এ অধ্যায়ে আমরা যে বিষয়গুলি আলোচনা করব।
- পর্যায়ক্রমিক অন্তরীকরণ
- অন্তরজের প্রতীক
- কতকগুলি বিশেষ ফাংশনের \(n\) তম অন্তরজ
- ম্যাকলরিনের উপপাদ্য
- সমাধানকৃত উদাহরণমালা
- অতি সংক্ষিপ্ত প্রশ্ন-উত্তর
- সংক্ষিপ্ত প্রশ্ন-উত্তর
- বর্ণনামূলক প্রশ্ন-উত্তর
পর্যায়ক্রমিক অন্তরীকরণ।
Successive Differentiation.
\(x\) এর সাপেক্ষে \(y=f(x)\) এর প্রথম অন্তরজকে \(\frac{dy}{dx}, \ f^{\prime}(x), \ y_{1}\) বা \(y^{\prime}\) প্রতীক দ্বারা প্রকাশ করা হয়। যদি প্রথম অন্তরজও সাধারণত \(x\) এর একটি ফাংশন হয়। তবে \(x\) এর এই নতুন ফাংশনের \(x\) এর সাপেক্ষে অন্তরজকে \(f(x)\) এর দ্বিতীয় অন্তরজ বলা হয়। \(f(x)\) এর দ্বিতীয় অন্তরজকে \(\frac{d}{dx}\left(\frac{dy}{dx}\right)\) বা সংক্ষেপে \(\frac{d^2y}{dx^2}, \ f^{\prime\prime}(x), \ y_{2}\) বা \(y^{\prime\prime}\) দ্বারা প্রকাশ করা হয়। অনুরূপভাবে, \(x\) এর সাপেক্ষে \(\frac{d^2y}{dx^2}\) এর অন্তরজকে \(f(x)\) এর তৃতীয় অন্তরজ বলা হয় এবং একে \(\frac{d^3y}{dx^3}, \ f^{\prime\prime\prime}(x), \ y_{3}\) বা \(y^{\prime\prime\prime}\) দ্বারা প্রকাশ করা হয় এবং এরূপভাবে \(f(x)\) এর \(n\) তম অন্তরজ \(\frac{d^ny}{dx^n}, \ f^{n}(x), \ y_{n}\) বা \(y^{n}\) দ্বারা প্রকাশ করা হয়।
এভাবে কোনো ফাংশনকে ধারাবাহিকভাবে অন্তরীকরণ করার প্রক্রিয়াকে পর্যায়ক্রমিক অন্তরীকরণ বলা হয়।
অন্তরজের প্রতীক।
\(\frac{dy}{dx}=y_{1}, \ \frac{dy_{1}}{dx}=y_{2}, \ \frac{dy_{2}}{dx}=y_{3}, \ ...\frac{dy_{n-1}}{dx}=y_{n}\)
\(\frac{d}{dx}\left(\frac{dy}{dx}\right)=y_{2}\)
\(\frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=y_{3}\)
\(\frac{d}{dx}\left(\frac{d^3y}{dx^3}\right)=y_{4}\)
\(............\)
\(\frac{d}{dx}\left(\frac{d^{n-1}y}{dx^{n-1}}\right)=y_{n}\)
\(\frac{d}{dx}\left(\frac{dy}{dx}\right)=y_{2}\)
\(\frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=y_{3}\)
\(\frac{d}{dx}\left(\frac{d^3y}{dx^3}\right)=y_{4}\)
\(............\)
\(\frac{d}{dx}\left(\frac{d^{n-1}y}{dx^{n-1}}\right)=y_{n}\)
কতকগুলি বিশেষ ফাংশনের \(n\) তম অন্তরজ। যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
The \(n^{th}\) derivative of some of spesial function.Where n is positive integer.
\((1)\) \(y=x^n\) হলে,
\(y_{n}=n!\)
\((2)\) \(y=e^x\) হলে,
\(y_{n}=e^x\)
\((3)\) \(y=e^{ax}\) হলে,
\(y_{n}=a^ne^{ax}\)
\((4)\) \(y=a^{x}\) হলে,
\(y_{n}=(\ln{a})^na^x\)
\((5)\) \(y=\ln{x}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!}{x^n}\)
\((6)\) \(y=\frac{1}{x}\) হলে,
\(y_{n}=\frac{(-1){n}(n)!}{x^{n+1}}\)
\((7)\) \(y=\ln{(x+a)}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!}{(x+a)^{n}}\)
\((8)\) \(y=\frac{1}{x+a}\) হলে,
\(y_{n}=\frac{(-1)^{n}n!}{(x+a)^{n+1}}\)
\((9)\) \(y=\frac{1}{x-a}\) হলে,
\(y_{n}=\frac{(-1)^{n}n!}{(x-a)^{n+1}}\)
\((10)\) \(y=\ln{(ax)}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!a^n}{(ax)^{n}}\)
\((11)\) \(y=\ln{(ax+b)}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!a^n}{(ax+b)^{n}}\)
\(y_{n}=n!\)
\((2)\) \(y=e^x\) হলে,
\(y_{n}=e^x\)
\((3)\) \(y=e^{ax}\) হলে,
\(y_{n}=a^ne^{ax}\)
\((4)\) \(y=a^{x}\) হলে,
\(y_{n}=(\ln{a})^na^x\)
\((5)\) \(y=\ln{x}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!}{x^n}\)
\((6)\) \(y=\frac{1}{x}\) হলে,
\(y_{n}=\frac{(-1){n}(n)!}{x^{n+1}}\)
\((7)\) \(y=\ln{(x+a)}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!}{(x+a)^{n}}\)
\((8)\) \(y=\frac{1}{x+a}\) হলে,
\(y_{n}=\frac{(-1)^{n}n!}{(x+a)^{n+1}}\)
\((9)\) \(y=\frac{1}{x-a}\) হলে,
\(y_{n}=\frac{(-1)^{n}n!}{(x-a)^{n+1}}\)
\((10)\) \(y=\ln{(ax)}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!a^n}{(ax)^{n}}\)
\((11)\) \(y=\ln{(ax+b)}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!a^n}{(ax+b)^{n}}\)
\((12)\) \(y=\sin{x}\) হলে,
\(y_{n}=\sin{\left(\frac{n\pi}{2}+x\right)}\)
\((13)\) \(y=\sin{(ax)}\) হলে,
\(y_{n}=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
\((14)\) \(y=\sin{(ax+b)}\) হলে,
\(y_{n}=a^n\sin{\{\frac{n\pi}{2}+(ax+b)\}}\)
\((15)\) \(y=\cos{x}\) হলে,
\(y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\)
\((16)\) \(y=\cos{(ax)}\) হলে,
\(y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\((17)\) \(y=\cos{(ax+b)}\) হলে,
\(y_{n}=a^n\cos{\{\frac{n\pi}{2}+(ax+b)\}}\)
\((18)\) \(y=e^{ax}\sin{(bx)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\((19)\) \(y=e^{ax}\cos{(bx)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\((20)\) \(y=e^{ax}\sin{(bx+c)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\((21)\) \(y=e^{ax}\cos{(bx+c)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\(y_{n}=\sin{\left(\frac{n\pi}{2}+x\right)}\)
\((13)\) \(y=\sin{(ax)}\) হলে,
\(y_{n}=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
\((14)\) \(y=\sin{(ax+b)}\) হলে,
\(y_{n}=a^n\sin{\{\frac{n\pi}{2}+(ax+b)\}}\)
\((15)\) \(y=\cos{x}\) হলে,
\(y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\)
\((16)\) \(y=\cos{(ax)}\) হলে,
\(y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\((17)\) \(y=\cos{(ax+b)}\) হলে,
\(y_{n}=a^n\cos{\{\frac{n\pi}{2}+(ax+b)\}}\)
\((18)\) \(y=e^{ax}\sin{(bx)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\((19)\) \(y=e^{ax}\cos{(bx)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\((20)\) \(y=e^{ax}\sin{(bx+c)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\((21)\) \(y=e^{ax}\cos{(bx+c)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
ম্যাকলরিনের উপপাদ্য।
Maclaurins theorem.
\(f(x)\) যদি \(x\) এর এমন একটি ফাংশন হয়, যাকে \(x\) এর ধনাত্মক পূর্ণ সাংখ্যিক, ক্রমবর্ধমান শক্তির একটি অসীম ধারায় বিস্তৃত করা যায় এবং ঐ বিস্তৃতির প্রতিটি পদ যে কোনো সংখ্যক বার অন্তরীকরণযোগ্য হয়, তাহলে,
\(f(x)=f(0)+\frac{x}{1!}f^{\prime}(0)+\frac{x^2}{2!}f^{\prime\prime}(0)+\frac{x^3}{3!}f^{\prime\prime\prime}(0)+ .......+\frac{x^n}{n!}f^{n}(0)+ ...\infty\)
\(f(x)=f(0)+\frac{x}{1!}f^{\prime}(0)+\frac{x^2}{2!}f^{\prime\prime}(0)+\frac{x^3}{3!}f^{\prime\prime\prime}(0)+ .......+\frac{x^n}{n!}f^{n}(0)+ ...\infty\)
\((1)\) \(y=x^n\) হলে,
\(y_{n}=n!\)
\(y_{n}=n!\)
Proof:
দেওয়া আছে,
\(y=x^n\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^n)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=nx^{n-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=n\frac{d}{dx}(x^{n-1})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=n(n-1)x^{n-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=n(n-1)\frac{d}{dx}(x^{n-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=n(n-1)(n-2)\frac{d}{dx}(x^{n-3})\)
...............
\(\Rightarrow y_{n}=n(n-1)(n-2)........3.2.1.x^{n-n}\)
\(\Rightarrow y_{n}=n(n-1)(n-2)........3.2.1.x^{0}\)
\(\Rightarrow y_{n}=n(n-1)(n-2)........3.2.1.1\) ➜ \(\because x^{0}=1\)
\(\Rightarrow y_{n}=n(n-1)(n-2)........3.2.1\)
\(\therefore y_{n}=n!\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because n(n-1)(n-2)........3.2.1=n!\)
\(y=x^n\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^n)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=nx^{n-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=n\frac{d}{dx}(x^{n-1})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=n(n-1)x^{n-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=n(n-1)\frac{d}{dx}(x^{n-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=n(n-1)(n-2)\frac{d}{dx}(x^{n-3})\)
...............
\(\Rightarrow y_{n}=n(n-1)(n-2)........3.2.1.x^{n-n}\)
\(\Rightarrow y_{n}=n(n-1)(n-2)........3.2.1.x^{0}\)
\(\Rightarrow y_{n}=n(n-1)(n-2)........3.2.1.1\) ➜ \(\because x^{0}=1\)
\(\Rightarrow y_{n}=n(n-1)(n-2)........3.2.1\)
\(\therefore y_{n}=n!\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because n(n-1)(n-2)........3.2.1=n!\)
\((2)\) \(y=e^x\) হলে,
\(y_{n}=e^x\)
\(y_{n}=e^x\)
Proof:
দেওয়া আছে,
\(y=e^x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^x)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^x\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(e^x)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=e^x\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}(e^x)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=e^x\)
\(...............\)
\(\Rightarrow y_{n}=e^x\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=e^x\)
\(y=e^x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^x)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^x\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(e^x)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=e^x\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}(e^x)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=e^x\)
\(...............\)
\(\Rightarrow y_{n}=e^x\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=e^x\)
\((3)\) \(y=e^{ax}\) হলে,
\(y_{n}=a^ne^{ax}\)
\(y_{n}=a^ne^{ax}\)
Proof:
দেওয়া আছে,
\(y=e^{ax}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{ax})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{1}=e^{ax}.a\)
\(\Rightarrow y_{1}=ae^{ax}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}(e^{ax})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=ae^{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{2}=ae^{ax}.a\)
\(\Rightarrow y_{2}=a^2e^{ax}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}(e^{ax})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2e^{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{3}=a^2e^{ax}.a\)
\(\Rightarrow y_{3}=a^3e^{ax}\)
\(...............\)
\(\Rightarrow y_{n}=a^{n}e^{ax}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^{n}e^{ax}\)
\(y=e^{ax}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{ax})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{1}=e^{ax}.a\)
\(\Rightarrow y_{1}=ae^{ax}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}(e^{ax})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=ae^{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{2}=ae^{ax}.a\)
\(\Rightarrow y_{2}=a^2e^{ax}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}(e^{ax})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2e^{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{3}=a^2e^{ax}.a\)
\(\Rightarrow y_{3}=a^3e^{ax}\)
\(...............\)
\(\Rightarrow y_{n}=a^{n}e^{ax}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^{n}e^{ax}\)
\((4)\) \(y=a^{x}\) হলে,
\(y_{n}=(\ln{a})^na^x\)
\(y_{n}=(\ln{a})^na^x\)
Proof:
দেওয়া আছে,
\(y=a^{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(a^{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a^{x}\ln{a}\)
\(\Rightarrow y_{1}=\ln{a}.a^{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\ln{a}\frac{d}{dx}(a^{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\ln{a}.a^{x}\ln{a}\)
\(\Rightarrow y_{2}=(\ln{a})^2a^{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(\ln{a})^2\frac{d}{dx}(a^{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(\ln{a})^2a^{x}\ln{a}\)
\(\Rightarrow y_{3}=(\ln{a})^3a^{x}\)
\(...............\)
\(\Rightarrow y_{n}=(\ln{a})^na^{x}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=(\ln{a})^na^{x}\)
\(y=a^{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(a^{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a^{x}\ln{a}\)
\(\Rightarrow y_{1}=\ln{a}.a^{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\ln{a}\frac{d}{dx}(a^{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\ln{a}.a^{x}\ln{a}\)
\(\Rightarrow y_{2}=(\ln{a})^2a^{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(\ln{a})^2\frac{d}{dx}(a^{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(\ln{a})^2a^{x}\ln{a}\)
\(\Rightarrow y_{3}=(\ln{a})^3a^{x}\)
\(...............\)
\(\Rightarrow y_{n}=(\ln{a})^na^{x}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=(\ln{a})^na^{x}\)
\((5)\) \(y=\ln{x}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!}{x^n}\)
\(y_{n}=\frac{(-1){n-1}(n-1)!}{x^n}\)
Proof:
দেওয়া আছে,
\(y=\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x}\)
\(\Rightarrow y_{1}=(x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(x^{-1})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-1(x)^{-1-1}\)
\(\Rightarrow y_{2}=(-1)^1.1(x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^1.1\frac{d}{dx}(x^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^1.1.(-2)x^{-2-1}\)
\(\Rightarrow y_{3}=(-1)^2.1.2.x^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^2.1.2\frac{d}{dx}(x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{4})=(-1)^2.1.2.(-3)x^{-3-1}\)
\(\Rightarrow \frac{d}{dx}(y_{4})=(-1)^3.1.2.3.x^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n-1}.1.2.3 .........(n-1)x^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!\frac{1}{(x)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!}{(x)^{n}}\)
\(y=\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x}\)
\(\Rightarrow y_{1}=(x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(x^{-1})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-1(x)^{-1-1}\)
\(\Rightarrow y_{2}=(-1)^1.1(x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^1.1\frac{d}{dx}(x^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^1.1.(-2)x^{-2-1}\)
\(\Rightarrow y_{3}=(-1)^2.1.2.x^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^2.1.2\frac{d}{dx}(x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{4})=(-1)^2.1.2.(-3)x^{-3-1}\)
\(\Rightarrow \frac{d}{dx}(y_{4})=(-1)^3.1.2.3.x^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n-1}.1.2.3 .........(n-1)x^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!\frac{1}{(x)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!}{(x)^{n}}\)
\((6)\) \(y=\frac{1}{x}\) হলে,
\(y_{n}=\frac{(-1){n}(n)!}{x^{n+1}}\)
\(y_{n}=\frac{(-1){n}(n)!}{x^{n+1}}\)
Proof:
দেওয়া আছে,
\(y=\frac{1}{x}\)
\(\Rightarrow y=(x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{-1})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(x)^{-1-1}\)
\(\Rightarrow y_{1}=(-1)^1.1(x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=(-1)^1.1\frac{d}{dx}(x^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)^1.1.(-2)x^{-2-1}\)
\(\Rightarrow y_{2}=(-1)^2.1.2.x^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^2.1.2\frac{d}{dx}(x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^2.1.2.(-3)x^{-3-1}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^3.1.2.3.x^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n}.1.2.3 .........nx^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n}n!\frac{1}{(x)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{(-1)^{n}n!}{(x)^{n+1}}\)
\(y=\frac{1}{x}\)
\(\Rightarrow y=(x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{-1})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(x)^{-1-1}\)
\(\Rightarrow y_{1}=(-1)^1.1(x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=(-1)^1.1\frac{d}{dx}(x^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)^1.1.(-2)x^{-2-1}\)
\(\Rightarrow y_{2}=(-1)^2.1.2.x^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^2.1.2\frac{d}{dx}(x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^2.1.2.(-3)x^{-3-1}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^3.1.2.3.x^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n}.1.2.3 .........nx^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n}n!\frac{1}{(x)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{(-1)^{n}n!}{(x)^{n+1}}\)
\((7)\) \(y=\ln{(x+a)}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!}{(x+a)^{n}}\)
\(y_{n}=\frac{(-1){n-1}(n-1)!}{(x+a)^{n}}\)
Proof:
দেওয়া আছে,
\(y=\ln{(x+a)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(x+a)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x+a}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{1}=\frac{1}{x+a}.1\)
\(\Rightarrow y_{1}=(x+a)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{(x+a)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-1(x+a)^{-1-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{2}=(-1)^1.1(x+a)^{-2}.1\)
\(\Rightarrow y_{2}=(-1)^1.1(x+a)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^1.1\frac{d}{dx}\{(x+a)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^1.1.(-2)(x+a)^{-2-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{3}=(-1)^2.1.2.(x+a)^{-3}.1\)
\(\Rightarrow y_{3}=(-1)^2.1.2.(x+a)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^2.1.2\frac{d}{dx}\{(x+a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=(-1)^2.1.2.(-3)(x+a)^{-3-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{4}=(-1)^3.1.2.3.(x+a)^{-4}.1\)
\(\Rightarrow y_{4}=(-1)^3.1.2.3.(x+a)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n-1}.1.2.3 .........(n-1)(x+a)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!\frac{1}{(x+a)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!}{(x+a)^{n}}\)
\(y=\ln{(x+a)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(x+a)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x+a}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{1}=\frac{1}{x+a}.1\)
\(\Rightarrow y_{1}=(x+a)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{(x+a)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-1(x+a)^{-1-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{2}=(-1)^1.1(x+a)^{-2}.1\)
\(\Rightarrow y_{2}=(-1)^1.1(x+a)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^1.1\frac{d}{dx}\{(x+a)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^1.1.(-2)(x+a)^{-2-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{3}=(-1)^2.1.2.(x+a)^{-3}.1\)
\(\Rightarrow y_{3}=(-1)^2.1.2.(x+a)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^2.1.2\frac{d}{dx}\{(x+a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=(-1)^2.1.2.(-3)(x+a)^{-3-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{4}=(-1)^3.1.2.3.(x+a)^{-4}.1\)
\(\Rightarrow y_{4}=(-1)^3.1.2.3.(x+a)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n-1}.1.2.3 .........(n-1)(x+a)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!\frac{1}{(x+a)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!}{(x+a)^{n}}\)
\((8)\) \(y=\frac{1}{x+a}\) হলে,
\(y_{n}=\frac{(-1)^{n}n!}{(x+a)^{n+1}}\)
\(y_{n}=\frac{(-1)^{n}n!}{(x+a)^{n+1}}\)
Proof:
দেওয়া আছে,
\(y=\frac{1}{x+a}\)
\(\Rightarrow y=(x+a)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x+a)^{-1}\}\) ➜\(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(x+a)^{-1-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{1}=(-1)^1.1(x+a)^{-2}.1\)
\(\Rightarrow y_{1}=(-1)^1.1(x+a)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=(-1)^1.1\frac{d}{dx}\{(x+a)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)^1.1.(-2)(x+a)^{-2-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{2}=(-1)^2.1.2.(x+a)^{-3}.1\)
\(\Rightarrow y_{2}=(-1)^2.1.2.(x+a)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^2.1.2\frac{d}{dx}\{(x+a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^2.1.2.(-3)(x+a)^{-3-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.(x+a)^{-(3+1)}.1\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.(x+a)^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n}.1.2.3 .........n(x+a)^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n}n!\frac{1}{(x+a)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{(-1)^{n}n!}{(x+a)^{n+1}}\)
\(y=\frac{1}{x+a}\)
\(\Rightarrow y=(x+a)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x+a)^{-1}\}\) ➜\(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(x+a)^{-1-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{1}=(-1)^1.1(x+a)^{-2}.1\)
\(\Rightarrow y_{1}=(-1)^1.1(x+a)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=(-1)^1.1\frac{d}{dx}\{(x+a)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)^1.1.(-2)(x+a)^{-2-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{2}=(-1)^2.1.2.(x+a)^{-3}.1\)
\(\Rightarrow y_{2}=(-1)^2.1.2.(x+a)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^2.1.2\frac{d}{dx}\{(x+a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^2.1.2.(-3)(x+a)^{-3-1}\frac{d}{dx}(x+a)\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.(x+a)^{-(3+1)}.1\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.(x+a)^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n}.1.2.3 .........n(x+a)^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n}n!\frac{1}{(x+a)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{(-1)^{n}n!}{(x+a)^{n+1}}\)
\((9)\) \(y=\frac{1}{x-a}\) হলে,
\(y_{n}=\frac{(-1)^{n}n!}{(x-a)^{n+1}}\)
\(y_{n}=\frac{(-1)^{n}n!}{(x-a)^{n+1}}\)
Proof:
দেওয়া আছে,
\(y=\frac{1}{x-a}\)
\(\Rightarrow y=(x-a)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x-a)^{-1}\}\) ➜\(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(x+a)^{-1-1}\frac{d}{dx}(x-a)\)
\(\Rightarrow y_{1}=(-1)^1.1(x-a)^{-2}.1\)
\(\Rightarrow y_{1}=(-1)^1.1(x-a)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=(-1)^1.1\frac{d}{dx}\{(x-a)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)^1.1.(-2)(x-a)^{-2-1}\frac{d}{dx}(x-a)\)
\(\Rightarrow y_{2}=(-1)^2.1.2.(x-a)^{-3}.1\)
\(\Rightarrow y_{2}=(-1)^2.1.2.(x-a)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^2.1.2\frac{d}{dx}\{(x-a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^2.1.2.(-3)(x-a)^{-3-1}\frac{d}{dx}(x-a)\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.(x-a)^{-(3+1)}.1\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.(x-a)^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n}.1.2.3 .........n(x-a)^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n}n!\frac{1}{(x-a)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{(-1)^{n}n!}{(x-a)^{n+1}}\)
\(y=\frac{1}{x-a}\)
\(\Rightarrow y=(x-a)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x-a)^{-1}\}\) ➜\(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(x+a)^{-1-1}\frac{d}{dx}(x-a)\)
\(\Rightarrow y_{1}=(-1)^1.1(x-a)^{-2}.1\)
\(\Rightarrow y_{1}=(-1)^1.1(x-a)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=(-1)^1.1\frac{d}{dx}\{(x-a)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)^1.1.(-2)(x-a)^{-2-1}\frac{d}{dx}(x-a)\)
\(\Rightarrow y_{2}=(-1)^2.1.2.(x-a)^{-3}.1\)
\(\Rightarrow y_{2}=(-1)^2.1.2.(x-a)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^2.1.2\frac{d}{dx}\{(x-a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^2.1.2.(-3)(x-a)^{-3-1}\frac{d}{dx}(x-a)\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.(x-a)^{-(3+1)}.1\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.(x-a)^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n}.1.2.3 .........n(x-a)^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n}n!\frac{1}{(x-a)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{(-1)^{n}n!}{(x-a)^{n+1}}\)
\((10)\) \(y=\ln{(ax)}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!a^n}{(ax)^{n}}\)
\(y_{n}=\frac{(-1){n-1}(n-1)!a^n}{(ax)^{n}}\)
Proof:
দেওয়া আছে,
\(y=\ln{xa}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(xa)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{xa}\frac{d}{dx}(xa)\)
\(\Rightarrow y_{1}=\frac{1}{xa}.a\)
\(\Rightarrow y_{1}=a(xa)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{(xa)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a(-1)(xa)^{-1-1}\frac{d}{dx}(xa)\)
\(\Rightarrow y_{2}=a(-1)^1.1(xa)^{-2}.a\)
\(\Rightarrow y_{2}=a^2(-1)^1.1(xa)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2(-1)^1.1\frac{d}{dx}\{(xa)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2(-1)^1.1.(-2)(xa)^{-2-1}.\frac{d}{dx}(xa)\)
\(\Rightarrow y_{3}=a^2(-1)^2.1.2.(xa)^{-3}.a\)
\(\Rightarrow y_{3}=a^3(-1)^2.1.2.(xa)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=a^3(-1)^2.1.2\frac{d}{dx}\{(x+a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=a^3(-1)^2.1.2.(-3)(xa)^{-3-1}.\frac{d}{dx}(xa)\)
\(\Rightarrow y_{4}=a^3(-1)^3.1.2.3.(xa)^{-4}.a\)
\(\Rightarrow y_{4}=a^4(-1)^3.1.2.3.(xa)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=a^n(-1)^{n-1}.1.2.3 .........(n-1)(xa)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=a^n(-1)^{n-1}(n-1)!\frac{1}{(xa)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!a^n}{(xa)^{n}}\)
\(y=\ln{xa}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(xa)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{xa}\frac{d}{dx}(xa)\)
\(\Rightarrow y_{1}=\frac{1}{xa}.a\)
\(\Rightarrow y_{1}=a(xa)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{(xa)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a(-1)(xa)^{-1-1}\frac{d}{dx}(xa)\)
\(\Rightarrow y_{2}=a(-1)^1.1(xa)^{-2}.a\)
\(\Rightarrow y_{2}=a^2(-1)^1.1(xa)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2(-1)^1.1\frac{d}{dx}\{(xa)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2(-1)^1.1.(-2)(xa)^{-2-1}.\frac{d}{dx}(xa)\)
\(\Rightarrow y_{3}=a^2(-1)^2.1.2.(xa)^{-3}.a\)
\(\Rightarrow y_{3}=a^3(-1)^2.1.2.(xa)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=a^3(-1)^2.1.2\frac{d}{dx}\{(x+a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=a^3(-1)^2.1.2.(-3)(xa)^{-3-1}.\frac{d}{dx}(xa)\)
\(\Rightarrow y_{4}=a^3(-1)^3.1.2.3.(xa)^{-4}.a\)
\(\Rightarrow y_{4}=a^4(-1)^3.1.2.3.(xa)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=a^n(-1)^{n-1}.1.2.3 .........(n-1)(xa)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=a^n(-1)^{n-1}(n-1)!\frac{1}{(xa)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!a^n}{(xa)^{n}}\)
\((11)\) \(y=\ln{(ax+b)}\) হলে,
\(y_{n}=\frac{(-1){n-1}(n-1)!a^n}{(ax+b)^{n}}\)
\(y_{n}=\frac{(-1){n-1}(n-1)!a^n}{(ax+b)^{n}}\)
Proof:
দেওয়া আছে,
\(y=\ln{(ax+b)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(ax+b)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{xa+b}\frac{d}{dx}(xa+b)\)
\(\Rightarrow y_{1}=\frac{1}{xa+b}.a\)
\(\Rightarrow y_{1}=a(xa+b)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{(xa+b)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a(-1)(xa+b)^{-1-1}\frac{d}{dx}(xa+b)\)
\(\Rightarrow y_{2}=a(-1)^1.1(xa+b)^{-2}.a\)
\(\Rightarrow y_{2}=a^2(-1)^1.1(xa+b)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2(-1)^1.1\frac{d}{dx}\{(xa+b)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2(-1)^1.1.(-2)(xa+b)^{-2-1}.\frac{d}{dx}(xa+b)\)
\(\Rightarrow y_{3}=a^2(-1)^2.1.2.(xa+b)^{-3}.a\)
\(\Rightarrow y_{3}=a^3(-1)^2.1.2.(xa+b)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=a^3(-1)^2.1.2\frac{d}{dx}\{(x+a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=a^3(-1)^2.1.2.(-3)(xa+b)^{-3-1}.\frac{d}{dx}(xa+b)\)
\(\Rightarrow y_{4}=a^3(-1)^3.1.2.3.(xa+b)^{-4}.a\)
\(\Rightarrow y_{4}=a^4(-1)^3.1.2.3.(xa+b)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=a^n(-1)^{n-1}.1.2.3 .........(n-1)(xa+b)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=a^n(-1)^{n-1}(n-1)!\frac{1}{(xa+b)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!a^n}{(xa+b)^{n}}\)
\(y=\ln{(ax+b)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(ax+b)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{xa+b}\frac{d}{dx}(xa+b)\)
\(\Rightarrow y_{1}=\frac{1}{xa+b}.a\)
\(\Rightarrow y_{1}=a(xa+b)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{(xa+b)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a(-1)(xa+b)^{-1-1}\frac{d}{dx}(xa+b)\)
\(\Rightarrow y_{2}=a(-1)^1.1(xa+b)^{-2}.a\)
\(\Rightarrow y_{2}=a^2(-1)^1.1(xa+b)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2(-1)^1.1\frac{d}{dx}\{(xa+b)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2(-1)^1.1.(-2)(xa+b)^{-2-1}.\frac{d}{dx}(xa+b)\)
\(\Rightarrow y_{3}=a^2(-1)^2.1.2.(xa+b)^{-3}.a\)
\(\Rightarrow y_{3}=a^3(-1)^2.1.2.(xa+b)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=a^3(-1)^2.1.2\frac{d}{dx}\{(x+a)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=a^3(-1)^2.1.2.(-3)(xa+b)^{-3-1}.\frac{d}{dx}(xa+b)\)
\(\Rightarrow y_{4}=a^3(-1)^3.1.2.3.(xa+b)^{-4}.a\)
\(\Rightarrow y_{4}=a^4(-1)^3.1.2.3.(xa+b)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=a^n(-1)^{n-1}.1.2.3 .........(n-1)(xa+b)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=a^n(-1)^{n-1}(n-1)!\frac{1}{(xa+b)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!a^n}{(xa+b)^{n}}\)
\((12)\) \(y=\sin{x}\) হলে,
\(y_{n}=\sin{\left(\frac{n\pi}{2}+x\right)}\)
\(y_{n}=\sin{\left(\frac{n\pi}{2}+x\right)}\)
Proof:
দেওয়া আছে,
\(y=\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{x}\)
\(\Rightarrow y_{1}=\sin{\left(\frac{\pi}{2}+x\right)}\) ➜ \(\because \cos{A}=\sin{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{\sin{\left(\frac{\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\cos{\left(\frac{\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{\pi}{2}+x\right)\)
\(\Rightarrow y_{2}=\cos{\left(\frac{\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{2}=\sin{\left(\frac{2\pi}{2}+x\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}\{\sin{\left(\frac{2\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=\cos{\left(\frac{2\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+x\right)\)
\(\Rightarrow y_{3}=\cos{\left(\frac{2\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{3}=\sin{\left(\frac{3\pi}{2}+x\right)}\)
\(...............\)
\(\Rightarrow y_{n}=\sin{\left(\frac{n\pi}{2}+x\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=\sin{\left(\frac{n\pi}{2}+x\right)}\)
\(y=\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{x}\)
\(\Rightarrow y_{1}=\sin{\left(\frac{\pi}{2}+x\right)}\) ➜ \(\because \cos{A}=\sin{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{\sin{\left(\frac{\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\cos{\left(\frac{\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{\pi}{2}+x\right)\)
\(\Rightarrow y_{2}=\cos{\left(\frac{\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{2}=\sin{\left(\frac{2\pi}{2}+x\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}\{\sin{\left(\frac{2\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=\cos{\left(\frac{2\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+x\right)\)
\(\Rightarrow y_{3}=\cos{\left(\frac{2\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{3}=\sin{\left(\frac{3\pi}{2}+x\right)}\)
\(...............\)
\(\Rightarrow y_{n}=\sin{\left(\frac{n\pi}{2}+x\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=\sin{\left(\frac{n\pi}{2}+x\right)}\)
\((13)\) \(y=\sin{(ax)}\) হলে,
\(y_{n}=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
\(y_{n}=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
Proof:
দেওয়া আছে,
\(y=\sin{(ax)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(ax)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{1}=\cos{ax}.a\)
\(\Rightarrow y_{1}=a\sin{\left(\frac{\pi}{2}+ax\right)}\) ➜ \(\because \cos{A}=\sin{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\sin{\left(\frac{\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a\cos{\left(\frac{\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{\pi}{2}+ax\right)\)
\(\Rightarrow y_{2}=a\cos{\left(\frac{\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{2}=a^2\sin{\left(\frac{2\pi}{2}+ax\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\sin{\left(\frac{2\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2\cos{\left(\frac{2\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+ax\right)\)
\(\Rightarrow y_{3}=a^2\cos{\left(\frac{2\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{3}=a^3\sin{\left(\frac{3\pi}{2}+ax\right)}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
\(y=\sin{(ax)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(ax)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{1}=\cos{ax}.a\)
\(\Rightarrow y_{1}=a\sin{\left(\frac{\pi}{2}+ax\right)}\) ➜ \(\because \cos{A}=\sin{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\sin{\left(\frac{\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a\cos{\left(\frac{\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{\pi}{2}+ax\right)\)
\(\Rightarrow y_{2}=a\cos{\left(\frac{\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{2}=a^2\sin{\left(\frac{2\pi}{2}+ax\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\sin{\left(\frac{2\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2\cos{\left(\frac{2\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+ax\right)\)
\(\Rightarrow y_{3}=a^2\cos{\left(\frac{2\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{3}=a^3\sin{\left(\frac{3\pi}{2}+ax\right)}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
\((14)\) \(y=\sin{(ax+b)}\) হলে,
\(y_{n}=a^n\sin{\{\frac{n\pi}{2}+(ax+b)\}}\)
\(y_{n}=a^n\sin{\{\frac{n\pi}{2}+(ax+b)\}}\)
Proof:
দেওয়া আছে,
\(y=\sin{(ax+b)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(ax+b)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{(ax+b)}\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{1}=\cos{(ax+b)}.a\)
\(\Rightarrow y_{1}=a\sin{\{\frac{\pi}{2}+(ax+b)\}}\) ➜ \(\because \cos{A}=\sin{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\sin{\{\frac{\pi}{2}+(ax+b)\}}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a\cos{\{\frac{\pi}{2}+(ax+b)\}}\frac{d}{dx}\{\frac{\pi}{2}+(ax+b)\}\)
\(\Rightarrow y_{2}=a\cos{\{\frac{\pi}{2}+(ax+b)\}}.a\)
\(\Rightarrow y_{2}=a^2\sin{\{\frac{2\pi}{2}+(ax+b)\}}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\sin{\{\frac{2\pi}{2}+(ax+b)\}}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2\cos{\{\frac{2\pi}{2}+(ax+b)\}}\frac{d}{dx}\{\frac{2\pi}{2}+(ax+b)\}\)
\(\Rightarrow y_{3}=a^2\cos{\{\frac{2\pi}{2}+(ax+b)\}}.a\)
\(\Rightarrow y_{3}=a^3\sin{\{\frac{3\pi}{2}+(ax+b)\}}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\sin{\{\frac{n\pi}{2}+(ax+b)\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\sin{\{\frac{n\pi}{2}+(ax+b)\}}\)
\(y=\sin{(ax+b)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(ax+b)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{(ax+b)}\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{1}=\cos{(ax+b)}.a\)
\(\Rightarrow y_{1}=a\sin{\{\frac{\pi}{2}+(ax+b)\}}\) ➜ \(\because \cos{A}=\sin{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\sin{\{\frac{\pi}{2}+(ax+b)\}}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a\cos{\{\frac{\pi}{2}+(ax+b)\}}\frac{d}{dx}\{\frac{\pi}{2}+(ax+b)\}\)
\(\Rightarrow y_{2}=a\cos{\{\frac{\pi}{2}+(ax+b)\}}.a\)
\(\Rightarrow y_{2}=a^2\sin{\{\frac{2\pi}{2}+(ax+b)\}}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\sin{\{\frac{2\pi}{2}+(ax+b)\}}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2\cos{\{\frac{2\pi}{2}+(ax+b)\}}\frac{d}{dx}\{\frac{2\pi}{2}+(ax+b)\}\)
\(\Rightarrow y_{3}=a^2\cos{\{\frac{2\pi}{2}+(ax+b)\}}.a\)
\(\Rightarrow y_{3}=a^3\sin{\{\frac{3\pi}{2}+(ax+b)\}}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\sin{\{\frac{n\pi}{2}+(ax+b)\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\sin{\{\frac{n\pi}{2}+(ax+b)\}}\)
\((15)\) \(y=\cos{x}\) হলে,
\(y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\)
\(y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\)
Proof:
দেওয়া আছে,
\(y=\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{x}\)
\(\Rightarrow y_{1}=\cos{\left(\frac{\pi}{2}+x\right)}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{\cos{\left(\frac{\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-\sin{\left(\frac{\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{\pi}{2}+x\right)\)
\(\Rightarrow y_{2}=-\sin{\left(\frac{\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{2}=\cos{\left(\frac{2\pi}{2}+x\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}\{\cos{\left(\frac{2\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-\sin{\left(\frac{2\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+x\right)\)
\(\Rightarrow y_{3}=-\sin{\left(\frac{2\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{3}=\cos{\left(\frac{3\pi}{2}+x\right)}\)
\(...............\)
\(\Rightarrow y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\)
\(y=\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{x}\)
\(\Rightarrow y_{1}=\cos{\left(\frac{\pi}{2}+x\right)}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{\cos{\left(\frac{\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-\sin{\left(\frac{\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{\pi}{2}+x\right)\)
\(\Rightarrow y_{2}=-\sin{\left(\frac{\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{2}=\cos{\left(\frac{2\pi}{2}+x\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}\{\cos{\left(\frac{2\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-\sin{\left(\frac{2\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+x\right)\)
\(\Rightarrow y_{3}=-\sin{\left(\frac{2\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{3}=\cos{\left(\frac{3\pi}{2}+x\right)}\)
\(...............\)
\(\Rightarrow y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\)
\((16)\) \(y=\cos{(ax)}\) হলে,
\(y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\(y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
Proof:
দেওয়া আছে,
\(y=\cos{(ax)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(ax)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{1}=-\sin{ax}.a\)
\(\Rightarrow y_{1}=a\cos{\left(\frac{\pi}{2}+ax\right)}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\cos{\left(\frac{\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-a\sin{\left(\frac{\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{\pi}{2}+ax\right)\)
\(\Rightarrow y_{2}=-a\sin{\left(\frac{\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{2}=a^2\cos{\left(\frac{2\pi}{2}+ax\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\cos{\left(\frac{2\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-a^2\sin{\left(\frac{2\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+ax\right)\)
\(\Rightarrow y_{3}=-a^2\sin{\left(\frac{2\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{3}=a^3\cos{\left(\frac{3\pi}{2}+ax\right)}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\(y=\cos{(ax)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(ax)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{ax}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{1}=-\sin{ax}.a\)
\(\Rightarrow y_{1}=a\cos{\left(\frac{\pi}{2}+ax\right)}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\cos{\left(\frac{\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-a\sin{\left(\frac{\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{\pi}{2}+ax\right)\)
\(\Rightarrow y_{2}=-a\sin{\left(\frac{\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{2}=a^2\cos{\left(\frac{2\pi}{2}+ax\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\cos{\left(\frac{2\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-a^2\sin{\left(\frac{2\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+ax\right)\)
\(\Rightarrow y_{3}=-a^2\sin{\left(\frac{2\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{3}=a^3\cos{\left(\frac{3\pi}{2}+ax\right)}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\((17)\) \(y=\cos{(ax+b)}\) হলে,
\(y_{n}=a^n\cos{\{\frac{n\pi}{2}+(ax+b)\}}\)
\(y_{n}=a^n\cos{\{\frac{n\pi}{2}+(ax+b)\}}\)
Proof:
দেওয়া আছে,
\(y=\cos{(ax+b)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(ax+b)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(ax+b)}\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{1}=-\sin{(ax+b)}.a\)
\(\Rightarrow y_{1}=a\cos{\{\frac{\pi}{2}+(ax+b)\}}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\cos{\{\frac{\pi}{2}+(ax+b)\}}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-a\sin{\{\frac{\pi}{2}+(ax+b)\}}\frac{d}{dx}\{\frac{\pi}{2}+(ax+b)\}\)
\(\Rightarrow y_{2}=-a\sin{\{\frac{\pi}{2}+(ax+b)\}}.a\)
\(\Rightarrow y_{2}=a^2\cos{\{\frac{2\pi}{2}+(ax+b)\}}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\cos{\{\frac{2\pi}{2}+(ax+b)\}}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-a^2\sin{\{\frac{2\pi}{2}+(ax+b)\}}\frac{d}{dx}\{\frac{2\pi}{2}+(ax+b)\}\)
\(\Rightarrow y_{3}=-a^2\sin{\{\frac{2\pi}{2}+(ax+b)\}}.a\)
\(\Rightarrow y_{3}=a^3\cos{\{\frac{3\pi}{2}+(ax+b)\}}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\cos{\{\frac{n\pi}{2}+(ax+b)\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\cos{\{\frac{n\pi}{2}+(ax+b)\}}\)
\(y=\cos{(ax+b)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(ax+b)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(ax+b)}\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{1}=-\sin{(ax+b)}.a\)
\(\Rightarrow y_{1}=a\cos{\{\frac{\pi}{2}+(ax+b)\}}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\cos{\{\frac{\pi}{2}+(ax+b)\}}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-a\sin{\{\frac{\pi}{2}+(ax+b)\}}\frac{d}{dx}\{\frac{\pi}{2}+(ax+b)\}\)
\(\Rightarrow y_{2}=-a\sin{\{\frac{\pi}{2}+(ax+b)\}}.a\)
\(\Rightarrow y_{2}=a^2\cos{\{\frac{2\pi}{2}+(ax+b)\}}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\cos{\{\frac{2\pi}{2}+(ax+b)\}}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-a^2\sin{\{\frac{2\pi}{2}+(ax+b)\}}\frac{d}{dx}\{\frac{2\pi}{2}+(ax+b)\}\)
\(\Rightarrow y_{3}=-a^2\sin{\{\frac{2\pi}{2}+(ax+b)\}}.a\)
\(\Rightarrow y_{3}=a^3\cos{\{\frac{3\pi}{2}+(ax+b)\}}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\cos{\{\frac{n\pi}{2}+(ax+b)\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\cos{\{\frac{n\pi}{2}+(ax+b)\}}\)
\((18)\) \(y=e^{ax}\sin{(bx)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
Proof:
দেওয়া আছে,
\(y=e^{ax}\sin{bx}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{e^{ax}\sin{(bx)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}\{\sin{(bx)}\}+\sin{(bx)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=e^{ax}\cos{(bx)}\frac{d}{dx}(bx)+e^{ax}\sin{(bx)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^{ax}\cos{(bx)}.b+e^{ax}\sin{(bx)}.a\)
\(\therefore y_{1}=e^{ax}\sin{(bx)}.a+e^{ax}\cos{(bx)}.b\)
ধরি,
\(a=r\cos{\theta} .....(1)\)
\(b=r\sin{\theta} ......(2)\)
\(\therefore a^2+b^2=r^2(\cos^2{\theta}+\sin^2{\theta})\) ➜ \((1)\) ও \((2)\) বর্গ করে যোগ করি।
\(\Rightarrow a^2+b^2=r^2.1\) ➜ \(\because \cos^2{\theta}+\sin^2{\theta}=1\)
\(\Rightarrow a^2+b^2=r^2\)
\(\Rightarrow r^2=a^2+b^2\)
\(\therefore r=(a^2+b^2)^{\frac{1}{2}}\)
আবার,
\(\frac{b}{a}=\frac{r\sin{\theta}}{r\cos{\theta}}\) ➜ \((2)\div{(1)}\)এর সাহায্যে।
\(\Rightarrow \frac{b}{a}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \frac{b}{a}=\tan{\theta}\) ➜ \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)
\(\Rightarrow \tan{\theta}=\frac{b}{a}\)
\(\therefore \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
এখন,
\(y_{1}=e^{ax}\sin{(bx)}.r\cos{\theta}+e^{ax}\cos{(bx)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{1}=re^{ax}\{\sin{(bx)}\cos{\theta}+\cos{(bx)}\sin{\theta}\}\)
\(\Rightarrow y_{1}=re^{ax}\sin{(bx+\theta)}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\therefore y_{1}=re^{ax}\sin{(bx+\theta)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=r\frac{d}{dx}\{e^{ax}\sin{(bx+\theta)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=re^{ax}\frac{d}{dx}\{\sin{(bx+\theta)}\}+r\sin{(bx+\theta)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=re^{ax}\cos{(bx+\theta)}\frac{d}{dx}(bx+\theta)+re^{ax}\sin{(bx+\theta)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=re^{ax}\cos{(bx+\theta)}.b+re^{ax}\sin{(bx+\theta)}.a\)
\(\therefore y_{2}=re^{ax}\sin{(bx+\theta)}.a+re^{ax}\cos{(bx+\theta)}.b\)
আবার,
\(y_{2}=re^{ax}\sin{(bx+\theta)}.r\cos{\theta}+re^{ax}\cos{(bx+\theta)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{2}=r^2e^{ax}\sin{(bx+\theta)}\cos{\theta}+r^2e^{ax}\cos{(bx+\theta)}\sin{\theta}\)
\(\Rightarrow y_{2}=r^2e^{ax}\{\sin{(bx+\theta)}\cos{\theta}+\cos{(bx+\theta)}\sin{\theta}\}\)
\(\Rightarrow y_{2}=r^2e^{ax}\sin{(bx+\theta+\theta)}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\therefore y_{2}=r^2e^{ax}\sin{(bx+2\theta)}\)
\(...............\)
\(...............\)
\(...............\)
\(\Rightarrow y_{n}=r^ne^{ax}\sin{(bx+n\theta)}\)
\(\therefore y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because r=(a^2+b^2)^{\frac{1}{2}}, \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
\(y=e^{ax}\sin{bx}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{e^{ax}\sin{(bx)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}\{\sin{(bx)}\}+\sin{(bx)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=e^{ax}\cos{(bx)}\frac{d}{dx}(bx)+e^{ax}\sin{(bx)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^{ax}\cos{(bx)}.b+e^{ax}\sin{(bx)}.a\)
\(\therefore y_{1}=e^{ax}\sin{(bx)}.a+e^{ax}\cos{(bx)}.b\)
ধরি,
\(a=r\cos{\theta} .....(1)\)
\(b=r\sin{\theta} ......(2)\)
\(\therefore a^2+b^2=r^2(\cos^2{\theta}+\sin^2{\theta})\) ➜ \((1)\) ও \((2)\) বর্গ করে যোগ করি।
\(\Rightarrow a^2+b^2=r^2.1\) ➜ \(\because \cos^2{\theta}+\sin^2{\theta}=1\)
\(\Rightarrow a^2+b^2=r^2\)
\(\Rightarrow r^2=a^2+b^2\)
\(\therefore r=(a^2+b^2)^{\frac{1}{2}}\)
আবার,
\(\frac{b}{a}=\frac{r\sin{\theta}}{r\cos{\theta}}\) ➜ \((2)\div{(1)}\)এর সাহায্যে।
\(\Rightarrow \frac{b}{a}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \frac{b}{a}=\tan{\theta}\) ➜ \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)
\(\Rightarrow \tan{\theta}=\frac{b}{a}\)
\(\therefore \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
এখন,
\(y_{1}=e^{ax}\sin{(bx)}.r\cos{\theta}+e^{ax}\cos{(bx)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{1}=re^{ax}\{\sin{(bx)}\cos{\theta}+\cos{(bx)}\sin{\theta}\}\)
\(\Rightarrow y_{1}=re^{ax}\sin{(bx+\theta)}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\therefore y_{1}=re^{ax}\sin{(bx+\theta)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=r\frac{d}{dx}\{e^{ax}\sin{(bx+\theta)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=re^{ax}\frac{d}{dx}\{\sin{(bx+\theta)}\}+r\sin{(bx+\theta)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=re^{ax}\cos{(bx+\theta)}\frac{d}{dx}(bx+\theta)+re^{ax}\sin{(bx+\theta)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=re^{ax}\cos{(bx+\theta)}.b+re^{ax}\sin{(bx+\theta)}.a\)
\(\therefore y_{2}=re^{ax}\sin{(bx+\theta)}.a+re^{ax}\cos{(bx+\theta)}.b\)
আবার,
\(y_{2}=re^{ax}\sin{(bx+\theta)}.r\cos{\theta}+re^{ax}\cos{(bx+\theta)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{2}=r^2e^{ax}\sin{(bx+\theta)}\cos{\theta}+r^2e^{ax}\cos{(bx+\theta)}\sin{\theta}\)
\(\Rightarrow y_{2}=r^2e^{ax}\{\sin{(bx+\theta)}\cos{\theta}+\cos{(bx+\theta)}\sin{\theta}\}\)
\(\Rightarrow y_{2}=r^2e^{ax}\sin{(bx+\theta+\theta)}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\therefore y_{2}=r^2e^{ax}\sin{(bx+2\theta)}\)
\(...............\)
\(...............\)
\(...............\)
\(\Rightarrow y_{n}=r^ne^{ax}\sin{(bx+n\theta)}\)
\(\therefore y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because r=(a^2+b^2)^{\frac{1}{2}}, \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
\((19)\) \(y=e^{ax}\cos{(bx)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
Proof:
দেওয়া আছে,
\(y=e^{ax}\cos{(bx)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{e^{ax}\cos{(bx)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}\{\cos{(bx)}\}+\cos{(bx)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-e^{ax}\sin{(bx)}\frac{d}{dx}(bx)+e^{ax}\cos{(bx)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=-e^{ax}\sin{(bx)}.b+e^{ax}\cos{(bx)}.a\)
\(\therefore y_{1}=e^{ax}\cos{(bx)}.a-e^{ax}\sin{(bx)}.b\)
ধরি,
\(a=r\cos{\theta} .....(1)\)
\(b=r\sin{\theta} ......(2)\)
\(\therefore a^2+b^2=r^2(\cos^2{\theta}+\sin^2{\theta})\) ➜ \((1)\) ও \((2)\) বর্গ করে যোগ করি।
\(\Rightarrow a^2+b^2=r^2.1\) ➜ \(\because \cos^2{\theta}+\sin^2{\theta}=1\)
\(\Rightarrow a^2+b^2=r^2\)
\(\Rightarrow r^2=a^2+b^2\)
\(\therefore r=(a^2+b^2)^{\frac{1}{2}}\)
আবার,
\(\frac{b}{a}=\frac{r\sin{\theta}}{r\cos{\theta}}\) ➜ \((2)\div{(1)}\)এর সাহায্যে।
\(\Rightarrow \frac{b}{a}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \frac{b}{a}=\tan{\theta}\) ➜ \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)
\(\Rightarrow \tan{\theta}=\frac{b}{a}\)
\(\therefore \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
এখন,
\(y_{1}=e^{ax}\cos{(bx)}.r\cos{\theta}-e^{ax}\sin{(bx)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{1}=re^{ax}\{\cos{(bx)}\cos{\theta}-\sin{(bx)}\sin{\theta}\}\)
\(\Rightarrow y_{1}=re^{ax}\cos{(bx+\theta)}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\therefore y_{1}=re^{ax}\cos{(bx+\theta)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=r\frac{d}{dx}\{e^{ax}\cos{(bx+\theta)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=re^{ax}\frac{d}{dx}\{\cos{(bx+\theta)}\}+r\cos{(bx+\theta)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-re^{ax}\sin{(bx+\theta)}\frac{d}{dx}(bx+\theta)+re^{ax}\cos{(bx+\theta)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=-re^{ax}\sin{(bx+\theta)}.b+re^{ax}\cos{(bx+\theta)}.a\)
\(\therefore y_{2}=re^{ax}\cos{(bx+\theta)}.a-re^{ax}\sin{(bx+\theta)}.b\)
আবার,
\(y_{2}=re^{ax}\cos{(bx+\theta)}.r\cos{\theta}-re^{ax}\sin{(bx+\theta)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{2}=r^2e^{ax}\cos{(bx+\theta)}\cos{\theta}-r^2e^{ax}\sin{(bx+\theta)}\sin{\theta}\)
\(\Rightarrow y_{2}=r^2e^{ax}\{\cos{(bx+\theta)}\cos{\theta}-\sin{(bx+\theta)}\sin{\theta}\}\)
\(\Rightarrow y_{2}=r^2e^{ax}\cos{(bx+\theta+\theta)}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\therefore y_{2}=r^2e^{ax}\cos{(bx+2\theta)}\)
\(...............\)
\(...............\)
\(...............\)
\(\Rightarrow y_{n}=r^ne^{ax}\cos{(bx+n\theta)}\)
\(\therefore y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because r=(a^2+b^2)^{\frac{1}{2}}, \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
\(y=e^{ax}\cos{(bx)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{e^{ax}\cos{(bx)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}\{\cos{(bx)}\}+\cos{(bx)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-e^{ax}\sin{(bx)}\frac{d}{dx}(bx)+e^{ax}\cos{(bx)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=-e^{ax}\sin{(bx)}.b+e^{ax}\cos{(bx)}.a\)
\(\therefore y_{1}=e^{ax}\cos{(bx)}.a-e^{ax}\sin{(bx)}.b\)
ধরি,
\(a=r\cos{\theta} .....(1)\)
\(b=r\sin{\theta} ......(2)\)
\(\therefore a^2+b^2=r^2(\cos^2{\theta}+\sin^2{\theta})\) ➜ \((1)\) ও \((2)\) বর্গ করে যোগ করি।
\(\Rightarrow a^2+b^2=r^2.1\) ➜ \(\because \cos^2{\theta}+\sin^2{\theta}=1\)
\(\Rightarrow a^2+b^2=r^2\)
\(\Rightarrow r^2=a^2+b^2\)
\(\therefore r=(a^2+b^2)^{\frac{1}{2}}\)
আবার,
\(\frac{b}{a}=\frac{r\sin{\theta}}{r\cos{\theta}}\) ➜ \((2)\div{(1)}\)এর সাহায্যে।
\(\Rightarrow \frac{b}{a}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \frac{b}{a}=\tan{\theta}\) ➜ \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)
\(\Rightarrow \tan{\theta}=\frac{b}{a}\)
\(\therefore \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
এখন,
\(y_{1}=e^{ax}\cos{(bx)}.r\cos{\theta}-e^{ax}\sin{(bx)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{1}=re^{ax}\{\cos{(bx)}\cos{\theta}-\sin{(bx)}\sin{\theta}\}\)
\(\Rightarrow y_{1}=re^{ax}\cos{(bx+\theta)}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\therefore y_{1}=re^{ax}\cos{(bx+\theta)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=r\frac{d}{dx}\{e^{ax}\cos{(bx+\theta)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=re^{ax}\frac{d}{dx}\{\cos{(bx+\theta)}\}+r\cos{(bx+\theta)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-re^{ax}\sin{(bx+\theta)}\frac{d}{dx}(bx+\theta)+re^{ax}\cos{(bx+\theta)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=-re^{ax}\sin{(bx+\theta)}.b+re^{ax}\cos{(bx+\theta)}.a\)
\(\therefore y_{2}=re^{ax}\cos{(bx+\theta)}.a-re^{ax}\sin{(bx+\theta)}.b\)
আবার,
\(y_{2}=re^{ax}\cos{(bx+\theta)}.r\cos{\theta}-re^{ax}\sin{(bx+\theta)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{2}=r^2e^{ax}\cos{(bx+\theta)}\cos{\theta}-r^2e^{ax}\sin{(bx+\theta)}\sin{\theta}\)
\(\Rightarrow y_{2}=r^2e^{ax}\{\cos{(bx+\theta)}\cos{\theta}-\sin{(bx+\theta)}\sin{\theta}\}\)
\(\Rightarrow y_{2}=r^2e^{ax}\cos{(bx+\theta+\theta)}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\therefore y_{2}=r^2e^{ax}\cos{(bx+2\theta)}\)
\(...............\)
\(...............\)
\(...............\)
\(\Rightarrow y_{n}=r^ne^{ax}\cos{(bx+n\theta)}\)
\(\therefore y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because r=(a^2+b^2)^{\frac{1}{2}}, \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
\((20)\) \(y=e^{ax}\sin{(bx+c)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
Proof:
দেওয়া আছে,
\(y=e^{ax}\sin{(bx+c)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{e^{ax}\sin{(bx+c)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}(\sin{(bx+c)})+\sin{(bx+c)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=e^{ax}\cos{(bx+c)}\frac{d}{dx}(bx+c)+e^{ax}\sin{(bx+c)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^{ax}\cos{(bx+c)}.b+e^{ax}\sin{(bx+c)}.a\)
\(\therefore y_{1}=e^{ax}\sin{(bx+c)}.a+e^{ax}\cos{(bx+c)}.b\)
ধরি,
\(a=r\cos{\theta} .....(1)\)
\(b=r\sin{\theta} ......(2)\)
\(\therefore a^2+b^2=r^2(\cos^2{\theta}+\sin^2{\theta})\) ➜ \((1)\) ও \((2)\) বর্গ করে যোগ করি।
\(\Rightarrow a^2+b^2=r^2.1\) ➜ \(\because \cos^2{\theta}+\sin^2{\theta}=1\)
\(\Rightarrow a^2+b^2=r^2\)
\(\Rightarrow r^2=a^2+b^2\)
\(\therefore r=(a^2+b^2)^{\frac{1}{2}}\)
আবার,
\(\frac{b}{a}=\frac{r\sin{\theta}}{r\cos{\theta}}\) ➜ \((2)\div{(1)}\)এর সাহায্যে।
\(\Rightarrow \frac{b}{a}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \frac{b}{a}=\tan{\theta}\) ➜ \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)
\(\Rightarrow \tan{\theta}=\frac{b}{a}\)
\(\therefore \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
এখন,
\(y_{1}=e^{ax}\sin{(bx+c)}.r\cos{\theta}+e^{ax}\cos{(bx+c)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{1}=re^{ax}\{\sin{(bx+c)}\cos{\theta}+\cos{(bx+c)}\sin{\theta}\}\)
\(\Rightarrow y_{1}=re^{ax}\sin{(bx+\theta)}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\therefore y_{1}=re^{ax}\sin{(bx+c+\theta)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=r\frac{d}{dx}\{e^{ax}\sin{(bx+c+\theta)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=re^{ax}\frac{d}{dx}\{\sin{(bx+c+\theta)}\}+r\sin{(bx+c+\theta)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=re^{ax}\cos{(bx+c+\theta)}\frac{d}{dx}(bx+c+\theta)+re^{ax}\sin{(bx+c+\theta)}\frac{d}{dx}(ax)\) ➜ \((bx+c)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=re^{ax}\cos{(bx+c+\theta)}.b+re^{ax}\sin{(bx+c+\theta)}.a\)
\(\therefore y_{2}=re^{ax}\sin{(bx+c+\theta)}.a+re^{ax}\cos{(bx+c+\theta)}.b\)
আবার,
\(y_{2}=re^{ax}\sin{(bx+c+\theta)}.r\cos{\theta}+re^{ax}\cos{(bx+c+\theta)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{2}=r^2e^{ax}\sin{(bx+c+\theta)}\cos{\theta}+r^2e^{ax}\cos{(bx+c+\theta)}\sin{\theta}\)
\(\Rightarrow y_{2}=r^2e^{ax}\{\sin{(bx+c+\theta)}\cos{\theta}+\cos{(bx+c+\theta)}\sin{\theta}\}\)
\(\Rightarrow y_{2}=r^2e^{ax}\sin{(bx+c+\theta+\theta)}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\therefore y_{2}=r^2e^{ax}\sin{(bx+c+2\theta)}\)
\(...............\)
\(...............\)
\(...............\)
\(\Rightarrow y_{n}=r^ne^{ax}\sin{(bx+c+n\theta)}\)
\(\therefore y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because r=(a^2+b^2)^{\frac{1}{2}}, \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
\(y=e^{ax}\sin{(bx+c)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{e^{ax}\sin{(bx+c)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}(\sin{(bx+c)})+\sin{(bx+c)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=e^{ax}\cos{(bx+c)}\frac{d}{dx}(bx+c)+e^{ax}\sin{(bx+c)}\frac{d}{dx}(ax)\) ➜ \((bx)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^{ax}\cos{(bx+c)}.b+e^{ax}\sin{(bx+c)}.a\)
\(\therefore y_{1}=e^{ax}\sin{(bx+c)}.a+e^{ax}\cos{(bx+c)}.b\)
ধরি,
\(a=r\cos{\theta} .....(1)\)
\(b=r\sin{\theta} ......(2)\)
\(\therefore a^2+b^2=r^2(\cos^2{\theta}+\sin^2{\theta})\) ➜ \((1)\) ও \((2)\) বর্গ করে যোগ করি।
\(\Rightarrow a^2+b^2=r^2.1\) ➜ \(\because \cos^2{\theta}+\sin^2{\theta}=1\)
\(\Rightarrow a^2+b^2=r^2\)
\(\Rightarrow r^2=a^2+b^2\)
\(\therefore r=(a^2+b^2)^{\frac{1}{2}}\)
আবার,
\(\frac{b}{a}=\frac{r\sin{\theta}}{r\cos{\theta}}\) ➜ \((2)\div{(1)}\)এর সাহায্যে।
\(\Rightarrow \frac{b}{a}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \frac{b}{a}=\tan{\theta}\) ➜ \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)
\(\Rightarrow \tan{\theta}=\frac{b}{a}\)
\(\therefore \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
এখন,
\(y_{1}=e^{ax}\sin{(bx+c)}.r\cos{\theta}+e^{ax}\cos{(bx+c)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{1}=re^{ax}\{\sin{(bx+c)}\cos{\theta}+\cos{(bx+c)}\sin{\theta}\}\)
\(\Rightarrow y_{1}=re^{ax}\sin{(bx+\theta)}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\therefore y_{1}=re^{ax}\sin{(bx+c+\theta)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=r\frac{d}{dx}\{e^{ax}\sin{(bx+c+\theta)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=re^{ax}\frac{d}{dx}\{\sin{(bx+c+\theta)}\}+r\sin{(bx+c+\theta)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=re^{ax}\cos{(bx+c+\theta)}\frac{d}{dx}(bx+c+\theta)+re^{ax}\sin{(bx+c+\theta)}\frac{d}{dx}(ax)\) ➜ \((bx+c)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=re^{ax}\cos{(bx+c+\theta)}.b+re^{ax}\sin{(bx+c+\theta)}.a\)
\(\therefore y_{2}=re^{ax}\sin{(bx+c+\theta)}.a+re^{ax}\cos{(bx+c+\theta)}.b\)
আবার,
\(y_{2}=re^{ax}\sin{(bx+c+\theta)}.r\cos{\theta}+re^{ax}\cos{(bx+c+\theta)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{2}=r^2e^{ax}\sin{(bx+c+\theta)}\cos{\theta}+r^2e^{ax}\cos{(bx+c+\theta)}\sin{\theta}\)
\(\Rightarrow y_{2}=r^2e^{ax}\{\sin{(bx+c+\theta)}\cos{\theta}+\cos{(bx+c+\theta)}\sin{\theta}\}\)
\(\Rightarrow y_{2}=r^2e^{ax}\sin{(bx+c+\theta+\theta)}\) ➜ \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(\therefore y_{2}=r^2e^{ax}\sin{(bx+c+2\theta)}\)
\(...............\)
\(...............\)
\(...............\)
\(\Rightarrow y_{n}=r^ne^{ax}\sin{(bx+c+n\theta)}\)
\(\therefore y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\sin{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because r=(a^2+b^2)^{\frac{1}{2}}, \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
\((21)\) \(y=e^{ax}\cos{(bx+c)}\) হলে,
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\(y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
Proof:
দেওয়া আছে,
\(y=e^{ax}\cos{(bx+c)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{e^{ax}\cos{(bx+c)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}(\cos{(bx+c)})+\cos{(bx+c)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-e^{ax}\sin{(bx+c)}\frac{d}{dx}(bx+c)+e^{ax}\cos{(bx+c)}\frac{d}{dx}(ax)\) ➜ \((bx+c)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=-e^{ax}\sin{(bx+c)}.b+e^{ax}\cos{(bx+c)}.a\)
\(\therefore y_{1}=e^{ax}\cos{(bx+c)}.a-e^{ax}\sin{(bx+c)}.b\)
ধরি,
\(a=r\cos{\theta} .....(1)\)
\(b=r\sin{\theta} ......(2)\)
\(\therefore a^2+b^2=r^2(\cos^2{\theta}+\sin^2{\theta})\) ➜ \((1)\) ও \((2)\) বর্গ করে যোগ করি।
\(\Rightarrow a^2+b^2=r^2.1\) ➜ \(\because \cos^2{\theta}+\sin^2{\theta}=1\)
\(\Rightarrow a^2+b^2=r^2\)
\(\Rightarrow r^2=a^2+b^2\)
\(\therefore r=(a^2+b^2)^{\frac{1}{2}}\)
আবার,
\(\frac{b}{a}=\frac{r\sin{\theta}}{r\cos{\theta}}\) ➜ \((2)\div{(1)}\)এর সাহায্যে।
\(\Rightarrow \frac{b}{a}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \frac{b}{a}=\tan{\theta}\) ➜ \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)
\(\Rightarrow \tan{\theta}=\frac{b}{a}\)
\(\therefore \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
এখন,
\(y_{1}=e^{ax}\cos{(bx+c)}.r\cos{\theta}-e^{ax}\sin{(bx+c)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{1}=re^{ax}\{\cos{(bx+c)}\cos{\theta}-\sin{(bx+c)}\sin{\theta}\}\)
\(\Rightarrow y_{1}=re^{ax}\cos{(bx+c+\theta)}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\therefore y_{1}=re^{ax}\cos{(bx+c+\theta)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=r\frac{d}{dx}\{e^{ax}\cos{(bx+c+\theta)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=re^{ax}\frac{d}{dx}\{\cos{(bx+c+\theta)}\}+r\cos{(bx+c+\theta)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-re^{ax}\sin{(bx+c+\theta)}\frac{d}{dx}(bx+c+\theta)+re^{ax}\cos{(bx+c+\theta)}\frac{d}{dx}(ax)\) ➜ \((bx+c)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=-re^{ax}\sin{(bx+c+\theta)}.b+re^{ax}\cos{(bx+c+\theta)}.a\)
\(\therefore y_{2}=re^{ax}\cos{(bx+c+\theta)}.a-re^{ax}\sin{(bx+c+\theta)}.b\)
আবার,
\(y_{2}=re^{ax}\cos{(bx+c+\theta)}.r\cos{\theta}-re^{ax}\sin{(bx+c+\theta)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{2}=r^2e^{ax}\cos{(bx+c+\theta)}\cos{\theta}-r^2e^{ax}\sin{(bx+c+\theta)}\sin{\theta}\)
\(\Rightarrow y_{2}=r^2e^{ax}\{\cos{(bx+c+\theta)}\cos{\theta}-\sin{(bx+c+\theta)}\sin{\theta}\}\)
\(\Rightarrow y_{2}=r^2e^{ax}\cos{(bx+c+\theta+\theta)}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\therefore y_{2}=r^2e^{ax}\cos{(bx+c+2\theta)}\)
\(...............\)
\(...............\)
\(...............\)
\(\Rightarrow y_{n}=r^ne^{ax}\cos{(bx+c+n\theta)}\)
\(\therefore y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because r=(a^2+b^2)^{\frac{1}{2}}, \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
\(y=e^{ax}\cos{(bx+c)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{e^{ax}\cos{(bx+c)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}(\cos{(bx+c)})+\cos{(bx+c)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-e^{ax}\sin{(bx+c)}\frac{d}{dx}(bx+c)+e^{ax}\cos{(bx+c)}\frac{d}{dx}(ax)\) ➜ \((bx+c)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=-e^{ax}\sin{(bx+c)}.b+e^{ax}\cos{(bx+c)}.a\)
\(\therefore y_{1}=e^{ax}\cos{(bx+c)}.a-e^{ax}\sin{(bx+c)}.b\)
ধরি,
\(a=r\cos{\theta} .....(1)\)
\(b=r\sin{\theta} ......(2)\)
\(\therefore a^2+b^2=r^2(\cos^2{\theta}+\sin^2{\theta})\) ➜ \((1)\) ও \((2)\) বর্গ করে যোগ করি।
\(\Rightarrow a^2+b^2=r^2.1\) ➜ \(\because \cos^2{\theta}+\sin^2{\theta}=1\)
\(\Rightarrow a^2+b^2=r^2\)
\(\Rightarrow r^2=a^2+b^2\)
\(\therefore r=(a^2+b^2)^{\frac{1}{2}}\)
আবার,
\(\frac{b}{a}=\frac{r\sin{\theta}}{r\cos{\theta}}\) ➜ \((2)\div{(1)}\)এর সাহায্যে।
\(\Rightarrow \frac{b}{a}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \frac{b}{a}=\tan{\theta}\) ➜ \(\because \frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}\)
\(\Rightarrow \tan{\theta}=\frac{b}{a}\)
\(\therefore \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
এখন,
\(y_{1}=e^{ax}\cos{(bx+c)}.r\cos{\theta}-e^{ax}\sin{(bx+c)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{1}=re^{ax}\{\cos{(bx+c)}\cos{\theta}-\sin{(bx+c)}\sin{\theta}\}\)
\(\Rightarrow y_{1}=re^{ax}\cos{(bx+c+\theta)}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\therefore y_{1}=re^{ax}\cos{(bx+c+\theta)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=r\frac{d}{dx}\{e^{ax}\cos{(bx+c+\theta)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=re^{ax}\frac{d}{dx}\{\cos{(bx+c+\theta)}\}+r\cos{(bx+c+\theta)}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-re^{ax}\sin{(bx+c+\theta)}\frac{d}{dx}(bx+c+\theta)+re^{ax}\cos{(bx+c+\theta)}\frac{d}{dx}(ax)\) ➜ \((bx+c)\) এবং \((ax)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=-re^{ax}\sin{(bx+c+\theta)}.b+re^{ax}\cos{(bx+c+\theta)}.a\)
\(\therefore y_{2}=re^{ax}\cos{(bx+c+\theta)}.a-re^{ax}\sin{(bx+c+\theta)}.b\)
আবার,
\(y_{2}=re^{ax}\cos{(bx+c+\theta)}.r\cos{\theta}-re^{ax}\sin{(bx+c+\theta)}.r\sin{\theta}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।।
\(\Rightarrow y_{2}=r^2e^{ax}\cos{(bx+c+\theta)}\cos{\theta}-r^2e^{ax}\sin{(bx+c+\theta)}\sin{\theta}\)
\(\Rightarrow y_{2}=r^2e^{ax}\{\cos{(bx+c+\theta)}\cos{\theta}-\sin{(bx+c+\theta)}\sin{\theta}\}\)
\(\Rightarrow y_{2}=r^2e^{ax}\cos{(bx+c+\theta+\theta)}\) ➜ \(\because \cos{A}\cos{B}-\sin{A}\sin{B}=\cos{(A+B)}\)
\(\therefore y_{2}=r^2e^{ax}\cos{(bx+c+2\theta)}\)
\(...............\)
\(...............\)
\(...............\)
\(\Rightarrow y_{n}=r^ne^{ax}\cos{(bx+c+n\theta)}\)
\(\therefore y_{n}=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+c+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা। ➜ \(\because r=(a^2+b^2)^{\frac{1}{2}}, \theta=\tan^{-1}{\left(\frac{b}{a}\right)}\)
\(f(x)\) যদি \(x\) এর এমন একটি ফাংশন হয়, যাকে \(x\) এর ধনাত্মক পূর্ণ সাংখ্যিক, ক্রমবর্ধমান শক্তির একটি অসীম ধারায় বিস্তৃত করা যায় এবং ঐ বিস্তৃতির প্রতিটি পদ যে কোনো সংখ্যক বার অন্তরীকরণযোগ্য হয়, তাহলে,
\(f(x)=f(0)+\frac{x}{1!}f^{'}(0)+\frac{x^2}{2!}f^{''}(0)+\frac{x^3}{3!}f^{'''}(0)+ .......+\frac{x^3}{3!}f^{'''}(0)+ ...\infty\)
\(f(x)=f(0)+\frac{x}{1!}f^{'}(0)+\frac{x^2}{2!}f^{''}(0)+\frac{x^3}{3!}f^{'''}(0)+ .......+\frac{x^3}{3!}f^{'''}(0)+ ...\infty\)
Proof:
মনে করি,
\(f(x)=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+ ..... (1)\)
\((1)\) নং কে পর্যায়ক্রম অন্তরীকরণ করে।
\(f^{'}(x)=a_{1}+2a_{2}x+3a_{3}x^2+4a_{4}x^3+ ..... \)
\(f^{''}(x)=2a_{2}+6a_{3}x+12a_{4}x^2+ ..... \)
\(f^{'''}(x)=6a_{3}+24a_{4}x+ ..... \)
\(...............\)
\(...............\)
\(x=0\) বসালে,
\(f(0)=a_{0}, \ f^{'}(0)=1!a_{1}, \ f^{''}(0)=2!a_{2}\), \( \ f^{'''}(0)=3!a_{3} ..... \ f^{n}(0)=n!a_{n}\)
\(\Rightarrow a_{0}=f(0), \ a_{1}=\frac{1}{1!}f^{'}(0), \ a_{2}=\frac{1}{2!}f^{''}(0)\), \( \ a_{3}=\frac{1}{3!}f^{'''}(0) ..... \ a_{n}=\frac{1}{n!}f^{n}(0)\)
এখন,
\(a_{0}, \ a_{1}, \ a_{2}, \ a_{3} ..... \ a_{n}\) এর মাণ \((1)\) নং সমীকরণে বসিয়ে,
\(f(x)=f(0)+\frac{x}{1!}f^{'}(0)+\frac{x^2}{2!}f^{''}(0)+\frac{x^3}{3!}f^{'''}(0)+ .......+\frac{x^3}{3!}f^{'''}(0)+ ...\infty\)
\(f(x)=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+ ..... (1)\)
\((1)\) নং কে পর্যায়ক্রম অন্তরীকরণ করে।
\(f^{'}(x)=a_{1}+2a_{2}x+3a_{3}x^2+4a_{4}x^3+ ..... \)
\(f^{''}(x)=2a_{2}+6a_{3}x+12a_{4}x^2+ ..... \)
\(f^{'''}(x)=6a_{3}+24a_{4}x+ ..... \)
\(...............\)
\(...............\)
\(x=0\) বসালে,
\(f(0)=a_{0}, \ f^{'}(0)=1!a_{1}, \ f^{''}(0)=2!a_{2}\), \( \ f^{'''}(0)=3!a_{3} ..... \ f^{n}(0)=n!a_{n}\)
\(\Rightarrow a_{0}=f(0), \ a_{1}=\frac{1}{1!}f^{'}(0), \ a_{2}=\frac{1}{2!}f^{''}(0)\), \( \ a_{3}=\frac{1}{3!}f^{'''}(0) ..... \ a_{n}=\frac{1}{n!}f^{n}(0)\)
এখন,
\(a_{0}, \ a_{1}, \ a_{2}, \ a_{3} ..... \ a_{n}\) এর মাণ \((1)\) নং সমীকরণে বসিয়ে,
\(f(x)=f(0)+\frac{x}{1!}f^{'}(0)+\frac{x^2}{2!}f^{''}(0)+\frac{x^3}{3!}f^{'''}(0)+ .......+\frac{x^3}{3!}f^{'''}(0)+ ...\infty\)
অনুশীলনী \(9.G\) উদাহরণ সমুহ
\((1)\) যদি \(y=a\sin{x}+b\cos{x}\) হয় তবে দেখাও যে, \(\frac{d^2y}{dx^2}+y=0\)
\((2)\) \(y=\sqrt{(4+3\sin{x})}\) হলে, দেখাও যে, \(2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=4\)
\((3)\) \(y=(p+qx)e^{-2x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\)
\((4)\) \(y=\sin{(\sin{x})}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+y\cos^2{x}=0\)
[ বঃ ২০১৪,২০০৯;ঢাঃ ২০১৪, ২০০১;কুঃ ২০১৩,২০০৭;সিঃ২০১১,২০০৬;যঃ২০০৫ ]
\((5)\) \(y=x^2+\frac{1}{x^2}\) হলে, দেখাও যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-4y=0\)
\((2)\) \(y=\sqrt{(4+3\sin{x})}\) হলে, দেখাও যে, \(2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=4\)
\((3)\) \(y=(p+qx)e^{-2x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\)
\((4)\) \(y=\sin{(\sin{x})}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+y\cos^2{x}=0\)
[ বঃ ২০১৪,২০০৯;ঢাঃ ২০১৪, ২০০১;কুঃ ২০১৩,২০০৭;সিঃ২০১১,২০০৬;যঃ২০০৫ ]
\((5)\) \(y=x^2+\frac{1}{x^2}\) হলে, দেখাও যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-4y=0\)
\((6)\) \(y=ax^2+\frac{b}{\sqrt{x}}\) হলে, দেখাও যে, \(2x^2y_{2}-xy_{1}-2y=0\)
[ ঢাঃ২০০৬;রাঃ২০১৩,২০০৬;কুঃ২০০৯,২০০৬;যঃ২০১৪,২০১২,২০০৫; দিঃ২০১৪,২০১১;চঃ২০১৩,২০১১;সিঃ২০১৩,২০১০;মাঃ২০১৩,২০০৯ ]
\((7)\) \(y=\sin{(2\cos^{-1}{x})}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}+4y=0\) অথবা, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+4y=0\)
\((8)\) \(y=\sin{(m\sin{x})}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+m^2y\cos^2{x}=0\)
\((9)\) \(y=x^3+5x^2+10x+14\) হলে, \(\frac{dy}{dx}, \frac{d^2y}{dx^2}, \frac{d^3y}{dx^3}\) এবং \(\frac{d^4y}{dx^4}\) নির্ণয় কর।
উত্তরঃ \(3x^2+10x+10, 6x+10, 6, 0\)
\((10)\) \(y=ax\sin{x}\) হলে, প্রমাণ কর যে, \(x^2y_{2}-2xy_{1}+(x^2+2)y=0\)
[ ঢাঃ২০০৬;রাঃ২০১৩,২০০৬;কুঃ২০০৯,২০০৬;যঃ২০১৪,২০১২,২০০৫; দিঃ২০১৪,২০১১;চঃ২০১৩,২০১১;সিঃ২০১৩,২০১০;মাঃ২০১৩,২০০৯ ]
\((7)\) \(y=\sin{(2\cos^{-1}{x})}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}+4y=0\) অথবা, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+4y=0\)
\((8)\) \(y=\sin{(m\sin{x})}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+m^2y\cos^2{x}=0\)
\((9)\) \(y=x^3+5x^2+10x+14\) হলে, \(\frac{dy}{dx}, \frac{d^2y}{dx^2}, \frac{d^3y}{dx^3}\) এবং \(\frac{d^4y}{dx^4}\) নির্ণয় কর।
উত্তরঃ \(3x^2+10x+10, 6x+10, 6, 0\)
\((10)\) \(y=ax\sin{x}\) হলে, প্রমাণ কর যে, \(x^2y_{2}-2xy_{1}+(x^2+2)y=0\)
\((1)\) যদি \(y=a\sin{x}+b\cos{x}\) হয় তবে দেখাও যে, \(\frac{d^2y}{dx^2}+y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=a\sin{x}+b\cos{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(a\sin{x}+b\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=a\frac{d}{dx}(\sin{x})+b\frac{d}{dx}(\cos{x})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=a\cos{x}-b\sin{x}\) ➜\(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(a\cos{x}-b\sin{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=a\frac{d}{dx}(\cos{x})-b\frac{d}{dx}(\sin{x})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=-a\sin{x}-b\cos{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=-(a\sin{x}+b\cos{x})\)
\(\Rightarrow \frac{d^2y}{dx^2}=-y\) ➜ \(\because a\sin{x}+b\cos{x}=y\)
\(\therefore \frac{d^2y}{dx^2}+y=0\)
(Showed)
\(y=a\sin{x}+b\cos{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(a\sin{x}+b\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=a\frac{d}{dx}(\sin{x})+b\frac{d}{dx}(\cos{x})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=a\cos{x}-b\sin{x}\) ➜\(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(a\cos{x}-b\sin{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=a\frac{d}{dx}(\cos{x})-b\frac{d}{dx}(\sin{x})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=-a\sin{x}-b\cos{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=-(a\sin{x}+b\cos{x})\)
\(\Rightarrow \frac{d^2y}{dx^2}=-y\) ➜ \(\because a\sin{x}+b\cos{x}=y\)
\(\therefore \frac{d^2y}{dx^2}+y=0\)
(Showed)
\((2)\) \(y=\sqrt{(4+3\sin{x})}\) হলে, দেখাও যে, \(2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=4\)
সমাধানঃ
দেওয়া আছে,
\(y=\sqrt{(4+3\sin{x})}\)
\(\Rightarrow y^2=4+3\sin{x}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(y^2)=\frac{d}{dx}(4+3\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y\frac{dy}{dx}=\frac{d}{dx}(4)+3\frac{d}{dx}(\sin{x})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow 2y\frac{dy}{dx}=0+3\cos{x}\) ➜\(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow 2\frac{d}{dx}\left(y\frac{dy}{dx}\right)=3\frac{d}{dx}(\cos{x})\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y\frac{d}{dx}\left(\frac{dy}{dx}\right)+2\frac{dy}{dx}.\frac{dy}{dx}=-3\sin{x}\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2=-4-3\sin{x}+4\)
\(\Rightarrow 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2=-(4+3\sin{x})+4\)
\(\Rightarrow 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2=-y^2+4\) ➜\(\because 4+3\sin{x}=y^2\)
\(\therefore 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=4\)
(Showed)
\(y=\sqrt{(4+3\sin{x})}\)
\(\Rightarrow y^2=4+3\sin{x}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(y^2)=\frac{d}{dx}(4+3\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y\frac{dy}{dx}=\frac{d}{dx}(4)+3\frac{d}{dx}(\sin{x})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow 2y\frac{dy}{dx}=0+3\cos{x}\) ➜\(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow 2\frac{d}{dx}\left(y\frac{dy}{dx}\right)=3\frac{d}{dx}(\cos{x})\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y\frac{d}{dx}\left(\frac{dy}{dx}\right)+2\frac{dy}{dx}.\frac{dy}{dx}=-3\sin{x}\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2=-4-3\sin{x}+4\)
\(\Rightarrow 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2=-(4+3\sin{x})+4\)
\(\Rightarrow 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2=-y^2+4\) ➜\(\because 4+3\sin{x}=y^2\)
\(\therefore 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=4\)
(Showed)
\((3)\) \(y=(p+qx)e^{-2x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=(p+qx)e^{-2x}\)
\(\Rightarrow ye^{2x}=p+qx\)
\(\Rightarrow \frac{d}{dx}(ye^{2x})=\frac{d}{dx}(p+qx)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y\frac{d}{dx}(e^{2x})+e^{2x}\frac{d}{dx}(y)=\frac{d}{dx}(p)+\frac{d}{dx}(qx)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow ye^{2x}\frac{d}{dx}(2x)+e^{2x}\frac{dy}{dx}=0+q\) ➜ \(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x)=1\)
\(\Rightarrow ye^{2x}.2+e^{2x}\frac{dy}{dx}=q\) ➜ \(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow 2ye^{2x}+e^{2x}y_{1}=q\) ➜ \(\because \frac{dy}{dx}=y_{1}\)
\(\Rightarrow e^{2x}(2y+y_{1})=q\)
\(\Rightarrow \frac{d}{dx}\{e^{2x}(2y+y_{1})\}=\frac{d}{dx}(q)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow e^{2x}\frac{d}{dx}(2y+y_{1})+(2y+y_{1})\frac{d}{dx}(e^{2x})=0\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(\Rightarrow e^{2x}(2y_{1}+y_{2})+(2y+y_{1})e^{2x}.\frac{d}{dx}(2x)=0\) ➜\(\because \frac{d}{dx}(2y+y_{1})=2y_{1}+y_{2}\)
\(\Rightarrow e^{2x}(2y_{1}+y_{2})+(2y+y_{1})e^{2x}.2=0\) ➜\(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow e^{2x}\{2y_{1}+y_{2}+2(2y+y_{1})\}=0\)
\(\Rightarrow 2y_{1}+y_{2}+2(2y+y_{1})=0, e^{2x}\ne{0}\)
\(\Rightarrow 2y_{1}+y_{2}+4y+2y_{1}=0\)
\(\Rightarrow y_{2}+4y_{1}+4y=0\)
\(\therefore \frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\) ➜\(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\(y=(p+qx)e^{-2x}\)
\(\Rightarrow ye^{2x}=p+qx\)
\(\Rightarrow \frac{d}{dx}(ye^{2x})=\frac{d}{dx}(p+qx)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y\frac{d}{dx}(e^{2x})+e^{2x}\frac{d}{dx}(y)=\frac{d}{dx}(p)+\frac{d}{dx}(qx)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow ye^{2x}\frac{d}{dx}(2x)+e^{2x}\frac{dy}{dx}=0+q\) ➜ \(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x)=1\)
\(\Rightarrow ye^{2x}.2+e^{2x}\frac{dy}{dx}=q\) ➜ \(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow 2ye^{2x}+e^{2x}y_{1}=q\) ➜ \(\because \frac{dy}{dx}=y_{1}\)
\(\Rightarrow e^{2x}(2y+y_{1})=q\)
\(\Rightarrow \frac{d}{dx}\{e^{2x}(2y+y_{1})\}=\frac{d}{dx}(q)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow e^{2x}\frac{d}{dx}(2y+y_{1})+(2y+y_{1})\frac{d}{dx}(e^{2x})=0\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(\Rightarrow e^{2x}(2y_{1}+y_{2})+(2y+y_{1})e^{2x}.\frac{d}{dx}(2x)=0\) ➜\(\because \frac{d}{dx}(2y+y_{1})=2y_{1}+y_{2}\)
\(\Rightarrow e^{2x}(2y_{1}+y_{2})+(2y+y_{1})e^{2x}.2=0\) ➜\(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow e^{2x}\{2y_{1}+y_{2}+2(2y+y_{1})\}=0\)
\(\Rightarrow 2y_{1}+y_{2}+2(2y+y_{1})=0, e^{2x}\ne{0}\)
\(\Rightarrow 2y_{1}+y_{2}+4y+2y_{1}=0\)
\(\Rightarrow y_{2}+4y_{1}+4y=0\)
\(\therefore \frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\) ➜\(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\((4)\) \(y=\sin{(\sin{x})}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+y\cos^2{x}=0\)
[ বঃ ২০১৪,২০০৯;ঢাঃ ২০১৪, ২০০১;কুঃ ২০১৩,২০০৭;সিঃ২০১১,২০০৬;যঃ২০০৫ ]
[ বঃ ২০১৪,২০০৯;ঢাঃ ২০১৪, ২০০১;কুঃ ২০১৩,২০০৭;সিঃ২০১১,২০০৬;যঃ২০০৫ ]
সমাধানঃ
দেওয়া আছে,
\(y=\sin{(\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(\sin{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\cos{(\sin{x})}\frac{d}{dx}(\sin{x})\) ➜ \(\sin{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{(\sin{x})}\cos{x}\) ➜ \(\because y_{1}=\frac{dy}{dx}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}\frac{1}{\cos{x}}=\cos{(\sin{x})}\)
\(\Rightarrow y_{1}\sec{x}=\cos{(\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y_{1}\sec{x})=\frac{d}{dx}\{\cos{(\sin{x})}\}\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}\frac{d}{dx}(\sec{x})+\sec{x}\frac{d}{dx}(y_{1})=-\sin{(\sin{x})}\frac{d}{dx}(\sin{x})\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(\cos{x})=-\sin{x} \)
\(\Rightarrow y_{1}\sec{x}\tan{x}+\sec{x}y_{2}=-\sin{(\sin{x})}\cos{x}\) ➜\(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}, \frac{d}{dx}(\sin{x})=\cos{x} \)
\(\Rightarrow y_{1}\frac{1}{\cos{x}}\tan{x}+\frac{1}{\cos{x}}y_{2}=-y\cos{x}\)
\(\Rightarrow y_{1}\tan{x}+y_{2}=-y\cos{x}\times{\cos{x}}\) ➜ \(\cos{x}\) দ্বারা গুণ করে।
\(\Rightarrow y_{2}+y_{1}\tan{x}=-y\cos^2{x}\)
\(\therefore \frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+y\cos^2{x}=0\)
(Showed)
\(y=\sin{(\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(\sin{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\cos{(\sin{x})}\frac{d}{dx}(\sin{x})\) ➜ \(\sin{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{(\sin{x})}\cos{x}\) ➜ \(\because y_{1}=\frac{dy}{dx}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}\frac{1}{\cos{x}}=\cos{(\sin{x})}\)
\(\Rightarrow y_{1}\sec{x}=\cos{(\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y_{1}\sec{x})=\frac{d}{dx}\{\cos{(\sin{x})}\}\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}\frac{d}{dx}(\sec{x})+\sec{x}\frac{d}{dx}(y_{1})=-\sin{(\sin{x})}\frac{d}{dx}(\sin{x})\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(\cos{x})=-\sin{x} \)
\(\Rightarrow y_{1}\sec{x}\tan{x}+\sec{x}y_{2}=-\sin{(\sin{x})}\cos{x}\) ➜\(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}, \frac{d}{dx}(\sin{x})=\cos{x} \)
\(\Rightarrow y_{1}\frac{1}{\cos{x}}\tan{x}+\frac{1}{\cos{x}}y_{2}=-y\cos{x}\)
\(\Rightarrow y_{1}\tan{x}+y_{2}=-y\cos{x}\times{\cos{x}}\) ➜ \(\cos{x}\) দ্বারা গুণ করে।
\(\Rightarrow y_{2}+y_{1}\tan{x}=-y\cos^2{x}\)
\(\therefore \frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+y\cos^2{x}=0\)
(Showed)
\((5)\) \(y=x^2+\frac{1}{x^2}\) হলে, দেখাও যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-4y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=x^2+\frac{1}{x^2}\)
\(\Rightarrow y=x^2+x^{-2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2+x^{-2})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^2)+\frac{d}{dx}(x^{-2})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-2-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-3}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(2x-2x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dx}(2x)-2\frac{d}{dx}(x^{-3})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2+6x^{-3-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-4}\times{x^2}\) ➜ \(x^2\) দ্বারা গুণ করে।
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-4+2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-2x^2+2x^{-2}+4x^2+4x^{-2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-x(2x-2x^{-3})+4(x^2+x^{-2})\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-x(\frac{dy}{dx})+4(y)\) ➜ \(\because \frac{dy}{dx}=2x-2x^{-3}, y=x^2+x^{-2}\)
\(\therefore x^2\frac{d^2y}{dx^2}+x(\frac{dy}{dx})-4(y)=0\)
(Showed)
\(y=x^2+\frac{1}{x^2}\)
\(\Rightarrow y=x^2+x^{-2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2+x^{-2})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^2)+\frac{d}{dx}(x^{-2})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-2-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-3}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(2x-2x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dx}(2x)-2\frac{d}{dx}(x^{-3})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2+6x^{-3-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-4}\times{x^2}\) ➜ \(x^2\) দ্বারা গুণ করে।
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-4+2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-2x^2+2x^{-2}+4x^2+4x^{-2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-x(2x-2x^{-3})+4(x^2+x^{-2})\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-x(\frac{dy}{dx})+4(y)\) ➜ \(\because \frac{dy}{dx}=2x-2x^{-3}, y=x^2+x^{-2}\)
\(\therefore x^2\frac{d^2y}{dx^2}+x(\frac{dy}{dx})-4(y)=0\)
(Showed)
\((6)\) \(y=ax^2+\frac{b}{\sqrt{x}}\) অথবা, \(y=ax^2+bx^{-\frac{1}{2}}\) হলে, দেখাও যে, \(2x^2y_{2}-xy_{1}-2y=0\)
[ ঢাঃ২০০৬;রাঃ২০১৩,২০০৬;কুঃ২০০৯,২০০৬;যঃ২০১৪,২০১২,২০০৫;দিঃ২০১৪,২০১১;চঃ২০১৩,২০১১;সিঃ২০১৩,২০১০;মাঃ২০১৩,২০০৯ ]
[ ঢাঃ২০০৬;রাঃ২০১৩,২০০৬;কুঃ২০০৯,২০০৬;যঃ২০১৪,২০১২,২০০৫;দিঃ২০১৪,২০১১;চঃ২০১৩,২০১১;সিঃ২০১৩,২০১০;মাঃ২০১৩,২০০৯ ]
সমাধানঃ
দেওয়া আছে,
\(y=ax^2+\frac{b}{\sqrt{x}}\)
\(\Rightarrow y=ax^2+\frac{b}{x^{\frac{1}{2}}}\)
\(\Rightarrow y=ax^2+bx^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(ax^2+bx^{-\frac{1}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=a\frac{d}{dx}(x^2)+b\frac{d}{dx}(x^{-\frac{1}{2}})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=a.2x-b\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=2ax-\frac{b}{2}x^{\frac{-1-2}{2}}\)
\(\Rightarrow \frac{dy}{dx}=2ax-\frac{b}{2}x^{-\frac{3}{2}}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(2ax-\frac{1}{2}x^{-\frac{3}{2}}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=2a\frac{d}{dx}(x)-\frac{b}{2}\frac{d}{dx}(x^{-\frac{3}{2}})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2a+\frac{b}{2}.\frac{3}{2}x^{-\frac{3}{2}-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2a+\frac{3b}{4}x^{\frac{-3-2}{2}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2a+\frac{3b}{4}x^{\frac{-5}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=4ax^2+\frac{3b}{2}x^{\frac{-5}{2}}\times{x^2}\) ➜ \(2x^2\) দ্বারা গুণ করে।
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=4ax^2+\frac{3b}{2}x^{\frac{-5}{2}+2}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=4ax^2+\frac{3b}{2}x^{\frac{-5+4}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=4ax^2+\frac{3b}{2}x^{\frac{-1}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=2ax^2-\frac{b}{2}x^{-\frac{1}{2}}+2ax^2+2bx^{-\frac{1}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=x(2ax-\frac{b}{2}x^{-\frac{3}{2}})+2(ax^2+bx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=x\frac{dy}{dx}+2y\) ➜ \(\because \frac{dy}{dx}=2ax-\frac{b}{2}x^{-\frac{3}{2}}, y=ax^2+bx^{-\frac{1}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2y=0\)
\(\therefore 2x^2y_{2}-xy_{1}-2y=0\)
(Showed)
\(y=ax^2+\frac{b}{\sqrt{x}}\)
\(\Rightarrow y=ax^2+\frac{b}{x^{\frac{1}{2}}}\)
\(\Rightarrow y=ax^2+bx^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(ax^2+bx^{-\frac{1}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=a\frac{d}{dx}(x^2)+b\frac{d}{dx}(x^{-\frac{1}{2}})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=a.2x-b\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=2ax-\frac{b}{2}x^{\frac{-1-2}{2}}\)
\(\Rightarrow \frac{dy}{dx}=2ax-\frac{b}{2}x^{-\frac{3}{2}}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(2ax-\frac{1}{2}x^{-\frac{3}{2}}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=2a\frac{d}{dx}(x)-\frac{b}{2}\frac{d}{dx}(x^{-\frac{3}{2}})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2a+\frac{b}{2}.\frac{3}{2}x^{-\frac{3}{2}-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2a+\frac{3b}{4}x^{\frac{-3-2}{2}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2a+\frac{3b}{4}x^{\frac{-5}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=4ax^2+\frac{3b}{2}x^{\frac{-5}{2}}\times{x^2}\) ➜ \(2x^2\) দ্বারা গুণ করে।
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=4ax^2+\frac{3b}{2}x^{\frac{-5}{2}+2}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=4ax^2+\frac{3b}{2}x^{\frac{-5+4}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=4ax^2+\frac{3b}{2}x^{\frac{-1}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=2ax^2-\frac{b}{2}x^{-\frac{1}{2}}+2ax^2+2bx^{-\frac{1}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=x(2ax-\frac{b}{2}x^{-\frac{3}{2}})+2(ax^2+bx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}=x\frac{dy}{dx}+2y\) ➜ \(\because \frac{dy}{dx}=2ax-\frac{b}{2}x^{-\frac{3}{2}}, y=ax^2+bx^{-\frac{1}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2y=0\)
\(\therefore 2x^2y_{2}-xy_{1}-2y=0\)
(Showed)
\((7)\) \(y=\sin{(2\cos^{-1}{x})}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}+4y=0\) অথবা, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+4y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=\sin{(2\cos^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(2\sin^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{(2\cos^{-1}{x})}.2\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(2\cos^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}=-2\cos{(2\cos^{-1}{x})}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-2\cos{(2\cos^{-1}{x})}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=2^2\cos^2{(2\cos^{-1}{x})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(1-\sin^2{(2\cos^{-1}{x})})\) ➜ \(\because \cos^2{x}=1-\sin^2{x}\)
\(\Rightarrow (1-x^2)y^2_{1}=4(1-y^2)\) ➜ \(\because y=\sin{(2\cos^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(1-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=1, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=-8yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}+8yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}+4y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}+4y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}+4y=0\)
অথবা,
\((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+4y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Proved)
\(y=\sin{(2\cos^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(2\sin^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{(2\cos^{-1}{x})}.2\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(2\cos^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}=-2\cos{(2\cos^{-1}{x})}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-2\cos{(2\cos^{-1}{x})}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=2^2\cos^2{(2\cos^{-1}{x})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(1-\sin^2{(2\cos^{-1}{x})})\) ➜ \(\because \cos^2{x}=1-\sin^2{x}\)
\(\Rightarrow (1-x^2)y^2_{1}=4(1-y^2)\) ➜ \(\because y=\sin{(2\cos^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(1-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=1, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=-8yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}+8yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}+4y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}+4y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}+4y=0\)
অথবা,
\((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+4y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Proved)
\((8)\) \(y=\sin{(m\sin{x})}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+m^2y\cos^2{x}=0\)
সমাধানঃ
দেওয়া আছে,
\(y=\sin{(m\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(m\sin{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\cos{(m\sin{x})}.m\frac{d}{dx}(\sin{x})\) ➜ \(m\sin{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=m\cos{(m\sin{x})}\cos{x}\) ➜ \(\because y_{1}=\frac{dy}{dx}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}\frac{1}{\cos{x}}=m\cos{(m\sin{x})}\)
\(\Rightarrow y_{1}\sec{x}=m\cos{(m\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y_{1}\sec{x})=\frac{d}{dx}\{\cos{(m\sin{x})}\}\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}\frac{d}{dx}(\sec{x})+\sec{x}\frac{d}{dx}(y_{1})=-m\sin{(m\sin{x})}.m\frac{d}{dx}(\sin{x})\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(\cos{x})=-\sin{x} \)
\(\Rightarrow y_{1}\sec{x}\tan{x}+\sec{x}y_{2}=-m^2\sin{(m\sin{x})}\cos{x}\) ➜\(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}, \frac{d}{dx}(\sin{x})=\cos{x} \)
\(\Rightarrow y_{1}\frac{1}{\cos{x}}\tan{x}+\frac{1}{\cos{x}}y_{2}=-m^2y\cos{x}\)
\(\Rightarrow y_{1}\tan{x}+y_{2}=-m^2y\cos{x}\times{\cos{x}}\) ➜ \(\cos{x}\) দ্বারা গুণ করে।
\(\Rightarrow y_{2}+y_{1}\tan{x}=-m^2y\cos^2{x}\)
\(\therefore \frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+m^2y\cos^2{x}=0\)
(Showed)
\(y=\sin{(m\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(m\sin{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\cos{(m\sin{x})}.m\frac{d}{dx}(\sin{x})\) ➜ \(m\sin{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=m\cos{(m\sin{x})}\cos{x}\) ➜ \(\because y_{1}=\frac{dy}{dx}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}\frac{1}{\cos{x}}=m\cos{(m\sin{x})}\)
\(\Rightarrow y_{1}\sec{x}=m\cos{(m\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y_{1}\sec{x})=\frac{d}{dx}\{\cos{(m\sin{x})}\}\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}\frac{d}{dx}(\sec{x})+\sec{x}\frac{d}{dx}(y_{1})=-m\sin{(m\sin{x})}.m\frac{d}{dx}(\sin{x})\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(\cos{x})=-\sin{x} \)
\(\Rightarrow y_{1}\sec{x}\tan{x}+\sec{x}y_{2}=-m^2\sin{(m\sin{x})}\cos{x}\) ➜\(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}, \frac{d}{dx}(\sin{x})=\cos{x} \)
\(\Rightarrow y_{1}\frac{1}{\cos{x}}\tan{x}+\frac{1}{\cos{x}}y_{2}=-m^2y\cos{x}\)
\(\Rightarrow y_{1}\tan{x}+y_{2}=-m^2y\cos{x}\times{\cos{x}}\) ➜ \(\cos{x}\) দ্বারা গুণ করে।
\(\Rightarrow y_{2}+y_{1}\tan{x}=-m^2y\cos^2{x}\)
\(\therefore \frac{d^2y}{dx^2}+\frac{dy}{dx}\tan{x}+m^2y\cos^2{x}=0\)
(Showed)
\((9)\) \(y=x^3+5x^2+10x+14\) হলে, \(\frac{dy}{dx}, \frac{d^2y}{dx^2}, \frac{d^3y}{dx^3}\) এবং \(\frac{d^4y}{dx^4}\) নির্ণয় কর।
উত্তরঃ \(3x^2+10x+10, 6x+10, 6, 0\)
উত্তরঃ \(3x^2+10x+10, 6x+10, 6, 0\)
সমাধানঃ
দেওয়া আছে,
\(y=x^3+5x^2+10x+14\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3+5x^2+10x+14)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^3)+5\frac{d}{dx}(x^2)+10\frac{d}{dx}(x)+\frac{d}{dx}(14)\) ➜ \(\because \frac{d}{dx}(u+v+w+s)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\) \(+\frac{d}{dx}(w)+\frac{d}{dx}(s)\)
\(\Rightarrow \frac{dy}{dx}=3x^2+5.2x+10.1+0\)
\(\therefore \frac{dy}{dx}=3x^2+10x+10\)
আবার,
\(\frac{dy}{dx}=3x^2+10x+10\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(3x^2+10x+10)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=3\frac{d}{dx}(x^2)+10\frac{d}{dx}(x)+\frac{d}{dx}(10)\) ➜\(\because \frac{d}{dx}(u+v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow \frac{d^2y}{dx^2}=3.2x+10.1+0\)
\(\therefore \frac{d^2y}{dx^2}=6x+10\)
আবার,
\(\frac{d^2y}{dx^2}=6x+10\)
\(\Rightarrow \frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=\frac{d}{dx}(6x+10)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^3y}{dx^3}=6\frac{d}{dx}(x)+\frac{d}{dx}(10)\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^3y}{dx^3}=6.1+0\)
\(\therefore \frac{d^3y}{dx^3}=6\)
আবার,
\(\Rightarrow \frac{d}{dx}\left(\frac{d^3y}{dx^3}\right)=\frac{d}{dx}(6)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\therefore \frac{d^4y}{dx^4}=0\)
\(y=x^3+5x^2+10x+14\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3+5x^2+10x+14)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^3)+5\frac{d}{dx}(x^2)+10\frac{d}{dx}(x)+\frac{d}{dx}(14)\) ➜ \(\because \frac{d}{dx}(u+v+w+s)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\) \(+\frac{d}{dx}(w)+\frac{d}{dx}(s)\)
\(\Rightarrow \frac{dy}{dx}=3x^2+5.2x+10.1+0\)
\(\therefore \frac{dy}{dx}=3x^2+10x+10\)
আবার,
\(\frac{dy}{dx}=3x^2+10x+10\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(3x^2+10x+10)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=3\frac{d}{dx}(x^2)+10\frac{d}{dx}(x)+\frac{d}{dx}(10)\) ➜\(\because \frac{d}{dx}(u+v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow \frac{d^2y}{dx^2}=3.2x+10.1+0\)
\(\therefore \frac{d^2y}{dx^2}=6x+10\)
আবার,
\(\frac{d^2y}{dx^2}=6x+10\)
\(\Rightarrow \frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=\frac{d}{dx}(6x+10)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^3y}{dx^3}=6\frac{d}{dx}(x)+\frac{d}{dx}(10)\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^3y}{dx^3}=6.1+0\)
\(\therefore \frac{d^3y}{dx^3}=6\)
আবার,
\(\Rightarrow \frac{d}{dx}\left(\frac{d^3y}{dx^3}\right)=\frac{d}{dx}(6)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\therefore \frac{d^4y}{dx^4}=0\)
\((10)\) \(y=ax\sin{x}\) হলে, প্রমাণ কর যে, \(x^2y_{2}-2xy_{1}+(x^2+2)y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=ax\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y)=a\frac{d}{dx}(x\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=ax\frac{d}{dx}(\sin{x})+a\sin{x}\frac{d}{dx}(x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=ax\cos{x}+a\sin{x}.1\) ➜ \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=ax\cos{x}+a\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(ax\cos{x}+a\sin{x})\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a\frac{d}{dx}(x\cos{x})+a\frac{d}{dx}(\sin{x})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=ax\frac{d}{dx}(\cos{x})+a\cos{x}\frac{d}{dx}(x)+a\cos{x}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{2}=-ax\sin{x}+a\cos{x}.1+a\cos{x}\)
\(\Rightarrow y_{2}=-ax\sin{x}+a\cos{x}+a\cos{x}\)
\(\Rightarrow y_{2}=-ax\sin{x}+2a\cos{x}\)
\(\Rightarrow x^2y_{2}=-ax^3\sin{x}+2ax^2\cos{x}\) ➜ উভয় পার্শে \(x^2\) গুণ করে।
\(\Rightarrow x^2y_{2}=-ax^3\sin{x}+2ax^2\cos{x}\)
\(\Rightarrow x^2y_{2}=2ax^2\cos{x}+2ax\sin{x}-ax^3\sin{x}-2ax\sin{x}\)
\(\Rightarrow x^2y_{2}=2x(ax\cos{x}+a\sin{x})-(x^2+2)ax\sin{x}\)
\(\Rightarrow x^2y_{2}=2xy_{1}-(x^2+2)y\)
\(\therefore x^2y_{2}-2xy_{1}+(x^2+2)y=0\)
(Proved)
\(y=ax\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y)=a\frac{d}{dx}(x\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=ax\frac{d}{dx}(\sin{x})+a\sin{x}\frac{d}{dx}(x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=ax\cos{x}+a\sin{x}.1\) ➜ \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=ax\cos{x}+a\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(ax\cos{x}+a\sin{x})\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a\frac{d}{dx}(x\cos{x})+a\frac{d}{dx}(\sin{x})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=ax\frac{d}{dx}(\cos{x})+a\cos{x}\frac{d}{dx}(x)+a\cos{x}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{2}=-ax\sin{x}+a\cos{x}.1+a\cos{x}\)
\(\Rightarrow y_{2}=-ax\sin{x}+a\cos{x}+a\cos{x}\)
\(\Rightarrow y_{2}=-ax\sin{x}+2a\cos{x}\)
\(\Rightarrow x^2y_{2}=-ax^3\sin{x}+2ax^2\cos{x}\) ➜ উভয় পার্শে \(x^2\) গুণ করে।
\(\Rightarrow x^2y_{2}=-ax^3\sin{x}+2ax^2\cos{x}\)
\(\Rightarrow x^2y_{2}=2ax^2\cos{x}+2ax\sin{x}-ax^3\sin{x}-2ax\sin{x}\)
\(\Rightarrow x^2y_{2}=2x(ax\cos{x}+a\sin{x})-(x^2+2)ax\sin{x}\)
\(\Rightarrow x^2y_{2}=2xy_{1}-(x^2+2)y\)
\(\therefore x^2y_{2}-2xy_{1}+(x^2+2)y=0\)
(Proved)
অনুশীলনী \(9.G / Q.1\)-এর অতি সংক্ষিপ্ত প্রশ্নসমুহ
\(Q.1.(i)\) \(y=x(x^2-5)\) হলে, \(\frac{d^3y}{dx^3}\) নির্ণয় কর।
উত্তরঃ \(6\)
\(Q.1.(ii)\) \(y=x^2-2+\frac{1}{x^2}\) হলে, \(\frac{d^2y}{dx^2}\) এবং \(\frac{d^3y}{dx^3}\) নির্ণয় কর।
উত্তরঃ \(2+\frac{6}{x^4}; -\frac{24}{x^5}\)
\(Q.1.(iii)\) \(y=x^2+\frac{1}{x^2}\) হলে, দেখাও যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-4y=0\)
\(Q.1.(iv)\) \(y=\sqrt{x}+\frac{1}{\sqrt{x}}\) হলে, দেখাও যে, \(2x\frac{dy}{dx}+y=2\sqrt{x}\)
[ কুঃ ২০০৮;ঢাঃ২০০৭,২০০৩;যঃ২০০৭;মাঃ২০০১ ]
\(Q.1.(v)\) \(y=ax^2+\frac{b}{\sqrt{x}}\) হলে, দেখাও যে, \(2x^2y_{2}-xy_{1}-2y=0\)
[ কুঃ ২০০৯;চঃ২০১২;ঢাঃ২০০৬ ]
\(Q.1.(vi)\) \(y=px^2+qx^{-\frac{1}{2}}\) হয়, তবে প্রমাণ কর যে, \(2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2y\)
[ যঃ ২০১২,২০০৫;চঃ২০১১;দিঃ২০১১;সিঃ২০১০,২০০৮; কুঃ২০০৬,রাঃ২০০৬,২০০০;বঃ২০০৩;মাঃ২০০৯ ]
\(Q.1.(vii)\) \(x\sqrt{1+y}+y\sqrt{1+x}=0\) হলে, দেখাও যে, \(\frac{dy}{dx}=-\frac{1}{(1+x)^2}\)
[বুয়েটঃ২০০৩-২০০৪ ]
\(Q.1.(viii)\) \(y=5x^4-3x^3+5x+2\) হলে, \(x=2\) বিন্দুতে \(y_{2}\) ও \(y_{3}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(204; 222 \)
উত্তরঃ \(6\)
\(Q.1.(ii)\) \(y=x^2-2+\frac{1}{x^2}\) হলে, \(\frac{d^2y}{dx^2}\) এবং \(\frac{d^3y}{dx^3}\) নির্ণয় কর।
উত্তরঃ \(2+\frac{6}{x^4}; -\frac{24}{x^5}\)
\(Q.1.(iii)\) \(y=x^2+\frac{1}{x^2}\) হলে, দেখাও যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-4y=0\)
\(Q.1.(iv)\) \(y=\sqrt{x}+\frac{1}{\sqrt{x}}\) হলে, দেখাও যে, \(2x\frac{dy}{dx}+y=2\sqrt{x}\)
[ কুঃ ২০০৮;ঢাঃ২০০৭,২০০৩;যঃ২০০৭;মাঃ২০০১ ]
\(Q.1.(v)\) \(y=ax^2+\frac{b}{\sqrt{x}}\) হলে, দেখাও যে, \(2x^2y_{2}-xy_{1}-2y=0\)
[ কুঃ ২০০৯;চঃ২০১২;ঢাঃ২০০৬ ]
\(Q.1.(vi)\) \(y=px^2+qx^{-\frac{1}{2}}\) হয়, তবে প্রমাণ কর যে, \(2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2y\)
[ যঃ ২০১২,২০০৫;চঃ২০১১;দিঃ২০১১;সিঃ২০১০,২০০৮; কুঃ২০০৬,রাঃ২০০৬,২০০০;বঃ২০০৩;মাঃ২০০৯ ]
\(Q.1.(vii)\) \(x\sqrt{1+y}+y\sqrt{1+x}=0\) হলে, দেখাও যে, \(\frac{dy}{dx}=-\frac{1}{(1+x)^2}\)
[বুয়েটঃ২০০৩-২০০৪ ]
\(Q.1.(viii)\) \(y=5x^4-3x^3+5x+2\) হলে, \(x=2\) বিন্দুতে \(y_{2}\) ও \(y_{3}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(204; 222 \)
\(Q.1.(ix)\) \(y=px+\frac{q}{x}\) হলে, দেখাও যে, \(x\frac{d^2y}{dx^2}+2\frac{dy}{dx}=2p\)
[ ঢাঃ২০০৯;যঃ২০০৯;চঃ২০০৫]
\(Q.1.(x)\) \(y=\sqrt{(1-x)(1+x)}\) হলে, দেখাও যে, \((1-x^2)\frac{dy}{dx}+xy=0\)
[ যঃ২০০৪ ]
\(Q.1.(xi)\) \(y=(x^2-1)^n\) হয়, তবে প্রমাণ কর যে, \((x^2-1)y_{2}-2(n-1)xy_{1}-2ny=0\)
\(Q.1.(xii)\) \(y=\frac{x^2}{1-x}\) হয়, তবে প্রমাণ কর যে, \((1-x)y_{2}-2y_{1}=2\)
\(Q.1.(xiii)\) \(y=x^{\frac{2}{3}}+x^{-\frac{2}{3}}\) হয়, তবে প্রমাণ কর যে, \(3x\frac{dy}{dx}+2y=4x^{\frac{2}{3}}\)
\(Q.1.(xiv)\) \(y=4x^{\frac{3}{2}}-3+2x^{\frac{1}{2}}\) হলে, \(y_{2}\) নির্ণয় কর এবং \(x=4\) হলে, \(y_{2}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(3x^{-\frac{1}{2}}-\frac{1}{2}x^{\frac{-3}{2}}; \frac{23}{16}\)
\(Q.1.(xv)\) \(y=\frac{x}{x+2}\) হলে, প্রমাণ কর যে, \(xy_{1}=y(1-y)\)
\(Q.1.(xvi)\) \(y=ax^{n+1}+bx^{-n}\) হলে, দেখাও যে, \(x^2y_{2}=n(n+1)y\)
\(Q.1.(xvii)\) \(y=\sqrt{ax^2+bx+c}\) হলে, দেখাও যে, \(4y^3y_{2}=4ac-b^2\)
[ ঢাঃ২০০৯;যঃ২০০৯;চঃ২০০৫]
\(Q.1.(x)\) \(y=\sqrt{(1-x)(1+x)}\) হলে, দেখাও যে, \((1-x^2)\frac{dy}{dx}+xy=0\)
[ যঃ২০০৪ ]
\(Q.1.(xi)\) \(y=(x^2-1)^n\) হয়, তবে প্রমাণ কর যে, \((x^2-1)y_{2}-2(n-1)xy_{1}-2ny=0\)
\(Q.1.(xii)\) \(y=\frac{x^2}{1-x}\) হয়, তবে প্রমাণ কর যে, \((1-x)y_{2}-2y_{1}=2\)
\(Q.1.(xiii)\) \(y=x^{\frac{2}{3}}+x^{-\frac{2}{3}}\) হয়, তবে প্রমাণ কর যে, \(3x\frac{dy}{dx}+2y=4x^{\frac{2}{3}}\)
\(Q.1.(xiv)\) \(y=4x^{\frac{3}{2}}-3+2x^{\frac{1}{2}}\) হলে, \(y_{2}\) নির্ণয় কর এবং \(x=4\) হলে, \(y_{2}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(3x^{-\frac{1}{2}}-\frac{1}{2}x^{\frac{-3}{2}}; \frac{23}{16}\)
\(Q.1.(xv)\) \(y=\frac{x}{x+2}\) হলে, প্রমাণ কর যে, \(xy_{1}=y(1-y)\)
\(Q.1.(xvi)\) \(y=ax^{n+1}+bx^{-n}\) হলে, দেখাও যে, \(x^2y_{2}=n(n+1)y\)
\(Q.1.(xvii)\) \(y=\sqrt{ax^2+bx+c}\) হলে, দেখাও যে, \(4y^3y_{2}=4ac-b^2\)
\(Q.1.(i)\) \(y=x(x^2-5)\) হলে, \(\frac{d^3y}{dx^3}\) নির্ণয় কর।
উত্তরঃ \(6\)
উত্তরঃ \(6\)
সমাধানঃ
দেওয়া আছে,
\(y=x(x^2-5)\)
\(\Rightarrow y=x^3-5x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3-5x)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^3)-\frac{d}{dx}(5x)\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=3x^2-5.1\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=3x^2-5\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(3x^2-5)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=3\frac{d}{dx}(x^2)-\frac{d}{dx}(5)\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=3.2x-0\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{d^2y}{dx^2}=6x\)
\(\Rightarrow \frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=6\frac{d}{dx}(x)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^3y}{dx^3}=6.1\) ➜ \(\because \frac{d}{dx}(x)=1\)
\(\therefore \frac{d^3y}{dx^3}=6\)
\(y=x(x^2-5)\)
\(\Rightarrow y=x^3-5x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3-5x)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^3)-\frac{d}{dx}(5x)\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=3x^2-5.1\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=3x^2-5\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(3x^2-5)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=3\frac{d}{dx}(x^2)-\frac{d}{dx}(5)\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=3.2x-0\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{d^2y}{dx^2}=6x\)
\(\Rightarrow \frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=6\frac{d}{dx}(x)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^3y}{dx^3}=6.1\) ➜ \(\because \frac{d}{dx}(x)=1\)
\(\therefore \frac{d^3y}{dx^3}=6\)
\(Q.1.(ii)\) \(y=x^2-2+\frac{1}{x^2}\) হলে, \(\frac{d^2y}{dx^2}\) এবং \(\frac{d^3y}{dx^3}\) নির্ণয় কর।
উত্তরঃ \(2+\frac{6}{x^4}; -\frac{24}{x^5}\)
উত্তরঃ \(2+\frac{6}{x^4}; -\frac{24}{x^5}\)
সমাধানঃ
দেওয়া আছে,
\(y=x^2-2+\frac{1}{x^2}\)
\(\Rightarrow y=x^2-2+x^{-2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2-2+x^{-2})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^2)-\frac{d}{dx}(2)+\frac{d}{dx}(x^{-2})\) ➜ \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow \frac{dy}{dx}=2x-0-2x^{-2-1}\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-3}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{dy}{dx}(2x-2x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=2\frac{dy}{dx}(x)-2\frac{dy}{dx}(x^{-3})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2.1-2(-3)x^{-3-1}\) ➜ \(\because \frac{d}{dx}(x)=1\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2+6x^{-4}=2+\frac{6}{x^4}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=\frac{d}{dx}(2+6x^{-4})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^3y}{dx^3}=\frac{d}{dx}(2)+6\frac{d}{dx}(x^{-4})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^3y}{dx^3}=0+6(-4)x^{-4-1}\) ➜ \(\because \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)\)
\(\Rightarrow \frac{d^3y}{dx^3}=-24x^{-5}\)
\(\therefore \frac{d^3y}{dx^3}=-\frac{24}{x^{5}}\)
\(y=x^2-2+\frac{1}{x^2}\)
\(\Rightarrow y=x^2-2+x^{-2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2-2+x^{-2})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^2)-\frac{d}{dx}(2)+\frac{d}{dx}(x^{-2})\) ➜ \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow \frac{dy}{dx}=2x-0-2x^{-2-1}\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-3}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{dy}{dx}(2x-2x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=2\frac{dy}{dx}(x)-2\frac{dy}{dx}(x^{-3})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2.1-2(-3)x^{-3-1}\) ➜ \(\because \frac{d}{dx}(x)=1\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2+6x^{-4}=2+\frac{6}{x^4}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=\frac{d}{dx}(2+6x^{-4})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^3y}{dx^3}=\frac{d}{dx}(2)+6\frac{d}{dx}(x^{-4})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^3y}{dx^3}=0+6(-4)x^{-4-1}\) ➜ \(\because \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)\)
\(\Rightarrow \frac{d^3y}{dx^3}=-24x^{-5}\)
\(\therefore \frac{d^3y}{dx^3}=-\frac{24}{x^{5}}\)
\(Q.1.(iii)\) \(y=x^2+\frac{1}{x^2}\) হলে, দেখাও যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-4y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=x^2+\frac{1}{x^2}\)
\(\Rightarrow y=x^2+x^{-2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2+x^{-2})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^2)+\frac{d}{dx}(x^{-2})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-2-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-3}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(2x-2x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dx}(2x)-2\frac{d}{dx}(x^{-3})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2+6x^{-3-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-4}\times{x^2}\) ➜ \(x^2\) দ্বারা গুণ করে।
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-4+2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-2x^2+2x^{-2}+4x^2+4x^{-2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-x(2x-2x^{-3})+4(x^2+x^{-2})\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-x\frac{dy}{dx}+4y\) ➜ \(\because \frac{dy}{dx}=2x-2x^{-3}, y=x^2+x^{-2}\)
\(\therefore x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-4y=0\)
(Showed)
\(y=x^2+\frac{1}{x^2}\)
\(\Rightarrow y=x^2+x^{-2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2+x^{-2})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^2)+\frac{d}{dx}(x^{-2})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-2-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=2x-2x^{-3}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(2x-2x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dx}(2x)-2\frac{d}{dx}(x^{-3})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2+6x^{-3-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-4}\times{x^2}\) ➜ \(x^2\) দ্বারা গুণ করে।
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-4+2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=2x^2+6x^{-2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-2x^2+2x^{-2}+4x^2+4x^{-2}\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-x(2x-2x^{-3})+4(x^2+x^{-2})\)
\(\Rightarrow x^2\frac{d^2y}{dx^2}=-x\frac{dy}{dx}+4y\) ➜ \(\because \frac{dy}{dx}=2x-2x^{-3}, y=x^2+x^{-2}\)
\(\therefore x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-4y=0\)
(Showed)
\(Q.1.(iv)\) \(y=\sqrt{x}+\frac{1}{\sqrt{x}}\) হলে, দেখাও যে, \(2x\frac{dy}{dx}+y=2\sqrt{x}\)
[ কুঃ ২০০৮;ঢাঃ২০০৭,২০০৩;যঃ২০০৭;মাঃ২০০১ ]
[ কুঃ ২০০৮;ঢাঃ২০০৭,২০০৩;যঃ২০০৭;মাঃ২০০১ ]
সমাধানঃ
দেওয়া আছে,
\(y=\sqrt{x}+\frac{1}{\sqrt{x}}\)
\(\Rightarrow y=\sqrt{x}+\frac{1}{x^{\frac{1}{2}}}\)
\(\Rightarrow y=\sqrt{x}+x^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt{x}+x^{-\frac{1}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\sqrt{x})+\frac{d}{dx}(x^{-\frac{1}{2}})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{\frac{-3}{2}}\)
\(\Rightarrow 2x\frac{dy}{dx}=\frac{2x}{2\sqrt{x}}-\frac{1}{2}x^{\frac{-3}{2}}\times{2x}\) ➜ উভয় পার্শে \(2x\) গুন করে।
\(\Rightarrow 2x\frac{dy}{dx}=\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}}-x^{\frac{-3}{2}+1}\)
\(\Rightarrow 2x\frac{dy}{dx}=\sqrt{x}-x^{\frac{-3+2}{2}}\)
\(\Rightarrow 2x\frac{dy}{dx}=\sqrt{x}-x^{\frac{-1}{2}}\)
\(\Rightarrow 2x\frac{dy}{dx}=\sqrt{x}-\frac{1}{x^{\frac{1}{2}}}\)
\(\Rightarrow 2x\frac{dy}{dx}=\sqrt{x}-\frac{1}{\sqrt{x}}\)
\(\Rightarrow 2x\frac{dy}{dx}=2\sqrt{x}-\sqrt{x}-\frac{1}{\sqrt{x}}\)
\(\Rightarrow 2x\frac{dy}{dx}=2\sqrt{x}-\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\)
\(\Rightarrow 2x\frac{dy}{dx}=2\sqrt{x}-y\) ➜ \(\because y=\sqrt{x}+\frac{1}{\sqrt{x}}\)
\(\therefore 2x\frac{dy}{dx}+y=2\sqrt{x}\)
(Showed)
\(y=\sqrt{x}+\frac{1}{\sqrt{x}}\)
\(\Rightarrow y=\sqrt{x}+\frac{1}{x^{\frac{1}{2}}}\)
\(\Rightarrow y=\sqrt{x}+x^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt{x}+x^{-\frac{1}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\sqrt{x})+\frac{d}{dx}(x^{-\frac{1}{2}})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{\frac{-3}{2}}\)
\(\Rightarrow 2x\frac{dy}{dx}=\frac{2x}{2\sqrt{x}}-\frac{1}{2}x^{\frac{-3}{2}}\times{2x}\) ➜ উভয় পার্শে \(2x\) গুন করে।
\(\Rightarrow 2x\frac{dy}{dx}=\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}}-x^{\frac{-3}{2}+1}\)
\(\Rightarrow 2x\frac{dy}{dx}=\sqrt{x}-x^{\frac{-3+2}{2}}\)
\(\Rightarrow 2x\frac{dy}{dx}=\sqrt{x}-x^{\frac{-1}{2}}\)
\(\Rightarrow 2x\frac{dy}{dx}=\sqrt{x}-\frac{1}{x^{\frac{1}{2}}}\)
\(\Rightarrow 2x\frac{dy}{dx}=\sqrt{x}-\frac{1}{\sqrt{x}}\)
\(\Rightarrow 2x\frac{dy}{dx}=2\sqrt{x}-\sqrt{x}-\frac{1}{\sqrt{x}}\)
\(\Rightarrow 2x\frac{dy}{dx}=2\sqrt{x}-\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\)
\(\Rightarrow 2x\frac{dy}{dx}=2\sqrt{x}-y\) ➜ \(\because y=\sqrt{x}+\frac{1}{\sqrt{x}}\)
\(\therefore 2x\frac{dy}{dx}+y=2\sqrt{x}\)
(Showed)
\(Q.1.(v)\) \(y=ax^2+\frac{b}{\sqrt{x}}\) হলে, দেখাও যে, \(2x^2y_{2}-xy_{1}-2y=0\)
[ কুঃ ২০০৯;চঃ২০১২;ঢাঃ২০০৬ ]
[ কুঃ ২০০৯;চঃ২০১২;ঢাঃ২০০৬ ]
সমাধানঃ
দেওয়া আছে,
\(y=ax^2+\frac{b}{\sqrt{x}}\)
\(\Rightarrow y=ax^2+\frac{b}{x^{\frac{1}{2}}}\)
\(\Rightarrow y=ax^2+bx^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(ax^2+bx^{-\frac{1}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=a\frac{d}{dx}(x^2)+b\frac{d}{dx}(x^{-\frac{1}{2}})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=a.2x-b\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{dy}{dx}=y_{1}\)
\(\Rightarrow y_{1}=2ax-b\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(\Rightarrow y_{1}=2ax-\frac{b}{2}x^{\frac{-3}{2}}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(2ax-b\frac{1}{2}x^{\frac{-3}{2}}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=2a\frac{d}{dx}(x)-b\frac{1}{2}\frac{d}{dx}\left(x^{\frac{-3}{2}}\right)\) ➜\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=2a.1-b\frac{1}{2}.\frac{-3}{2}x^{\frac{-3}{2}-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-3-2}{2}}\times{x^2}\) ➜ উভয় পার্শে \(2x^2\) গুন করে।
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-5}{2}}\times{x^2}\) ➜ উভয় পার্শে \(2x^2\) গুন করে।
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-5}{2}+2}\)
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-5+4}{2}}\)
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-1}{2}}\)
\(\Rightarrow 2x^2y_{2}=2ax^2-\frac{b}{2}x^{\frac{-1}{2}}+2ax^2+2bx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2y_{2}=x\left(2ax-\frac{b}{2}x^{\frac{-3}{2}}\right)+2(ax^2+bx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2y_{2}=xy_{1}+2y\) ➜ \(\because y_{1}=2ax-\frac{b}{2}x^{\frac{-3}{2}}, y=ax^2+bx^{-\frac{1}{2}}\)
\(\therefore 2x^2y_{2}-xy_{1}-2y=0\)
(Showed)
\(y=ax^2+\frac{b}{\sqrt{x}}\)
\(\Rightarrow y=ax^2+\frac{b}{x^{\frac{1}{2}}}\)
\(\Rightarrow y=ax^2+bx^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(ax^2+bx^{-\frac{1}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=a\frac{d}{dx}(x^2)+b\frac{d}{dx}(x^{-\frac{1}{2}})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=a.2x-b\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{dy}{dx}=y_{1}\)
\(\Rightarrow y_{1}=2ax-b\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(\Rightarrow y_{1}=2ax-\frac{b}{2}x^{\frac{-3}{2}}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(2ax-b\frac{1}{2}x^{\frac{-3}{2}}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=2a\frac{d}{dx}(x)-b\frac{1}{2}\frac{d}{dx}\left(x^{\frac{-3}{2}}\right)\) ➜\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=2a.1-b\frac{1}{2}.\frac{-3}{2}x^{\frac{-3}{2}-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-3-2}{2}}\times{x^2}\) ➜ উভয় পার্শে \(2x^2\) গুন করে।
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-5}{2}}\times{x^2}\) ➜ উভয় পার্শে \(2x^2\) গুন করে।
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-5}{2}+2}\)
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-5+4}{2}}\)
\(\Rightarrow 2x^2y_{2}=4ax^2+\frac{3b}{2}x^{\frac{-1}{2}}\)
\(\Rightarrow 2x^2y_{2}=2ax^2-\frac{b}{2}x^{\frac{-1}{2}}+2ax^2+2bx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2y_{2}=x\left(2ax-\frac{b}{2}x^{\frac{-3}{2}}\right)+2(ax^2+bx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2y_{2}=xy_{1}+2y\) ➜ \(\because y_{1}=2ax-\frac{b}{2}x^{\frac{-3}{2}}, y=ax^2+bx^{-\frac{1}{2}}\)
\(\therefore 2x^2y_{2}-xy_{1}-2y=0\)
(Showed)
\(Q.1.(vi)\) \(y=px^2+qx^{-\frac{1}{2}}\) হয়, তবে প্রমাণ কর যে, \(2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2y\)
[ যঃ ২০১২,২০০৫;চঃ২০১১;দিঃ২০১১;সিঃ২০১০,২০০৮;কুঃ২০০৬,রাঃ২০০৬,২০০০;বঃ২০০৩;মাঃ২০০৯ ]
[ যঃ ২০১২,২০০৫;চঃ২০১১;দিঃ২০১১;সিঃ২০১০,২০০৮;কুঃ২০০৬,রাঃ২০০৬,২০০০;বঃ২০০৩;মাঃ২০০৯ ]
সমাধানঃ
দেওয়া আছে,
\(y=px^2+qx^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(px^2+qx^{-\frac{1}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=p\frac{d}{dx}(x^2)+q\frac{d}{dx}(x^{-\frac{1}{2}})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=p.2x-q\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{dy}{dx}=y_{1}\)
\(\Rightarrow y_{1}=2px-q\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(\Rightarrow y_{1}=2px-\frac{q}{2}x^{\frac{-3}{2}}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(2ax-b\frac{1}{2}x^{\frac{-3}{2}}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=2p\frac{d}{dx}(x)-q\frac{1}{2}\frac{d}{dx}\left(x^{\frac{-3}{2}}\right)\) ➜\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=2p.1-q\frac{1}{2}.\frac{-3}{2}x^{\frac{-3}{2}-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-3-2}{2}}\times{x^2}\) ➜ উভয় পার্শে \(2x^2\) গুন করে।
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-5}{2}}\times{x^2}\) ➜ উভয় পার্শে \(2x^2\) গুন করে।
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-5}{2}+2}\)
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-5+4}{2}}\)
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-1}{2}}\)
\(\Rightarrow 2x^2y_{2}=2px^2-\frac{q}{2}x^{\frac{-1}{2}}+2px^2+2qx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2y_{2}=x\left(2px-\frac{q}{2}x^{\frac{-3}{2}}\right)+2(px^2+qx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2y_{2}=xy_{1}+2y\) ➜ \(\because y_{1}=2px-\frac{q}{2}x^{\frac{-3}{2}}, y=px^2+qx^{-\frac{1}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2y\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
\(\therefore 2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2y\)
(Showed)
\(y=px^2+qx^{-\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(px^2+qx^{-\frac{1}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=p\frac{d}{dx}(x^2)+q\frac{d}{dx}(x^{-\frac{1}{2}})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=p.2x-q\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{dy}{dx}=y_{1}\)
\(\Rightarrow y_{1}=2px-q\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(\Rightarrow y_{1}=2px-\frac{q}{2}x^{\frac{-3}{2}}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(2ax-b\frac{1}{2}x^{\frac{-3}{2}}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=2p\frac{d}{dx}(x)-q\frac{1}{2}\frac{d}{dx}\left(x^{\frac{-3}{2}}\right)\) ➜\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=2p.1-q\frac{1}{2}.\frac{-3}{2}x^{\frac{-3}{2}-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-3-2}{2}}\times{x^2}\) ➜ উভয় পার্শে \(2x^2\) গুন করে।
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-5}{2}}\times{x^2}\) ➜ উভয় পার্শে \(2x^2\) গুন করে।
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-5}{2}+2}\)
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-5+4}{2}}\)
\(\Rightarrow 2x^2y_{2}=4px^2+\frac{3q}{2}x^{\frac{-1}{2}}\)
\(\Rightarrow 2x^2y_{2}=2px^2-\frac{q}{2}x^{\frac{-1}{2}}+2px^2+2qx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2y_{2}=x\left(2px-\frac{q}{2}x^{\frac{-3}{2}}\right)+2(px^2+qx^{-\frac{1}{2}})\)
\(\Rightarrow 2x^2y_{2}=xy_{1}+2y\) ➜ \(\because y_{1}=2px-\frac{q}{2}x^{\frac{-3}{2}}, y=px^2+qx^{-\frac{1}{2}}\)
\(\Rightarrow 2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2y\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
\(\therefore 2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2y\)
(Showed)
\(Q.1.(vii)\) \(x\sqrt{1+y}+y\sqrt{1+x}=0\) হলে, দেখাও যে, \(\frac{dy}{dx}=-\frac{1}{(1+x)^2}\)
[বুয়েটঃ২০০৩-২০০৪ ]
[বুয়েটঃ২০০৩-২০০৪ ]
সমাধানঃ
দেওয়া আছে,
\(x\sqrt{(1+y)}+y\sqrt{(1+x)}=0\)
\(\Rightarrow x\sqrt{(1+y)}=-y\sqrt{(1+x)}\)
\(\Rightarrow x^2(1+y)=y^2(1+x)\)
\(\Rightarrow x^2+x^2y=y^2+xy^2\)
\(\Rightarrow x^2+x^2y-y^2-xy^2=0\)
\(\Rightarrow x^2-y^2+x^2y-xy^2=0\)
\(\Rightarrow (x+y)(x-y)+xy(x-y)=0\)
\(\Rightarrow (x-y)(x+y+xy)=0\)
\(\Rightarrow (x-y)\ne{0}, (x+y+xy)=0\)
\(\Rightarrow x+y+xy=0\)
\(\Rightarrow y(1+x)=-x\)
\(\Rightarrow y=-\frac{x}{1+x}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{d}{dx}(\frac{x}{1+x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=-\frac{(1+x)\frac{d}{dx}(x)-x\frac{d}{dx}(1+x)}{(1+x)^2}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{(1+x).1-x(0+1)}{(1+x)^2}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1+x-x}{(1+x)^2}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{(1+x)^2}\)
(Showed)
\(x\sqrt{(1+y)}+y\sqrt{(1+x)}=0\)
\(\Rightarrow x\sqrt{(1+y)}=-y\sqrt{(1+x)}\)
\(\Rightarrow x^2(1+y)=y^2(1+x)\)
\(\Rightarrow x^2+x^2y=y^2+xy^2\)
\(\Rightarrow x^2+x^2y-y^2-xy^2=0\)
\(\Rightarrow x^2-y^2+x^2y-xy^2=0\)
\(\Rightarrow (x+y)(x-y)+xy(x-y)=0\)
\(\Rightarrow (x-y)(x+y+xy)=0\)
\(\Rightarrow (x-y)\ne{0}, (x+y+xy)=0\)
\(\Rightarrow x+y+xy=0\)
\(\Rightarrow y(1+x)=-x\)
\(\Rightarrow y=-\frac{x}{1+x}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{d}{dx}(\frac{x}{1+x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=-\frac{(1+x)\frac{d}{dx}(x)-x\frac{d}{dx}(1+x)}{(1+x)^2}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{(1+x).1-x(0+1)}{(1+x)^2}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1+x-x}{(1+x)^2}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{(1+x)^2}\)
(Showed)
\(Q.1.(viii)\) \(y=5x^4-3x^3+5x+2\) হলে, \(x=2\) বিন্দুতে \(y_{2}\) ও \(y_{3}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(204; 222 \)
উত্তরঃ \(204; 222 \)
সমাধানঃ
দেওয়া আছে,
\(y=5x^4-3x^3+5x+2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(5x^4-3x^3+5x+2)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=5\frac{d}{dx}(x^4)-3\frac{d}{dx}(x^3)+5\frac{d}{dx}(x)+\frac{d}{dx}(2)\) ➜ \(\because \frac{d}{dx}(u-v+w+s)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)+\frac{d}{dx}(s)\)
\(\Rightarrow y_{1}=5.4x^3-3.3x^2+5.1+.0\) ➜ \(\because \frac{d}{dx}(x^4)=4x^3,\frac{d}{dx}(x^3)=3x^2\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{1}=20x^3-9x^2+5\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(20x^3-9x^2+5)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=20\frac{d}{dx}(x^3)-9\frac{d}{dx}(x^2)+\frac{d}{dx}(5)\) ➜ \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow y_{2}=20.3x^2-9.2x+0\) ➜ \(\because \frac{d}{dx}(x^3)=3x^2\), \(\frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{2}=60x^2-18x\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}(60x^2-18x)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=60\frac{d}{dx}(x^2)-18\frac{d}{dx}(x)\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{3}=60.2x-18.1\) ➜\(\because \frac{d}{dx}(x^2)=2x\), \(\frac{d}{dx}(x)=1\)
\(\Rightarrow y_{3}=120x-18\)
এখন,
\(x=2\) বিন্দুতে
\(y_{2}=60.2^2-18.2\)
\(=60.4-36\)
\(=240-36\)
\(=204\)
আবার,
\(x=2\) বিন্দুতে
\(y_{3}=120.2-18\)
\(=240-18\)
\(=222\)
\(y=5x^4-3x^3+5x+2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(5x^4-3x^3+5x+2)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=5\frac{d}{dx}(x^4)-3\frac{d}{dx}(x^3)+5\frac{d}{dx}(x)+\frac{d}{dx}(2)\) ➜ \(\because \frac{d}{dx}(u-v+w+s)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)+\frac{d}{dx}(s)\)
\(\Rightarrow y_{1}=5.4x^3-3.3x^2+5.1+.0\) ➜ \(\because \frac{d}{dx}(x^4)=4x^3,\frac{d}{dx}(x^3)=3x^2\), \(\frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{1}=20x^3-9x^2+5\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(20x^3-9x^2+5)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=20\frac{d}{dx}(x^3)-9\frac{d}{dx}(x^2)+\frac{d}{dx}(5)\) ➜ \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow y_{2}=20.3x^2-9.2x+0\) ➜ \(\because \frac{d}{dx}(x^3)=3x^2\), \(\frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{2}=60x^2-18x\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}(60x^2-18x)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=60\frac{d}{dx}(x^2)-18\frac{d}{dx}(x)\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{3}=60.2x-18.1\) ➜\(\because \frac{d}{dx}(x^2)=2x\), \(\frac{d}{dx}(x)=1\)
\(\Rightarrow y_{3}=120x-18\)
এখন,
\(x=2\) বিন্দুতে
\(y_{2}=60.2^2-18.2\)
\(=60.4-36\)
\(=240-36\)
\(=204\)
আবার,
\(x=2\) বিন্দুতে
\(y_{3}=120.2-18\)
\(=240-18\)
\(=222\)
\(Q.1.(ix)\) \(y=px+\frac{q}{x}\) হলে, দেখাও যে, \(x\frac{d^2y}{dx^2}+2\frac{dy}{dx}=2p\)
[ ঢাঃ২০০৯;যঃ২০০৯;চঃ২০০৫]
[ ঢাঃ২০০৯;যঃ২০০৯;চঃ২০০৫]
সমাধানঃ
দেওয়া আছে,
\(y=px+\frac{q}{x}\)
\(\Rightarrow y=px+qx^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(px+qx^{-1})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=p\frac{d}{dx}(x)+q\frac{d}{dx}(x^{-1})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=p.1-q.x^{-1-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{1}=p-qx^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(p-qx^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\frac{d}{dx}(p)-q\frac{d}{dx}(x^{-2})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=0-q.(-2)x^{-2-1}\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{2}=2qx^{-3}\)
\(\Rightarrow xy_{2}=2qx^{-3}\times{x}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow xy_{2}=2qx^{-3+1}\)
\(\Rightarrow xy_{2}=2qx^{-2}\)
\(\Rightarrow xy_{2}=2p-2p+2qx^{-2}\)
\(\Rightarrow xy_{2}=2p-2(p-qx^{-2})\)
\(\Rightarrow xy_{2}=2p-2y_{1}\) ➜ \(\because y_{1}=p-qx^{-2}\)
\(\Rightarrow xy_{2}+2y_{1}=2p\)
\(\Rightarrow x\frac{d^2y}{dx^2}+2\frac{dy}{dx}=2p\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\(y=px+\frac{q}{x}\)
\(\Rightarrow y=px+qx^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(px+qx^{-1})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=p\frac{d}{dx}(x)+q\frac{d}{dx}(x^{-1})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=p.1-q.x^{-1-1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{1}=p-qx^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(p-qx^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\frac{d}{dx}(p)-q\frac{d}{dx}(x^{-2})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=0-q.(-2)x^{-2-1}\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{2}=2qx^{-3}\)
\(\Rightarrow xy_{2}=2qx^{-3}\times{x}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow xy_{2}=2qx^{-3+1}\)
\(\Rightarrow xy_{2}=2qx^{-2}\)
\(\Rightarrow xy_{2}=2p-2p+2qx^{-2}\)
\(\Rightarrow xy_{2}=2p-2(p-qx^{-2})\)
\(\Rightarrow xy_{2}=2p-2y_{1}\) ➜ \(\because y_{1}=p-qx^{-2}\)
\(\Rightarrow xy_{2}+2y_{1}=2p\)
\(\Rightarrow x\frac{d^2y}{dx^2}+2\frac{dy}{dx}=2p\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\(Q.1.(x)\) \(y=\sqrt{(1-x)(1+x)}\) হলে, দেখাও যে, \((1-x^2)\frac{dy}{dx}+xy=0\)
[ যঃ২০০৪ ]
[ যঃ২০০৪ ]
সমাধানঃ
দেওয়া আছে,
\(y=\sqrt{(1-x)(1+x)}\)
\(\Rightarrow y=\sqrt{(1-x^2)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sqrt{(1-x^2)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{(1-x^2)}}\frac{d}{dx}(1-x^2)\) ➜ \((1-x^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{(1-x^2)}}(0-2x)\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{dy}{dx}=-\frac{x}{\sqrt{(1-x^2)}}\)
\(\Rightarrow (1-x^2)\frac{dy}{dx}=-\frac{x(1-x^2)}{\sqrt{(1-x^2)}}\) ➜ উভয় পার্শে \((1-x^2)\)গুণ করে।
\(\Rightarrow (1-x^2)\frac{dy}{dx}=-\frac{x.\sqrt{(1-x^2)}.\sqrt{(1-x^2)}}{\sqrt{(1-x^2)}}\)
\(\Rightarrow (1-x^2)\frac{dy}{dx}=-x.\sqrt{(1-x^2)}\)
\(\Rightarrow (1-x^2)\frac{dy}{dx}=-xy\) ➜ \(\because y=\sqrt{(1-x^2)}\)
\(\Rightarrow (1-x^2)\frac{dy}{dx}+xy=0\)
(Showed)
\(y=\sqrt{(1-x)(1+x)}\)
\(\Rightarrow y=\sqrt{(1-x^2)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sqrt{(1-x^2)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{(1-x^2)}}\frac{d}{dx}(1-x^2)\) ➜ \((1-x^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{(1-x^2)}}(0-2x)\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{dy}{dx}=-\frac{x}{\sqrt{(1-x^2)}}\)
\(\Rightarrow (1-x^2)\frac{dy}{dx}=-\frac{x(1-x^2)}{\sqrt{(1-x^2)}}\) ➜ উভয় পার্শে \((1-x^2)\)গুণ করে।
\(\Rightarrow (1-x^2)\frac{dy}{dx}=-\frac{x.\sqrt{(1-x^2)}.\sqrt{(1-x^2)}}{\sqrt{(1-x^2)}}\)
\(\Rightarrow (1-x^2)\frac{dy}{dx}=-x.\sqrt{(1-x^2)}\)
\(\Rightarrow (1-x^2)\frac{dy}{dx}=-xy\) ➜ \(\because y=\sqrt{(1-x^2)}\)
\(\Rightarrow (1-x^2)\frac{dy}{dx}+xy=0\)
(Showed)
\(Q.1.(xi)\) \(y=(x^2-1)^n\) হয়, তবে প্রমাণ কর যে, \((x^2-1)y_{2}-2(n-1)xy_{1}-2ny=0\)
সমাধানঃ
দেওয়া আছে,
\(y=(x^2-1)^n\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x^2-1)^n\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=n(x^2-1)^{n-1}\frac{d}{dx}(x^2-1)\) ➜ \((x^2-1)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{1}=n(x^2-1)^{n-1}(2x-0)\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{1}=2nx(x^2-1)^{n-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=2n\frac{d}{dx}\{x(x^2-1)^{n-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=2n\{x\frac{d}{dx}(x^2-1)^{n-1}+(x^2-1)^{n-1}\frac{d}{dx}(x)\}\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=2n\{(n-1)x(x^2-1)^{n-1-1}\frac{d}{dx}(x^2-1)+(x^2-1)^{n-1}.1\}\) ➜\(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{2}=2n(n-1)x(x^2-1)^{n-2}(2x-0)+2n(x^2-1)^{n-1}\)
\(\Rightarrow y_{2}=4n(n-1)x^2(x^2-1)^{n-2}+2n(x^2-1)^{n-1}\)
\(\Rightarrow (x^2-1)y_{2}=4n(n-1)x^2(x^2-1)^{n-2}\times{(x^2-1)}+2n(x^2-1)^{n-1}\times{(x^2-1)}\) ➜ উভয় পার্শে \((x^2-1)\)গুণ করে।
\(\Rightarrow (x^2-1)y_{2}=4n(n-1)x^2(x^2-1)^{n-2+1}+2n(x^2-1)^{n-1+1}\)
\(\Rightarrow (x^2-1)y_{2}=4n(n-1)x^2(x^2-1)^{n-1}+2n(x^2-1)^{n}\)
\(\Rightarrow (x^2-1)y_{2}=2(n-1)xy_{1}+2ny\) ➜ \(\because y_{1}=2nx(x^2-1)^{n-1}, y=(x^2-1)^{n}\)
\(\Rightarrow (x^2-1)y_{2}-2(n-1)xy_{1}-2ny=0\)
(Proved)
\(y=(x^2-1)^n\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x^2-1)^n\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=n(x^2-1)^{n-1}\frac{d}{dx}(x^2-1)\) ➜ \((x^2-1)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{1}=n(x^2-1)^{n-1}(2x-0)\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{1}=2nx(x^2-1)^{n-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=2n\frac{d}{dx}\{x(x^2-1)^{n-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=2n\{x\frac{d}{dx}(x^2-1)^{n-1}+(x^2-1)^{n-1}\frac{d}{dx}(x)\}\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=2n\{(n-1)x(x^2-1)^{n-1-1}\frac{d}{dx}(x^2-1)+(x^2-1)^{n-1}.1\}\) ➜\(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{2}=2n(n-1)x(x^2-1)^{n-2}(2x-0)+2n(x^2-1)^{n-1}\)
\(\Rightarrow y_{2}=4n(n-1)x^2(x^2-1)^{n-2}+2n(x^2-1)^{n-1}\)
\(\Rightarrow (x^2-1)y_{2}=4n(n-1)x^2(x^2-1)^{n-2}\times{(x^2-1)}+2n(x^2-1)^{n-1}\times{(x^2-1)}\) ➜ উভয় পার্শে \((x^2-1)\)গুণ করে।
\(\Rightarrow (x^2-1)y_{2}=4n(n-1)x^2(x^2-1)^{n-2+1}+2n(x^2-1)^{n-1+1}\)
\(\Rightarrow (x^2-1)y_{2}=4n(n-1)x^2(x^2-1)^{n-1}+2n(x^2-1)^{n}\)
\(\Rightarrow (x^2-1)y_{2}=2(n-1)xy_{1}+2ny\) ➜ \(\because y_{1}=2nx(x^2-1)^{n-1}, y=(x^2-1)^{n}\)
\(\Rightarrow (x^2-1)y_{2}-2(n-1)xy_{1}-2ny=0\)
(Proved)
\(Q.1.(xii)\) \(y=\frac{x^2}{1-x}\) হয়, তবে প্রমাণ কর যে, \((1-x)y_{2}-2y_{1}=2\)
সমাধানঃ
দেওয়া আছে,
\(y=\frac{x^2}{1-x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x^2}{1-x}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{(1-x)\frac{d}{dx}(x^2)-x^2\frac{d}{dx}(1-x)}{(1-x)^2}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow y_{1}=\frac{(1-x).2x-x^2(0-1)}{(1-x)^2}\) ➜ \(\because \frac{d}{dx}(x^2)=2x\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=\frac{2x-2x^2+x^2}{(1-x)^2}\)
\(\Rightarrow y_{1}=\frac{2x-x^2}{(1-x)^2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(\frac{2x-x^2}{(1-x)^2}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{(1-x)^2\frac{d}{dx}(2x-x^2)-(2x-x^2)\frac{d}{dx}(1-x)^2}{(1-x)^4}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow y_{2}=\frac{(1-x)^2(2.1-2x)-(2x-x^2).2(1-x).(-1)}{(1-x)^4}\)
\(\Rightarrow y_{2}=\frac{(1-x)^2(2-2x)+2(2x-x^2)(1-x)}{(1-x)^4}\)
\(\Rightarrow y_{2}=\frac{2(1-x)^2(1-x)+2(2x-x^2)(1-x)}{(1-x)^4}\)
\(\Rightarrow y_{2}=\frac{2(1-x)\{(1-x)^2+(2x-x^2)\}}{(1-x)^4}\)
\(\Rightarrow y_{2}=\frac{2\{(1-x)^2+(2x-x^2)\}}{(1-x)^3}\)
\(\Rightarrow y_{2}=\frac{2\{1-2x+x^2+2x-x^2\}}{(1-x)^3}\)
\(\Rightarrow y_{2}=\frac{2.1}{(1-x)^3}\)
\(\Rightarrow (1-x)y_{2}=\frac{2}{(1-x)^3}\times{(1-x)}\)
\(\Rightarrow (1-x)y_{2}=\frac{2}{(1-x)^2}\)
\(\Rightarrow (1-x)y_{2}=\frac{4x-2x^2+2-4x+2x^2)}{(1-x)^2}\)
\(\Rightarrow (1-x)y_{2}=\frac{4x-2x^2+2(1-2x+x^2)}{(1-x)^2}\)
\(\Rightarrow (1-x)y_{2}=\frac{4x-2x^2+2(1-x)^2}{(1-x)^2}\)
\(\Rightarrow (1-x)y_{2}=\frac{2(2x-x^2)}{(1-x)^2}+2\)
\(\Rightarrow (1-x)y_{2}=2\left(\frac{2x-x^2}{(1-x)^2}\right)+2\)
\(\Rightarrow (1-x)y_{2}=2y_{1}+2\)
\(\therefore (1-x)y_{2}-2y_{1}=2\)
(Proved)
\(y=\frac{x^2}{1-x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x^2}{1-x}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{(1-x)\frac{d}{dx}(x^2)-x^2\frac{d}{dx}(1-x)}{(1-x)^2}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow y_{1}=\frac{(1-x).2x-x^2(0-1)}{(1-x)^2}\) ➜ \(\because \frac{d}{dx}(x^2)=2x\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=\frac{2x-2x^2+x^2}{(1-x)^2}\)
\(\Rightarrow y_{1}=\frac{2x-x^2}{(1-x)^2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(\frac{2x-x^2}{(1-x)^2}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{(1-x)^2\frac{d}{dx}(2x-x^2)-(2x-x^2)\frac{d}{dx}(1-x)^2}{(1-x)^4}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow y_{2}=\frac{(1-x)^2(2.1-2x)-(2x-x^2).2(1-x).(-1)}{(1-x)^4}\)
\(\Rightarrow y_{2}=\frac{(1-x)^2(2-2x)+2(2x-x^2)(1-x)}{(1-x)^4}\)
\(\Rightarrow y_{2}=\frac{2(1-x)^2(1-x)+2(2x-x^2)(1-x)}{(1-x)^4}\)
\(\Rightarrow y_{2}=\frac{2(1-x)\{(1-x)^2+(2x-x^2)\}}{(1-x)^4}\)
\(\Rightarrow y_{2}=\frac{2\{(1-x)^2+(2x-x^2)\}}{(1-x)^3}\)
\(\Rightarrow y_{2}=\frac{2\{1-2x+x^2+2x-x^2\}}{(1-x)^3}\)
\(\Rightarrow y_{2}=\frac{2.1}{(1-x)^3}\)
\(\Rightarrow (1-x)y_{2}=\frac{2}{(1-x)^3}\times{(1-x)}\)
\(\Rightarrow (1-x)y_{2}=\frac{2}{(1-x)^2}\)
\(\Rightarrow (1-x)y_{2}=\frac{4x-2x^2+2-4x+2x^2)}{(1-x)^2}\)
\(\Rightarrow (1-x)y_{2}=\frac{4x-2x^2+2(1-2x+x^2)}{(1-x)^2}\)
\(\Rightarrow (1-x)y_{2}=\frac{4x-2x^2+2(1-x)^2}{(1-x)^2}\)
\(\Rightarrow (1-x)y_{2}=\frac{2(2x-x^2)}{(1-x)^2}+2\)
\(\Rightarrow (1-x)y_{2}=2\left(\frac{2x-x^2}{(1-x)^2}\right)+2\)
\(\Rightarrow (1-x)y_{2}=2y_{1}+2\)
\(\therefore (1-x)y_{2}-2y_{1}=2\)
(Proved)
\(Q.1.(xiii)\) \(y=x^{\frac{2}{3}}+x^{-\frac{2}{3}}\) হয়, তবে প্রমাণ কর যে, \(3x\frac{dy}{dx}+2y=4x^{\frac{2}{3}}\)
সমাধানঃ
দেওয়া আছে,
\(y=x^{\frac{2}{3}}+x^{-\frac{2}{3}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(x^{\frac{2}{3}}+x^{-\frac{2}{3}}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(x^{\frac{2}{3}}\right)+\frac{d}{dx}\left(x^{-\frac{2}{3}}\right)\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}x^{\frac{2}{3}-1}-\frac{2}{3}x^{-\frac{2}{3}-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}x^{\frac{2-3}{3}}-\frac{2}{3}x^{\frac{-2-3}{3}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}x^{\frac{-1}{3}}-\frac{2}{3}x^{\frac{-5}{3}}\)
\(\Rightarrow 3x\frac{dy}{dx}=\frac{2}{3}x^{\frac{-1}{3}}\times{3x}-\frac{2}{3}x^{\frac{-5}{3}}\times{3x}\) ➜ উভয় পার্শে \(3x\)গুণ করে।
\(\Rightarrow 3x\frac{dy}{dx}=\frac{2}{3}x^{\frac{-1}{3}}\times{3x}-\frac{2}{3}x^{\frac{-5}{3}}\times{3x}\) ➜ উভয় পার্শে \(3x\)গুণ করে।
\(\Rightarrow 3x\frac{dy}{dx}=2x^{\frac{-1}{3}+1}-2x^{\frac{-5}{3}+1}\)
\(\Rightarrow 3x\frac{dy}{dx}=2x^{\frac{-1+3}{3}}-2x^{\frac{-5+3}{3}}\)
\(\Rightarrow 3x\frac{dy}{dx}=2x^{\frac{2}{3}}-2x^{\frac{-2}{3}}\)
\(\Rightarrow 3x\frac{dy}{dx}=2x^{\frac{2}{3}}-2x^{\frac{-2}{3}}\)
\(\Rightarrow 3x\frac{dy}{dx}=4x^{\frac{2}{3}}-2x^{\frac{2}{3}}-2x^{-\frac{2}{3}})\)
\(\Rightarrow 3x\frac{dy}{dx}=4x^{\frac{2}{3}}-2(x^{\frac{2}{3}}+x^{-\frac{2}{3}})\)
\(\Rightarrow 3x\frac{dy}{dx}=4x^{\frac{2}{3}}-2y\)
\(\therefore 3x\frac{dy}{dx}+2y=4x^{\frac{2}{3}}\)
(Proved)
\(y=x^{\frac{2}{3}}+x^{-\frac{2}{3}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(x^{\frac{2}{3}}+x^{-\frac{2}{3}}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(x^{\frac{2}{3}}\right)+\frac{d}{dx}\left(x^{-\frac{2}{3}}\right)\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}x^{\frac{2}{3}-1}-\frac{2}{3}x^{-\frac{2}{3}-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}x^{\frac{2-3}{3}}-\frac{2}{3}x^{\frac{-2-3}{3}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}x^{\frac{-1}{3}}-\frac{2}{3}x^{\frac{-5}{3}}\)
\(\Rightarrow 3x\frac{dy}{dx}=\frac{2}{3}x^{\frac{-1}{3}}\times{3x}-\frac{2}{3}x^{\frac{-5}{3}}\times{3x}\) ➜ উভয় পার্শে \(3x\)গুণ করে।
\(\Rightarrow 3x\frac{dy}{dx}=\frac{2}{3}x^{\frac{-1}{3}}\times{3x}-\frac{2}{3}x^{\frac{-5}{3}}\times{3x}\) ➜ উভয় পার্শে \(3x\)গুণ করে।
\(\Rightarrow 3x\frac{dy}{dx}=2x^{\frac{-1}{3}+1}-2x^{\frac{-5}{3}+1}\)
\(\Rightarrow 3x\frac{dy}{dx}=2x^{\frac{-1+3}{3}}-2x^{\frac{-5+3}{3}}\)
\(\Rightarrow 3x\frac{dy}{dx}=2x^{\frac{2}{3}}-2x^{\frac{-2}{3}}\)
\(\Rightarrow 3x\frac{dy}{dx}=2x^{\frac{2}{3}}-2x^{\frac{-2}{3}}\)
\(\Rightarrow 3x\frac{dy}{dx}=4x^{\frac{2}{3}}-2x^{\frac{2}{3}}-2x^{-\frac{2}{3}})\)
\(\Rightarrow 3x\frac{dy}{dx}=4x^{\frac{2}{3}}-2(x^{\frac{2}{3}}+x^{-\frac{2}{3}})\)
\(\Rightarrow 3x\frac{dy}{dx}=4x^{\frac{2}{3}}-2y\)
\(\therefore 3x\frac{dy}{dx}+2y=4x^{\frac{2}{3}}\)
(Proved)
\(Q.1.(xiv)\) \(y=4x^{\frac{3}{2}}-3+2x^{\frac{1}{2}}\)হলে, \(y_{2}\) নির্ণয় কর এবং \(x=4\) হলে, \(y_{2}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(3x^{-\frac{1}{2}}-\frac{1}{2}x^{\frac{-3}{2}}; \frac{23}{16}\)
উত্তরঃ \(3x^{-\frac{1}{2}}-\frac{1}{2}x^{\frac{-3}{2}}; \frac{23}{16}\)
সমাধানঃ
দেওয়া আছে,
\(y=4x^{\frac{3}{2}}-3+2x^{\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(4x^{\frac{3}{2}}-3+2x^{\frac{1}{2}}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=4\frac{d}{dx}\left(x^{\frac{3}{2}}\right)-\frac{d}{dx}(3)+2\frac{d}{dx}\left(x^{\frac{1}{2}}\right)\) ➜ \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow y_{1}=4.\frac{3}{2}x^{\frac{3}{2}-1}-0+2.\frac{1}{2}x^{\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{1}=6x^{\frac{3-2}{2}}+x^{\frac{1-2}{2}}\)
\(\Rightarrow y_{1}=6x^{\frac{1}{2}}+x^{\frac{-1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(6x^{\frac{1}{2}}+x^{\frac{-1}{2}}\right)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=6\frac{d}{dx}\left(x^{\frac{1}{2}}\right)+\frac{d}{dx}\left(x^{\frac{-1}{2}}\right)\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=6.\frac{1}{2}x^{\frac{1}{2}-1}-\frac{1}{2}x^{\frac{-1}{2}-1}\) ➜\(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{2}=3x^{\frac{1-2}{2}}-\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(\therefore y_{2}=3x^{-\frac{1}{2}}-\frac{1}{2}x^{\frac{-3}{2}}\)
আবার,
\(x=4\) হলে,
\(y_{2}=3.4^{-\frac{1}{2}}-\frac{1}{2}4^{\frac{-3}{2}}\)
\(=\frac{3}{4^{\frac{1}{2}}}-\frac{1}{2.4^{\frac{3}{2}}}\)
\(=\frac{3}{2}-\frac{1}{2.4^{\frac{1+2}{2}}}\)
\(=\frac{3}{2}-\frac{1}{2.4^{\frac{1}{2}+1}}\)
\(=\frac{3}{2}-\frac{1}{2.4^{\frac{1}{2}}.4}\)
\(=\frac{3}{2}-\frac{1}{2.2.4}\)
\(=\frac{3}{2}-\frac{1}{16}\)
\(=\frac{24-1}{16}\)
\(=\frac{23}{16}\)
\(y=4x^{\frac{3}{2}}-3+2x^{\frac{1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(4x^{\frac{3}{2}}-3+2x^{\frac{1}{2}}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=4\frac{d}{dx}\left(x^{\frac{3}{2}}\right)-\frac{d}{dx}(3)+2\frac{d}{dx}\left(x^{\frac{1}{2}}\right)\) ➜ \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow y_{1}=4.\frac{3}{2}x^{\frac{3}{2}-1}-0+2.\frac{1}{2}x^{\frac{1}{2}-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{1}=6x^{\frac{3-2}{2}}+x^{\frac{1-2}{2}}\)
\(\Rightarrow y_{1}=6x^{\frac{1}{2}}+x^{\frac{-1}{2}}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(6x^{\frac{1}{2}}+x^{\frac{-1}{2}}\right)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=6\frac{d}{dx}\left(x^{\frac{1}{2}}\right)+\frac{d}{dx}\left(x^{\frac{-1}{2}}\right)\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=6.\frac{1}{2}x^{\frac{1}{2}-1}-\frac{1}{2}x^{\frac{-1}{2}-1}\) ➜\(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{2}=3x^{\frac{1-2}{2}}-\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(\therefore y_{2}=3x^{-\frac{1}{2}}-\frac{1}{2}x^{\frac{-3}{2}}\)
আবার,
\(x=4\) হলে,
\(y_{2}=3.4^{-\frac{1}{2}}-\frac{1}{2}4^{\frac{-3}{2}}\)
\(=\frac{3}{4^{\frac{1}{2}}}-\frac{1}{2.4^{\frac{3}{2}}}\)
\(=\frac{3}{2}-\frac{1}{2.4^{\frac{1+2}{2}}}\)
\(=\frac{3}{2}-\frac{1}{2.4^{\frac{1}{2}+1}}\)
\(=\frac{3}{2}-\frac{1}{2.4^{\frac{1}{2}}.4}\)
\(=\frac{3}{2}-\frac{1}{2.2.4}\)
\(=\frac{3}{2}-\frac{1}{16}\)
\(=\frac{24-1}{16}\)
\(=\frac{23}{16}\)
\(Q.1.(xv)\) \(y=\frac{x}{x+2}\) হলে, প্রমাণ কর যে, \(xy_{1}=y(1-y)\)
সমাধানঃ
দেওয়া আছে,
\(y=\frac{x}{x+2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x}{x+2}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{(x+2)\frac{d}{dx}(x)-x\frac{d}{dx}(x+2)}{(x+2)^2}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow y_{1}=\frac{(x+2).1-x(1+0)}{(x+2)^2}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{1}=\frac{x+2-x}{(x+2)^2}\)
\(\Rightarrow xy_{1}=\frac{2x}{(x+2)^2}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow xy_{1}=\frac{x^2+2x-x^2}{(x+2)^2}\)
\(\Rightarrow xy_{1}=\frac{x(x+2)-x^2}{(x+2)^2}\)
\(\Rightarrow xy_{1}=\frac{x}{x+2}-\frac{x^2}{(x+2)^2}\)
\(\Rightarrow xy_{1}=\frac{x}{x+2}-\left(\frac{x}{x+2}\right)^2\)
\(\Rightarrow xy_{1}=y-y^2\) ➜\(\because y=\frac{x}{x+2}\)
\(\therefore xy_{1}=y(1-y)\)
(Proved)
\(y=\frac{x}{x+2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x}{x+2}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{(x+2)\frac{d}{dx}(x)-x\frac{d}{dx}(x+2)}{(x+2)^2}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow y_{1}=\frac{(x+2).1-x(1+0)}{(x+2)^2}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow y_{1}=\frac{x+2-x}{(x+2)^2}\)
\(\Rightarrow xy_{1}=\frac{2x}{(x+2)^2}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow xy_{1}=\frac{x^2+2x-x^2}{(x+2)^2}\)
\(\Rightarrow xy_{1}=\frac{x(x+2)-x^2}{(x+2)^2}\)
\(\Rightarrow xy_{1}=\frac{x}{x+2}-\frac{x^2}{(x+2)^2}\)
\(\Rightarrow xy_{1}=\frac{x}{x+2}-\left(\frac{x}{x+2}\right)^2\)
\(\Rightarrow xy_{1}=y-y^2\) ➜\(\because y=\frac{x}{x+2}\)
\(\therefore xy_{1}=y(1-y)\)
(Proved)
\(Q.1.(xvi)\) \(y=ax^{n+1}+bx^{-n}\) হলে, দেখাও যে, \(x^2y_{2}=n(n+1)y\)
সমাধানঃ
দেওয়া আছে,
\(y=ax^{n+1}+bx^{-n}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(ax^{n+1}+bx^{-n}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a\frac{d}{dx}(x^{n+1})+b\frac{d}{dx}(x^{-n})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=a(n+1)x^{n+1-1}+b(-n)x^{-n-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{1}=a(n+1)x^{n}-bnx^{-n-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(a(n+1)x^{n}-bnx^{-n-1}\right)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a(n+1)\frac{d}{dx}(x^{n})-bn\frac{d}{dx}(x^{-n-1})\) ➜\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=a(n+1)nx^{n-1}-bn(-n-1)x^{-n-1-1}\) ➜\(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{2}=an(n+1)x^{n-1}+bn(n+1)x^{-n-2}\)
\(\Rightarrow x^2y_{2}=an(n+1)x^{n-1}\times{x^2}+bn(n+1)x^{-n-2}\times{x^2}\) ➜ উভয় পার্শে \(x^2\)গুণ করে।
\(\Rightarrow x^2y_{2}=an(n+1)x^{n-1+2}+bn(n+1)x^{-n-2+2}\)
\(\Rightarrow x^2y_{2}=an(n+1)x^{n+1}+bn(n+1)x^{-n}\)
\(\Rightarrow x^2y_{2}=n(n+1)(ax^{n+1}+bx^{-n})\)
\(\therefore x^2y_{2}=n(n+1)y\) ➜\(\because y=ax^{n+1}+bx^{-n}\)
(Showed)
\(y=ax^{n+1}+bx^{-n}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(ax^{n+1}+bx^{-n}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a\frac{d}{dx}(x^{n+1})+b\frac{d}{dx}(x^{-n})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=a(n+1)x^{n+1-1}+b(-n)x^{-n-1}\) ➜ \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{1}=a(n+1)x^{n}-bnx^{-n-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\left(a(n+1)x^{n}-bnx^{-n-1}\right)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a(n+1)\frac{d}{dx}(x^{n})-bn\frac{d}{dx}(x^{-n-1})\) ➜\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=a(n+1)nx^{n-1}-bn(-n-1)x^{-n-1-1}\) ➜\(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{2}=an(n+1)x^{n-1}+bn(n+1)x^{-n-2}\)
\(\Rightarrow x^2y_{2}=an(n+1)x^{n-1}\times{x^2}+bn(n+1)x^{-n-2}\times{x^2}\) ➜ উভয় পার্শে \(x^2\)গুণ করে।
\(\Rightarrow x^2y_{2}=an(n+1)x^{n-1+2}+bn(n+1)x^{-n-2+2}\)
\(\Rightarrow x^2y_{2}=an(n+1)x^{n+1}+bn(n+1)x^{-n}\)
\(\Rightarrow x^2y_{2}=n(n+1)(ax^{n+1}+bx^{-n})\)
\(\therefore x^2y_{2}=n(n+1)y\) ➜\(\because y=ax^{n+1}+bx^{-n}\)
(Showed)
\(Q.1.(xvii)\) \(y=\sqrt{ax^2+bx+c}\) হলে, দেখাও যে, \(4y^3y_{2}=4ac-b^2\)
সমাধানঃ
দেওয়া আছে,
\(y=\sqrt{ax^2+bx+c}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt{ax^2+bx+c})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{2\sqrt{ax^2+bx+c}}\frac{d}{dx}(ax^2+bx+c)\) ➜ \((ax^2+bx+c)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow y_{1}=\frac{1}{2\sqrt{ax^2+bx+c}}(2ax+b)\)
\(\Rightarrow y_{1}=\frac{2ax+b}{2\sqrt{ax^2+bx+c}}\)
\(\Rightarrow y_{1}=\frac{2ax+b}{2y}\) ➜\(\because y=\sqrt{ax^2+bx+c}\)
\(\Rightarrow 2y_{1}=\frac{2ax+b}{y}\) ➜ উভয় পার্শে \(2\)গুণ করে।
\(\Rightarrow 2\frac{d}{dx}(y_{1})=\frac{d}{dx}\left(\frac{2ax+b}{y}\right)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y_{2}=\frac{y\frac{d}{dx}(2ax+b)-(2ax+b)y_{1}}{y^2}\) ➜\(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow 2y_{2}=\frac{y(2a.1+0)-(2ax+b)y_{1}}{y^2}\)
\(\Rightarrow 2y_{2}=\frac{2ay-(2ax+b)\frac{2ax+b}{2y}}{y^2}\) ➜\(\because y_{1}=\frac{2ax+b}{2y}\)
\(\Rightarrow 2y_{2}=\frac{2ay-(2ax+b)\frac{2ax+b}{2y}}{y^2}\)
\(\Rightarrow 2y_{2}=\frac{\frac{4ay^2-(2ax+b)^2}{2y}}{y^2}\)
\(\Rightarrow 2y_{2}=\frac{4ay^2-4a^2x^2-4abx-b^2}{2y^3}\)
\(\Rightarrow 2y_{2}=\frac{4a(ax^2+bx+c)-4a^2x^2-4abx-b^2}{2y^3}\) ➜\(\because y=\sqrt{ax^2+bx+c}\Rightarrow y^2=ax^2+bx+c\)
\(\Rightarrow 2y_{2}=\frac{4a^2x^2+4abx+4ac-4a^2x^2-4abx-b^2}{2y^3}\)
\(\Rightarrow 2y_{2}=\frac{4ac-b^2}{2y^3}\)
\(\therefore 4y^3y_{2}=4ac-b^2\) ➜ উভয় পার্শে \(2y^3\) গুণ করে।
(Showed)
\(y=\sqrt{ax^2+bx+c}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt{ax^2+bx+c})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{2\sqrt{ax^2+bx+c}}\frac{d}{dx}(ax^2+bx+c)\) ➜ \((ax^2+bx+c)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow y_{1}=\frac{1}{2\sqrt{ax^2+bx+c}}(2ax+b)\)
\(\Rightarrow y_{1}=\frac{2ax+b}{2\sqrt{ax^2+bx+c}}\)
\(\Rightarrow y_{1}=\frac{2ax+b}{2y}\) ➜\(\because y=\sqrt{ax^2+bx+c}\)
\(\Rightarrow 2y_{1}=\frac{2ax+b}{y}\) ➜ উভয় পার্শে \(2\)গুণ করে।
\(\Rightarrow 2\frac{d}{dx}(y_{1})=\frac{d}{dx}\left(\frac{2ax+b}{y}\right)\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y_{2}=\frac{y\frac{d}{dx}(2ax+b)-(2ax+b)y_{1}}{y^2}\) ➜\(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow 2y_{2}=\frac{y(2a.1+0)-(2ax+b)y_{1}}{y^2}\)
\(\Rightarrow 2y_{2}=\frac{2ay-(2ax+b)\frac{2ax+b}{2y}}{y^2}\) ➜\(\because y_{1}=\frac{2ax+b}{2y}\)
\(\Rightarrow 2y_{2}=\frac{2ay-(2ax+b)\frac{2ax+b}{2y}}{y^2}\)
\(\Rightarrow 2y_{2}=\frac{\frac{4ay^2-(2ax+b)^2}{2y}}{y^2}\)
\(\Rightarrow 2y_{2}=\frac{4ay^2-4a^2x^2-4abx-b^2}{2y^3}\)
\(\Rightarrow 2y_{2}=\frac{4a(ax^2+bx+c)-4a^2x^2-4abx-b^2}{2y^3}\) ➜\(\because y=\sqrt{ax^2+bx+c}\Rightarrow y^2=ax^2+bx+c\)
\(\Rightarrow 2y_{2}=\frac{4a^2x^2+4abx+4ac-4a^2x^2-4abx-b^2}{2y^3}\)
\(\Rightarrow 2y_{2}=\frac{4ac-b^2}{2y^3}\)
\(\therefore 4y^3y_{2}=4ac-b^2\) ➜ উভয় পার্শে \(2y^3\) গুণ করে।
(Showed)
অনুশীলনী \(9.G / Q.2\)-এর সংক্ষিপ্ত প্রশ্নসমুহ
\(Q.2.(i)\) \(y=\tan{x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}=2y(1+y^2)\)
[ মাঃ২০০২ ]
\(Q.2.(ii)\) \(y=\sin{x}\) হলে, দেখাও যে, \(y_{4}-y=0\)
[ রাঃ২০০৪;বঃ২০০৪ ]
\(Q.2.(iii)\) \(y=x^3\ln{x}\) হলে, প্রমাণ কর যে, \(y_{4}=\frac{6}{x}\)
\(Q.2.(iv)\) \(y=\frac{\ln{x}}{x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}=\frac{2\ln{x}-3}{x^3}\)
[ দিঃ২০০৯;বঃ২০০৭;ঢাঃ২০০৬;চঃ২০০৪,২০০১]
\(Q.2.(v)\) \(y=\tan{x}+\sec{x}\) হলে, প্রমাণ কর যে, \(\frac{d^2y}{dx^2}=\frac{\cos{x}}{(1-\sin{x})^2}\)
[ রাঃ২০১০,২০০৫,২০০১]
\(Q.2.(vi)\) \(y=\sec{x}\) হলে, দেখাও যে, \(y_{2}=y(2y^2-1)\)
[ যঃ২০১১,২০০৮;কুঃ২০১০,২০০২,২০০০;চঃ২০০৮,২০০৬,২০০৪;রাঃ২০০৭,২০০২; সিঃ২০০৭;বঃ২০০৬;ঢাঃ২০১৭,২০০০;মাঃ২০১২,২০০৫,২০০১ ]
\(Q.2.(vii)\) \(y=a\cos{(\ln{x})}+b\sin{(\ln{x})}\) হলে, দেখাও যে, \(x^2y_{2}+xy_{1}+y=0\)
[ ঢাঃ২০০৯ ]
\(Q.2.(viii)\) \(y=x^m\ln{x}\) হলে, প্রমাণ কর যে, \(xy_{1}=my+x^m\)
\(Q.2.(ix)\) \(y=x^{n-1}\ln{x}\) হলে, প্রমাণ কর যে, \(x^2y_{2}+(3-2n)xy_{1}+(n-1)^2y=0\)
[ বুয়েটঃ২০০৯-২০১০]
\(Q.2.(x)\) \(y=\sin^{-1}{x}\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=0\)
\(Q.2.(xi)\) \(y=(\sin^{-1}{x})^2\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}-2=0\)
[ রাঃ২০১১,২০০১;বঃ২০০৮;কুঃ২০০০;মাঃ২০১০,২০০৩]
\(Q.2.(xii)\) \(f(u)=\sin^{-1}{u}, y=\{f(2x)\}^2\) হলে, দেখাও যে, \((1-4x^2)y_{2}-4xy_{1}-8=0\)
[ সিঃ২০১৭ ]
\(Q.2.(xiii)\) \(x=\sin{\sqrt{y}}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}-2=0\)
[ বঃ২০১২,২০০৫,২০০৩;চঃ২০১১;ঢাঃ২০০৮;কুঃ২০০৮,২০০৩ ]
\(Q.2.(xiv)\) \(y=\frac{1}{2}(\sin^{-1}{x})^2\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}-1=0\)
[ বুয়েটঃ২০০৫-২০০৬]
\(Q.2.(xv)\) \(y=\cos^{-1}{x}\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=0\)
\(Q.2.(xvi)\) \(\ln{y}=bz\) এবং \(\cos{z}=x\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=b^2y\)
[ বঃ২০১৭]
\(Q.2.(xvii)\) \(y=(\cos^{-1}{x})^2\) হলে, দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2\)
[ ঢাঃ২০১২,২০০৪;দিঃ২০১২;সিঃ২০০৮;বঃ২০০৬;চঃ২০০৩;যঃ২০০৩; কুঃ২০০১;রাঃ২০০৪;মাঃ২০০৮ ]
\(Q.2.(xviii)\) যদি \(\cos{\sqrt{y}}=x\) হয়, তবে দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2=0\)
[ যঃ২০১২,২০০৪,২০০৬;ঢাঃ২০১১,২০০১;সিঃ২০০১০,২০০৪; রাঃ২০০৯,২০০৭;চঃ২০০৬ ]
\(Q.2.(xix)\) \(y=\tan^{-1}{x}\) হয়, তবে প্রমাণ কর যে, \((1+x^2)y_{2}+2xy_{1}=0\)
[ মাঃ২০০৭,২০০৫;ঢাঃ,কুঃ২০০৫]
\(Q.2.(xx)\) \(y=(\tan^{-1}{x})^2\) হয়, তবে প্রমাণ কর যে, \((1+x^2)^2y_{2}+2(1+x^2)xy_{1}=2\)
\(Q.2.(xxi)\) \(y=\sqrt{a+b\cos{x}}\) হলে, দেখাও যে, \(2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=a\)
[ যঃ২০১৭]
\(Q.2.(xxii)\) \(y=\ln{(x+\sqrt{a^2+x^2})}\) হলে, দেখাও যে, \((a^2+x^2)y_{2}+xy_{1}=0\)
[ চঃ,মাঃ২০১০;যঃ২০০৩,২০০১]
\(Q.2.(xxiii)\) \(y=(x+\sqrt{1+x^2})^m\) হয়, তবে প্রমাণ কর যে, \((1+x^2)y_{2}+xy_{1}-m^2y=0\)
[সিঃ২০১১;যঃ,বঃ২০০১০;চঃ২০০৯]
[ মাঃ২০০২ ]
\(Q.2.(ii)\) \(y=\sin{x}\) হলে, দেখাও যে, \(y_{4}-y=0\)
[ রাঃ২০০৪;বঃ২০০৪ ]
\(Q.2.(iii)\) \(y=x^3\ln{x}\) হলে, প্রমাণ কর যে, \(y_{4}=\frac{6}{x}\)
\(Q.2.(iv)\) \(y=\frac{\ln{x}}{x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}=\frac{2\ln{x}-3}{x^3}\)
[ দিঃ২০০৯;বঃ২০০৭;ঢাঃ২০০৬;চঃ২০০৪,২০০১]
\(Q.2.(v)\) \(y=\tan{x}+\sec{x}\) হলে, প্রমাণ কর যে, \(\frac{d^2y}{dx^2}=\frac{\cos{x}}{(1-\sin{x})^2}\)
[ রাঃ২০১০,২০০৫,২০০১]
\(Q.2.(vi)\) \(y=\sec{x}\) হলে, দেখাও যে, \(y_{2}=y(2y^2-1)\)
[ যঃ২০১১,২০০৮;কুঃ২০১০,২০০২,২০০০;চঃ২০০৮,২০০৬,২০০৪;রাঃ২০০৭,২০০২; সিঃ২০০৭;বঃ২০০৬;ঢাঃ২০১৭,২০০০;মাঃ২০১২,২০০৫,২০০১ ]
\(Q.2.(vii)\) \(y=a\cos{(\ln{x})}+b\sin{(\ln{x})}\) হলে, দেখাও যে, \(x^2y_{2}+xy_{1}+y=0\)
[ ঢাঃ২০০৯ ]
\(Q.2.(viii)\) \(y=x^m\ln{x}\) হলে, প্রমাণ কর যে, \(xy_{1}=my+x^m\)
\(Q.2.(ix)\) \(y=x^{n-1}\ln{x}\) হলে, প্রমাণ কর যে, \(x^2y_{2}+(3-2n)xy_{1}+(n-1)^2y=0\)
[ বুয়েটঃ২০০৯-২০১০]
\(Q.2.(x)\) \(y=\sin^{-1}{x}\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=0\)
\(Q.2.(xi)\) \(y=(\sin^{-1}{x})^2\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}-2=0\)
[ রাঃ২০১১,২০০১;বঃ২০০৮;কুঃ২০০০;মাঃ২০১০,২০০৩]
\(Q.2.(xii)\) \(f(u)=\sin^{-1}{u}, y=\{f(2x)\}^2\) হলে, দেখাও যে, \((1-4x^2)y_{2}-4xy_{1}-8=0\)
[ সিঃ২০১৭ ]
\(Q.2.(xiii)\) \(x=\sin{\sqrt{y}}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}-2=0\)
[ বঃ২০১২,২০০৫,২০০৩;চঃ২০১১;ঢাঃ২০০৮;কুঃ২০০৮,২০০৩ ]
\(Q.2.(xiv)\) \(y=\frac{1}{2}(\sin^{-1}{x})^2\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}-1=0\)
[ বুয়েটঃ২০০৫-২০০৬]
\(Q.2.(xv)\) \(y=\cos^{-1}{x}\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=0\)
\(Q.2.(xvi)\) \(\ln{y}=bz\) এবং \(\cos{z}=x\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=b^2y\)
[ বঃ২০১৭]
\(Q.2.(xvii)\) \(y=(\cos^{-1}{x})^2\) হলে, দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2\)
[ ঢাঃ২০১২,২০০৪;দিঃ২০১২;সিঃ২০০৮;বঃ২০০৬;চঃ২০০৩;যঃ২০০৩; কুঃ২০০১;রাঃ২০০৪;মাঃ২০০৮ ]
\(Q.2.(xviii)\) যদি \(\cos{\sqrt{y}}=x\) হয়, তবে দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2=0\)
[ যঃ২০১২,২০০৪,২০০৬;ঢাঃ২০১১,২০০১;সিঃ২০০১০,২০০৪; রাঃ২০০৯,২০০৭;চঃ২০০৬ ]
\(Q.2.(xix)\) \(y=\tan^{-1}{x}\) হয়, তবে প্রমাণ কর যে, \((1+x^2)y_{2}+2xy_{1}=0\)
[ মাঃ২০০৭,২০০৫;ঢাঃ,কুঃ২০০৫]
\(Q.2.(xx)\) \(y=(\tan^{-1}{x})^2\) হয়, তবে প্রমাণ কর যে, \((1+x^2)^2y_{2}+2(1+x^2)xy_{1}=2\)
\(Q.2.(xxi)\) \(y=\sqrt{a+b\cos{x}}\) হলে, দেখাও যে, \(2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=a\)
[ যঃ২০১৭]
\(Q.2.(xxii)\) \(y=\ln{(x+\sqrt{a^2+x^2})}\) হলে, দেখাও যে, \((a^2+x^2)y_{2}+xy_{1}=0\)
[ চঃ,মাঃ২০১০;যঃ২০০৩,২০০১]
\(Q.2.(xxiii)\) \(y=(x+\sqrt{1+x^2})^m\) হয়, তবে প্রমাণ কর যে, \((1+x^2)y_{2}+xy_{1}-m^2y=0\)
[সিঃ২০১১;যঃ,বঃ২০০১০;চঃ২০০৯]
\(Q.2.(xxiv)\) \(y=e^{a\sin^{-1}{x}}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}=a^2y\)
[কুঃ২০১২,২০০১;ঢাঃ২০১১,২০০৩;বঃ২০১১;যঃ২০০৯,২০০২ সিঃ২০০৯,২০০৭,২০০৪;চঃ২০০৫;রাঃ২০০০;মাঃ২০১২,২০০৯,২০০৬]
\(Q.2.(xxv)\) \(\ln{y}=a\sin^{-1}{x}\) হলে, দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-a^2y=0\)
[ ঢাঃ২০০৭ ]
\(Q.2.(xxvi)\) যদি \(y=e^{\tan^{-1}{x}}\) অথবা \(\ln{y}=\tan^{-1}{x}\) হয়, তবে প্রমাণ কর যে, \((1+x^2)y_{2}+(2x-1)y_{1}=0\)
[ ঢাঃ,বঃ২০১২;কুঃ২০১৭,২০১১;রাঃ২০১০,২০০৮,২০০৫,২০০২;যঃ২০১০ ]
\(Q.2.(xxvii)\) \(y=e^{4\sin^{-1}{x}}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}=16y\)
[ চঃ২০০২ ]
\(Q.2.(xxviii)\) \(y=\sin{(m\sin^{-1}{x})}\) হলে, দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+m^2y=0\)
[ বঃ২০১১,২০০৩,২০০১;ঢাঃ২০১০,২০০৫;রাঃ২০০৯; চঃ২০০৮;সিঃ২০০৩;মাঃ২০০৪ ]
\(Q.2.(xxix)\) \(y=\cos{(m\sin^{-1}{p})}\) হলে, প্রমাণ কর যে, \((1-p^2)y_{2}-py_{1}+m^2y=0\)
[ চঃ২০১৭ ]
\(Q.2.(xxx)\) \(y=\cos{(2\sin^{-1}{x})}\) হলে, প্রমাণ কর যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+4y=0\)
[ বুয়েটঃ২০০৬-২০০৭ ]
\(Q.2.(xxxi)\) \(y=\tan{(m\tan^{-1}{x})}\) হলে, দেখাও যে, \((1+x^2)y_{1}=m(1+y^2)\)
[ চঃ২০১২;কুঃ২০০৭,২০০৪]
\(Q.2.(xxxii)\) \(y=\tan{(m\tan^{-1}{x})}\) হলে, দেখাও যে, \((1+x^2)y_{2}-2(my-x)y_{1}=0\)
[ ঢাঃ২০১৩,২০০০;কুঃ২০১২;যঃ২০১১;সিঃ২০০৬;মাঃ২০০৬,২০০৪]
\(Q.2.(xxxiii)\) \(y=x^2\ln{x}\) হলে, \(y_{3}\) নির্ণয় কর।
উত্তরঃ \(\frac{2}{x}\)
\(Q.2.(xxxiv)\) \(y=\ln{(\sin{x})}\) হলে, দেখাও যে, \(\frac{d^3y}{dx^3}=\frac{2\cos{x}}{\sin^3{x}}\)
\(Q.2.(xxxv)\) \(y=a\cos{\{\ln{(1+x)}\}}\) হলে, দেখাও যে, \((1+x)^2y_{2}+(1+x)y_{1}+y=0\)
\(Q.2.(xxxvi)\) \(y=Ae^{mx}+Be^{-mx}\) হলে, দেখাও যে, \(y_{2}-m^2y=0\)
[ যঃ২০০৭;দিঃ২০১০;বঃ২০১৩,২০০৮;সিঃ২০১১ ]
\(Q.2.(xxxvii)\) \(y=\frac{1}{2}(e^{x}+e^{-x})\) হলে, দেখাও যে, \(\left(\frac{dy}{dx}\right)^2=y^2-1\)
[ চঃ২০০৩ ]
\(Q.2.(xxxviii)\) \(y=a\sin^{-1}{x}+b\cos^{-1}{x}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}=0\)
[ ঢাঃ২০০৮;রাঃ২০১১,২০০৩;কুঃ২০১৩,২০০৮,২০০৩; বঃ২০১২,২০০৮,২০০৫;চঃ২০১১;মাঃ২০১০ ]
\(Q.2.(xxxix)\) \(x=\cos{\sqrt{y}} \) অথবা, \(y=(\cos^{-1}{x})^2 \) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}-2=0\)
[ বুয়েটঃ২০০৮-২০০৯; কুয়েটঃ২০১৩-২০১৪;ঢাঃ২০১২,২০১১,২০০৪; রাঃ২০০৯,২০০৭,২০০৪;কুঃ২০১০; দিঃ২০১২,২০১৩;যঃ২০১২,২০০৮,২০০৬,২০০; চঃ২০১৩,২০০৬,২০০৩; সিঃ২০১৪,২০১০,২০০৮,২০০৪; বঃ২০১০,২০০৬]
\(Q.2.(xL)\) \(2y=(\tan^{-1}{x})^2\) হলে, দেখাও যে, \((1+x^2)^2y_{2}+2(1+x^2)xy_{1}=1\)
\(Q.2.(xLi)\) \(y=e^{m\cos^{-1}{x}}\) অথবা, \(\ln{y}=m\cos^{-1}{x}\) অথবা, \(x=\cos\left(\frac{1}{m}\ln{y}\right)\)হলে,
প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=m^2y\)
\(Q.2.(xLii)\) \(y=\cos{(m\sin^{-1}{x})}\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}+m^2y=0\)
\(Q.2.(xLiii)\) \(y=a\sin{(\ln{x})}+b\cos{(\ln{x})}\) হলে, প্রমাণ কর যে, \(x^2y_{2}+xy_{1}+y=0\)
\(Q.2.(xLiv)\) \(y=\sec{2x}\) হলে, প্রমাণ কর যে, \(y_{2}+y=3y^5\)
\(Q.2.(xLv)\) \(y=\cot{x}+cosec{x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}=\frac{\sin{x}}{(1-\cos{x})^2}\)
\(Q.2.(i)\) \(y=\tan{x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}=2y(1+y^2)\)
[ মাঃ২০০২ ]
[ মাঃ২০০২ ]
সমাধানঃ
দেওয়া আছে,
\(y=\tan{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\sec^2{x}\) ➜ \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(\sec^2{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec{x}\frac{d}{dx}(\sec{x})\) ➜ \(\sec{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec{x}\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\tan{x}\sec^2{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\tan{x}(1+\tan^2{x})\) ➜ \(\because \sec^2{x}=1+\tan^2{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2y(1+y^2)\) ➜ \(\because y=\tan{x}\)
\(\therefore \frac{d^2y}{dx^2}=2y(1+y^2)\)
(Showed)
\(y=\tan{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\sec^2{x}\) ➜ \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(\sec^2{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec{x}\frac{d}{dx}(\sec{x})\) ➜ \(\sec{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec{x}\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\tan{x}\sec^2{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\tan{x}(1+\tan^2{x})\) ➜ \(\because \sec^2{x}=1+\tan^2{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2y(1+y^2)\) ➜ \(\because y=\tan{x}\)
\(\therefore \frac{d^2y}{dx^2}=2y(1+y^2)\)
(Showed)
\(Q.2.(ii)\) \(y=\sin{x}\) হলে, দেখাও যে, \(y_{4}-y=0\)
[ রাঃ২০০৪;বঃ২০০৪ ]
[ রাঃ২০০৪;বঃ২০০৪ ]
সমাধানঃ
দেওয়া আছে,
\(y=\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\cos{x}\) ➜ \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}=\cos{x}\) ➜ \(\because \frac{dy}{dx}=y_{1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(\cos{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-\sin{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=-\frac{d}{dx}(\sin{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-\cos{x}\) ➜ \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=-\frac{d}{dx}(\cos{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=\sin{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{4}=y\) ➜ \(\because y=\sin{x}\)
\(\therefore y_{4}-y=0\)
(Showed)
\(y=\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\cos{x}\) ➜ \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}=\cos{x}\) ➜ \(\because \frac{dy}{dx}=y_{1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(\cos{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-\sin{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=-\frac{d}{dx}(\sin{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-\cos{x}\) ➜ \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=-\frac{d}{dx}(\cos{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=\sin{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{4}=y\) ➜ \(\because y=\sin{x}\)
\(\therefore y_{4}-y=0\)
(Showed)
\(Q.2.(iii)\) \(y=x^3\ln{x}\) হলে, প্রমাণ কর যে, \(y_{4}=\frac{6}{x}\)
সমাধানঃ
দেওয়া আছে,
\(y=x^3\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^3\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^3)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^3\frac{1}{x}+\ln{x}.3x^2\) ➜ \(\because \frac{dy}{dx}=y_{1}, \frac{d}{dx}(\ln{x})=\frac{1}{x}\), \(\frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow y_{1}=x^2+3x^2\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(x^2+3x^2\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\frac{d}{dx}(x^2)+3\frac{d}{dx}(x^2\ln{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=2x+3x^2\frac{d}{dx}(\ln{x})+3\ln{x}\frac{d}{dx}(x^2)\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=2x+3x^2\frac{1}{x}+3\ln{x}.2x\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{2}=2x+3x+6x\ln{x}\)
\(\Rightarrow y_{2}=5x+6x\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}(5x+6x\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=5\frac{d}{dx}(x)+6\frac{d}{dx}(x\ln{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{3}=5.1+6x\frac{d}{dx}(\ln{x})+6\ln{x}\frac{d}{dx}(x)\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{3}=5+6x\frac{1}{x}+6\ln{x}.1\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{3}=5+6+6\ln{x}\)
\(\Rightarrow y_{3}=11+6\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=\frac{d}{dx}(11+6\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=\frac{d}{dx}(11)+6\frac{d}{dx}(\ln{x})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{4}=0+6\frac{1}{x}\) ➜\(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\therefore y_{4}=\frac{6}{x}\)
(Proved)
\(y=x^3\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^3\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^3)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^3\frac{1}{x}+\ln{x}.3x^2\) ➜ \(\because \frac{dy}{dx}=y_{1}, \frac{d}{dx}(\ln{x})=\frac{1}{x}\), \(\frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow y_{1}=x^2+3x^2\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(x^2+3x^2\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\frac{d}{dx}(x^2)+3\frac{d}{dx}(x^2\ln{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=2x+3x^2\frac{d}{dx}(\ln{x})+3\ln{x}\frac{d}{dx}(x^2)\) ➜ \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=2x+3x^2\frac{1}{x}+3\ln{x}.2x\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{2}=2x+3x+6x\ln{x}\)
\(\Rightarrow y_{2}=5x+6x\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}(5x+6x\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=5\frac{d}{dx}(x)+6\frac{d}{dx}(x\ln{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{3}=5.1+6x\frac{d}{dx}(\ln{x})+6\ln{x}\frac{d}{dx}(x)\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{3}=5+6x\frac{1}{x}+6\ln{x}.1\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{3}=5+6+6\ln{x}\)
\(\Rightarrow y_{3}=11+6\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=\frac{d}{dx}(11+6\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=\frac{d}{dx}(11)+6\frac{d}{dx}(\ln{x})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{4}=0+6\frac{1}{x}\) ➜\(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\therefore y_{4}=\frac{6}{x}\)
(Proved)
\(Q.2.(iv)\) \(y=\frac{\ln{x}}{x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}=\frac{2\ln{x}-3}{x^3}\)
[ দিঃ২০০৯;বঃ২০০৭;ঢাঃ২০০৬;চঃ২০০৪,২০০১]
[ দিঃ২০০৯;বঃ২০০৭;ঢাঃ২০০৬;চঃ২০০৪,২০০১]
সমাধানঃ
দেওয়া আছে,
\(y=\frac{\ln{x}}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\ln{x}}{x}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{x\frac{d}{dx}(\ln{x})-\ln{x}\frac{d}{dx}(x)}{x^2}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{x\frac{1}{x}-\ln{x}.1}{x^2}\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=\frac{1-\ln{x}}{x^2}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{1-\ln{x}}{x^2}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{x^2\frac{d}{dx}(1-\ln{x})-(1-\ln{x})\frac{d}{dx}(x^2)}{x^4}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{x^2\times{-\frac{1}{x}}-(1-\ln{x}).2x}{x^4}\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\ln{x})=\frac{1}{x}\), \(\frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{-x-2x+2x\ln{x}}{x^4}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{-3x+2x\ln{x}}{x^4}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{x(2\ln{x}-3)}{x^4}\)
\(\therefore \frac{d^2y}{dx^2}=\frac{2\ln{x}-3}{x^3}\)
(Showed)
\(y=\frac{\ln{x}}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\ln{x}}{x}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{x\frac{d}{dx}(\ln{x})-\ln{x}\frac{d}{dx}(x)}{x^2}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{x\frac{1}{x}-\ln{x}.1}{x^2}\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=\frac{1-\ln{x}}{x^2}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{1-\ln{x}}{x^2}\right)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{x^2\frac{d}{dx}(1-\ln{x})-(1-\ln{x})\frac{d}{dx}(x^2)}{x^4}\) ➜ \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{x^2\times{-\frac{1}{x}}-(1-\ln{x}).2x}{x^4}\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\ln{x})=\frac{1}{x}\), \(\frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{-x-2x+2x\ln{x}}{x^4}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{-3x+2x\ln{x}}{x^4}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{x(2\ln{x}-3)}{x^4}\)
\(\therefore \frac{d^2y}{dx^2}=\frac{2\ln{x}-3}{x^3}\)
(Showed)
\(Q.2.(v)\) \(y=\tan{x}+\sec{x}\) হলে, প্রমাণ কর যে, \(\frac{d^2y}{dx^2}=\frac{\cos{x}}{(1-\sin{x})^2}\)
[ রাঃ২০১০,২০০৫,২০০১]
[ রাঃ২০১০,২০০৫,২০০১]
সমাধানঃ
দেওয়া আছে,
\(y=\tan{x}+\sec{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan{x}+\sec{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\tan{x})+\frac{d}{dx}(\sec{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\sec^2{x}+\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(tan{x})=\sec^2{x}, \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(\sec^2{x}+\sec{x}\tan{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dx}(\sec^2{x})+\frac{d}{dx}(\sec{x}\tan{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec{x}\frac{d}{dx}(\sec{x})+\sec{x}\frac{d}{dx}(\tan{x})+\tan{x}\frac{d}{dx}(\sec{x})\) ➜ \(\sec{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec{x}\sec{x}\tan{x}+\sec{x}\sec^2{x}+\tan{x}\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}, \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec^2{x}\tan{x}+\sec^3{x}+\tan^2{x}\sec{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\times{\frac{1}{\cos^2{x}}}\times{\frac{\sin{x}}{\cos{x}}}+\frac{1}{\cos^3{x}}+\frac{\sin^2{x}}{\cos^2{x}}\times{\frac{1}{\cos{x}}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{2\sin{x}}{\cos^3{x}}+\frac{1}{\cos^3{x}}+\frac{\sin^2{x}}{\cos^3{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{2\sin{x}+1+\sin^2{x}}{\cos^3{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1+2\sin{x}+\sin^2{x}}{\cos{x}\cos^2{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\sin{x})^2}{\cos{x}(1-\sin^2{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\sin{x})^2}{\cos{x}(1+\sin{x})(1-\sin{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1+\sin{x}}{\cos{x}(1-\sin{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\sin{x})(1-\sin{x})}{\cos{x}(1-\sin{x})^2}\) ➜ লব ও হরের সহিত \(1-\sin{x}\) গুণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1-\sin^2{x}}{\cos{x}(1-\sin{x})^2}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{\cos^2{x}}{\cos{x}(1-\sin{x})^2}\)
\(\therefore \frac{d^2y}{dx^2}=\frac{\cos{x}}{(1-\sin{x})^2}\)
(Proved)
\(y=\tan{x}+\sec{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan{x}+\sec{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\tan{x})+\frac{d}{dx}(\sec{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\sec^2{x}+\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(tan{x})=\sec^2{x}, \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(\sec^2{x}+\sec{x}\tan{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dx}(\sec^2{x})+\frac{d}{dx}(\sec{x}\tan{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec{x}\frac{d}{dx}(\sec{x})+\sec{x}\frac{d}{dx}(\tan{x})+\tan{x}\frac{d}{dx}(\sec{x})\) ➜ \(\sec{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec{x}\sec{x}\tan{x}+\sec{x}\sec^2{x}+\tan{x}\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}, \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\sec^2{x}\tan{x}+\sec^3{x}+\tan^2{x}\sec{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\times{\frac{1}{\cos^2{x}}}\times{\frac{\sin{x}}{\cos{x}}}+\frac{1}{\cos^3{x}}+\frac{\sin^2{x}}{\cos^2{x}}\times{\frac{1}{\cos{x}}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{2\sin{x}}{\cos^3{x}}+\frac{1}{\cos^3{x}}+\frac{\sin^2{x}}{\cos^3{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{2\sin{x}+1+\sin^2{x}}{\cos^3{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1+2\sin{x}+\sin^2{x}}{\cos{x}\cos^2{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\sin{x})^2}{\cos{x}(1-\sin^2{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\sin{x})^2}{\cos{x}(1+\sin{x})(1-\sin{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1+\sin{x}}{\cos{x}(1-\sin{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\sin{x})(1-\sin{x})}{\cos{x}(1-\sin{x})^2}\) ➜ লব ও হরের সহিত \(1-\sin{x}\) গুণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1-\sin^2{x}}{\cos{x}(1-\sin{x})^2}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{\cos^2{x}}{\cos{x}(1-\sin{x})^2}\)
\(\therefore \frac{d^2y}{dx^2}=\frac{\cos{x}}{(1-\sin{x})^2}\)
(Proved)
\(Q.2.(vi)\) \(y=\sec{x}\) হলে, দেখাও যে, \(y_{2}=y(2y^2-1)\)
[ যঃ২০১১,২০০৮;কুঃ২০১০,২০০২,২০০০;চঃ২০০৮,২০০৬,২০০৪;রাঃ২০০৭,২০০২; সিঃ২০০৭;বঃ২০০৬;ঢাঃ২০১৭,২০০০;মাঃ২০১২,২০০৫,২০০১ ]
[ যঃ২০১১,২০০৮;কুঃ২০১০,২০০২,২০০০;চঃ২০০৮,২০০৬,২০০৪;রাঃ২০০৭,২০০২; সিঃ২০০৭;বঃ২০০৬;ঢাঃ২০১৭,২০০০;মাঃ২০১২,২০০৫,২০০১ ]
সমাধানঃ
দেওয়া আছে,
\(y=\sec{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sec{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(\sec{x}\tan{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\sec{x}\frac{d}{dx}(\tan{x})+\tan{x}\frac{d}{dx}(\sec{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=\sec{x}\sec^2{x}+\tan{x}\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}, \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow y_{2}=\sec^3{x}+\tan^2{x}\sec{x}\)
\(\Rightarrow y_{2}=\sec^3{x}+(\sec^2{x}-1)\sec{x}\) ➜ \(\because \tan^2{x}=\sec^2{x}-1\)
\(\Rightarrow y_{2}=y^3+(y^2-1)y\) ➜ \(\because y=\sec{x}\)
\(\Rightarrow y_{2}=y(y^2+y^2-1)\)
\(\therefore y_{2}=y(2y^2-1)\)
(Showed)
\(y=\sec{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sec{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(\sec{x}\tan{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\sec{x}\frac{d}{dx}(\tan{x})+\tan{x}\frac{d}{dx}(\sec{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=\sec{x}\sec^2{x}+\tan{x}\sec{x}\tan{x}\) ➜ \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}, \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow y_{2}=\sec^3{x}+\tan^2{x}\sec{x}\)
\(\Rightarrow y_{2}=\sec^3{x}+(\sec^2{x}-1)\sec{x}\) ➜ \(\because \tan^2{x}=\sec^2{x}-1\)
\(\Rightarrow y_{2}=y^3+(y^2-1)y\) ➜ \(\because y=\sec{x}\)
\(\Rightarrow y_{2}=y(y^2+y^2-1)\)
\(\therefore y_{2}=y(2y^2-1)\)
(Showed)
\(Q.2.(vii)\) \(y=a\cos{(\ln{x})}+b\sin{(\ln{x})}\) হলে, দেখাও যে, \(x^2y_{2}+xy_{1}+y=0\)
[ ঢাঃ২০০৯ ]
[ ঢাঃ২০০৯ ]
সমাধানঃ
দেওয়া আছে,
\(y=a\cos{(\ln{x})}+b\sin{(\ln{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(a\cos{(\ln{x})}+b\sin{(\ln{x})})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a\frac{d}{dx}(\cos{(\ln{x})})+b\frac{d}{dx}(\sin{(\ln{x})})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=-a\sin{(\ln{x})}\frac{d}{dx}(\ln{x})+b\cos{(\ln{x})}\frac{d}{dx}(\ln{x})\) ➜\(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}=-a\sin{(\ln{x})}\frac{1}{x}+b\cos{(\ln{x})}\frac{1}{x}\) ➜\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow xy_{1}=-a\sin{(\ln{x})}+b\cos{(\ln{x})}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow \frac{d}{dx}(xy_{1})=\frac{d}{dx}(-a\sin{(\ln{x})}+b\cos{(\ln{x})})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(x)=-a\frac{d}{dx}(\sin{(\ln{x})})+b\frac{d}{dx}(\cos{(\ln{x})})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow xy_{2}+y_{1}.1=-a\cos{(\ln{x})}\frac{d}{dx}(\ln{x})-b\sin{(\ln{x})}\frac{d}{dx}(\ln{x})\) ➜ \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow xy_{2}+y_{1}=-a\cos{(\ln{x})}\frac{1}{x}-b\sin{(\ln{x})}\frac{1}{x}\) ➜\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow x^2y_{2}+xy_{1}=-a\cos{(\ln{x})}-b\sin{(\ln{x})}\) ➜উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow x^2y_{2}+xy_{1}=-(a\cos{(\ln{x})}+b\sin{(\ln{x})})\)
\(\Rightarrow x^2y_{2}+xy_{1}=-y\) ➜\(\because y=a\cos{(\ln{x})}+b\sin{(\ln{x})}\)
\(\therefore x^2y_{2}+xy_{1}+y=0\)
(Showed)
\(y=a\cos{(\ln{x})}+b\sin{(\ln{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(a\cos{(\ln{x})}+b\sin{(\ln{x})})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a\frac{d}{dx}(\cos{(\ln{x})})+b\frac{d}{dx}(\sin{(\ln{x})})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=-a\sin{(\ln{x})}\frac{d}{dx}(\ln{x})+b\cos{(\ln{x})}\frac{d}{dx}(\ln{x})\) ➜\(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}=-a\sin{(\ln{x})}\frac{1}{x}+b\cos{(\ln{x})}\frac{1}{x}\) ➜\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow xy_{1}=-a\sin{(\ln{x})}+b\cos{(\ln{x})}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow \frac{d}{dx}(xy_{1})=\frac{d}{dx}(-a\sin{(\ln{x})}+b\cos{(\ln{x})})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(x)=-a\frac{d}{dx}(\sin{(\ln{x})})+b\frac{d}{dx}(\cos{(\ln{x})})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow xy_{2}+y_{1}.1=-a\cos{(\ln{x})}\frac{d}{dx}(\ln{x})-b\sin{(\ln{x})}\frac{d}{dx}(\ln{x})\) ➜ \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow xy_{2}+y_{1}=-a\cos{(\ln{x})}\frac{1}{x}-b\sin{(\ln{x})}\frac{1}{x}\) ➜\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow x^2y_{2}+xy_{1}=-a\cos{(\ln{x})}-b\sin{(\ln{x})}\) ➜উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow x^2y_{2}+xy_{1}=-(a\cos{(\ln{x})}+b\sin{(\ln{x})})\)
\(\Rightarrow x^2y_{2}+xy_{1}=-y\) ➜\(\because y=a\cos{(\ln{x})}+b\sin{(\ln{x})}\)
\(\therefore x^2y_{2}+xy_{1}+y=0\)
(Showed)
\(Q.2.(viii)\) \(y=x^m\ln{x}\) হলে, প্রমাণ কর যে, \(xy_{1}=my+x^m\)
সমাধানঃ
দেওয়া আছে,
\(y=x^m\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^m\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=x^m\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^m)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^m\frac{1}{x}+\ln{x}mx^{m-1}\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow xy_{1}=x^m+\ln{x}mx^{m-1}\times{x}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow xy_{1}=x^m+mx^{m-1+1}\ln{x}\)
\(\Rightarrow xy_{1}=x^m+mx^{m}\ln{x}\)
\(\Rightarrow xy_{1}=x^m+my\) ➜ \(\because y=x^m\ln{x}\)
\(\therefore xy_{1}=my+x^m\)
(Proved)
\(y=x^m\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^m\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=x^m\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^m)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^m\frac{1}{x}+\ln{x}mx^{m-1}\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow xy_{1}=x^m+\ln{x}mx^{m-1}\times{x}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow xy_{1}=x^m+mx^{m-1+1}\ln{x}\)
\(\Rightarrow xy_{1}=x^m+mx^{m}\ln{x}\)
\(\Rightarrow xy_{1}=x^m+my\) ➜ \(\because y=x^m\ln{x}\)
\(\therefore xy_{1}=my+x^m\)
(Proved)
\(Q.2.(ix)\) \(y=x^{n-1}\ln{x}\) হলে, প্রমাণ কর যে, \(x^2y_{2}+(3-2n)xy_{1}+(n-1)^2y=0\)
[ বুয়েটঃ২০০৯-২০১০]
[ বুয়েটঃ২০০৯-২০১০]
সমাধানঃ
দেওয়া আছে,
\(y=x^{n-1}\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{n-1}\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=x^{n-1}\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^{n-1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^{n-1}\frac{1}{x}+\ln{x}(n-1)x^{n-1-1}\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow xy_{1}=x^{n-1}+\ln{x}(n-1)x^{n-2}\times{x}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow xy_{1}=x^{n-1}+(n-1)x^{n-2+1}\ln{x}\)
\(\Rightarrow xy_{1}=x^{n-1}+(n-1)x^{n-1}\ln{x}\)
\(\Rightarrow xy_{1}=x^{n-1}+(n-1)y\) ➜ \(\because y=x^m\ln{x}\)
\(\Rightarrow \frac{d}{dx}(xy_{1})=\frac{d}{dx}\{x^{n-1}+(n-1)y\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(xy_{1})=\frac{d}{dx}\{x^{n-1}+(n-1)y\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(x)=\frac{d}{dx}(x^{n-1})+(n-1)\frac{d}{dx}(y)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow xy_{2}+y_{1}.1=(n-1)x^{n-1-1}+(n-1)y_{1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)\)
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)x^{n-2}\times{x}+(n-1)xy_{1}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)x^{n-2+1}+(n-1)xy_{1}\)
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)x^{n-1}+(n-1)xy_{1}\)
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)\{xy_{1}-(n-1)y\}+(n-1)xy_{1}\) ➜ \(\because xy_{1}=x^{n-1}+(n-1)y \Rightarrow xy_{1}-(n-1)y=x^{n-1}\)
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)xy_{1}-(n-1)^2y+(n-1)xy_{1}\)
\(\Rightarrow x^2y_{2}+xy_{1}=2(n-1)xy_{1}-(n-1)^2y\)
\(\Rightarrow x^2y_{2}+xy_{1}-2(n-1)xy_{1}+(n-1)^2y=0\)
\(\Rightarrow x^2y_{2}+(1-2n+2)xy_{1}+(n-1)^2y=0\)
\(\therefore x^2y_{2}+(3-2n)xy_{1}+(n-1)^2y=0\)
(Proved)
\(y=x^{n-1}\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{n-1}\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=x^{n-1}\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^{n-1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^{n-1}\frac{1}{x}+\ln{x}(n-1)x^{n-1-1}\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow xy_{1}=x^{n-1}+\ln{x}(n-1)x^{n-2}\times{x}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow xy_{1}=x^{n-1}+(n-1)x^{n-2+1}\ln{x}\)
\(\Rightarrow xy_{1}=x^{n-1}+(n-1)x^{n-1}\ln{x}\)
\(\Rightarrow xy_{1}=x^{n-1}+(n-1)y\) ➜ \(\because y=x^m\ln{x}\)
\(\Rightarrow \frac{d}{dx}(xy_{1})=\frac{d}{dx}\{x^{n-1}+(n-1)y\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(xy_{1})=\frac{d}{dx}\{x^{n-1}+(n-1)y\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(x)=\frac{d}{dx}(x^{n-1})+(n-1)\frac{d}{dx}(y)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow xy_{2}+y_{1}.1=(n-1)x^{n-1-1}+(n-1)y_{1}\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(x^n)=nx^{n-1}\)\)
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)x^{n-2}\times{x}+(n-1)xy_{1}\) ➜ উভয় পার্শে \(x\)গুণ করে।
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)x^{n-2+1}+(n-1)xy_{1}\)
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)x^{n-1}+(n-1)xy_{1}\)
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)\{xy_{1}-(n-1)y\}+(n-1)xy_{1}\) ➜ \(\because xy_{1}=x^{n-1}+(n-1)y \Rightarrow xy_{1}-(n-1)y=x^{n-1}\)
\(\Rightarrow x^2y_{2}+xy_{1}=(n-1)xy_{1}-(n-1)^2y+(n-1)xy_{1}\)
\(\Rightarrow x^2y_{2}+xy_{1}=2(n-1)xy_{1}-(n-1)^2y\)
\(\Rightarrow x^2y_{2}+xy_{1}-2(n-1)xy_{1}+(n-1)^2y=0\)
\(\Rightarrow x^2y_{2}+(1-2n+2)xy_{1}+(n-1)^2y=0\)
\(\therefore x^2y_{2}+(3-2n)xy_{1}+(n-1)^2y=0\)
(Proved)
\(Q.2.(x)\) \(y=\sin^{-1}{x}\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=0\)
সমাধানঃ
দেওয়া আছে,
\(y=\sin^{-1}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=1\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=1\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=\frac{d}{dx}(1)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=0\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=0\)
(Proved)
\(y=\sin^{-1}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=1\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=1\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=\frac{d}{dx}(1)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=0\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=0\)
(Proved)
\(Q.2.(xi)\) \(y=(\sin^{-1}{x})^2\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}-2=0\)
[ রাঃ২০১১,২০০১;বঃ২০০৮;কুঃ২০০০;মাঃ২০১০,২০০৩]
[ রাঃ২০১১,২০০১;বঃ২০০৮;কুঃ২০০০;মাঃ২০১০,২০০৩]
সমাধানঃ
দেওয়া আছে,
\(y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{x}\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=2\sin^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\sin^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-2=0\)
(Proved)
\(y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{x}\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=2\sin^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\sin^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-2=0\)
(Proved)
\(Q.2.(xii)\) \(f(u)=\sin^{-1}{u}, y=\{f(2x)\}^2\) হলে, দেখাও যে, \((1-4x^2)y_{2}-4xy_{1}-8=0\)
[ সিঃ২০১৭ ]
[ সিঃ২০১৭ ]
সমাধানঃ
দেওয়া আছে,
\(f(u)=\sin^{-1}{u}, y=\{f(2x)\}^2\)
ধরি,
\(f(u)=\sin^{-1}{u} ....(1)\) এবং
\(y=\{f(2x)\}^2 ......(2)\)
\((1)\) হতে,
\(f(2x)=\sin^{-1}{2x} ....(3)\)
\((2)\) ও \((3)\) হতে,
\(y=(\sin^{-1}{2x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{2x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{2x}\frac{d}{dx}(\sin^{-1}{2x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{2x}.\frac{1}{\sqrt{1-4x^2}}\frac{d}{dx}(2x)\) ➜ \(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{2x}.\frac{1}{\sqrt{1-4x^2}}.2\) ➜ \(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \sqrt{1-4x^2}y_{1}=4\sin^{-1}{2x}\) ➜ উভয় পার্শে \(\sqrt{1-4x^2}\) গুণ করে।
\(\Rightarrow (1-4x^2)y^2_{1}=16(\sin^{-1}{2x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-4x^2)y^2_{1}=16y\) ➜ \(\because y=(\sin^{-1}{2x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-4x^2)y^2_{1}\}=16\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-4x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-4x^2)=16y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-4x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-8x)=16y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-4x^2)y_{1}y_{2}-8xy^2_{1}=16y_{1}\)
\(\Rightarrow 2(1-4x^2)y_{1}y_{2}-8xy^2_{1}-16y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-4x^2)y_{2}-4xy_{1}-8\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-4x^2)y_{2}-4xy_{1}-8=0\)
\(\therefore (1-4x^2)y_{2}-4xy_{1}-8=0\)
(Proved)
\(f(u)=\sin^{-1}{u}, y=\{f(2x)\}^2\)
ধরি,
\(f(u)=\sin^{-1}{u} ....(1)\) এবং
\(y=\{f(2x)\}^2 ......(2)\)
\((1)\) হতে,
\(f(2x)=\sin^{-1}{2x} ....(3)\)
\((2)\) ও \((3)\) হতে,
\(y=(\sin^{-1}{2x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{2x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{2x}\frac{d}{dx}(\sin^{-1}{2x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{2x}.\frac{1}{\sqrt{1-4x^2}}\frac{d}{dx}(2x)\) ➜ \(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{2x}.\frac{1}{\sqrt{1-4x^2}}.2\) ➜ \(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \sqrt{1-4x^2}y_{1}=4\sin^{-1}{2x}\) ➜ উভয় পার্শে \(\sqrt{1-4x^2}\) গুণ করে।
\(\Rightarrow (1-4x^2)y^2_{1}=16(\sin^{-1}{2x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-4x^2)y^2_{1}=16y\) ➜ \(\because y=(\sin^{-1}{2x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-4x^2)y^2_{1}\}=16\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-4x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-4x^2)=16y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-4x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-8x)=16y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-4x^2)y_{1}y_{2}-8xy^2_{1}=16y_{1}\)
\(\Rightarrow 2(1-4x^2)y_{1}y_{2}-8xy^2_{1}-16y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-4x^2)y_{2}-4xy_{1}-8\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-4x^2)y_{2}-4xy_{1}-8=0\)
\(\therefore (1-4x^2)y_{2}-4xy_{1}-8=0\)
(Proved)
\(Q.2.(xiii)\) \(x=\sin{\sqrt{y}}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}-2=0\)
[ বঃ২০১২,২০০৫,২০০৩;চঃ২০১১;ঢাঃ২০০৮;কুঃ২০০৮,২০০৩ ]
[ বঃ২০১২,২০০৫,২০০৩;চঃ২০১১;ঢাঃ২০০৮;কুঃ২০০৮,২০০৩ ]
সমাধানঃ
দেওয়া আছে,
\(x=\sin{\sqrt{y}}\)
\(\Rightarrow \sin{\sqrt{y}}=x\)
\(\Rightarrow \sqrt{y}=\sin^{-1}{x}\)
\(\Rightarrow y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{x}\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=2\sin^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\sin^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-2=0\)
(Showed)
\(x=\sin{\sqrt{y}}\)
\(\Rightarrow \sin{\sqrt{y}}=x\)
\(\Rightarrow \sqrt{y}=\sin^{-1}{x}\)
\(\Rightarrow y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{x}\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\sin^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=2\sin^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\sin^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-2=0\)
(Showed)
\(Q.2.(xiv)\) \(y=\frac{1}{2}(\sin^{-1}{x})^2\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}-1=0\)
[ বুয়েটঃ২০০৫-২০০৬]
[ বুয়েটঃ২০০৫-২০০৬]
সমাধানঃ
দেওয়া আছে,
\(y=\frac{1}{2}(\sin^{-1}{x})^2\)
\(\Rightarrow 2y=(\sin^{-1}{x})^2\)
\(\Rightarrow 2\frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y_{1}=2\sin^{-1}{x}\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\sin^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=\sin^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=(\sin^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=2y\) ➜ \(\because 2y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=2\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=2y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=2y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=2y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-2y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-1\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-1=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-1=0\)
(Showed)
\(y=\frac{1}{2}(\sin^{-1}{x})^2\)
\(\Rightarrow 2y=(\sin^{-1}{x})^2\)
\(\Rightarrow 2\frac{d}{dx}(y)=\frac{d}{dx}(\sin^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y_{1}=2\sin^{-1}{x}\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\sin^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=\sin^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=(\sin^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=2y\) ➜ \(\because 2y=(\sin^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=2\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=2y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=2y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=2y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-2y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-1\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-1=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-1=0\)
(Showed)
\(Q.2.(xv)\) \(y=\cos^{-1}{x}\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=0\)
সমাধানঃ
দেওয়া আছে,
\(y=\cos^{-1}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-1\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=1\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=\frac{d}{dx}(1)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=0\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=0\)
(Proved)
\(y=\cos^{-1}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-1\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=1\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=\frac{d}{dx}(1)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=0\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=0\)
(Proved)
\(Q.2.(xvi)\) \(\ln{y}=bz\) এবং \(\cos{z}=x\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=b^2y\)
[ বঃ২০১৭]
[ বঃ২০১৭]
সমাধানঃ
দেওয়া আছে,
\(\ln{y}=bz\) এবং \(\cos{z}=x\)
ধরি,
\(\ln{y}=bz ....(1)\) এবং
\(\cos{z}=x ......(2)\)
\((2)\) হতে,
\(z=\cos^{-1}{x} ....(3)\)
\((1)\) ও \((3)\) হতে,
\(\ln{y}=b\cos^{-1}{x}\)
\(\Rightarrow y=e^{b\cos^{-1}{x}}\) ➜ \(\because \ln{x}=y\Rightarrow x=e^{y}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{b\cos^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{b\cos^{-1}{x}}.b\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(b\cos^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-be^{b\cos^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-by\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে এবং \(\because y=e^{b\cos^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=b^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=b^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=b^2.2yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=2b^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=2b^2yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-2b^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-b^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-b^2y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}-b^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=b^2y\)
(Proved)
\(\ln{y}=bz\) এবং \(\cos{z}=x\)
ধরি,
\(\ln{y}=bz ....(1)\) এবং
\(\cos{z}=x ......(2)\)
\((2)\) হতে,
\(z=\cos^{-1}{x} ....(3)\)
\((1)\) ও \((3)\) হতে,
\(\ln{y}=b\cos^{-1}{x}\)
\(\Rightarrow y=e^{b\cos^{-1}{x}}\) ➜ \(\because \ln{x}=y\Rightarrow x=e^{y}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{b\cos^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{b\cos^{-1}{x}}.b\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(b\cos^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-be^{b\cos^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-by\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে এবং \(\because y=e^{b\cos^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=b^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=b^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=b^2.2yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=2b^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=2b^2yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-2b^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-b^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-b^2y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}-b^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=b^2y\)
(Proved)
\(Q.2.(xvii)\) \(y=(\cos^{-1}{x})^2\) হলে, দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2\)
[ ঢাঃ২০১২,২০০৪;দিঃ২০১২;সিঃ২০০৮;বঃ২০০৬;চঃ২০০৩; যঃ২০০৩;কুঃ২০০১;রাঃ২০০৪;মাঃ২০০৮ ]
[ ঢাঃ২০১২,২০০৪;দিঃ২০১২;সিঃ২০০৮;বঃ২০০৬;চঃ২০০৩; যঃ২০০৩;কুঃ২০০১;রাঃ২০০৪;মাঃ২০০৮ ]
সমাধানঃ
দেওয়া আছে,
\(y=(\cos^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\cos^{-1}{x}\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-2\cos^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-2\cos^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\cos^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\cos^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\(y=(\cos^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\cos^{-1}{x}\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-2\cos^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-2\cos^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\cos^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\cos^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=2\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\(Q.2.(xviii)\) যদি \(\cos{\sqrt{y}}=x\) হয়, তবে দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2=0\)
[ যঃ২০১২,২০০৪,২০০৬;ঢাঃ২০১১,২০০১;সিঃ২০০১০,২০০৪; রাঃ২০০৯,২০০৭;চঃ২০০৬ ]
[ যঃ২০১২,২০০৪,২০০৬;ঢাঃ২০১১,২০০১;সিঃ২০০১০,২০০৪; রাঃ২০০৯,২০০৭;চঃ২০০৬ ]
সমাধানঃ
দেওয়া আছে,
\(\cos{\sqrt{y}}=x\)
\(\Rightarrow \sqrt{y}=\cos^{-1}{x}\)
\(\Rightarrow y=(\cos^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\cos^{-1}{x}\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-2\cos^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-2\cos^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\cos^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\cos^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\(\cos{\sqrt{y}}=x\)
\(\Rightarrow \sqrt{y}=\cos^{-1}{x}\)
\(\Rightarrow y=(\cos^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\cos^{-1}{x}\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-2\cos^{-1}{x}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-2\cos^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\cos^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\cos^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\(Q.2.(xix)\) \(y=\tan^{-1}{x}\) হয়, তবে প্রমাণ কর যে, \((1+x^2)y_{2}+2xy_{1}=0\)
[ মাঃ২০০৭,২০০৫;ঢাঃ,কুঃ২০০৫]
[ মাঃ২০০৭,২০০৫;ঢাঃ,কুঃ২০০৫]
সমাধানঃ
দেওয়া আছে,
\(y=\tan^{-1}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=1\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y^2_{1}\}=\frac{d}{dx}(1)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1+x^2)=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow (1+x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0+2x)=0\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1+x^2)y_{1}y_{2}+2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(1+x^2)y_{2}+xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1+x^2)y_{2}+xy_{1}=0\)
\(\therefore (1+x^2)y_{2}+xy_{1}=0\)
(Proved)
\(y=\tan^{-1}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=1\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y^2_{1}\}=\frac{d}{dx}(1)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1+x^2)=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow (1+x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0+2x)=0\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1+x^2)y_{1}y_{2}+2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(1+x^2)y_{2}+xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1+x^2)y_{2}+xy_{1}=0\)
\(\therefore (1+x^2)y_{2}+xy_{1}=0\)
(Proved)
\(Q.2.(xx)\) \(y=(\tan^{-1}{x})^2\) হয়, তবে প্রমাণ কর যে, \((1+x^2)^2y_{2}+2(1+x^2)xy_{1}=2\)
সমাধানঃ
দেওয়া আছে,
\(y=(\tan^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\tan^{-1}{x}\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\tan^{-1}{x}.\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=2\tan^{-1}{x}\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow (1+x^2)^2y^2_{1}=4(\tan^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1+x^2)^2y^2_{1}=4y\) ➜ \(\because y=(\tan^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1+x^2)^2y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)^2\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1+x^2)^2=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1+x^2)^2.2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}2(1+x^2)(0+2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1+x^2)^2y_{1}y_{2}+4x(1+x^2)y^2_{1}=4y_{1}\)
\(\Rightarrow 2(1+x^2)^2y_{1}y_{2}+4x(1+x^2)y^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1+x^2)^2y_{2}+2x(1+x^2)y_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1+x^2)^2y_{2}+2x(1+x^2)y_{1}-2=0\)
\(\Rightarrow (1+x^2)^2y_{2}+2x(1+x^2)y_{1}-2=0\)
\(\therefore (1+x^2)^2y_{2}+2(1+x^2)xy_{1}=2\)
(Proved)
\(y=(\tan^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\tan^{-1}{x}\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\tan^{-1}{x}.\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=2\tan^{-1}{x}\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow (1+x^2)^2y^2_{1}=4(\tan^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1+x^2)^2y^2_{1}=4y\) ➜ \(\because y=(\tan^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1+x^2)^2y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)^2\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1+x^2)^2=4y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1+x^2)^2.2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}2(1+x^2)(0+2x)=4y_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1+x^2)^2y_{1}y_{2}+4x(1+x^2)y^2_{1}=4y_{1}\)
\(\Rightarrow 2(1+x^2)^2y_{1}y_{2}+4x(1+x^2)y^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1+x^2)^2y_{2}+2x(1+x^2)y_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1+x^2)^2y_{2}+2x(1+x^2)y_{1}-2=0\)
\(\Rightarrow (1+x^2)^2y_{2}+2x(1+x^2)y_{1}-2=0\)
\(\therefore (1+x^2)^2y_{2}+2(1+x^2)xy_{1}=2\)
(Proved)
\(Q.2.(xxi)\) \(y=\sqrt{a+b\cos{x}}\) হলে, দেখাও যে, \(2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=a\)
[ যঃ২০১৭]
[ যঃ২০১৭]
সমাধানঃ
দেওয়া আছে,
\(y=\sqrt{a+b\cos{x}}\)
\(\Rightarrow y^2=a+b\cos{x}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(y^2)=\frac{d}{dx}(a+b\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2yy_{1}=\frac{d}{dx}(a)+b\frac{d}{dx}(\cos{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow 2yy_{1}=0-b\sin{x}\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow 2yy_{1}=-b\sin{x}\)
\(\Rightarrow 2\frac{d}{dx}(yy_{1})=-b\frac{d}{dx}(\sin{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y\frac{d}{dx}(y_{1})+2y_{1}\frac{d}{dx}(y)=-b\cos{x}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow 2yy_{2}+2y_{1}y_{1}=-b\cos{x}\)
\(\Rightarrow 2yy_{2}+2(y_{1})^2=a-a-b\cos{x}\)
\(\Rightarrow 2yy_{2}+2(y_{1})^2=a-(a+b\cos{x})\)
\(\Rightarrow 2yy_{2}+2(y_{1})^2=a-y^2\) ➜ \(\because y^2=a+b\cos{x}\)
\(\therefore 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=a\)| \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\(y=\sqrt{a+b\cos{x}}\)
\(\Rightarrow y^2=a+b\cos{x}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(y^2)=\frac{d}{dx}(a+b\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2yy_{1}=\frac{d}{dx}(a)+b\frac{d}{dx}(\cos{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow 2yy_{1}=0-b\sin{x}\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow 2yy_{1}=-b\sin{x}\)
\(\Rightarrow 2\frac{d}{dx}(yy_{1})=-b\frac{d}{dx}(\sin{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y\frac{d}{dx}(y_{1})+2y_{1}\frac{d}{dx}(y)=-b\cos{x}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow 2yy_{2}+2y_{1}y_{1}=-b\cos{x}\)
\(\Rightarrow 2yy_{2}+2(y_{1})^2=a-a-b\cos{x}\)
\(\Rightarrow 2yy_{2}+2(y_{1})^2=a-(a+b\cos{x})\)
\(\Rightarrow 2yy_{2}+2(y_{1})^2=a-y^2\) ➜ \(\because y^2=a+b\cos{x}\)
\(\therefore 2y\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+y^2=a\)| \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Showed)
\(Q.2.(xxii)\) \(y=\ln{(x+\sqrt{a^2+x^2})}\) হলে, দেখাও যে, \((a^2+x^2)y_{2}+xy_{1}=0\)
[ চঃ,মাঃ২০১০;যঃ২০০৩,২০০১]
[ চঃ,মাঃ২০১০;যঃ২০০৩,২০০১]
সমাধানঃ
দেওয়া আছে,
\(y=\ln{(x+\sqrt{a^2+x^2})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(x+\sqrt{a^2+x^2})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\frac{d}{dx}(x+\sqrt{a^2+x^2})\) ➜\((x+\sqrt{a^2+x^2})\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\left[\frac{d}{dx}(x)+\frac{d}{dx}(\sqrt{a^2+x^2})\right]\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\left[1+\frac{1}{2\sqrt{a^2+x^2}}\frac{d}{dx}(a^2+x^2)\right]\) ➜ \((a^2+x^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\left[1+\frac{1}{2\sqrt{a^2+x^2}}(0+2x)\right]\) ➜\(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\left[1+\frac{x}{\sqrt{a^2+x^2}}\right]\)
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\times{\frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}}}\)
\(\Rightarrow y_{1}=\frac{1}{\sqrt{a^2+x^2}}\)
\(\Rightarrow (\sqrt{a^2+x^2})y_{1}=1\) ➜ উভয় পার্শে \(\sqrt{a^2+x^2}\) গুণ করে।
\(\Rightarrow (a^2+x^2)y^2_{1}=1\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(a^2+x^2)y^2_{1}\}=\frac{d}{dx}(1)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (a^2+x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(a^2+x^2)=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(\Rightarrow (a^2+x^2).2y_{1}y_{2}+y^2_{1}(0+2x)=0\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(a^2+x^2)y_{1}y_{2}+2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(a^2+x^2)y_{2}+xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (a^2+x^2)y_{2}+xy_{1}=0\)
\(\therefore (a^2+x^2)y_{2}+xy_{1}=0\)
(Showed)
\(y=\ln{(x+\sqrt{a^2+x^2})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(x+\sqrt{a^2+x^2})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\frac{d}{dx}(x+\sqrt{a^2+x^2})\) ➜\((x+\sqrt{a^2+x^2})\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\left[\frac{d}{dx}(x)+\frac{d}{dx}(\sqrt{a^2+x^2})\right]\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\left[1+\frac{1}{2\sqrt{a^2+x^2}}\frac{d}{dx}(a^2+x^2)\right]\) ➜ \((a^2+x^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\left[1+\frac{1}{2\sqrt{a^2+x^2}}(0+2x)\right]\) ➜\(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\left[1+\frac{x}{\sqrt{a^2+x^2}}\right]\)
\(\Rightarrow y_{1}=\frac{1}{x+\sqrt{a^2+x^2}}\times{\frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}}}\)
\(\Rightarrow y_{1}=\frac{1}{\sqrt{a^2+x^2}}\)
\(\Rightarrow (\sqrt{a^2+x^2})y_{1}=1\) ➜ উভয় পার্শে \(\sqrt{a^2+x^2}\) গুণ করে।
\(\Rightarrow (a^2+x^2)y^2_{1}=1\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(a^2+x^2)y^2_{1}\}=\frac{d}{dx}(1)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (a^2+x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(a^2+x^2)=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(\Rightarrow (a^2+x^2).2y_{1}y_{2}+y^2_{1}(0+2x)=0\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(a^2+x^2)y_{1}y_{2}+2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(a^2+x^2)y_{2}+xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (a^2+x^2)y_{2}+xy_{1}=0\)
\(\therefore (a^2+x^2)y_{2}+xy_{1}=0\)
(Showed)
\(Q.2.(xxiii)\) \(y=(x+\sqrt{1+x^2})^m\) হয়, তবে প্রমাণ কর যে, \((1+x^2)y_{2}+xy_{1}-m^2y=0\)
[সিঃ২০১১;যঃ,বঃ২০০১০;চঃ২০০৯]
[সিঃ২০১১;যঃ,বঃ২০০১০;চঃ২০০৯]
সমাধানঃ
দেওয়া আছে,
\(y=(x+\sqrt{1+x^2})^m\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x+\sqrt{1+x^2})^m\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\frac{d}{dx}(x+\sqrt{1+x^2})\) ➜\((x+\sqrt{1+x^2})\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\left[\frac{d}{dx}(x)+\frac{d}{dx}(\sqrt{1+x^2})\right]\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\left[1+\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}(1+x^2)\right]\) ➜ \((1+x^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\left[1+\frac{1}{2\sqrt{1+x^2}}(0+2x)\right]\) ➜\(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\left[1+\frac{x}{\sqrt{1+x^2}}\right]\)
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\times{\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}}\)
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m}\frac{1}{\sqrt{1+x^2}}\)
\(\Rightarrow (\sqrt{1+x^2})y_{1}=m(x+\sqrt{1+x^2})^{m}\) ➜ উভয় পার্শে \(\sqrt{1+x^2}\) গুণ করে।
\(\Rightarrow (\sqrt{1+x^2})y_{1}=my\) ➜\(\because y=(x+\sqrt{1+x^2})^{m}\)
\(\Rightarrow (1+x^2)y^2_{1}=m^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y^2_{1}\}=m^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1+x^2)=m^22yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1+x^2).2y_{1}y_{2}+y^2_{1}(0+2x)=m^22yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1+x^2)y_{1}y_{2}+2xy^2_{1}-m^22yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1+x^2)y_{2}+xy_{1}-m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1+x^2)y_{2}+xy_{1}-m^2y=0\)
\(\therefore (1+x^2)y_{2}+xy_{1}-m^2y=0\)
(Showed)
\(y=(x+\sqrt{1+x^2})^m\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x+\sqrt{1+x^2})^m\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\frac{d}{dx}(x+\sqrt{1+x^2})\) ➜\((x+\sqrt{1+x^2})\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\left[\frac{d}{dx}(x)+\frac{d}{dx}(\sqrt{1+x^2})\right]\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\left[1+\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}(1+x^2)\right]\) ➜ \((1+x^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\left[1+\frac{1}{2\sqrt{1+x^2}}(0+2x)\right]\) ➜\(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\left[1+\frac{x}{\sqrt{1+x^2}}\right]\)
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m-1}\times{\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}}\)
\(\Rightarrow y_{1}=m(x+\sqrt{1+x^2})^{m}\frac{1}{\sqrt{1+x^2}}\)
\(\Rightarrow (\sqrt{1+x^2})y_{1}=m(x+\sqrt{1+x^2})^{m}\) ➜ উভয় পার্শে \(\sqrt{1+x^2}\) গুণ করে।
\(\Rightarrow (\sqrt{1+x^2})y_{1}=my\) ➜\(\because y=(x+\sqrt{1+x^2})^{m}\)
\(\Rightarrow (1+x^2)y^2_{1}=m^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y^2_{1}\}=m^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1+x^2)=m^22yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1+x^2).2y_{1}y_{2}+y^2_{1}(0+2x)=m^22yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1+x^2)y_{1}y_{2}+2xy^2_{1}-m^22yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1+x^2)y_{2}+xy_{1}-m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1+x^2)y_{2}+xy_{1}-m^2y=0\)
\(\therefore (1+x^2)y_{2}+xy_{1}-m^2y=0\)
(Showed)
\(Q.2.(xxiv)\) \(y=e^{a\sin^{-1}{x}}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}=a^2y\)
[কুঃ২০১২,২০০১;ঢাঃ২০১১,২০০৩;বঃ২০১১;যঃ২০০৯,২০০২; সিঃ২০০৯,২০০৭,২০০৪;চঃ২০০৫;রাঃ২০০০;মাঃ২০১২,২০০৯,২০০৬]
[কুঃ২০১২,২০০১;ঢাঃ২০১১,২০০৩;বঃ২০১১;যঃ২০০৯,২০০২; সিঃ২০০৯,২০০৭,২০০৪;চঃ২০০৫;রাঃ২০০০;মাঃ২০১২,২০০৯,২০০৬]
সমাধানঃ
দেওয়া আছে,
\(y=e^{a\sin^{-1}{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{a\sin^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{a\sin^{-1}{x}}.a\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(a\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=ae^{a\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=ay\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে এবং \(\because y=e^{a\sin^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=a^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=a^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=a^2.2yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=2a^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=2a^2yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-2a^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-a^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-a^2y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}-a^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=a^2y\)
(Showed)
\(y=e^{a\sin^{-1}{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{a\sin^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{a\sin^{-1}{x}}.a\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(a\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=ae^{a\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=ay\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে এবং \(\because y=e^{a\sin^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=a^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=a^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=a^2.2yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=2a^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=2a^2yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-2a^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-a^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-a^2y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}-a^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=a^2y\)
(Showed)
\(Q.2.(xxv)\) \(\ln{y}=a\sin^{-1}{x}\) হলে, দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-a^2y=0\)
[ ঢাঃ২০০৭ ]
[ ঢাঃ২০০৭ ]
সমাধানঃ
দেওয়া আছে,
\(\ln{y}=a\sin^{-1}{x}\)
\(\Rightarrow y=e^{a\sin^{-1}{x}}\) ➜ \(\because \ln{x}=y\Rightarrow x=e^{y}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{a\sin^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{a\sin^{-1}{x}}.a\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(a\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=ae^{a\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=ay\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে এবং \(\because y=e^{a\sin^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=a^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=a^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=a^2.2yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=2a^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=2a^2yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-2a^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-a^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-a^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-a^2y=0\)
(Showed)
\(\ln{y}=a\sin^{-1}{x}\)
\(\Rightarrow y=e^{a\sin^{-1}{x}}\) ➜ \(\because \ln{x}=y\Rightarrow x=e^{y}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{a\sin^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{a\sin^{-1}{x}}.a\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(a\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=ae^{a\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=ay\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে এবং \(\because y=e^{a\sin^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=a^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=a^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=a^2.2yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=2a^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=2a^2yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-2a^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-a^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-a^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-a^2y=0\)
(Showed)
\(Q.2.(xxvi)\) যদি \(y=e^{\tan^{-1}{x}}\) অথবা \(\ln{y}=\tan^{-1}{x}\) হয়, তবে প্রমাণ কর যে, \((1+x^2)y_{2}+(2x-1)y_{1}=0\)
[ ঢাঃ,বঃ২০১২;কুঃ২০১৭,২০১১;রাঃ২০১০,২০০৮,২০০৫,২০০২;যঃ২০১০ ]
[ ঢাঃ,বঃ২০১২;কুঃ২০১৭,২০১১;রাঃ২০১০,২০০৮,২০০৫,২০০২;যঃ২০১০ ]
সমাধানঃ
দেওয়া আছে,
\(y=e^{\tan^{-1}{x}}\) অথবা, \(\ln{y}=\tan^{-1}{x}\)
\(\Rightarrow y=e^{\tan^{-1}{x}}\) ➜ \(\because \ln{x}=y\Rightarrow x=e^{y}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{\tan^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{\tan^{-1}{x}}.\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=e^{\tan^{-1}{x}}\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=y\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে এবং \(\because y=e^{\tan^{-1}{x}}\)
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y_{1}\}=\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(1+x^2)=y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1+x^2)y_{2}+y_{1}(0+2x)=y_{1}\) ➜ \(\because \frac{d}{dx}(y_{1})=y_{2}\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow (1+x^2)y_{2}+2xy_{1}-y_{1}=0\)
\(\therefore (1+x^2)y_{2}+(2x-1)y_{1}=0\)
(Proved)
\(y=e^{\tan^{-1}{x}}\) অথবা, \(\ln{y}=\tan^{-1}{x}\)
\(\Rightarrow y=e^{\tan^{-1}{x}}\) ➜ \(\because \ln{x}=y\Rightarrow x=e^{y}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{\tan^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{\tan^{-1}{x}}.\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=e^{\tan^{-1}{x}}\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=y\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে এবং \(\because y=e^{\tan^{-1}{x}}\)
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y_{1}\}=\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(1+x^2)=y_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow (1+x^2)y_{2}+y_{1}(0+2x)=y_{1}\) ➜ \(\because \frac{d}{dx}(y_{1})=y_{2}\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow (1+x^2)y_{2}+2xy_{1}-y_{1}=0\)
\(\therefore (1+x^2)y_{2}+(2x-1)y_{1}=0\)
(Proved)
\(Q.2.(xxvii)\) \(y=e^{4\sin^{-1}{x}}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}=16y\)
[ চঃ২০০২ ]
[ চঃ২০০২ ]
সমাধানঃ
দেওয়া আছে,
\(y=e^{4\sin^{-1}{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{4\sin^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{4\sin^{-1}{x}}.4\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(4\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=4e^{4\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=4y\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে এবং \(\because y=e^{4\sin^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=16y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=16\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=16.2yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=32yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=32yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-32yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-16y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-16y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}-16y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=16y\)
(Showed)
\(y=e^{4\sin^{-1}{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{4\sin^{-1}{x}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{4\sin^{-1}{x}}.4\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(4\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=4e^{4\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=4y\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে এবং \(\because y=e^{4\sin^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=16y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=16\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=16.2yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2).2y_{1}\frac{d}{dx}(y_{1})+y^2_{1}(0-2x)=32yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}\frac{d}{dx}(y_{1})\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=32yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-32yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-16y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-16y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}-16y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=16y\)
(Showed)
\(Q.2.(xxviii)\) \(y=\sin{(m\sin^{-1}{x})}\) হলে, দেখাও যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+m^2y=0\)
[ বঃ২০১১,২০০৩,২০০১;ঢাঃ২০১০,২০০৫;রাঃ২০০৯; চঃ২০০৮;সিঃ২০০৩;মাঃ২০০৪ ]
[ বঃ২০১১,২০০৩,২০০১;ঢাঃ২০১০,২০০৫;রাঃ২০০৯; চঃ২০০৮;সিঃ২০০৩;মাঃ২০০৪ ]
সমাধানঃ
দেওয়া আছে,
\(y=\sin{(m\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(m\sin^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{(m\sin^{-1}{x})}.m\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(m\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m\cos{(m\sin^{-1}{x})}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=m\cos{(m\sin^{-1}{x})}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=m^2\cos^2{(m\sin^{-1}{x})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=m^2\{1-\sin^2{(m\sin^{-1}{x})}\}\) ➜\(\because \cos^2{x}=1-\sin^2{x}\)
\(\Rightarrow (1-x^2)y^2_{1}=m^2(1-y^2)\) ➜\(\because y=\sin{(m\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=m^2\frac{d}{dx}(1-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=m^2(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=-2m^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=-2m^2yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}+2m^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}+m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}+m^2y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}+m^2y=0\)
\(\therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+m^2y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}(c)\)
(Showed)
\(y=\sin{(m\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\sin{(m\sin^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\cos{(m\sin^{-1}{x})}.m\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(m\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m\cos{(m\sin^{-1}{x})}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=m\cos{(m\sin^{-1}{x})}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=m^2\cos^2{(m\sin^{-1}{x})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=m^2\{1-\sin^2{(m\sin^{-1}{x})}\}\) ➜\(\because \cos^2{x}=1-\sin^2{x}\)
\(\Rightarrow (1-x^2)y^2_{1}=m^2(1-y^2)\) ➜\(\because y=\sin{(m\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=m^2\frac{d}{dx}(1-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=m^2(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=-2m^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=-2m^2yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}+2m^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}+m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}+m^2y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}+m^2y=0\)
\(\therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+m^2y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}(c)\)
(Showed)
\(Q.2.(xxix)\) \(y=\cos{(m\sin^{-1}{p})}\) হলে, প্রমাণ কর যে, \((1-p^2)y_{2}-py_{1}+m^2y=0\)
[ চঃ২০১৭ ]
[ চঃ২০১৭ ]
সমাধানঃ
দেওয়া আছে,
\(y=\cos{(m\sin^{-1}{p})}\)
\(\Rightarrow \frac{d}{dp}(y)=\frac{d}{dp}\{\cos{(m\sin^{-1}{p})}\}\) ➜ \(p\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(m\sin^{-1}{p})}.m\frac{d}{dp}(\sin^{-1}{p})\) ➜ \(m\sin^{-1}{p}\) কে \(p\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-m\sin{(m\sin^{-1}{p})}\frac{1}{\sqrt{1-p^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-p^2}y_{1}=m\sin{(m\sin^{-1}{p})}\) ➜ উভয় পার্শে \(\sqrt{1-p^2}\) গুণ করে।
\(\Rightarrow (1-p^2)y^2_{1}=m^2\sin^2{(m\sin^{-1}{p})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-p^2)y^2_{1}=m^2\{1-\cos^2{(m\sin^{-1}{p})}\}\) ➜\(\because \sin^2{x}=1-\cos^2{x}\)
\(\Rightarrow (1-p^2)y^2_{1}=m^2(1-y^2)\) ➜\(\because y=\cos{(m\sin^{-1}{p})}\)
\(\Rightarrow \frac{d}{dp}\{(1-p^2)y^2_{1}\}=m^2\frac{d}{dp}(1-y^2)\) ➜ আবার, \(p\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-p^2)\frac{d}{dp}(y^2_{1})+y^2_{1}\frac{d}{dp}(1-p^2)=m^2(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-p^2)2y_{1}y_{2}+y^2_{1}(0-2p)=-2m^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-p^2)y_{1}y_{2}-2py^2_{1}=-2m^2yy_{1}\)
\(\Rightarrow 2(1-p^2)y_{1}y_{2}-2py^2_{1}+2m^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-p^2)y_{2}-py_{1}+m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-p^2)y_{2}-py_{1}+m^2y=0\)
\(\therefore (1-p^2)y_{2}-py_{1}+m^2y=0\)
(Showed)
\(y=\cos{(m\sin^{-1}{p})}\)
\(\Rightarrow \frac{d}{dp}(y)=\frac{d}{dp}\{\cos{(m\sin^{-1}{p})}\}\) ➜ \(p\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(m\sin^{-1}{p})}.m\frac{d}{dp}(\sin^{-1}{p})\) ➜ \(m\sin^{-1}{p}\) কে \(p\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-m\sin{(m\sin^{-1}{p})}\frac{1}{\sqrt{1-p^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-p^2}y_{1}=m\sin{(m\sin^{-1}{p})}\) ➜ উভয় পার্শে \(\sqrt{1-p^2}\) গুণ করে।
\(\Rightarrow (1-p^2)y^2_{1}=m^2\sin^2{(m\sin^{-1}{p})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-p^2)y^2_{1}=m^2\{1-\cos^2{(m\sin^{-1}{p})}\}\) ➜\(\because \sin^2{x}=1-\cos^2{x}\)
\(\Rightarrow (1-p^2)y^2_{1}=m^2(1-y^2)\) ➜\(\because y=\cos{(m\sin^{-1}{p})}\)
\(\Rightarrow \frac{d}{dp}\{(1-p^2)y^2_{1}\}=m^2\frac{d}{dp}(1-y^2)\) ➜ আবার, \(p\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-p^2)\frac{d}{dp}(y^2_{1})+y^2_{1}\frac{d}{dp}(1-p^2)=m^2(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-p^2)2y_{1}y_{2}+y^2_{1}(0-2p)=-2m^2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-p^2)y_{1}y_{2}-2py^2_{1}=-2m^2yy_{1}\)
\(\Rightarrow 2(1-p^2)y_{1}y_{2}-2py^2_{1}+2m^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-p^2)y_{2}-py_{1}+m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-p^2)y_{2}-py_{1}+m^2y=0\)
\(\therefore (1-p^2)y_{2}-py_{1}+m^2y=0\)
(Showed)
\(Q.2.(xxx)\) \(y=\cos{(2\sin^{-1}{x})}\) হলে, প্রমাণ কর যে, \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+4y=0\)
[ বুয়েটঃ২০০৬-২০০৭ ]
[ বুয়েটঃ২০০৬-২০০৭ ]
সমাধানঃ
দেওয়া আছে,
\(y=\cos{(2\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(2\sin^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(2\sin^{-1}{x})}.2\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(2\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-2\sin{(2\sin^{-1}{x})}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=2\sin{(2\sin^{-1}{x})}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4\sin^2{(2\sin^{-1}{x})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4\{1-\cos^2{(2\sin^{-1}{x})}\}\) ➜\(\because \sin^2{x}=1-\cos^2{x}\)
\(\Rightarrow (1-x^2)y^2_{1}=4(1-y^2)\) ➜\(\because y=\cos{(2\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(1-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=-8yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=-8yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}+8yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}+4y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}+4y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}+4y=0\)
\(\therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+4y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}(c)\)
(Showed)
\(y=\cos{(2\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(2\sin^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(2\sin^{-1}{x})}.2\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(2\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-2\sin{(2\sin^{-1}{x})}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=2\sin{(2\sin^{-1}{x})}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4\sin^2{(2\sin^{-1}{x})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4\{1-\cos^2{(2\sin^{-1}{x})}\}\) ➜\(\because \sin^2{x}=1-\cos^2{x}\)
\(\Rightarrow (1-x^2)y^2_{1}=4(1-y^2)\) ➜\(\because y=\cos{(2\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(1-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=-8yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=-8yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}+8yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}+4y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}+4y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}+4y=0\)
\(\therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+4y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}(c)\)
(Showed)
\(Q.2.(xxxi)\) \(y=\tan{(m\tan^{-1}{x})}\) হলে, দেখাও যে, \((1+x^2)y_{1}=m(1+y^2)\)
[ চঃ২০১২;কুঃ২০০৭,২০০৪]
[ চঃ২০১২;কুঃ২০০৭,২০০৪]
সমাধানঃ
দেওয়া আছে,
\(y=\tan{(m\tan^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\tan{(m\tan^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\sec^2{(m\tan^{-1}{x})}.m\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(m\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m\sec^2{(m\tan^{-1}{x})}\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=m\sec^2{(m\tan^{-1}{x})}\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow (1+x^2)y_{1}=m\{1+\tan^2{(m\tan^{-1}{x})}\}\) ➜\(\because \sec^2{x}=1+\tan^2{x}\)
\(\therefore (1+x^2)y_{1}=m(1+y^2)\) ➜\(\because y=\tan{(m\tan^{-1}{x})}\)
(Showed)
\(y=\tan{(m\tan^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\tan{(m\tan^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\sec^2{(m\tan^{-1}{x})}.m\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(m\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m\sec^2{(m\tan^{-1}{x})}\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=m\sec^2{(m\tan^{-1}{x})}\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow (1+x^2)y_{1}=m\{1+\tan^2{(m\tan^{-1}{x})}\}\) ➜\(\because \sec^2{x}=1+\tan^2{x}\)
\(\therefore (1+x^2)y_{1}=m(1+y^2)\) ➜\(\because y=\tan{(m\tan^{-1}{x})}\)
(Showed)
\(Q.2.(xxxii)\) \(y=\tan{(m\tan^{-1}{x})}\) হলে, দেখাও যে, \((1+x^2)y_{2}-2(my-x)y_{1}=0\)
[ ঢাঃ২০১৩,২০০০;কুঃ২০১২;যঃ২০১১;সিঃ২০০৬;মাঃ২০০৬,২০০৪]
[ ঢাঃ২০১৩,২০০০;কুঃ২০১২;যঃ২০১১;সিঃ২০০৬;মাঃ২০০৬,২০০৪]
সমাধানঃ
দেওয়া আছে,
\(y=\tan{(m\tan^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan{(m\tan^{-1}{x})})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\sec^2{(m\tan^{-1}{x})}.m\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(m\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m\sec^2{(m\tan^{-1}{x})}\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=m\sec^2{(m\tan^{-1}{x})}\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow (1+x^2)y_{1}=m\{1+\tan^2{(m\tan^{-1}{x})}\}\) ➜\(\because \sec^2{x}=1+\tan^2{x}\)
\(\Rightarrow (1+x^2)y_{1}=m(1+y^2)\) ➜\(\because y=\tan{(m\tan^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y_{1}\}=m\frac{d}{dx}(1+y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(1+x^2)=m(0+2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1+x^2)y_{2}+y_{1}(0+2x)=2myy_{1}\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow (1+x^2)y_{2}+2xy_{1}=2myy_{1}\)
\(\Rightarrow (1+x^2)y_{2}+2xy_{1}-2myy_{1}=0\)
\(\Rightarrow (1+x^2)y_{2}-2(my-x)y_{1}=0\)
(Showed)
\(y=\tan{(m\tan^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\tan{(m\tan^{-1}{x})})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\sec^2{(m\tan^{-1}{x})}.m\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(m\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=m\sec^2{(m\tan^{-1}{x})}\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=m\sec^2{(m\tan^{-1}{x})}\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow (1+x^2)y_{1}=m\{1+\tan^2{(m\tan^{-1}{x})}\}\) ➜\(\because \sec^2{x}=1+\tan^2{x}\)
\(\Rightarrow (1+x^2)y_{1}=m(1+y^2)\) ➜\(\because y=\tan{(m\tan^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y_{1}\}=m\frac{d}{dx}(1+y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(1+x^2)=m(0+2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1+x^2)y_{2}+y_{1}(0+2x)=2myy_{1}\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow (1+x^2)y_{2}+2xy_{1}=2myy_{1}\)
\(\Rightarrow (1+x^2)y_{2}+2xy_{1}-2myy_{1}=0\)
\(\Rightarrow (1+x^2)y_{2}-2(my-x)y_{1}=0\)
(Showed)
\(Q.2.(xxxiii)\) \(y=x^2\ln{x}\) হলে, \(y_{3}\) নির্ণয় কর।
উত্তরঃ \(\frac{2}{x}\)
উত্তরঃ \(\frac{2}{x}\)
সমাধানঃ
দেওয়া আছে,
\(y=x^2\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=x^2\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^2)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^2\frac{1}{x}+\ln{x}.2x\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{1}=x+2x\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(x+2x\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\frac{d}{dx}(x)+2\frac{d}{dx}(x\ln{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=1+2x\frac{d}{dx}(\ln{x})+2\ln{x}\frac{d}{dx}(x)\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=1+2x\frac{1}{x}+2\ln{x}.1\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{2}=1+2+2\ln{x}\)
\(\Rightarrow y_{2}=3+2\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}(3+2\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=\frac{d}{dx}(3)+2\frac{d}{dx}(\ln{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{3}=0+2.\frac{1}{x}\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\therefore y_{3}=\frac{2}{x}\)
\(y=x^2\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=x^2\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^2)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^2\frac{1}{x}+\ln{x}.2x\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{1}=x+2x\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(x+2x\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=\frac{d}{dx}(x)+2\frac{d}{dx}(x\ln{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=1+2x\frac{d}{dx}(\ln{x})+2\ln{x}\frac{d}{dx}(x)\) ➜ \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=1+2x\frac{1}{x}+2\ln{x}.1\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{2}=1+2+2\ln{x}\)
\(\Rightarrow y_{2}=3+2\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}(3+2\ln{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=\frac{d}{dx}(3)+2\frac{d}{dx}(\ln{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{3}=0+2.\frac{1}{x}\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\therefore y_{3}=\frac{2}{x}\)
\(Q.2.(xxxiv)\) \(y=\ln{(\sin{x})}\) হলে, দেখাও যে, \(\frac{d^3y}{dx^3}=\frac{2\cos{x}}{\sin^3{x}}\)
সমাধানঃ
দেওয়া আছে,
\(y=\ln{(\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(\sin{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{\sin{x}}\frac{d}{dx}(\sin{x})\) ➜ \(\sin{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow y_{1}=\frac{1}{\sin{x}}(\cos{x})\) ➜\(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}=\frac{\cos{x}}{\sin{x}}\)
\(\Rightarrow y_{1}=\cot{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(\cot{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-cosec^2{x}\) ➜\(\because \frac{d}{dx}(\cot{x})=-cosec^2{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=-\frac{d}{dx}(cosec^2{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-2cosec{x}\frac{d}{dx}(cosec{x})\) ➜ \(cosec{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{3}=-2cosec{x}.(-cosec{x}\cot{x})\) ➜ \(\because \frac{d}{dx}(cosec{x})=-cosec{x}\cot{x}\)
\(\Rightarrow y_{3}=2cosec^2{x}\cot{x}\)
\(\Rightarrow y_{3}=2\frac{1}{\sin^2{x}}.\frac{\cos{x}}{\sin{x}}\)
\(\therefore \frac{d^3y}{dx^3}=\frac{2\cos{x}}{\sin^3{x}}\)
(Showed)
\(y=\ln{(\sin{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\ln{(\sin{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{\sin{x}}\frac{d}{dx}(\sin{x})\) ➜ \(\sin{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow y_{1}=\frac{1}{\sin{x}}(\cos{x})\) ➜\(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow y_{1}=\frac{\cos{x}}{\sin{x}}\)
\(\Rightarrow y_{1}=\cot{x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(\cot{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-cosec^2{x}\) ➜\(\because \frac{d}{dx}(\cot{x})=-cosec^2{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=-\frac{d}{dx}(cosec^2{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-2cosec{x}\frac{d}{dx}(cosec{x})\) ➜ \(cosec{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow y_{3}=-2cosec{x}.(-cosec{x}\cot{x})\) ➜ \(\because \frac{d}{dx}(cosec{x})=-cosec{x}\cot{x}\)
\(\Rightarrow y_{3}=2cosec^2{x}\cot{x}\)
\(\Rightarrow y_{3}=2\frac{1}{\sin^2{x}}.\frac{\cos{x}}{\sin{x}}\)
\(\therefore \frac{d^3y}{dx^3}=\frac{2\cos{x}}{\sin^3{x}}\)
(Showed)
\(Q.2.(xxxv)\) \(y=a\cos{\{\ln{(1+x)}\}}\) হলে, দেখাও যে, \((1+x)^2y_{2}+(1+x)y_{1}+y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=a\cos{\{\ln{(1+x)}\}}\)
\(\Rightarrow \frac{d}{dx}(y)=a\frac{d}{dx}\cos{\{\ln{(1+x)}\}}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-a\sin{\{\ln{(1+x)}\}}\frac{d}{dx}\{\ln{(1+x)}\}\) ➜ \(\ln{(1+x)}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{1}=-a\sin{\{\ln{(1+x)}\}}.\frac{1}{1+x}\frac{d}{dx}(1+x)\) ➜ \((1+x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow y_{1}=-a\sin{\{\ln{(1+x)}\}}.\frac{1}{1+x}(0+1)\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=-a\sin{\{\ln{(1+x)}\}}.\frac{1}{1+x}\)
\(\Rightarrow (1+x)y_{1}=-a\sin{\{\ln{(1+x)}\}}\) ➜ উভয় পার্শে \((1+x)\) গুণ করে।
\(\Rightarrow \frac{d}{dx}\{(1+x)y_{1}\}=-a\frac{d}{dx}\sin{\{\ln{(1+x)}\}}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x)\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(1+x)=-a\cos{\{\ln{(1+x)}\}}\frac{d}{dx}\{\ln{(1+x)}\}\) ➜ \(\ln{(1+x)}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow (1+x)y_{2}+y_{1}(0+1)=-a\cos{\{\ln{(1+x)}\}}.\frac{1}{1+x}\frac{d}{dx}(1+x)\) ➜ \((1+x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow (1+x)y_{2}+y_{1}=-a\cos{\{\ln{(1+x)}\}}.\frac{1}{1+x}\)
\(\Rightarrow (1+x)^2y_{2}+(1+x)y_{1}=-a\cos{\{\ln{(1+x)}\}}\) ➜ উভয় পার্শে \((1+x)\) গুণ করে।
\(\Rightarrow (1+x)^2y_{2}+(1+x)y_{1}=-y\) ➜ \(\because y=a\cos{\{\ln{(1+x)}\}}\)
\(\therefore (1+x)^2y_{2}+(1+x)y_{1}+y=0\)
(Showed)
\(y=a\cos{\{\ln{(1+x)}\}}\)
\(\Rightarrow \frac{d}{dx}(y)=a\frac{d}{dx}\cos{\{\ln{(1+x)}\}}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-a\sin{\{\ln{(1+x)}\}}\frac{d}{dx}\{\ln{(1+x)}\}\) ➜ \(\ln{(1+x)}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{1}=-a\sin{\{\ln{(1+x)}\}}.\frac{1}{1+x}\frac{d}{dx}(1+x)\) ➜ \((1+x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow y_{1}=-a\sin{\{\ln{(1+x)}\}}.\frac{1}{1+x}(0+1)\) ➜ \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=-a\sin{\{\ln{(1+x)}\}}.\frac{1}{1+x}\)
\(\Rightarrow (1+x)y_{1}=-a\sin{\{\ln{(1+x)}\}}\) ➜ উভয় পার্শে \((1+x)\) গুণ করে।
\(\Rightarrow \frac{d}{dx}\{(1+x)y_{1}\}=-a\frac{d}{dx}\sin{\{\ln{(1+x)}\}}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x)\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(1+x)=-a\cos{\{\ln{(1+x)}\}}\frac{d}{dx}\{\ln{(1+x)}\}\) ➜ \(\ln{(1+x)}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow (1+x)y_{2}+y_{1}(0+1)=-a\cos{\{\ln{(1+x)}\}}.\frac{1}{1+x}\frac{d}{dx}(1+x)\) ➜ \((1+x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow (1+x)y_{2}+y_{1}=-a\cos{\{\ln{(1+x)}\}}.\frac{1}{1+x}\)
\(\Rightarrow (1+x)^2y_{2}+(1+x)y_{1}=-a\cos{\{\ln{(1+x)}\}}\) ➜ উভয় পার্শে \((1+x)\) গুণ করে।
\(\Rightarrow (1+x)^2y_{2}+(1+x)y_{1}=-y\) ➜ \(\because y=a\cos{\{\ln{(1+x)}\}}\)
\(\therefore (1+x)^2y_{2}+(1+x)y_{1}+y=0\)
(Showed)
\(Q.2.(xxxvi)\) \(y=Ae^{mx}+Be^{-mx}\) হলে, দেখাও যে, \(y_{2}-m^2y=0\)
[ যঃ২০০৭;দিঃ২০১০;বঃ২০১৩,২০০৮;সিঃ২০১১ ]
[ যঃ২০০৭;দিঃ২০১০;বঃ২০১৩,২০০৮;সিঃ২০১১ ]
সমাধানঃ
দেওয়া আছে,
\(y=Ae^{mx}+Be^{-mx}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(Ae^{mx}+Be^{-mx})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=A\frac{d}{dx}(e^{mx})+B\frac{d}{dx}(e^{-mx})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=Ae^{mx}\frac{d}{dx}(mx)+Be^{-mx}\frac{d}{dx}(-mx)\) ➜ \(mx\) ও \(-mx\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^{x})=e^{x}\)
\(\Rightarrow y_{1}=Ae^{mx}.m+Be^{-mx}.(-m)\)
\(\Rightarrow y_{1}=Ame^{mx}-Bme^{-mx}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(Ame^{mx}-Bme^{-mx})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=Am\frac{d}{dx}(e^{mx})-Bm\frac{d}{dx}(e^{-mx})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=Ame^{mx}\frac{d}{dx}(mx)-Bme^{-mx}\frac{d}{dx}(-mx)\) ➜ \(mx\) ও \(-mx\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^{x})=e^{x}\)
\(\Rightarrow y_{2}=Ame^{mx}.m-Bme^{-mx}.(-m)\)
\(\Rightarrow y_{2}=Am^2e^{mx}+Bm^2e^{-mx}\)
\(\Rightarrow y_{2}=m^2(Ae^{mx}+Be^{-mx})\)
\(\Rightarrow y_{2}=m^2y\) ➜ \(\because y=Ae^{mx}+Be^{-mx}\)
\(\therefore y_{2}-m^2y=0\)
(Showed)
\(y=Ae^{mx}+Be^{-mx}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(Ae^{mx}+Be^{-mx})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=A\frac{d}{dx}(e^{mx})+B\frac{d}{dx}(e^{-mx})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=Ae^{mx}\frac{d}{dx}(mx)+Be^{-mx}\frac{d}{dx}(-mx)\) ➜ \(mx\) ও \(-mx\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^{x})=e^{x}\)
\(\Rightarrow y_{1}=Ae^{mx}.m+Be^{-mx}.(-m)\)
\(\Rightarrow y_{1}=Ame^{mx}-Bme^{-mx}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(Ame^{mx}-Bme^{-mx})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=Am\frac{d}{dx}(e^{mx})-Bm\frac{d}{dx}(e^{-mx})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=Ame^{mx}\frac{d}{dx}(mx)-Bme^{-mx}\frac{d}{dx}(-mx)\) ➜ \(mx\) ও \(-mx\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^{x})=e^{x}\)
\(\Rightarrow y_{2}=Ame^{mx}.m-Bme^{-mx}.(-m)\)
\(\Rightarrow y_{2}=Am^2e^{mx}+Bm^2e^{-mx}\)
\(\Rightarrow y_{2}=m^2(Ae^{mx}+Be^{-mx})\)
\(\Rightarrow y_{2}=m^2y\) ➜ \(\because y=Ae^{mx}+Be^{-mx}\)
\(\therefore y_{2}-m^2y=0\)
(Showed)
\(Q.2.(xxxvii)\) \(y=\frac{1}{2}(e^{x}+e^{-x})\) হলে, দেখাও যে, \(\left(\frac{dy}{dx}\right)^2=y^2-1\)
[ চঃ২০০৩ ]
[ চঃ২০০৩ ]
সমাধানঃ
দেওয়া আছে,
\(y=\frac{1}{2}(e^{x}+e^{-x})\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{2}\frac{d}{dx}(e^{x}+e^{-x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(e^{x})+\frac{1}{2}\frac{d}{dx}(e^{-x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}e^{x}+\frac{1}{2}e^{-x}\frac{d}{dx}(-x)\) ➜ \(-x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^{x})=e^{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}e^{x}+\frac{1}{2}e^{-x}.(-1)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}e^{x}-\frac{1}{2}e^{-x}\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}(e^{x}-e^{-x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4e^{x}e^{-x}\}\) ➜\(\because (a-b)^2=(a+b)^2-4ab\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4e^{x-x}\}\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4e^{0}\}\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4.1\}\) ➜\(\because e^0=1\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4\}\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}(e^{x}+e^{-x})^2-1\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\{\frac{1}{2}(e^{x}+e^{-x})\}^2-1\)
\(\therefore \left(\frac{dy}{dx}\right)^2=y^2-1\) ➜\(\because y^2=\frac{1}{2}(e^{x}+e^{-x})\)
(Showed)
\(y=\frac{1}{2}(e^{x}+e^{-x})\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{2}\frac{d}{dx}(e^{x}+e^{-x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(e^{x})+\frac{1}{2}\frac{d}{dx}(e^{-x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}e^{x}+\frac{1}{2}e^{-x}\frac{d}{dx}(-x)\) ➜ \(-x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^{x})=e^{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}e^{x}+\frac{1}{2}e^{-x}.(-1)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}e^{x}-\frac{1}{2}e^{-x}\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}(e^{x}-e^{-x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4e^{x}e^{-x}\}\) ➜\(\because (a-b)^2=(a+b)^2-4ab\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4e^{x-x}\}\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4e^{0}\}\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4.1\}\) ➜\(\because e^0=1\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}\{(e^{x}+e^{-x})^2-4\}\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{1}{4}(e^{x}+e^{-x})^2-1\)
\(\Rightarrow \left(\frac{dy}{dx}\right)^2=\{\frac{1}{2}(e^{x}+e^{-x})\}^2-1\)
\(\therefore \left(\frac{dy}{dx}\right)^2=y^2-1\) ➜\(\because y^2=\frac{1}{2}(e^{x}+e^{-x})\)
(Showed)
\(Q.2.(xxxviii)\) \(y=a\sin^{-1}{x}+b\cos^{-1}{x}\) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}=0\)
[ ঢাঃ২০০৮;রাঃ২০১১,২০০৩;কুঃ২০১৩,২০০৮,২০০৩; বঃ২০১২,২০০৮,২০০৫;চঃ২০১১;মাঃ২০১০ ]
[ ঢাঃ২০০৮;রাঃ২০১১,২০০৩;কুঃ২০১৩,২০০৮,২০০৩; বঃ২০১২,২০০৮,২০০৫;চঃ২০১১;মাঃ২০১০ ]
সমাধানঃ
দেওয়া আছে,
\(y=a\sin^{-1}{x}+b\cos^{-1}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(a\sin^{-1}{x}+b\cos^{-1}{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a\frac{d}{dx}(\sin^{-1}{x})+b\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=a\frac{1}{\sqrt{1-x^2}}-b\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}, \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=a-b\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=(a-b)^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=\frac{d}{dx}(a-b)^2\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=0\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=0\) ➜\(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=0\)
(Showed)
\(y=a\sin^{-1}{x}+b\cos^{-1}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(a\sin^{-1}{x}+b\cos^{-1}{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a\frac{d}{dx}(\sin^{-1}{x})+b\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=a\frac{1}{\sqrt{1-x^2}}-b\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}, \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=a-b\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=(a-b)^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=\frac{d}{dx}(a-b)^2\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=0\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=0\) ➜\(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=0\)
(Showed)
\(Q.2.(xxxix)\) \(x=\cos{\sqrt{y}} \) অথবা, \(y=(\cos^{-1}{x})^2 \) হলে, দেখাও যে, \((1-x^2)y_{2}-xy_{1}-2=0\)
[ বুয়েটঃ২০০৮-২০০৯; কুয়েটঃ২০১৩-২০১৪;ঢাঃ২০১২,২০১১,২০০৪; রাঃ২০০৯,২০০৭,২০০৪;কুঃ২০১০; দিঃ২০১২,২০১৩;যঃ২০১২,২০০৮,২০০৬,২০০; চঃ২০১৩,২০০৬,২০০৩; সিঃ২০১৪,২০১০,২০০৮,২০০৪; বঃ২০১০,২০০৬]
[ বুয়েটঃ২০০৮-২০০৯; কুয়েটঃ২০১৩-২০১৪;ঢাঃ২০১২,২০১১,২০০৪; রাঃ২০০৯,২০০৭,২০০৪;কুঃ২০১০; দিঃ২০১২,২০১৩;যঃ২০১২,২০০৮,২০০৬,২০০; চঃ২০১৩,২০০৬,২০০৩; সিঃ২০১৪,২০১০,২০০৮,২০০৪; বঃ২০১০,২০০৬]
সমাধানঃ
দেওয়া আছে,
\(x=\cos{\sqrt{y}} \)
\(\Rightarrow \cos^{-1}{x}=\sqrt{y} \)
\(\Rightarrow (\cos^{-1}{x})^2=y \)
\(\Rightarrow y=(\cos^{-1}{x})^2 \)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\cos^{-1}{x}\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(\cos^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-2\cos^{-1}{x}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-2\cos^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\cos^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\cos^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=4y_{1}\) ➜\(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-2=0\)
(Showed)
\(x=\cos{\sqrt{y}} \)
\(\Rightarrow \cos^{-1}{x}=\sqrt{y} \)
\(\Rightarrow (\cos^{-1}{x})^2=y \)
\(\Rightarrow y=(\cos^{-1}{x})^2 \)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=2\cos^{-1}{x}\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(\cos^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-2\cos^{-1}{x}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-2\cos^{-1}{x}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4(\cos^{-1}{x})^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=4y\) ➜ \(\because y=(\cos^{-1}{x})^2\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=4\frac{d}{dx}(y)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=4y_{1}\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=4y_{1}\) ➜\(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}=4y_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-4y_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-2\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-2=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}-2=0\)
(Showed)
\(Q.2.(xL)\) \(2y=(\tan^{-1}{x})^2\) হলে, দেখাও যে, \((1+x^2)^2y_{2}+2(1+x^2)xy_{1}=1\)
সমাধানঃ
দেওয়া আছে,
\(2y=(\tan^{-1}{x})^2\)
\(\Rightarrow 2\frac{d}{dx}(y)=\frac{d}{dx}(\tan^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y_{1}=2\tan^{-1}{x}\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=\tan^{-1}{x}.\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=\tan^{-1}{x}\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y_{1}\}=\frac{d}{dx}(\tan^{-1}{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(1+x^2)=\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{2}+y_{1}(0+2x)=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{2}+2xy_{1}=\frac{1}{1+x^2}\)
\(\therefore (1+x^2)^2y_{2}+2(1+x^2)xy_{1}=1\)
(Showed)
\(2y=(\tan^{-1}{x})^2\)
\(\Rightarrow 2\frac{d}{dx}(y)=\frac{d}{dx}(\tan^{-1}{x})^2\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2y_{1}=2\tan^{-1}{x}\frac{d}{dx}(\tan^{-1}{x})\) ➜ \(\tan^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=\tan^{-1}{x}.\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{1}=\tan^{-1}{x}\) ➜ উভয় পার্শে \(1+x^2\) গুণ করে।
\(\Rightarrow \frac{d}{dx}\{(1+x^2)y_{1}\}=\frac{d}{dx}(\tan^{-1}{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1+x^2)\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(1+x^2)=\frac{1}{1+x^2}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{2}+y_{1}(0+2x)=\frac{1}{1+x^2}\)
\(\Rightarrow (1+x^2)y_{2}+2xy_{1}=\frac{1}{1+x^2}\)
\(\therefore (1+x^2)^2y_{2}+2(1+x^2)xy_{1}=1\)
(Showed)
\(Q.2.(xLi)\) \(y=e^{m\cos^{-1}{x}}\) অথবা, \(\ln{y}=m\cos^{-1}{x}\) অথবা, \(x=\cos\left(\frac{1}{m}\ln{y}\right)\)হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}=m^2y\)
সমাধানঃ
দেওয়া আছে,
\(x=\cos\left(\frac{1}{m}\ln{y}\right)\)
\(\Rightarrow \cos^{-1}{x}=\frac{1}{m}\ln{y}\)
\(\Rightarrow m\cos^{-1}{x}=\ln{y}\)
\(\Rightarrow \ln{y}=m\cos^{-1}{x}\)
\(\Rightarrow y=e^{m\cos^{-1}{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(e^{m\cos^{-1}{x}}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{m\cos^{-1}{x}}.m\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(m\cos^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-me^{m\cos^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-me^{m\cos^{-1}{x}}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow \sqrt{1-x^2}y_{1}=-my\) ➜ \(\because y=e^{m\cos^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=m^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=m^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=m^22yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=m^22yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-m^22yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-m^2y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}-m^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=m^2y\)
(Proved)
\(x=\cos\left(\frac{1}{m}\ln{y}\right)\)
\(\Rightarrow \cos^{-1}{x}=\frac{1}{m}\ln{y}\)
\(\Rightarrow m\cos^{-1}{x}=\ln{y}\)
\(\Rightarrow \ln{y}=m\cos^{-1}{x}\)
\(\Rightarrow y=e^{m\cos^{-1}{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(e^{m\cos^{-1}{x}}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{m\cos^{-1}{x}}.m\frac{d}{dx}(\cos^{-1}{x})\) ➜ \(m\cos^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow y_{1}=-me^{m\cos^{-1}{x}}\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-me^{m\cos^{-1}{x}}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow \sqrt{1-x^2}y_{1}=-my\) ➜ \(\because y=e^{m\cos^{-1}{x}}\)
\(\Rightarrow (1-x^2)y^2_{1}=m^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=m^2\frac{d}{dx}(y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=m^22yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=m^22yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}-m^22yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}-m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}-m^2y=0\)
\(\Rightarrow (1-x^2)y_{2}-xy_{1}-m^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}=m^2y\)
(Proved)
\(Q.2.(xLii)\) \(y=\cos{(m\sin^{-1}{x})}\) হলে, প্রমাণ কর যে, \((1-x^2)y_{2}-xy_{1}+m^2y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=\cos{(m\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(m\sin^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(m\sin^{-1}{x})}.m\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(m\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{1}=-m\sin{(m\sin^{-1}{x})}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-m\sin{(m\sin^{-1}{x})}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=m^2\sin^2{(m\sin^{-1}{x})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=m^2\{1-\cos^2{(m\sin^{-1}{x})}\}\) ➜ \(\because \sin^2{x}=1-\cos^2{x}\)
\(\Rightarrow (1-x^2)y^2_{1}=m^2(1-y^2)\) ➜ \(\because y=\cos{(m\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=m^2\frac{d}{dx}(1-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=m^2(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=1, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=m^2(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=1, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=-m^22yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}+m^22yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}+m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}+m^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}+m^2y=0\)
(Proved)
\(y=\cos{(m\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(m\sin^{-1}{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(m\sin^{-1}{x})}.m\frac{d}{dx}(\sin^{-1}{x})\) ➜ \(m\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{1}=-m\sin{(m\sin^{-1}{x})}.\frac{1}{\sqrt{1-x^2}}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \sqrt{1-x^2}y_{1}=-m\sin{(m\sin^{-1}{x})}\) ➜ উভয় পার্শে \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow (1-x^2)y^2_{1}=m^2\sin^2{(m\sin^{-1}{x})}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (1-x^2)y^2_{1}=m^2\{1-\cos^2{(m\sin^{-1}{x})}\}\) ➜ \(\because \sin^2{x}=1-\cos^2{x}\)
\(\Rightarrow (1-x^2)y^2_{1}=m^2(1-y^2)\) ➜ \(\because y=\cos{(m\sin^{-1}{x})}\)
\(\Rightarrow \frac{d}{dx}\{(1-x^2)y^2_{1}\}=m^2\frac{d}{dx}(1-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=m^2(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=1, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(1-x^2)=m^2(0-2yy_{1})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=1, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow (1-x^2)2y_{1}y_{2}+y^2_{1}(0-2x)=-m^22yy_{1}\)
\(\Rightarrow 2(1-x^2)y_{1}y_{2}-2xy^2_{1}+m^22yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-x^2)y_{2}-xy_{1}+m^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-x^2)y_{2}-xy_{1}+m^2y=0\)
\(\therefore (1-x^2)y_{2}-xy_{1}+m^2y=0\)
(Proved)
\(Q.2.(xLiii)\) \(y=a\sin{(\ln{x})}+b\cos{(\ln{x})}\) হলে, প্রমাণ কর যে, \(x^2y_{2}+xy_{1}+y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=a\sin{(\ln{x})}+b\cos{(\ln{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{a\sin{(\ln{x})}+b\cos{(\ln{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a\frac{d}{dx}\{\sin{(\ln{x})}\}+b\frac{d}{dx}\{\cos{(\ln{x})}\}\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=a\cos{(\ln{x})}\frac{d}{dx}(\ln{x})-b\sin{(\ln{x})}\frac{d}{dx}(\ln{x})\) ➜\(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{1}=a\cos{(\ln{x})}\frac{1}{x}-b\sin{(\ln{x})}\frac{1}{x}\) ➜\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow xy_{1}=a\cos{(\ln{x})}-b\sin{(\ln{x})}\) ➜ উভয় পার্শে \(x\) গুণ করে।
\(\Rightarrow \frac{d}{dx}(xy_{1})=\frac{d}{dx}\{a\cos{(\ln{x})}-b\sin{(\ln{x})}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(x)=a\frac{d}{dx}\{\cos{(\ln{x})}\}-b\frac{d}{dx}\{\sin{(\ln{x})}\}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow xy_{2}+y_{1}.1=-a\sin{(\ln{x})}\frac{d}{dx}(\ln{x})-b\cos{(\ln{x})}\frac{d}{dx}(\ln{x})\) ➜ \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow xy_{2}+y_{1}=-a\sin{(\ln{x})}\frac{1}{x}-b\cos{(\ln{x})}\frac{1}{x}\) ➜\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow x^2y_{2}+xy_{1}=-a\sin{(\ln{x})}-b\cos{(\ln{x})}\) ➜উভয় পার্শে \(x\) গুণ করে।
\(\Rightarrow x^2y_{2}+xy_{1}=-\{a\sin{(\ln{x})}+b\cos{(\ln{x})}\}\)
\(\Rightarrow x^2y_{2}+xy_{1}=-y\) ➜\(\because y=a\cos{(\ln{x})}+b\sin{(\ln{x})}\)
\(\therefore x^2y_{2}+xy_{1}+y=0\) (Showed)
\(y=a\sin{(\ln{x})}+b\cos{(\ln{x})}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{a\sin{(\ln{x})}+b\cos{(\ln{x})}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=a\frac{d}{dx}\{\sin{(\ln{x})}\}+b\frac{d}{dx}\{\cos{(\ln{x})}\}\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=a\cos{(\ln{x})}\frac{d}{dx}(\ln{x})-b\sin{(\ln{x})}\frac{d}{dx}(\ln{x})\) ➜\(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{1}=a\cos{(\ln{x})}\frac{1}{x}-b\sin{(\ln{x})}\frac{1}{x}\) ➜\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow xy_{1}=a\cos{(\ln{x})}-b\sin{(\ln{x})}\) ➜ উভয় পার্শে \(x\) গুণ করে।
\(\Rightarrow \frac{d}{dx}(xy_{1})=\frac{d}{dx}\{a\cos{(\ln{x})}-b\sin{(\ln{x})}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(x)=a\frac{d}{dx}\{\cos{(\ln{x})}\}-b\frac{d}{dx}\{\sin{(\ln{x})}\}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow xy_{2}+y_{1}.1=-a\sin{(\ln{x})}\frac{d}{dx}(\ln{x})-b\cos{(\ln{x})}\frac{d}{dx}(\ln{x})\) ➜ \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow xy_{2}+y_{1}=-a\sin{(\ln{x})}\frac{1}{x}-b\cos{(\ln{x})}\frac{1}{x}\) ➜\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow x^2y_{2}+xy_{1}=-a\sin{(\ln{x})}-b\cos{(\ln{x})}\) ➜উভয় পার্শে \(x\) গুণ করে।
\(\Rightarrow x^2y_{2}+xy_{1}=-\{a\sin{(\ln{x})}+b\cos{(\ln{x})}\}\)
\(\Rightarrow x^2y_{2}+xy_{1}=-y\) ➜\(\because y=a\cos{(\ln{x})}+b\sin{(\ln{x})}\)
\(\therefore x^2y_{2}+xy_{1}+y=0\) (Showed)
\(Q.2.(xLiv)\) \(y^2=\sec{2x}\) হলে, প্রমাণ কর যে, \(y_{2}+y=3y^5\)
সমাধানঃ
দেওয়া আছে,
\(y^2=\sec{2x}\)
\(\Rightarrow \frac{d}{dx}(y^2)=\frac{d}{dx}(\sec{2x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2yy_{1}=\sec{2x}\tan{2x}\frac{d}{dx}(2x)\) ➜ \(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow 2yy_{1}=\sec{2x}\tan{2x}.2\)
\(\Rightarrow 2yy_{1}=2\sec{2x}\tan{2x}\)
\(\Rightarrow 2yy_{1}=2y^2\tan{2x}\) ➜\(\because y^2=\sec{2x}\)
\(\Rightarrow y_{1}=y\tan{2x}\) ➜\(\because 2y\ne{0}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(y\tan{2x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=y\frac{d}{dx}(\tan{2x})+\tan{2x}\frac{d}{dx}(y)\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=y\sec^2{2x}\frac{d}{dx}(2x)+\tan{2x}y_{1}\) ➜\(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow y_{2}=y\sec^2{2x}.2+\tan{2x}y_{1}\)
\(\Rightarrow y_{2}=2y(y^2)^2+\tan{2x}.y\tan{2x}\) ➜\(\because y^2=\sec{2x}, y_{1}=y\tan{2x}\)
\(\Rightarrow y_{2}=2yy^4+y\tan^2{2x}\)
\(\Rightarrow y_{2}=2y^5+y(\sec^2{2x}-1)\) ➜\(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow y_{2}=2y^5+y(y^4-1)\) ➜\(\because y^2=\sec{2x}\)
\(\Rightarrow y_{2}=2y^5+y^5-y\)
\(\therefore y_{2}+y=3y^5\)
(Proved)
\(y^2=\sec{2x}\)
\(\Rightarrow \frac{d}{dx}(y^2)=\frac{d}{dx}(\sec{2x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2yy_{1}=\sec{2x}\tan{2x}\frac{d}{dx}(2x)\) ➜ \(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow 2yy_{1}=\sec{2x}\tan{2x}.2\)
\(\Rightarrow 2yy_{1}=2\sec{2x}\tan{2x}\)
\(\Rightarrow 2yy_{1}=2y^2\tan{2x}\) ➜\(\because y^2=\sec{2x}\)
\(\Rightarrow y_{1}=y\tan{2x}\) ➜\(\because 2y\ne{0}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(y\tan{2x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=y\frac{d}{dx}(\tan{2x})+\tan{2x}\frac{d}{dx}(y)\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=y\sec^2{2x}\frac{d}{dx}(2x)+\tan{2x}y_{1}\) ➜\(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow y_{2}=y\sec^2{2x}.2+\tan{2x}y_{1}\)
\(\Rightarrow y_{2}=2y(y^2)^2+\tan{2x}.y\tan{2x}\) ➜\(\because y^2=\sec{2x}, y_{1}=y\tan{2x}\)
\(\Rightarrow y_{2}=2yy^4+y\tan^2{2x}\)
\(\Rightarrow y_{2}=2y^5+y(\sec^2{2x}-1)\) ➜\(\because \tan^2{A}=\sec^2{A}-1\)
\(\Rightarrow y_{2}=2y^5+y(y^4-1)\) ➜\(\because y^2=\sec{2x}\)
\(\Rightarrow y_{2}=2y^5+y^5-y\)
\(\therefore y_{2}+y=3y^5\)
(Proved)
\(Q.2.(xLv)\) \(y=\cot{x}+cosec{x}\) হলে, দেখাও যে, \(\frac{d^2y}{dx^2}=\frac{\sin{x}}{(1-\cos{x})^2}\)
সমাধানঃ
দেওয়া আছে,
\(y=\cot{x}+cosec{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cot{x}+cosec{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\cot{x})+\frac{d}{dx}(cosec{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=-cosec^2{x}-cosec{x}\cot{x}\) ➜ \(\because \frac{d}{dx}(cot{x})=-cosec^2{x}, \frac{d}{dx}(cosec{x})=-cosec{x}\cot{x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(-cosec^2{x}-cosec{x}\cot{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=-\frac{d}{dx}(cosec^2{x})-\frac{d}{dx}(cosec{x}\cot{x})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=-2 \ cosec{x}\frac{d}{dx}(cosec{x})-cosec{x}\frac{d}{dx}(\cot{x})-\cot{x}\frac{d}{dx}(cosec{x})\) ➜ \(\sec{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2 \ cosec{x} \ cosec{x}\cot{x}+cosec{x} \ cosec^2{x}+\cot{x} \ cosec{x}\cot{x}\) ➜ \(\because \frac{d}{dx}(cosec{x})=-cosec{x}\cot{x}, \frac{d}{dx}(\cot{x})=-cosec^2{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2 \ cosec^2{x}\cot{x}+cosec^3{x}+\cot^2{x} \ cosec{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\times{\frac{1}{\sin^2{x}}}\times{\frac{\cos{x}}{\sin{x}}}+\frac{1}{\sin^3{x}}+\frac{\cos^2{x}}{\sin^2{x}}\times{\frac{1}{\sin{x}}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{2\cos{x}}{\sin^3{x}}+\frac{1}{\sin^3{x}}+\frac{\cos^2{x}}{\sin^3{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{2\cos{x}+1+\cos^2{x}}{\sin^3{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1+2\cos{x}+\cos^2{x}}{\sin{x}\sin^2{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\cos{x})^2}{\sin{x}(1-\cos^2{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\cos{x})^2}{\sin{x}(1+\cos{x})(1-\cos{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1+\cos{x}}{\sin{x}(1-\cos{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\cos{x})(1-\cos{x})}{\sin{x}(1-\cos{x})^2}\) ➜ লব ও হরের সহিত \(1-\cos{x}\) গুণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1-\cos^2{x}}{\sin{x}(1-\cos{x})^2}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{\sin^2{x}}{\sin{x}(1-\cos{x})^2}\)
\(\therefore \frac{d^2y}{dx^2}=\frac{\sin{x}}{(1-\cos{x})^2}\)
(Proved)
\(y=\cot{x}+cosec{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cot{x}+cosec{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\cot{x})+\frac{d}{dx}(cosec{x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=-cosec^2{x}-cosec{x}\cot{x}\) ➜ \(\because \frac{d}{dx}(cot{x})=-cosec^2{x}, \frac{d}{dx}(cosec{x})=-cosec{x}\cot{x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(-cosec^2{x}-cosec{x}\cot{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=-\frac{d}{dx}(cosec^2{x})-\frac{d}{dx}(cosec{x}\cot{x})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d^2y}{dx^2}=-2 \ cosec{x}\frac{d}{dx}(cosec{x})-cosec{x}\frac{d}{dx}(\cot{x})-\cot{x}\frac{d}{dx}(cosec{x})\) ➜ \(\sec{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{d^2y}{dx^2}=2 \ cosec{x} \ cosec{x}\cot{x}+cosec{x} \ cosec^2{x}+\cot{x} \ cosec{x}\cot{x}\) ➜ \(\because \frac{d}{dx}(cosec{x})=-cosec{x}\cot{x}, \frac{d}{dx}(\cot{x})=-cosec^2{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2 \ cosec^2{x}\cot{x}+cosec^3{x}+\cot^2{x} \ cosec{x}\)
\(\Rightarrow \frac{d^2y}{dx^2}=2\times{\frac{1}{\sin^2{x}}}\times{\frac{\cos{x}}{\sin{x}}}+\frac{1}{\sin^3{x}}+\frac{\cos^2{x}}{\sin^2{x}}\times{\frac{1}{\sin{x}}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{2\cos{x}}{\sin^3{x}}+\frac{1}{\sin^3{x}}+\frac{\cos^2{x}}{\sin^3{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{2\cos{x}+1+\cos^2{x}}{\sin^3{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1+2\cos{x}+\cos^2{x}}{\sin{x}\sin^2{x}}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\cos{x})^2}{\sin{x}(1-\cos^2{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\cos{x})^2}{\sin{x}(1+\cos{x})(1-\cos{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1+\cos{x}}{\sin{x}(1-\cos{x})}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(1+\cos{x})(1-\cos{x})}{\sin{x}(1-\cos{x})^2}\) ➜ লব ও হরের সহিত \(1-\cos{x}\) গুণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{1-\cos^2{x}}{\sin{x}(1-\cos{x})^2}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{\sin^2{x}}{\sin{x}(1-\cos{x})^2}\)
\(\therefore \frac{d^2y}{dx^2}=\frac{\sin{x}}{(1-\cos{x})^2}\)
(Proved)
অনুশীলনী \(9.G / Q.3\)-এর বর্ণনামূলক প্রশ্নসমুহ
\(Q.3.(i)\) \(y=e^x\cos{x}\) হলে, প্রমাণ কর যে, \(y_{2}-2y_{1}+2y=0\)
[ যঃ২০১৩;দিঃ২০১০;মাঃ২০০৮;সিঃ২০০৫,২০০৩;রাঃ২০০৩;ঢাঃ২০০২]
\(Q.3.(ii)\) \(y=e^{x}\cos{x}\) হলে, দেখাও যে, \(y_{4}+4y=0\)
\(Q.3.(iii)\) \(y=e^{ax}\sin{bx}\) হলে, প্রমাণ কর যে, \(y_{2}-2ay_{1}+(a^2+b^2)y=0\)
[ কুয়েটঃ২০০৩-২০০৪]
\(Q.3.(iv)\) \(y=(e^{x}+e^{-x})\sin{x}\) হলে, প্রমাণ কর যে, \(\frac{d^4y}{dx^4}+4y=0\)
[ দিঃ২০১২;ঢাঃ২০১০,২০০৪;রাঃ২০০৬]
\(Q.3.(v)\) \(\cos^{-1}\left(\frac{y}{b}\right)=\ln{\left(\frac{x}{n}\right)^n}\) হলে,
প্রমাণ কর যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+n^2y=0\)
[ বুয়েটঃ২০১০-২০১১]
\(Q.3.(vi)\) \(y=\frac{\sin{x}}{\sqrt{x}}\) হলে,
প্রমাণ কর যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+\left(x^2-\frac{1}{4}\right)y=0\)
[ বুয়েটঃ২০০৪-২০০৫]
\(Q.3.(vii)\) \(y=A\sin{mx}+B\cos{mx}\) হলে,
প্রমাণ কর যে, \(y_{2}+m^2y=0\) এবং \(y_{4}-m^4y=0\)
[ বুয়েটঃ২০০৪-২০০৫]
\(Q.3.(viii)\) \(y=x^m\log_{e}x\) হলে, প্রমাণ কর যে, \(xy_{1}=my+x^m\)
[ বুয়েটঃ২০০৪-২০০৫]
\(Q.3.(ix)\) \(y=(a+bx)e^{2x}\) হলে, প্রমাণ কর যে, \(y_{2}-2y_{1}-2be^{2x}=0\)
\(Q.3.(x)\) \(y=(a+bx)e^{-2x}\) হলে, প্রমাণ কর যে, \(\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\)
\(Q.3.(xi)\) \(y=(p+qx)e^{-2x}\) হলে, প্রমাণ কর যে, \(y_{2}+4y_{1}+4y=0\)
\(Q.3.(xii)\) \(y=e^{a\sin{t}}\) হলে, দেখাও যে, \((1-t^2)y_{2}-ty_{1}=a^2y\)
[ ঢাঃ ২০১১ ]
\(Q.3.(xiii)\) \(x=a(\theta+\sin{\theta})\) ও \(y=a(1-\cos{\theta})\) হলে,
\(\frac{\theta}{2}\) এর মাধ্যমে \(\frac{dy}{dx}\) ও \(\frac{d^2y}{dx^2}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(\tan{\frac{\theta}{2}}; \frac{1}{4a}\sec^4{\frac{\theta}{2}}\)
\(Q.3.(xiv)\) \(2x=t+t^{-1}\) এবং \(2y=t-t^{-1} \) হলে,
দেখাও যে, \(\frac{dy}{dx}=\frac{t^2+1}{t^2-1}\)এবং \(\frac{d^2y}{dx^2}=-\frac{8t^3}{(t^2-1)^3}\)
\(Q.3.(xv)\) \(y=\sqrt{\cos{2x}}\) হলে, দেখাও যে, \((yy_{1})^2=1-y^4\)
\(Q.3.(xvi)\) \(y=\tan{\sqrt{1-x}}\) হলে, দেখাও যে, \(2y_{1}\sqrt{1-x}+(1+y^2)=0\)
[ যঃ২০১৩;দিঃ২০১০;মাঃ২০০৮;সিঃ২০০৫,২০০৩;রাঃ২০০৩;ঢাঃ২০০২]
\(Q.3.(ii)\) \(y=e^{x}\cos{x}\) হলে, দেখাও যে, \(y_{4}+4y=0\)
\(Q.3.(iii)\) \(y=e^{ax}\sin{bx}\) হলে, প্রমাণ কর যে, \(y_{2}-2ay_{1}+(a^2+b^2)y=0\)
[ কুয়েটঃ২০০৩-২০০৪]
\(Q.3.(iv)\) \(y=(e^{x}+e^{-x})\sin{x}\) হলে, প্রমাণ কর যে, \(\frac{d^4y}{dx^4}+4y=0\)
[ দিঃ২০১২;ঢাঃ২০১০,২০০৪;রাঃ২০০৬]
\(Q.3.(v)\) \(\cos^{-1}\left(\frac{y}{b}\right)=\ln{\left(\frac{x}{n}\right)^n}\) হলে,
প্রমাণ কর যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+n^2y=0\)
[ বুয়েটঃ২০১০-২০১১]
\(Q.3.(vi)\) \(y=\frac{\sin{x}}{\sqrt{x}}\) হলে,
প্রমাণ কর যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+\left(x^2-\frac{1}{4}\right)y=0\)
[ বুয়েটঃ২০০৪-২০০৫]
\(Q.3.(vii)\) \(y=A\sin{mx}+B\cos{mx}\) হলে,
প্রমাণ কর যে, \(y_{2}+m^2y=0\) এবং \(y_{4}-m^4y=0\)
[ বুয়েটঃ২০০৪-২০০৫]
\(Q.3.(viii)\) \(y=x^m\log_{e}x\) হলে, প্রমাণ কর যে, \(xy_{1}=my+x^m\)
[ বুয়েটঃ২০০৪-২০০৫]
\(Q.3.(ix)\) \(y=(a+bx)e^{2x}\) হলে, প্রমাণ কর যে, \(y_{2}-2y_{1}-2be^{2x}=0\)
\(Q.3.(x)\) \(y=(a+bx)e^{-2x}\) হলে, প্রমাণ কর যে, \(\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\)
\(Q.3.(xi)\) \(y=(p+qx)e^{-2x}\) হলে, প্রমাণ কর যে, \(y_{2}+4y_{1}+4y=0\)
\(Q.3.(xii)\) \(y=e^{a\sin{t}}\) হলে, দেখাও যে, \((1-t^2)y_{2}-ty_{1}=a^2y\)
[ ঢাঃ ২০১১ ]
\(Q.3.(xiii)\) \(x=a(\theta+\sin{\theta})\) ও \(y=a(1-\cos{\theta})\) হলে,
\(\frac{\theta}{2}\) এর মাধ্যমে \(\frac{dy}{dx}\) ও \(\frac{d^2y}{dx^2}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(\tan{\frac{\theta}{2}}; \frac{1}{4a}\sec^4{\frac{\theta}{2}}\)
\(Q.3.(xiv)\) \(2x=t+t^{-1}\) এবং \(2y=t-t^{-1} \) হলে,
দেখাও যে, \(\frac{dy}{dx}=\frac{t^2+1}{t^2-1}\)এবং \(\frac{d^2y}{dx^2}=-\frac{8t^3}{(t^2-1)^3}\)
\(Q.3.(xv)\) \(y=\sqrt{\cos{2x}}\) হলে, দেখাও যে, \((yy_{1})^2=1-y^4\)
\(Q.3.(xvi)\) \(y=\tan{\sqrt{1-x}}\) হলে, দেখাও যে, \(2y_{1}\sqrt{1-x}+(1+y^2)=0\)
\(Q.3.(xvii)\) \(y=\frac{4}{\sqrt{\sec{x}}}\) হলে, দেখাও যে, \(2\cot{x}\frac{dy}{dx}+y=0\)
উত্তরঃ \(\frac{n!}{(a-x)^{n+1}}\)
\(Q.3.(xix)\) \(y=\ln{\left(\frac{a-x}{a+x}\right)}\)
উত্তরঃ \((n-1)!\left[\frac{(-1)^n}{(a+x)^{n}}-\frac{1}{(a-x)^{n}}\right]\)
\(Q.3.(xx)\) \(y=\ln{(ax+x^2)}\)
উত্তরঃ \((-1)^{n-1}(n-1)!\left[\frac{1}{(x)^{n}}+\frac{1}{(a+x)^{n}}\right]\)
\(Q.3.(xxi)\) \(y=\cos{2x}\cos{x}\)
উত্তরঃ \(\frac{1}{2}\left[3^n\cos{\left(\frac{n\pi}{2}+3x\right)}+\cos{\left(\frac{n\pi}{2}+x\right)}\right]\)
\(Q.3.(xxii)\) \(y=\cos^3{x}\)
উত্তরঃ \(\frac{1}{4}\left[3\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\right]\)
\(Q.3.(xxiii)\) \(y=\sin{x}\sin{3x}\)
উত্তরঃ \(2^{n-1}\cos{\left(2x+\frac{n\pi}{2}\right)}-2^{2n-1}\cos{\left(4x+\frac{n\pi}{2}\right)}\)
\(Q.3.(xxiv)\) \(y=\ln{x}\)
উত্তরঃ \(\frac{(-1){n-1}(n-1)!}{x^n}\)
\(Q.3.(xxv)\) \(y=\frac{1}{x}\)
উত্তরঃ \(\frac{(-1){n}(n)!}{x^{n+1}}\)
\(Q.3.(xxvi)\) \(y=\frac{x^2+1}{(x-1)(x-2)(x-3)}\)
উত্তরঃ \((-1)^nn!\left[\frac{1}{(x-1)^{n+1}}-\frac{5}{(x-2)^{n+1}}+\frac{5}{(x-3)^{n+1}}\right]\)
\(Q.3.(xxvii)\) \(y=\cos{(ax)}\)
উত্তরঃ \(a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\(Q.3.(xxviii)\) \(y=2\sin{3x}\cos{2x}\)
উত্তরঃ \(5^n\sin{\left(\frac{n\pi}{2}+5x\right)}+\sin{\left(\frac{n\pi}{2}+x\right)}\)
\(Q.3.(xxix)\) \(y=4\sin^3{x}\)
উত্তরঃ \(3\sin{\left(\frac{n\pi}{2}+x\right)}-3^n\sin{\left(\frac{n\pi}{2}+3x\right)}\)
\(Q.3.(xxx)\) \(y=(ax+b)^{-m}\)
উত্তরঃ \(\frac{(-1)^n(m+n-1)!a^n}{(m-1)!(ax+b)^{m+n}}\)
\(Q.3.(xxxi)\) \(y=e^{3x}\sin^2{x}\)
উত্তরঃ \(\frac{1}{2}e^{3x}\left[3^{n}-(\sqrt{13})^n\cos{(2x+n\tan^{-1}{\frac{2}{3}})}\right]\)
\(Q.3.(xxxii)\) \(y=\cos{x}\) হলে, \(y_{n}\) নির্ণয় কর এবং \((y_{n})_{0}\) এর মাণ নির্ণয় কর।
নিচের ফাংশনগুলির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xviii)\) \(\frac{1}{a-x}\)উত্তরঃ \(\frac{n!}{(a-x)^{n+1}}\)
\(Q.3.(xix)\) \(y=\ln{\left(\frac{a-x}{a+x}\right)}\)
উত্তরঃ \((n-1)!\left[\frac{(-1)^n}{(a+x)^{n}}-\frac{1}{(a-x)^{n}}\right]\)
\(Q.3.(xx)\) \(y=\ln{(ax+x^2)}\)
উত্তরঃ \((-1)^{n-1}(n-1)!\left[\frac{1}{(x)^{n}}+\frac{1}{(a+x)^{n}}\right]\)
\(Q.3.(xxi)\) \(y=\cos{2x}\cos{x}\)
উত্তরঃ \(\frac{1}{2}\left[3^n\cos{\left(\frac{n\pi}{2}+3x\right)}+\cos{\left(\frac{n\pi}{2}+x\right)}\right]\)
\(Q.3.(xxii)\) \(y=\cos^3{x}\)
উত্তরঃ \(\frac{1}{4}\left[3\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\right]\)
\(Q.3.(xxiii)\) \(y=\sin{x}\sin{3x}\)
উত্তরঃ \(2^{n-1}\cos{\left(2x+\frac{n\pi}{2}\right)}-2^{2n-1}\cos{\left(4x+\frac{n\pi}{2}\right)}\)
\(Q.3.(xxiv)\) \(y=\ln{x}\)
উত্তরঃ \(\frac{(-1){n-1}(n-1)!}{x^n}\)
\(Q.3.(xxv)\) \(y=\frac{1}{x}\)
উত্তরঃ \(\frac{(-1){n}(n)!}{x^{n+1}}\)
\(Q.3.(xxvi)\) \(y=\frac{x^2+1}{(x-1)(x-2)(x-3)}\)
উত্তরঃ \((-1)^nn!\left[\frac{1}{(x-1)^{n+1}}-\frac{5}{(x-2)^{n+1}}+\frac{5}{(x-3)^{n+1}}\right]\)
\(Q.3.(xxvii)\) \(y=\cos{(ax)}\)
উত্তরঃ \(a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\(Q.3.(xxviii)\) \(y=2\sin{3x}\cos{2x}\)
উত্তরঃ \(5^n\sin{\left(\frac{n\pi}{2}+5x\right)}+\sin{\left(\frac{n\pi}{2}+x\right)}\)
\(Q.3.(xxix)\) \(y=4\sin^3{x}\)
উত্তরঃ \(3\sin{\left(\frac{n\pi}{2}+x\right)}-3^n\sin{\left(\frac{n\pi}{2}+3x\right)}\)
\(Q.3.(xxx)\) \(y=(ax+b)^{-m}\)
উত্তরঃ \(\frac{(-1)^n(m+n-1)!a^n}{(m-1)!(ax+b)^{m+n}}\)
\(Q.3.(xxxi)\) \(y=e^{3x}\sin^2{x}\)
উত্তরঃ \(\frac{1}{2}e^{3x}\left[3^{n}-(\sqrt{13})^n\cos{(2x+n\tan^{-1}{\frac{2}{3}})}\right]\)
\(Q.3.(xxxii)\) \(y=\cos{x}\) হলে, \(y_{n}\) নির্ণয় কর এবং \((y_{n})_{0}\) এর মাণ নির্ণয় কর।
\(Q.3.(i)\) \(y=e^x\cos{x}\) হলে, প্রমাণ কর যে, \(y_{2}-2y_{1}+2y=0\)
[ যঃ২০১৩;দিঃ২০১০;মাঃ২০০৮;সিঃ২০০৫,২০০৩;রাঃ২০০৩;ঢাঃ২০০২]
[ যঃ২০১৩;দিঃ২০১০;মাঃ২০০৮;সিঃ২০০৫,২০০৩;রাঃ২০০৩;ঢাঃ২০০২]
সমাধানঃ
দেওয়া আছে,
\(y=e^x\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^x\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^x\frac{d}{dx}(\cos{x})+\cos{x}\frac{d}{dx}(e^x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-e^x\sin{x}+e^x\cos{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \(\frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^x(\cos{x}-\sin{x})\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{e^x(\cos{x}-\sin{x})\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=e^x\frac{d}{dx}(\cos{x}-\sin{x})+(\cos{x}-\sin{x})\frac{d}{dx}(e^x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=e^x(-\sin{x}-\cos{x})+(\cos{x}-\sin{x})e^x\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\), \(\frac{d}{dx}(e^{x})=e^{x}\)
\(\Rightarrow y_{2}=e^x(-\sin{x}-\cos{x}+\cos{x}-\sin{x})\)
\(\Rightarrow y_{2}=-2e^x\sin{x}\)
\(\Rightarrow y_{2}=-2e^x\cos{x}+2e^x\cos{x}-2e^x\sin{x}\)
\(\Rightarrow y_{2}=-2e^x\cos{x}+2e^x(\cos{x}-\sin{x})\)
\(\Rightarrow y_{2}=-2y+2y_{1}\) ➜ \(\because y=e^x\cos{x}, y_{1}=e^x(\cos{x}-\sin{x})\)
\(\therefore y_{2}-2y_{1}+2y=0\)
(Proved)
\(y=e^x\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^x\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^x\frac{d}{dx}(\cos{x})+\cos{x}\frac{d}{dx}(e^x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-e^x\sin{x}+e^x\cos{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \(\frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^x(\cos{x}-\sin{x})\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{e^x(\cos{x}-\sin{x})\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=e^x\frac{d}{dx}(\cos{x}-\sin{x})+(\cos{x}-\sin{x})\frac{d}{dx}(e^x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=e^x(-\sin{x}-\cos{x})+(\cos{x}-\sin{x})e^x\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\), \(\frac{d}{dx}(e^{x})=e^{x}\)
\(\Rightarrow y_{2}=e^x(-\sin{x}-\cos{x}+\cos{x}-\sin{x})\)
\(\Rightarrow y_{2}=-2e^x\sin{x}\)
\(\Rightarrow y_{2}=-2e^x\cos{x}+2e^x\cos{x}-2e^x\sin{x}\)
\(\Rightarrow y_{2}=-2e^x\cos{x}+2e^x(\cos{x}-\sin{x})\)
\(\Rightarrow y_{2}=-2y+2y_{1}\) ➜ \(\because y=e^x\cos{x}, y_{1}=e^x(\cos{x}-\sin{x})\)
\(\therefore y_{2}-2y_{1}+2y=0\)
(Proved)
\(Q.3.(ii)\) \(y=e^{x}\cos{x}\) হলে, দেখাও যে, \(y_{4}+4y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=e^{x}\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{x}\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{x}\frac{d}{dx}(\cos{x})+\cos{x}\frac{d}{dx}(e^{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-e^{x}\sin{x}+\cos{x}e^{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=-e^{x}\sin{x}+e^{x}\cos{x}\)
\(\Rightarrow y_{1}=e^{x}(\cos{x}-\sin{x})\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{e^{x}(\cos{x}-\sin{x})\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=e^{x}\frac{d}{dx}(\cos{x}-\sin{x})+(\cos{x}-\sin{x})\frac{d}{dx}(e^{x})\}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=e^{x}(-\sin{x}-\cos{x})+(\cos{x}-\sin{x})e^{x}\}\) ➜\(\because \frac{d}{dx}(\cos{x})=-\sin{x}\), \(\frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=e^{x}(-\sin{x}-\cos{x}+\cos{x}-\sin{x})\)
\(\Rightarrow y_{2}=e^{x}\{-2\sin{x}\}\)
\(\Rightarrow y_{2}=-2e^{x}\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=-2\frac{d}{dx}(e^{x}\sin{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-2e^{x}\frac{d}{dx}(\sin{x})-2\sin{x}\frac{d}{dx}(e^{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{3}=-2e^{x}\cos{x}-2\sin{x}e^{x}\)
\(\Rightarrow y_{3}=-2e^{x}(\cos{x}+\sin{x})\)
\(\Rightarrow \frac{d}{dx}(y_{3})=-2\frac{d}{dx}\{e^{x}(\cos{x}+\sin{x})\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=-2e^{x}\frac{d}{dx}(\cos{x}+\sin{x})-2(\cos{x}+\sin{x})\frac{d}{dx}(e^{x})\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{4}=-2e^{x}(-\sin{x}+\cos{x})-2(\cos{x}+\sin{x})e^{x}\) ➜\(\because \frac{d}{dx}(\cos{x})=-\sin{x}\), \(\frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{4}=-2e^{x}(-\sin{x}+\cos{x}+\cos{x}+\sin{x})\)
\(\Rightarrow y_{4}=-2e^{x}.2\cos{x}\)
\(\Rightarrow y_{4}=-4e^{x}\cos{x}\)
\(\Rightarrow y_{4}=-4y\) ➜\(\because y=e^{x}\cos{x}\)
\(\therefore y_{4}+4y=0\)
(Showed)
\(y=e^{x}\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{x}\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{x}\frac{d}{dx}(\cos{x})+\cos{x}\frac{d}{dx}(e^{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=-e^{x}\sin{x}+\cos{x}e^{x}\) ➜ \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=-e^{x}\sin{x}+e^{x}\cos{x}\)
\(\Rightarrow y_{1}=e^{x}(\cos{x}-\sin{x})\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{e^{x}(\cos{x}-\sin{x})\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=e^{x}\frac{d}{dx}(\cos{x}-\sin{x})+(\cos{x}-\sin{x})\frac{d}{dx}(e^{x})\}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=e^{x}(-\sin{x}-\cos{x})+(\cos{x}-\sin{x})e^{x}\}\) ➜\(\because \frac{d}{dx}(\cos{x})=-\sin{x}\), \(\frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{2}=e^{x}(-\sin{x}-\cos{x}+\cos{x}-\sin{x})\)
\(\Rightarrow y_{2}=e^{x}\{-2\sin{x}\}\)
\(\Rightarrow y_{2}=-2e^{x}\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=-2\frac{d}{dx}(e^{x}\sin{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-2e^{x}\frac{d}{dx}(\sin{x})-2\sin{x}\frac{d}{dx}(e^{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{3}=-2e^{x}\cos{x}-2\sin{x}e^{x}\)
\(\Rightarrow y_{3}=-2e^{x}(\cos{x}+\sin{x})\)
\(\Rightarrow \frac{d}{dx}(y_{3})=-2\frac{d}{dx}\{e^{x}(\cos{x}+\sin{x})\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=-2e^{x}\frac{d}{dx}(\cos{x}+\sin{x})-2(\cos{x}+\sin{x})\frac{d}{dx}(e^{x})\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{4}=-2e^{x}(-\sin{x}+\cos{x})-2(\cos{x}+\sin{x})e^{x}\) ➜\(\because \frac{d}{dx}(\cos{x})=-\sin{x}\), \(\frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{4}=-2e^{x}(-\sin{x}+\cos{x}+\cos{x}+\sin{x})\)
\(\Rightarrow y_{4}=-2e^{x}.2\cos{x}\)
\(\Rightarrow y_{4}=-4e^{x}\cos{x}\)
\(\Rightarrow y_{4}=-4y\) ➜\(\because y=e^{x}\cos{x}\)
\(\therefore y_{4}+4y=0\)
(Showed)
\(Q.3.(iii)\) \(y=e^{ax}\sin{bx}\) হলে, প্রমাণ কর যে, \(y_{2}-2ay_{1}+(a^2+b^2)y=0\)
[ কুয়েটঃ২০০৩-২০০৪]
[ কুয়েটঃ২০০৩-২০০৪]
সমাধানঃ
দেওয়া আছে,
\(y=e^{ax}\sin{bx}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{ax}\sin{bx})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}(\sin{bx})+\sin{bx}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=e^{ax}\cos{bx}\frac{d}{dx}(bx)+\sin{bx}e^{ax}\frac{d}{dx}(ax)\) ➜\((ax)\) ও \((bx)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^{ax}\cos{bx}.b+\sin{bx}e^{ax}.a\)
\(\Rightarrow y_{1}=be^{ax}\cos{bx}+ae^{ax}\sin{bx}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=b\frac{d}{dx}(e^{ax}\cos{bx})+a\frac{d}{dx}(e^{ax}\sin{bx})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=be^{ax}\frac{d}{dx}(\cos{bx})+b\cos{bx}\frac{d}{dx}(e^{ax})\)\(+ae^{ax}\frac{d}{dx}(\sin{bx})+a\sin{bx}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=-be^{ax}\sin{bx}.b+b\cos{bx}e^{ax}.a\)\(+ae^{ax}\cos{bx}.b+a\sin{bx}e^{ax}.a\)
\(\Rightarrow y_{2}=-b^2e^{ax}\sin{bx}+abe^{ax}\cos{bx}+abe^{ax}\cos{bx}+a^2e^{ax}\sin{bx}\)
\(\Rightarrow y_{2}=(a^2-b^2)e^{ax}\sin{bx}+2abe^{ax}\cos{bx}\)
\(\Rightarrow y_{2}=(-a^2-b^2)e^{ax}\sin{bx}+2abe^{ax}\cos{bx}+2a^2e^{ax}\sin{bx}\)
\(\Rightarrow y_{2}=-(a^2+b^2)e^{ax}\sin{bx}+2a(be^{ax}\cos{bx}+ae^{ax}\sin{bx})\)
\(\Rightarrow y_{2}=-(a^2+b^2)y+2ay_{1}\) ➜ \(\because y=e^{ax}\sin{bx}, y_{1}=be^{ax}\cos{bx}+ae^{ax}\sin{bx}\)
\(\therefore y_{2}-2ay_{1}+(a^2+b^2)y=0\)
(Proved)
\(y=e^{ax}\sin{bx}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{ax}\sin{bx})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{ax}\frac{d}{dx}(\sin{bx})+\sin{bx}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=e^{ax}\cos{bx}\frac{d}{dx}(bx)+\sin{bx}e^{ax}\frac{d}{dx}(ax)\) ➜\((ax)\) ও \((bx)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^{ax}\cos{bx}.b+\sin{bx}e^{ax}.a\)
\(\Rightarrow y_{1}=be^{ax}\cos{bx}+ae^{ax}\sin{bx}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=b\frac{d}{dx}(e^{ax}\cos{bx})+a\frac{d}{dx}(e^{ax}\sin{bx})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=be^{ax}\frac{d}{dx}(\cos{bx})+b\cos{bx}\frac{d}{dx}(e^{ax})\)\(+ae^{ax}\frac{d}{dx}(\sin{bx})+a\sin{bx}\frac{d}{dx}(e^{ax})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=-be^{ax}\sin{bx}.b+b\cos{bx}e^{ax}.a\)\(+ae^{ax}\cos{bx}.b+a\sin{bx}e^{ax}.a\)
\(\Rightarrow y_{2}=-b^2e^{ax}\sin{bx}+abe^{ax}\cos{bx}+abe^{ax}\cos{bx}+a^2e^{ax}\sin{bx}\)
\(\Rightarrow y_{2}=(a^2-b^2)e^{ax}\sin{bx}+2abe^{ax}\cos{bx}\)
\(\Rightarrow y_{2}=(-a^2-b^2)e^{ax}\sin{bx}+2abe^{ax}\cos{bx}+2a^2e^{ax}\sin{bx}\)
\(\Rightarrow y_{2}=-(a^2+b^2)e^{ax}\sin{bx}+2a(be^{ax}\cos{bx}+ae^{ax}\sin{bx})\)
\(\Rightarrow y_{2}=-(a^2+b^2)y+2ay_{1}\) ➜ \(\because y=e^{ax}\sin{bx}, y_{1}=be^{ax}\cos{bx}+ae^{ax}\sin{bx}\)
\(\therefore y_{2}-2ay_{1}+(a^2+b^2)y=0\)
(Proved)
\(Q.3.(iv)\) \(y=(e^{x}+e^{-x})\sin{x}\) হলে, প্রমাণ কর যে, \(\frac{d^4y}{dx^4}+4y=0\)
[ দিঃ২০১২;ঢাঃ২০১০,২০০৪;রাঃ২০০৬]
[ দিঃ২০১২;ঢাঃ২০১০,২০০৪;রাঃ২০০৬]
সমাধানঃ
দেওয়া আছে,
\(y=(e^{x}+e^{-x})\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(e^{x}+e^{-x})\sin{x}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(e^{x}+e^{-x})\frac{d}{dx}(\sin{x})+\sin{x}\frac{d}{dx}(e^{x}+e^{-x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=(e^{x}+e^{-x})\cos{x}+\sin{x}\{e^{x}+e^{-x}\frac{d}{dx}(-x)\}\) ➜ \((-x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=(e^{x}+e^{-x})\cos{x}+\sin{x}\{e^{x}+e^{-x}(-1)\}\)
\(\Rightarrow y_{1}=(e^{x}+e^{-x})\cos{x}+\sin{x}(e^{x}-e^{-x})\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{(e^{x}+e^{-x})\cos{x}\}+\frac{d}{dx}\{\sin{x}(e^{x}-e^{-x})\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(e^{x}+e^{-x})\frac{d}{dx}(\cos{x})+\cos{x}\frac{d}{dx}(e^{x}+e^{-x})\)\(+\sin{x}\frac{d}{dx}(e^{x}-e^{-x})+(e^{x}-e^{-x})\frac{d}{dx}(\sin{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=-(e^{x}+e^{-x})\sin{x}+(e^{x}-e^{-x})\cos{x}\)\(+(e^{x}+e^{-x})\sin{x}+(e^{x}-e^{-x})\cos{x}\)
\(\Rightarrow y_{2}=2(e^{x}-e^{-x})\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=2\frac{d}{dx}\{(e^{x}-e^{-x})\cos{x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=2(e^{x}-e^{-x})\frac{d}{dx}(\cos{x})+2\cos{x}\frac{d}{dx}(e^{x}-e^{-x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{3}=-2(e^{x}-e^{-x})\sin{x})+2\cos{x}(e^{x}+e^{-x})\)
\(\Rightarrow y_{3}=-2(e^{x}-e^{-x})\sin{x})+2(e^{x}+e^{-x})\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=-2\frac{d}{dx}\{(e^{x}-e^{-x})\sin{x}\}+2\frac{d}{dx}\{(e^{x}+e^{-x})\cos{x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (y_{4})=-2(e^{x}-e^{-x})\frac{d}{dx}(\sin{x})-2\sin{x}\frac{d}{dx}(e^{x}-e^{-x})\)\(+2(e^{x}+e^{-x})\frac{d}{dx}(\cos{x})+2\cos{x}\frac{d}{dx}(e^{x}+e^{-x})\)
\(\Rightarrow y_{4}=-2(e^{x}-e^{-x})\cos{x}-2\sin{x}(e^{x}+e^{-x})\)\(-2(e^{x}+e^{-x})\sin{x}+2\cos{x}(e^{x}-e^{-x})\)
\(\Rightarrow y_{4}=-2(e^{x}-e^{-x})\cos{x}-2(e^{x}+e^{-x})\sin{x}\)\(-2(e^{x}+e^{-x})\sin{x}+2(e^{x}-e^{-x})\cos{x}\)
\(\Rightarrow y_{4}=-4(e^{x}+e^{-x})\sin{x}\)
\(\Rightarrow \frac{d^4y}{dx^4}=-4y\) ➜ \(\because y_{4}=\frac{d^4y}{dx^4}, y=(e^{x}+e^{-x})\sin{x}\)
\(\therefore \frac{d^4y}{dx^4}+4y=0\)
(Proved)
\(y=(e^{x}+e^{-x})\sin{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(e^{x}+e^{-x})\sin{x}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(e^{x}+e^{-x})\frac{d}{dx}(\sin{x})+\sin{x}\frac{d}{dx}(e^{x}+e^{-x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=(e^{x}+e^{-x})\cos{x}+\sin{x}\{e^{x}+e^{-x}\frac{d}{dx}(-x)\}\) ➜ \((-x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=(e^{x}+e^{-x})\cos{x}+\sin{x}\{e^{x}+e^{-x}(-1)\}\)
\(\Rightarrow y_{1}=(e^{x}+e^{-x})\cos{x}+\sin{x}(e^{x}-e^{-x})\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{(e^{x}+e^{-x})\cos{x}\}+\frac{d}{dx}\{\sin{x}(e^{x}-e^{-x})\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(e^{x}+e^{-x})\frac{d}{dx}(\cos{x})+\cos{x}\frac{d}{dx}(e^{x}+e^{-x})\)\(+\sin{x}\frac{d}{dx}(e^{x}-e^{-x})+(e^{x}-e^{-x})\frac{d}{dx}(\sin{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=-(e^{x}+e^{-x})\sin{x}+(e^{x}-e^{-x})\cos{x}\)\(+(e^{x}+e^{-x})\sin{x}+(e^{x}-e^{-x})\cos{x}\)
\(\Rightarrow y_{2}=2(e^{x}-e^{-x})\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=2\frac{d}{dx}\{(e^{x}-e^{-x})\cos{x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=2(e^{x}-e^{-x})\frac{d}{dx}(\cos{x})+2\cos{x}\frac{d}{dx}(e^{x}-e^{-x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{3}=-2(e^{x}-e^{-x})\sin{x})+2\cos{x}(e^{x}+e^{-x})\)
\(\Rightarrow y_{3}=-2(e^{x}-e^{-x})\sin{x})+2(e^{x}+e^{-x})\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=-2\frac{d}{dx}\{(e^{x}-e^{-x})\sin{x}\}+2\frac{d}{dx}\{(e^{x}+e^{-x})\cos{x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (y_{4})=-2(e^{x}-e^{-x})\frac{d}{dx}(\sin{x})-2\sin{x}\frac{d}{dx}(e^{x}-e^{-x})\)\(+2(e^{x}+e^{-x})\frac{d}{dx}(\cos{x})+2\cos{x}\frac{d}{dx}(e^{x}+e^{-x})\)
\(\Rightarrow y_{4}=-2(e^{x}-e^{-x})\cos{x}-2\sin{x}(e^{x}+e^{-x})\)\(-2(e^{x}+e^{-x})\sin{x}+2\cos{x}(e^{x}-e^{-x})\)
\(\Rightarrow y_{4}=-2(e^{x}-e^{-x})\cos{x}-2(e^{x}+e^{-x})\sin{x}\)\(-2(e^{x}+e^{-x})\sin{x}+2(e^{x}-e^{-x})\cos{x}\)
\(\Rightarrow y_{4}=-4(e^{x}+e^{-x})\sin{x}\)
\(\Rightarrow \frac{d^4y}{dx^4}=-4y\) ➜ \(\because y_{4}=\frac{d^4y}{dx^4}, y=(e^{x}+e^{-x})\sin{x}\)
\(\therefore \frac{d^4y}{dx^4}+4y=0\)
(Proved)
\(Q.3.(v)\) \(\cos^{-1}\left(\frac{y}{b}\right)=\ln{\left(\frac{x}{n}\right)^n}\) হলে,
প্রমাণ কর যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+n^2y=0\)
[ বুয়েটঃ২০১০-২০১১]
প্রমাণ কর যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+n^2y=0\)
[ বুয়েটঃ২০১০-২০১১]
সমাধানঃ
দেওয়া আছে,
\(\cos^{-1}\left(\frac{y}{b}\right)=\ln{\left(\frac{x}{n}\right)^n}\)
\(\Rightarrow \cos^{-1}\left(\frac{y}{b}\right)=n\ln{\left(\frac{x}{n}\right)}\)
\(\Rightarrow \cos^{-1}\left(\frac{y}{b}\right)=n\ln{x}-n\ln{n}\)
\(\Rightarrow \frac{d}{dx}\{\cos^{-1}\left(\frac{y}{b}\right)\}=\frac{d}{dx}(n\ln{x}-n\ln{n})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow -\frac{1}{\sqrt{1-\left(\frac{y}{b}\right)^2}}\frac{d}{dx}\left(\frac{y}{b}\right)=n\frac{d}{dx}(\ln{x})-n\frac{d}{dx}(\ln{n})\) ➜\(\left(\frac{y}{b}\right)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\), \(\frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow -\frac{1}{\sqrt{1-\frac{y^2}{b^2}}}.\frac{1}{b}.y_{1}=n\frac{1}{x}-n.0\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(c)=0\)
\(\Rightarrow -\frac{1}{\sqrt{\frac{b^2-y^2}{b^2}}}.\frac{1}{b}.y_{1}=n\frac{1}{x}\)
\(\Rightarrow -\frac{b}{\sqrt{b^2-y^2}}.\frac{1}{b}.y_{1}=n\frac{1}{x}\)
\(\Rightarrow -\frac{1}{\sqrt{b^2-y^2}}y_{1}=n\frac{1}{x}\)
\(\Rightarrow -y_{1}=n\frac{\sqrt{b^2-y^2}}{x}\)
\(\Rightarrow y^2_{1}=n^2\frac{b^2-y^2}{x^2}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow x^2y^2_{1}=n^2(b^2-y^2)\) ➜ উভয় পার্শে \(x^2\) গুণ করে।
\(\Rightarrow \frac{d}{dx}(x^2y^2_{1})=n^2\frac{d}{dx}(b^2-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x^2\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(x^2)=n^2\frac{d}{dx}(b^2)-n^2\frac{d}{dx}(y^2)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow x^2.2y_{1}y_{2}+y^2_{1}.2x=n^2.0-n^2.2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow 2x^2y_{1}y_{2}+2xy^2_{1}=-2n^2yy_{1}\)
\(\Rightarrow 2x^2y_{1}y_{2}+2xy^2_{1}+2n^2yy_{1}=0\)
\(\Rightarrow 2y_{1}(x^2y_{2}+xy_{1}+n^2y)=0\)
\(\Rightarrow 2y_{1}\ne{0}, x^2y_{2}+xy_{1}+n^2y=0\)
\(\Rightarrow x^2y_{2}+xy_{1}+n^2y=0\)
\(\therefore x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+n^2y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Proved)
\(\cos^{-1}\left(\frac{y}{b}\right)=\ln{\left(\frac{x}{n}\right)^n}\)
\(\Rightarrow \cos^{-1}\left(\frac{y}{b}\right)=n\ln{\left(\frac{x}{n}\right)}\)
\(\Rightarrow \cos^{-1}\left(\frac{y}{b}\right)=n\ln{x}-n\ln{n}\)
\(\Rightarrow \frac{d}{dx}\{\cos^{-1}\left(\frac{y}{b}\right)\}=\frac{d}{dx}(n\ln{x}-n\ln{n})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow -\frac{1}{\sqrt{1-\left(\frac{y}{b}\right)^2}}\frac{d}{dx}\left(\frac{y}{b}\right)=n\frac{d}{dx}(\ln{x})-n\frac{d}{dx}(\ln{n})\) ➜\(\left(\frac{y}{b}\right)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\), \(\frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow -\frac{1}{\sqrt{1-\frac{y^2}{b^2}}}.\frac{1}{b}.y_{1}=n\frac{1}{x}-n.0\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(c)=0\)
\(\Rightarrow -\frac{1}{\sqrt{\frac{b^2-y^2}{b^2}}}.\frac{1}{b}.y_{1}=n\frac{1}{x}\)
\(\Rightarrow -\frac{b}{\sqrt{b^2-y^2}}.\frac{1}{b}.y_{1}=n\frac{1}{x}\)
\(\Rightarrow -\frac{1}{\sqrt{b^2-y^2}}y_{1}=n\frac{1}{x}\)
\(\Rightarrow -y_{1}=n\frac{\sqrt{b^2-y^2}}{x}\)
\(\Rightarrow y^2_{1}=n^2\frac{b^2-y^2}{x^2}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow x^2y^2_{1}=n^2(b^2-y^2)\) ➜ উভয় পার্শে \(x^2\) গুণ করে।
\(\Rightarrow \frac{d}{dx}(x^2y^2_{1})=n^2\frac{d}{dx}(b^2-y^2)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x^2\frac{d}{dx}(y^2_{1})+y^2_{1}\frac{d}{dx}(x^2)=n^2\frac{d}{dx}(b^2)-n^2\frac{d}{dx}(y^2)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow x^2.2y_{1}y_{2}+y^2_{1}.2x=n^2.0-n^2.2yy_{1}\) ➜ \(\because \frac{d}{dx}(y^2_{1})=2y_{1}y_{2}, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow 2x^2y_{1}y_{2}+2xy^2_{1}=-2n^2yy_{1}\)
\(\Rightarrow 2x^2y_{1}y_{2}+2xy^2_{1}+2n^2yy_{1}=0\)
\(\Rightarrow 2y_{1}(x^2y_{2}+xy_{1}+n^2y)=0\)
\(\Rightarrow 2y_{1}\ne{0}, x^2y_{2}+xy_{1}+n^2y=0\)
\(\Rightarrow x^2y_{2}+xy_{1}+n^2y=0\)
\(\therefore x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+n^2y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Proved)
\(Q.3.(vi)\) \(y=\frac{\sin{x}}{\sqrt{x}}\) হলে, প্রমাণ কর যে, \(x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+\left(x^2-\frac{1}{4}\right)y=0\)
[ বুয়েটঃ২০০৪-২০০৫]
[ বুয়েটঃ২০০৪-২০০৫]
সমাধানঃ
দেওয়া আছে,
\(y=\frac{\sin{x}}{\sqrt{x}}\)
\(\Rightarrow \sqrt{x}y=\sin{x}\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x}y)=\frac{d}{dx}(\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \sqrt{x}\frac{d}{dx}(y)+y\frac{d}{dx}(\sqrt{x})=\cos{x}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u) , \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \sqrt{x}y_{1}+y.\frac{1}{2\sqrt{x}}=\cos{x}\) ➜ \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \sqrt{x}y_{1}+\frac{y}{2\sqrt{x}}=\cos{x}\)
\(\Rightarrow xy_{1}+\frac{y}{2}=\sqrt{x}\cos{x}\) ➜ উভয় পার্শে \(\sqrt{x}\) গুণ করে।
\(\Rightarrow \frac{d}{dx}(xy_{1})+\frac{d}{dx}\left(\frac{y}{2}\right)=\frac{d}{dx}(\sqrt{x}\cos{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(x)+\frac{y_{1}}{2}=\sqrt{x}\frac{d}{dx}(\cos{x})+\cos{x}\frac{d}{dx}(\sqrt{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow xy_{2}+y_{1}.1+\frac{y_{1}}{2}=-\sqrt{x}\sin{x}+\cos{x}.\frac{1}{2\sqrt{x}}\) ➜\(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow xy_{2}+y_{1}+\frac{y_{1}}{2}=-\sqrt{x}\sin{x}+\frac{1}{2\sqrt{x}}\cos{x}\)
\(\Rightarrow xy_{2}+y_{1}+\frac{y_{1}}{2}=-\sqrt{x}\times{\sqrt{x}y}+\frac{1}{2\sqrt{x}}\left(\sqrt{x}y_{1}+\frac{y}{2\sqrt{x}}\right)\) ➜\(\because \sin{x}=\sqrt{x}y, \cos{x}=\sqrt{x}y_{1}+\frac{y}{2\sqrt{x}}\)
\(\Rightarrow xy_{2}+y_{1}+\frac{y_{1}}{2}=-xy+\frac{y_{1}}{2}+\frac{y}{4x}\)
\(\Rightarrow xy_{2}+y_{1}=-xy+\frac{y}{4x}\)
\(\Rightarrow x^2y_{2}+xy_{1}=-x^2y+\frac{y}{4}\) ➜ উভয় পার্শে \(x\) গুণ করে।
\(\Rightarrow x^2y_{2}+xy_{1}=-(x^2-\frac{1}{4})y\)
\(\Rightarrow x^2y_{2}+xy_{1}+(x^2-\frac{1}{4})y=0\)
\(\therefore x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+(x^2-\frac{1}{4})y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Proved)
\(y=\frac{\sin{x}}{\sqrt{x}}\)
\(\Rightarrow \sqrt{x}y=\sin{x}\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x}y)=\frac{d}{dx}(\sin{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \sqrt{x}\frac{d}{dx}(y)+y\frac{d}{dx}(\sqrt{x})=\cos{x}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u) , \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \sqrt{x}y_{1}+y.\frac{1}{2\sqrt{x}}=\cos{x}\) ➜ \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \sqrt{x}y_{1}+\frac{y}{2\sqrt{x}}=\cos{x}\)
\(\Rightarrow xy_{1}+\frac{y}{2}=\sqrt{x}\cos{x}\) ➜ উভয় পার্শে \(\sqrt{x}\) গুণ করে।
\(\Rightarrow \frac{d}{dx}(xy_{1})+\frac{d}{dx}\left(\frac{y}{2}\right)=\frac{d}{dx}(\sqrt{x}\cos{x})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow x\frac{d}{dx}(y_{1})+y_{1}\frac{d}{dx}(x)+\frac{y_{1}}{2}=\sqrt{x}\frac{d}{dx}(\cos{x})+\cos{x}\frac{d}{dx}(\sqrt{x})\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow xy_{2}+y_{1}.1+\frac{y_{1}}{2}=-\sqrt{x}\sin{x}+\cos{x}.\frac{1}{2\sqrt{x}}\) ➜\(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow xy_{2}+y_{1}+\frac{y_{1}}{2}=-\sqrt{x}\sin{x}+\frac{1}{2\sqrt{x}}\cos{x}\)
\(\Rightarrow xy_{2}+y_{1}+\frac{y_{1}}{2}=-\sqrt{x}\times{\sqrt{x}y}+\frac{1}{2\sqrt{x}}\left(\sqrt{x}y_{1}+\frac{y}{2\sqrt{x}}\right)\) ➜\(\because \sin{x}=\sqrt{x}y, \cos{x}=\sqrt{x}y_{1}+\frac{y}{2\sqrt{x}}\)
\(\Rightarrow xy_{2}+y_{1}+\frac{y_{1}}{2}=-xy+\frac{y_{1}}{2}+\frac{y}{4x}\)
\(\Rightarrow xy_{2}+y_{1}=-xy+\frac{y}{4x}\)
\(\Rightarrow x^2y_{2}+xy_{1}=-x^2y+\frac{y}{4}\) ➜ উভয় পার্শে \(x\) গুণ করে।
\(\Rightarrow x^2y_{2}+xy_{1}=-(x^2-\frac{1}{4})y\)
\(\Rightarrow x^2y_{2}+xy_{1}+(x^2-\frac{1}{4})y=0\)
\(\therefore x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+(x^2-\frac{1}{4})y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Proved)
\(Q.3.(vii)\) \(y=A\sin{mx}+B\cos{mx}\) হলে, প্রমাণ কর যে, \(y_{2}+m^2y=0\) এবং \(y_{4}-m^4y=0\)
[ বুয়েটঃ২০০৪-২০০৫]
[ বুয়েটঃ২০০৪-২০০৫]
সমাধানঃ
দেওয়া আছে,
\(y=A\sin{mx}+B\cos{mx}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(A\sin{mx}+B\cos{mx})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=A\frac{d}{dx}(\sin{mx})+B\frac{d}{dx}(\cos{mx})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=A\cos{mx}\frac{d}{dx}(mx)-B\sin{mx}\frac{d}{dx}(mx)\) ➜ \((mx)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{1}=A\cos{mx}.m-B\sin{mx}.m\)
\(\Rightarrow y_{1}=Am\cos{mx}-Bm\sin{mx}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(Am\cos{mx}-Bm\sin{mx})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=Am\frac{d}{dx}(\cos{mx})-Bm\frac{d}{dx}(\sin{mx})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=-Am\sin{mx}\frac{d}{dx}(mx)-Bm\cos{mx}\frac{d}{dx}(mx)\) ➜ \((mx)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{2}=-Am\sin{mx}.m-Bm\cos{mx}.m\)
\(\Rightarrow y_{2}=-Am^2\sin{mx}-Bm^2\cos{mx}\)
\(\Rightarrow y_{2}=-m^2(A\sin{mx}+B\cos{mx})\)
\(\Rightarrow y_{2}=-m^2y\) ➜ \(\because y=A\sin{mx}+B\cos{mx}\)
\(\therefore y_{2}+m^2y=0\)
(Proved)
আবার,
\(y_{2}+m^2y=0\)
\(\Rightarrow \frac{d}{dx}(y_{2}+m^2y)=\frac{d}{dx}(0)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{2})+m^2\frac{d}{dx}(y)=0\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{3}+m^2y_{1}=0\)
\(\Rightarrow \frac{d}{dx}(y_{3}+m^2y_{1})=0\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{3})+m^2\frac{d}{dx}(y_{1})=0\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{4}+m^2y_{2}=0\)
\(\Rightarrow y_{4}+m^2(-m^2y)=0\) ➜\(\because y_{2}+m^2y=0\Rightarrow y_{2}=-m^2y\)
\(\therefore y_{4}-m^4y=0\)
(Proved)
\(y=A\sin{mx}+B\cos{mx}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(A\sin{mx}+B\cos{mx})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=A\frac{d}{dx}(\sin{mx})+B\frac{d}{dx}(\cos{mx})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{1}=A\cos{mx}\frac{d}{dx}(mx)-B\sin{mx}\frac{d}{dx}(mx)\) ➜ \((mx)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{1}=A\cos{mx}.m-B\sin{mx}.m\)
\(\Rightarrow y_{1}=Am\cos{mx}-Bm\sin{mx}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(Am\cos{mx}-Bm\sin{mx})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=Am\frac{d}{dx}(\cos{mx})-Bm\frac{d}{dx}(\sin{mx})\) ➜ \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=-Am\sin{mx}\frac{d}{dx}(mx)-Bm\cos{mx}\frac{d}{dx}(mx)\) ➜ \((mx)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow y_{2}=-Am\sin{mx}.m-Bm\cos{mx}.m\)
\(\Rightarrow y_{2}=-Am^2\sin{mx}-Bm^2\cos{mx}\)
\(\Rightarrow y_{2}=-m^2(A\sin{mx}+B\cos{mx})\)
\(\Rightarrow y_{2}=-m^2y\) ➜ \(\because y=A\sin{mx}+B\cos{mx}\)
\(\therefore y_{2}+m^2y=0\)
(Proved)
আবার,
\(y_{2}+m^2y=0\)
\(\Rightarrow \frac{d}{dx}(y_{2}+m^2y)=\frac{d}{dx}(0)\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{2})+m^2\frac{d}{dx}(y)=0\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{3}+m^2y_{1}=0\)
\(\Rightarrow \frac{d}{dx}(y_{3}+m^2y_{1})=0\) ➜আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y_{3})+m^2\frac{d}{dx}(y_{1})=0\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{4}+m^2y_{2}=0\)
\(\Rightarrow y_{4}+m^2(-m^2y)=0\) ➜\(\because y_{2}+m^2y=0\Rightarrow y_{2}=-m^2y\)
\(\therefore y_{4}-m^4y=0\)
(Proved)
\(Q.3.(viii)\) \(y=x^m\log_{e}x\) হলে, প্রমাণ কর যে, \(xy_{1}=my+x^m\)
[ বুয়েটঃ২০০৪-২০০৫]
[ বুয়েটঃ২০০৪-২০০৫]
সমাধানঃ
দেওয়া আছে,
\(y=x^m\log_{e}x\)
\(\Rightarrow y=x^m\ln{x}\) ➜ \(\because \log_{e}x=\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^m\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=x^m\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^m)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^m\frac{1}{x}+\ln{x}.mx^{m-1}\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{1}=x^{m-1}+m\ln{x}x^{m-1}\)
\(\Rightarrow xy_{1}=x^{m-1}\times{x}+m\ln{x}x^{m-1}\times{x}\) ➜ উভয় পার্শে \(x\) গুণ করে।
\(\Rightarrow xy_{1}=x^{m}+mx^{m}\ln{x}\)
\(\Rightarrow xy_{1}=x^{m}+my\) ➜ \(\because y=x^m\ln{x}\)
\(\therefore xy_{1}=my+x^{m}\)
(Proved)
\(y=x^m\log_{e}x\)
\(\Rightarrow y=x^m\ln{x}\) ➜ \(\because \log_{e}x=\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^m\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=x^m\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^m)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=x^m\frac{1}{x}+\ln{x}.mx^{m-1}\) ➜ \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow y_{1}=x^{m-1}+m\ln{x}x^{m-1}\)
\(\Rightarrow xy_{1}=x^{m-1}\times{x}+m\ln{x}x^{m-1}\times{x}\) ➜ উভয় পার্শে \(x\) গুণ করে।
\(\Rightarrow xy_{1}=x^{m}+mx^{m}\ln{x}\)
\(\Rightarrow xy_{1}=x^{m}+my\) ➜ \(\because y=x^m\ln{x}\)
\(\therefore xy_{1}=my+x^{m}\)
(Proved)
\(Q.3.(ix)\) \(y=(a+bx)e^{2x}\) হলে, প্রমাণ কর যে, \(y_{2}-2y_{1}-2be^{2x}=0\)
সমাধানঃ
দেওয়া আছে,
\(y=(a+bx)e^{2x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(a+bx)e^{2x}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(a+bx)\frac{d}{dx}(e^{2x})+e^{2x}\frac{d}{dx}(a+bx)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=(a+bx)e^{2x}\frac{d}{dx}(2x)+e^{2x}(0+b.1)\) ➜ \((2x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=(a+bx)e^{2x}.2+be^{2x}\)
\(\Rightarrow y_{1}=2(a+bx)e^{2x}+be^{2x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{2(a+bx)e^{2x}+be^{2x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=2\frac{d}{dx}\{(a+bx)e^{2x}\}+b\frac{d}{dx}(e^{2x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=2(a+bx)\frac{d}{dx}(e^{2x})+2e^{2x}\frac{d}{dx}(a+bx)+be^{2x}\frac{d}{dx}(2x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=2(a+bx)e^{2x}\frac{d}{dx}(2x)+2e^{2x}(0+b.1)+be^{2x}.2\)
\(\Rightarrow y_{2}=2(a+bx)e^{2x}.2+2be^{2x}+2be^{2x}\)
\(\Rightarrow y_{2}=4(a+bx)e^{2x}+2be^{2x})+2be^{2x}\)
\(\Rightarrow y_{2}=2(2(a+bx)e^{2x}+be^{2x})+2be^{2x}\)
\(\Rightarrow y_{2}=2y_{1}+2be^{2x}\)| \(\because y_{1}=2(a+bx)e^{2x}+be^{2x}\)
\(\therefore y_{2}-2y_{1}-2be^{2x}=0\)
(Proved)
\(y=(a+bx)e^{2x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(a+bx)e^{2x}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(a+bx)\frac{d}{dx}(e^{2x})+e^{2x}\frac{d}{dx}(a+bx)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=(a+bx)e^{2x}\frac{d}{dx}(2x)+e^{2x}(0+b.1)\) ➜ \((2x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=(a+bx)e^{2x}.2+be^{2x}\)
\(\Rightarrow y_{1}=2(a+bx)e^{2x}+be^{2x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{2(a+bx)e^{2x}+be^{2x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=2\frac{d}{dx}\{(a+bx)e^{2x}\}+b\frac{d}{dx}(e^{2x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=2(a+bx)\frac{d}{dx}(e^{2x})+2e^{2x}\frac{d}{dx}(a+bx)+be^{2x}\frac{d}{dx}(2x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=2(a+bx)e^{2x}\frac{d}{dx}(2x)+2e^{2x}(0+b.1)+be^{2x}.2\)
\(\Rightarrow y_{2}=2(a+bx)e^{2x}.2+2be^{2x}+2be^{2x}\)
\(\Rightarrow y_{2}=4(a+bx)e^{2x}+2be^{2x})+2be^{2x}\)
\(\Rightarrow y_{2}=2(2(a+bx)e^{2x}+be^{2x})+2be^{2x}\)
\(\Rightarrow y_{2}=2y_{1}+2be^{2x}\)| \(\because y_{1}=2(a+bx)e^{2x}+be^{2x}\)
\(\therefore y_{2}-2y_{1}-2be^{2x}=0\)
(Proved)
\(Q.3.(x)\) \(y=(a+bx)e^{-2x}\) হলে, প্রমাণ কর যে, \(\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=(a+bx)e^{-2x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(a+bx)e^{-2x}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(a+bx)\frac{d}{dx}(e^{-2x})+e^{-2x}\frac{d}{dx}(a+bx)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=(a+bx)e^{-2x}\frac{d}{dx}(-2x)+e^{-2x}(0+b.1)\) ➜ \((2x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=(a+bx)e^{-2x}.(-2)+be^{-2x}\)
\(\Rightarrow y_{1}=-2(a+bx)e^{-2x}+be^{-2x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{-2(a+bx)e^{-2x}+be^{-2x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-2\frac{d}{dx}\{(a+bx)e^{-2x}\}+b\frac{d}{dx}(e^{-2x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=-2(a+bx)\frac{d}{dx}(e^{-2x})-2e^{-2x}\frac{d}{dx}(a+bx)+be^{-2x}\frac{d}{dx}(-2x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=-2(a+bx)e^{-2x}\frac{d}{dx}(-2x)-2e^{-2x}(0+b.1)+be^{-2x}.(-2)\)
\(\Rightarrow y_{2}=-2(a+bx)e^{-2x}.(-2)-2be^{-2x}-2be^{-2x}\)
\(\Rightarrow y_{2}=4(a+bx)e^{2x}-4be^{2x})\)
\(\Rightarrow y_{2}=8(a+bx)e^{-2x}-4be^{-2x})-4(a+bx)e^{-2x}\)
\(\Rightarrow y_{2}=-4(-2(a+bx)e^{-2x}+be^{-2x})-4(a+bx)e^{-2x}\)
\(\Rightarrow y_{2}=-4y_{1}-4y\)| \(\because y_{1}=-2(a+bx)e^{-2x}+be^{-2x}, y=(a+bx)e^{-2x}\)
\(\Rightarrow y_{2}+4y_{1}+4y=0\)
\(\therefore \frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Proved)
\(y=(a+bx)e^{-2x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(a+bx)e^{-2x}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(a+bx)\frac{d}{dx}(e^{-2x})+e^{-2x}\frac{d}{dx}(a+bx)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=(a+bx)e^{-2x}\frac{d}{dx}(-2x)+e^{-2x}(0+b.1)\) ➜ \((2x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=(a+bx)e^{-2x}.(-2)+be^{-2x}\)
\(\Rightarrow y_{1}=-2(a+bx)e^{-2x}+be^{-2x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{-2(a+bx)e^{-2x}+be^{-2x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-2\frac{d}{dx}\{(a+bx)e^{-2x}\}+b\frac{d}{dx}(e^{-2x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=-2(a+bx)\frac{d}{dx}(e^{-2x})-2e^{-2x}\frac{d}{dx}(a+bx)+be^{-2x}\frac{d}{dx}(-2x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=-2(a+bx)e^{-2x}\frac{d}{dx}(-2x)-2e^{-2x}(0+b.1)+be^{-2x}.(-2)\)
\(\Rightarrow y_{2}=-2(a+bx)e^{-2x}.(-2)-2be^{-2x}-2be^{-2x}\)
\(\Rightarrow y_{2}=4(a+bx)e^{2x}-4be^{2x})\)
\(\Rightarrow y_{2}=8(a+bx)e^{-2x}-4be^{-2x})-4(a+bx)e^{-2x}\)
\(\Rightarrow y_{2}=-4(-2(a+bx)e^{-2x}+be^{-2x})-4(a+bx)e^{-2x}\)
\(\Rightarrow y_{2}=-4y_{1}-4y\)| \(\because y_{1}=-2(a+bx)e^{-2x}+be^{-2x}, y=(a+bx)e^{-2x}\)
\(\Rightarrow y_{2}+4y_{1}+4y=0\)
\(\therefore \frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0\) ➜ \(\because y_{2}=\frac{d^2y}{dx^2}, y_{1}=\frac{dy}{dx}\)
(Proved)
\(Q.3.(xi)\) \(y=(p+qx)e^{-2x}\) হলে, প্রমাণ কর যে, \(y_{2}+4y_{1}+4y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=(p+qx)e^{-2x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(p+qx)e^{-2x}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(p+qx)\frac{d}{dx}(e^{-2x})+e^{-2x}\frac{d}{dx}(p+qx)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=(p+qx)e^{-2x}\frac{d}{dx}(-2x)+e^{-2x}(0+q.1)\) ➜ \((2x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=(p+qx)e^{-2x}.(-2)+qe^{-2x}\)
\(\Rightarrow y_{1}=-2(p+qx)e^{-2x}+qe^{-2x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{-2(p+qx)e^{-2x}+qe^{-2x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-2\frac{d}{dx}\{(p+qx)e^{-2x}\}+q\frac{d}{dx}(e^{-2x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=-2(p+qx)\frac{d}{dx}(e^{-2x})-2e^{-2x}\frac{d}{dx}(p+qx)+qe^{-2x}\frac{d}{dx}(-2x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=-2(p+qx)e^{-2x}\frac{d}{dx}(-2x)-2e^{-2x}(0+q.1)+qe^{-2x}.(-2)\)
\(\Rightarrow y_{2}=-2(p+qx)e^{-2x}.(-2)-2qe^{-2x}-2qe^{-2x}\)
\(\Rightarrow y_{2}=4(p+qx)e^{2x}-4qe^{2x})\)
\(\Rightarrow y_{2}=8(p+qx)e^{-2x}-4qe^{-2x})-4(p+qx)e^{-2x}\)
\(\Rightarrow y_{2}=-4(-2(p+qx)e^{-2x}+qe^{-2x})-4(p+qx)e^{-2x}\)
\(\Rightarrow y_{2}=-4y_{1}-4y\)| \(\because y_{1}=-2(a+bx)e^{-2x}+be^{-2x}, y=(a+bx)e^{-2x}\)
\(\therefore y_{2}+4y_{1}+4y=0\)
(Proved)
\(y=(p+qx)e^{-2x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(p+qx)e^{-2x}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(p+qx)\frac{d}{dx}(e^{-2x})+e^{-2x}\frac{d}{dx}(p+qx)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{1}=(p+qx)e^{-2x}\frac{d}{dx}(-2x)+e^{-2x}(0+q.1)\) ➜ \((2x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow y_{1}=(p+qx)e^{-2x}.(-2)+qe^{-2x}\)
\(\Rightarrow y_{1}=-2(p+qx)e^{-2x}+qe^{-2x}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{-2(p+qx)e^{-2x}+qe^{-2x}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-2\frac{d}{dx}\{(p+qx)e^{-2x}\}+q\frac{d}{dx}(e^{-2x})\) ➜ \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y_{2}=-2(p+qx)\frac{d}{dx}(e^{-2x})-2e^{-2x}\frac{d}{dx}(p+qx)+qe^{-2x}\frac{d}{dx}(-2x)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y_{2}=-2(p+qx)e^{-2x}\frac{d}{dx}(-2x)-2e^{-2x}(0+q.1)+qe^{-2x}.(-2)\)
\(\Rightarrow y_{2}=-2(p+qx)e^{-2x}.(-2)-2qe^{-2x}-2qe^{-2x}\)
\(\Rightarrow y_{2}=4(p+qx)e^{2x}-4qe^{2x})\)
\(\Rightarrow y_{2}=8(p+qx)e^{-2x}-4qe^{-2x})-4(p+qx)e^{-2x}\)
\(\Rightarrow y_{2}=-4(-2(p+qx)e^{-2x}+qe^{-2x})-4(p+qx)e^{-2x}\)
\(\Rightarrow y_{2}=-4y_{1}-4y\)| \(\because y_{1}=-2(a+bx)e^{-2x}+be^{-2x}, y=(a+bx)e^{-2x}\)
\(\therefore y_{2}+4y_{1}+4y=0\)
(Proved)
\(Q.3.(xii)\) \(y=e^{a\sin^{-1}{t}}\) হলে, দেখাও যে, \((1-t^2)y_{2}-ty_{1}=a^2y\)
[ ঢাঃ ২০১১ ]
[ ঢাঃ ২০১১ ]
সমাধানঃ
দেওয়া আছে,
\(y=e^{a\sin^{-1}{t}}\)
\(\Rightarrow \frac{d}{dt}(y)=\frac{d}{dt}(e^{a\sin^{-1}{t}})\) ➜ \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{a\sin^{-1}{t}}.a\frac{d}{dt}(\sin^{-1}{t})\) ➜ \((a\sin^{-1}{t})\) কে \(t\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^{a\sin^{-1}{t}}.a\frac{1}{\sqrt{1-t^2}}\)
\(\Rightarrow \sqrt{1-t^2}y_{1}=ae^{a\sin^{-1}{t}}\) ➜ উভয় পার্শে \(\sqrt{1-t^2}\) গুণ করে।
\(\Rightarrow \sqrt{1-t^2}y_{1}=ay\)| \(\because y=e^{a\sin^{-1}{t}}\)
\(\Rightarrow (1-t^2)y^2_{1}=a^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dt}\{(1-t^2)y^2_{1}\}=a^2\frac{d}{dt}(y^2)\) ➜ আবার, \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-t^2)\frac{d}{dt}(y^2_{1})+y^2_{1}\frac{d}{dt}(1-t^2)=a^22yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dt}(y^2)=2yy_{1}\)
\(\Rightarrow (1-t^2)2y_{1}y_{2}+y^2_{1}(0-2t)=2a^2yy_{1}\) ➜ \(\because \frac{d}{dt}(y^2_{1})=2y_{1}y_{2}\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-t^2)y_{1}y_{2}-2ty^2_{1}-2a^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-t^2)y_{2}-ty_{1}-a^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-t^2)y_{2}-ty_{1}-a^2y=0\)
\(\Rightarrow (1-t^2)y_{2}-ty_{1}-a^2y=0\)
\(\therefore (1-t^2)y_{2}-ty_{1}=a^2y\)
(Showed)
\(y=e^{a\sin^{-1}{t}}\)
\(\Rightarrow \frac{d}{dt}(y)=\frac{d}{dt}(e^{a\sin^{-1}{t}})\) ➜ \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=e^{a\sin^{-1}{t}}.a\frac{d}{dt}(\sin^{-1}{t})\) ➜ \((a\sin^{-1}{t})\) কে \(t\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow y_{1}=e^{a\sin^{-1}{t}}.a\frac{1}{\sqrt{1-t^2}}\)
\(\Rightarrow \sqrt{1-t^2}y_{1}=ae^{a\sin^{-1}{t}}\) ➜ উভয় পার্শে \(\sqrt{1-t^2}\) গুণ করে।
\(\Rightarrow \sqrt{1-t^2}y_{1}=ay\)| \(\because y=e^{a\sin^{-1}{t}}\)
\(\Rightarrow (1-t^2)y^2_{1}=a^2y^2\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dt}\{(1-t^2)y^2_{1}\}=a^2\frac{d}{dt}(y^2)\) ➜ আবার, \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow (1-t^2)\frac{d}{dt}(y^2_{1})+y^2_{1}\frac{d}{dt}(1-t^2)=a^22yy_{1}\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dt}(y^2)=2yy_{1}\)
\(\Rightarrow (1-t^2)2y_{1}y_{2}+y^2_{1}(0-2t)=2a^2yy_{1}\) ➜ \(\because \frac{d}{dt}(y^2_{1})=2y_{1}y_{2}\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2(1-t^2)y_{1}y_{2}-2ty^2_{1}-2a^2yy_{1}=0\)
\(\Rightarrow 2y_{1}\{(1-t^2)y_{2}-ty_{1}-a^2y\}=0\)
\(\Rightarrow 2y_{1}\ne{0}, (1-t^2)y_{2}-ty_{1}-a^2y=0\)
\(\Rightarrow (1-t^2)y_{2}-ty_{1}-a^2y=0\)
\(\therefore (1-t^2)y_{2}-ty_{1}=a^2y\)
(Showed)
\(Q.3.(xiii)\) \(x=a(\theta+\sin{\theta})\) ও \(y=a(1-\cos{\theta})\) হলে, \(\frac{\theta}{2}\) এর মাধ্যমে \(\frac{dy}{dx}\) ও \(\frac{d^2y}{dx^2}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(\tan{\frac{\theta}{2}}; \frac{1}{4a}\sec^4{\frac{\theta}{2}}\)
উত্তরঃ \(\tan{\frac{\theta}{2}}; \frac{1}{4a}\sec^4{\frac{\theta}{2}}\)
সমাধানঃ
দেওয়া আছে,
\(x=a(\theta+\sin{\theta})\) এবং \(y=a(1-\cos{\theta})\)
\(\Rightarrow x=a(\theta+\sin{\theta})\)
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\theta+\sin{\theta})\) ➜ \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\theta)+a\frac{d}{d\theta}(\sin{\theta})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{d\theta}=a.1+a\cos{\theta}\) ➜\(\because \frac{d}{d\theta}(\theta)=1, \frac{d}{d\theta}(\sin{\theta})=\cos{\theta}\)
\(\therefore \frac{dx}{d\theta}=a(1+\cos{\theta}) ....(1)\)
আবার,
\(\Rightarrow y=a(1-\cos{\theta})\)
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(1-\cos{\theta})\) ➜ \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(1)-a\frac{d}{d\theta}(\cos{\theta})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{d\theta}=0+a\sin{\theta}\) ➜\(\because \frac{d}{d\theta}(c)=0, \frac{d}{d\theta}(\cos{\theta})=\sin{\theta}\)
\(\therefore \frac{dy}{d\theta}=a\sin{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=a\sin{\theta}\times{\frac{1}{a(1+\cos{\theta})}}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{\sin{\theta}}{1+\cos{\theta}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{2\cos^2{\frac{\theta}{2}}}\) ➜\(\because \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}, 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}\)
\(\therefore \frac{dy}{dx}=\tan{\frac{\theta}{2}}\)
আবার,
\(\Rightarrow \frac{dy}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(\tan{\frac{\theta}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\sec^2{\frac{\theta}{2}}.\frac{1}{2}\frac{d}{dx}(\theta)\) ➜ \(\frac{\theta}{2}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\frac{d\theta}{dx}\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\times{\frac{1}{a(1+\cos{\theta})}}\) ➜\(\because \frac{dx}{d\theta}=a(1+\cos{\theta})\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\times{\frac{1}{2a\cos^2{\frac{\theta}{2}}}}\) ➜\(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\times{\frac{1}{2a}\sec^2{\frac{\theta}{2}}}\)
\(=\frac{1}{4a}\sec^4{\frac{\theta}{2}}\)
\(x=a(\theta+\sin{\theta})\) এবং \(y=a(1-\cos{\theta})\)
\(\Rightarrow x=a(\theta+\sin{\theta})\)
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\theta+\sin{\theta})\) ➜ \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\theta)+a\frac{d}{d\theta}(\sin{\theta})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{d\theta}=a.1+a\cos{\theta}\) ➜\(\because \frac{d}{d\theta}(\theta)=1, \frac{d}{d\theta}(\sin{\theta})=\cos{\theta}\)
\(\therefore \frac{dx}{d\theta}=a(1+\cos{\theta}) ....(1)\)
আবার,
\(\Rightarrow y=a(1-\cos{\theta})\)
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(1-\cos{\theta})\) ➜ \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(1)-a\frac{d}{d\theta}(\cos{\theta})\) ➜\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{d\theta}=0+a\sin{\theta}\) ➜\(\because \frac{d}{d\theta}(c)=0, \frac{d}{d\theta}(\cos{\theta})=\sin{\theta}\)
\(\therefore \frac{dy}{d\theta}=a\sin{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=a\sin{\theta}\times{\frac{1}{a(1+\cos{\theta})}}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{\sin{\theta}}{1+\cos{\theta}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{2\cos^2{\frac{\theta}{2}}}\) ➜\(\because \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}, 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}\)
\(\therefore \frac{dy}{dx}=\tan{\frac{\theta}{2}}\)
আবার,
\(\Rightarrow \frac{dy}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(\tan{\frac{\theta}{2}})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\sec^2{\frac{\theta}{2}}.\frac{1}{2}\frac{d}{dx}(\theta)\) ➜ \(\frac{\theta}{2}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\frac{d\theta}{dx}\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\times{\frac{1}{a(1+\cos{\theta})}}\) ➜\(\because \frac{dx}{d\theta}=a(1+\cos{\theta})\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\times{\frac{1}{2a\cos^2{\frac{\theta}{2}}}}\) ➜\(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(=\frac{1}{2}\sec^2{\frac{\theta}{2}}\times{\frac{1}{2a}\sec^2{\frac{\theta}{2}}}\)
\(=\frac{1}{4a}\sec^4{\frac{\theta}{2}}\)
\(Q.3.(xiv)\) \(2x=t+t^{-1}\) এবং \(2y=t-t^{-1}\) হলে,
দেখাও যে, \(\frac{dy}{dx}=\frac{t^2+1}{t^2-1}\)এবং \(\frac{d^2y}{dx^2}=-\frac{8t^3}{(t^2-1)^3}\)
দেখাও যে, \(\frac{dy}{dx}=\frac{t^2+1}{t^2-1}\)এবং \(\frac{d^2y}{dx^2}=-\frac{8t^3}{(t^2-1)^3}\)
সমাধানঃ
দেওয়া আছে,
\(2x=t+t^{-1}\) এবং \(2y=t-t^{-1}\)
\(\Rightarrow 2x=t+t^{-1}\)
\(\Rightarrow 2\frac{dx}{dt}=\frac{d}{dt}(t+t^{-1})\) ➜ \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2\frac{dx}{dt}=1+(-1)t^{-1-1}\) ➜ \(\because \frac{d}{dt}(t)=1, \frac{d}{dt}(t^n)=nt^{n-1}\)
\(\therefore \frac{dx}{dt}=\frac{1}{2}(1-t^{-2}) ....(1)\)
আবার,
\(\Rightarrow 2\frac{dy}{dt}=\frac{d}{dt}(t-t^{-1})\) ➜ \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2\frac{dy}{dt}=1-(-1)t^{-1-1}\) ➜ \(\because \frac{d}{dt}(t)=1, \frac{d}{dt}(t^n)=nt^{n-1}\)
\(\therefore \frac{dy}{dt}=\frac{1}{2}(1+t^{-2}) ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(=\frac{1}{2}(1+t^{-2})\times{\frac{1}{\frac{1}{2}(1-t^{-2})}}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।
\(=(1+t^{-2})\times{\frac{1}{(1-t^{-2})}}\)
\(=\frac{1+t^{-2}}{1-t^{-2}}\)
\(=\frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}\)
\(=\frac{\frac{t^2+1}{t^2}}{\frac{t^2-1}{t^2}}\)
\(=\frac{t^2+1}{t^2}\times{\frac{t^2}{t^2-1}}\)
\(\therefore \frac{dy}{dx}=\frac{t^2+1}{t^2-1}\)
(Showed)
আবার,
\(\Rightarrow \frac{dy}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{t^2+1}{t^2-1}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(t^2-1)\frac{d}{dx}(t^2+1)-(t^2+1)\frac{d}{dx}(t^2-1)}{(t^2-1)^2}\) ➜\(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(t^2-1).2t\frac{dt}{dx}-(t^2+1).2t\frac{dt}{dx}}{(t^2-1)^2}\) ➜ \(t\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x\)
\(=2t\frac{dt}{dx}\frac{t^2-1-t^2-1}{(t^2-1)^2}\)
\(=2t\frac{dt}{dx}\frac{-2}{(t^2-1)^2}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{dt}{dx}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{1}{\frac{dx}{dt}}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{1}{\frac{1}{2}(1-t^{-2})}}\) ➜ \(\because \frac{dx}{dt}=\frac{1}{2}(1-t^{-2})\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{2}{1-t^{-2}}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{2}{1-\frac{1}{t^2}}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{2}{\frac{t^2-1}{t^2}}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{2t^2}{t^2-1}}\)
\(\therefore \frac{d^2y}{dx^2}=-\frac{8t^3}{(t^2-1)^3}\)
\(2x=t+t^{-1}\) এবং \(2y=t-t^{-1}\)
\(\Rightarrow 2x=t+t^{-1}\)
\(\Rightarrow 2\frac{dx}{dt}=\frac{d}{dt}(t+t^{-1})\) ➜ \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2\frac{dx}{dt}=1+(-1)t^{-1-1}\) ➜ \(\because \frac{d}{dt}(t)=1, \frac{d}{dt}(t^n)=nt^{n-1}\)
\(\therefore \frac{dx}{dt}=\frac{1}{2}(1-t^{-2}) ....(1)\)
আবার,
\(\Rightarrow 2\frac{dy}{dt}=\frac{d}{dt}(t-t^{-1})\) ➜ \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2\frac{dy}{dt}=1-(-1)t^{-1-1}\) ➜ \(\because \frac{d}{dt}(t)=1, \frac{d}{dt}(t^n)=nt^{n-1}\)
\(\therefore \frac{dy}{dt}=\frac{1}{2}(1+t^{-2}) ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(=\frac{1}{2}(1+t^{-2})\times{\frac{1}{\frac{1}{2}(1-t^{-2})}}\) ➜ \((1)\) ও \((2)\) এর সাহায্যে।
\(=(1+t^{-2})\times{\frac{1}{(1-t^{-2})}}\)
\(=\frac{1+t^{-2}}{1-t^{-2}}\)
\(=\frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}\)
\(=\frac{\frac{t^2+1}{t^2}}{\frac{t^2-1}{t^2}}\)
\(=\frac{t^2+1}{t^2}\times{\frac{t^2}{t^2-1}}\)
\(\therefore \frac{dy}{dx}=\frac{t^2+1}{t^2-1}\)
(Showed)
আবার,
\(\Rightarrow \frac{dy}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{t^2+1}{t^2-1}\right)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(t^2-1)\frac{d}{dx}(t^2+1)-(t^2+1)\frac{d}{dx}(t^2-1)}{(t^2-1)^2}\) ➜\(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{d^2y}{dx^2}=\frac{(t^2-1).2t\frac{dt}{dx}-(t^2+1).2t\frac{dt}{dx}}{(t^2-1)^2}\) ➜ \(t\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x\)
\(=2t\frac{dt}{dx}\frac{t^2-1-t^2-1}{(t^2-1)^2}\)
\(=2t\frac{dt}{dx}\frac{-2}{(t^2-1)^2}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{dt}{dx}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{1}{\frac{dx}{dt}}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{1}{\frac{1}{2}(1-t^{-2})}}\) ➜ \(\because \frac{dx}{dt}=\frac{1}{2}(1-t^{-2})\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{2}{1-t^{-2}}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{2}{1-\frac{1}{t^2}}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{2}{\frac{t^2-1}{t^2}}}\)
\(=-4\frac{t}{(t^2-1)^2}\times{\frac{2t^2}{t^2-1}}\)
\(\therefore \frac{d^2y}{dx^2}=-\frac{8t^3}{(t^2-1)^3}\)
\(Q.3.(xv)\) \(y=\sqrt{\cos{2x}}\) হলে, দেখাও যে, \((yy_{1})^2=1-y^4\)
সমাধানঃ
দেওয়া আছে,
\(y=\sqrt{\cos{2x}}\)
\(\Rightarrow y^2=\cos{2x}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(y^2)=\frac{d}{dx}(\cos{2x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2yy_{1}=-\sin{2x}\frac{d}{dx}(2x)\) ➜ \((2x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow 2yy_{1}=-\sin{2x}.2\)
\(\Rightarrow yy_{1}=-\sin{2x}\)
\(\Rightarrow (yy_{1})^2=\sin^2{2x}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (yy_{1})^2=1-\cos^2{2x}\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow (yy_{1})^2=1-(y^2)^2\) ➜ \(\because y^2=\cos{2x}\)
\(\therefore (yy_{1})^2=1-y^4\)
(Showed)
\(y=\sqrt{\cos{2x}}\)
\(\Rightarrow y^2=\cos{2x}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(y^2)=\frac{d}{dx}(\cos{2x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2yy_{1}=-\sin{2x}\frac{d}{dx}(2x)\) ➜ \((2x)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(y^2)=2yy_{1}\)
\(\Rightarrow 2yy_{1}=-\sin{2x}.2\)
\(\Rightarrow yy_{1}=-\sin{2x}\)
\(\Rightarrow (yy_{1})^2=\sin^2{2x}\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow (yy_{1})^2=1-\cos^2{2x}\) ➜ \(\because \sin^2{A}=1-\cos^2{A}\)
\(\Rightarrow (yy_{1})^2=1-(y^2)^2\) ➜ \(\because y^2=\cos{2x}\)
\(\therefore (yy_{1})^2=1-y^4\)
(Showed)
\(Q.3.(xvi)\) \(y=\tan{\sqrt{1-x}}\) হলে, দেখাও যে, \(2y_{1}\sqrt{1-x}+(1+y^2)=0\)
সমাধানঃ
দেওয়া আছে,
\(y=\tan{\sqrt{1-x}}\) \(\Rightarrow \tan^{-1}{y}=\sqrt{1-x}\) \(\Rightarrow (\tan^{-1}{y})^2=1-x\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{y})^2=\frac{d}{dx}(1-x)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2\tan^{-1}{y}\frac{d}{dx}(\tan^{-1}{y})=(0-1)\) ➜ \((\tan^{-1}{y})\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x)=1\)
\(\Rightarrow 2\tan^{-1}{y}\frac{1}{1+y^2}y_{1}=-1\)
\(\Rightarrow 2\tan^{-1}{y}y_{1}=-(1+y^2)\) ➜ উভয় পার্শে \(1+y^2\) গুণ করে।
\(\Rightarrow 2\sqrt{1-x}y_{1}=-(1+y^2)\) ➜ \(\because \tan^{-1}{y}=\sqrt{1-x}\)
\(\therefore 2\sqrt{1-x}y_{1}+(1+y^2)=0\)
(Showed)
\(y=\tan{\sqrt{1-x}}\) \(\Rightarrow \tan^{-1}{y}=\sqrt{1-x}\) \(\Rightarrow (\tan^{-1}{y})^2=1-x\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{y})^2=\frac{d}{dx}(1-x)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 2\tan^{-1}{y}\frac{d}{dx}(\tan^{-1}{y})=(0-1)\) ➜ \((\tan^{-1}{y})\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x)=1\)
\(\Rightarrow 2\tan^{-1}{y}\frac{1}{1+y^2}y_{1}=-1\)
\(\Rightarrow 2\tan^{-1}{y}y_{1}=-(1+y^2)\) ➜ উভয় পার্শে \(1+y^2\) গুণ করে।
\(\Rightarrow 2\sqrt{1-x}y_{1}=-(1+y^2)\) ➜ \(\because \tan^{-1}{y}=\sqrt{1-x}\)
\(\therefore 2\sqrt{1-x}y_{1}+(1+y^2)=0\)
(Showed)
\(Q.3.(xvii)\) \(y=\frac{4}{\sqrt{\sec{x}}}\) হলে, দেখাও যে, \(2\cot{x}\frac{dy}{dx}+y=0\)
সমাধানঃ
দেওয়া আছে,
\(y=\frac{4}{\sqrt{\sec{x}}}\)
\(\Rightarrow \sqrt{\sec{x}}y=4\)
\(\Rightarrow \sec{x}y^2=16\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(\sec{x}y^2)=\frac{d}{dx}(4)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \sec{x}\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(\sec{x})=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow \sec{x}2yy_{1}+y^2\sec{x}\tan{x}=0\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}\), \(\frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow 2y\sec{x}y_{1}+y^2\sec{x}\tan{x}=0\)
\(\Rightarrow y\sec{x}(2y_{1}+y\tan{x})=0\)
\(\Rightarrow y\sec{x}\ne{0}, 2y_{1}+y\tan{x}=0\)
\(\Rightarrow 2y_{1}+y\tan{x}=0\)
\(\Rightarrow 2\frac{1}{\tan{x}}y_{1}+y=0\) ➜ উভয় পার্শে \(\tan{x}\) ভাগ করে।
\(\Rightarrow 2\cot{x}y_{1}+y=0\) ➜ \(\because \frac{1}{\tan{x}}=\cot{x}\)
\(\therefore 2\cot{x}\frac{dy}{dx}+y=0\) ➜ \(\because y_{1}=\frac{dy}{dx}\)
(Showed)
\(y=\frac{4}{\sqrt{\sec{x}}}\)
\(\Rightarrow \sqrt{\sec{x}}y=4\)
\(\Rightarrow \sec{x}y^2=16\) ➜ উভয় পার্শে বর্গ করে।
\(\Rightarrow \frac{d}{dx}(\sec{x}y^2)=\frac{d}{dx}(4)\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \sec{x}\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(\sec{x})=0\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow \sec{x}2yy_{1}+y^2\sec{x}\tan{x}=0\) ➜ \(\because \frac{d}{dx}(y^2)=2yy_{1}\), \(\frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow 2y\sec{x}y_{1}+y^2\sec{x}\tan{x}=0\)
\(\Rightarrow y\sec{x}(2y_{1}+y\tan{x})=0\)
\(\Rightarrow y\sec{x}\ne{0}, 2y_{1}+y\tan{x}=0\)
\(\Rightarrow 2y_{1}+y\tan{x}=0\)
\(\Rightarrow 2\frac{1}{\tan{x}}y_{1}+y=0\) ➜ উভয় পার্শে \(\tan{x}\) ভাগ করে।
\(\Rightarrow 2\cot{x}y_{1}+y=0\) ➜ \(\because \frac{1}{\tan{x}}=\cot{x}\)
\(\therefore 2\cot{x}\frac{dy}{dx}+y=0\) ➜ \(\because y_{1}=\frac{dy}{dx}\)
(Showed)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xviii)\) \(\frac{1}{a-x}\)
উত্তরঃ \(\frac{n!}{(a-x)^{n+1}}\)
\(Q.3.(xviii)\) \(\frac{1}{a-x}\)
উত্তরঃ \(\frac{n!}{(a-x)^{n+1}}\)
সমাধানঃ
দেওয়া আছে,
\(y=\frac{1}{a-x}\)
\(\Rightarrow y=(a-x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(a-x)^{-1}\}\) ➜\(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(a-x)^{-1-1}\frac{d}{dx}(a-x)\)
\(\Rightarrow y_{1}=(-1).1(a-x)^{-2}.(-1)\)
\(\Rightarrow y_{1}=1(a-x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=1\frac{d}{dx}\{(a-x)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=1.(-2)(a-x)^{-2-1}\frac{d}{dx}(a-x)\)
\(\Rightarrow y_{2}=1.(-2).(a-x)^{-3}.(-1)\)
\(\Rightarrow y_{2}=1.2.(a-x)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=1.2\frac{d}{dx}\{(a-x)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=1.2.(-3)(a-x)^{-3-1}\frac{d}{dx}(a-x)\)
\(\Rightarrow y_{3}=1.2.(-3).(a-x)^{-(3+1)}.(-1)\)
\(\Rightarrow y_{3}=1.2.3.(a-x)^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=1.2.3 .........n(a-x)^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=n!\frac{1}{(a-x)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{n!}{(a-x)^{n+1}}\)
\(y=\frac{1}{a-x}\)
\(\Rightarrow y=(a-x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(a-x)^{-1}\}\) ➜\(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(a-x)^{-1-1}\frac{d}{dx}(a-x)\)
\(\Rightarrow y_{1}=(-1).1(a-x)^{-2}.(-1)\)
\(\Rightarrow y_{1}=1(a-x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=1\frac{d}{dx}\{(a-x)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=1.(-2)(a-x)^{-2-1}\frac{d}{dx}(a-x)\)
\(\Rightarrow y_{2}=1.(-2).(a-x)^{-3}.(-1)\)
\(\Rightarrow y_{2}=1.2.(a-x)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=1.2\frac{d}{dx}\{(a-x)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=1.2.(-3)(a-x)^{-3-1}\frac{d}{dx}(a-x)\)
\(\Rightarrow y_{3}=1.2.(-3).(a-x)^{-(3+1)}.(-1)\)
\(\Rightarrow y_{3}=1.2.3.(a-x)^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=1.2.3 .........n(a-x)^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=n!\frac{1}{(a-x)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{n!}{(a-x)^{n+1}}\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xix)\) \(y=\ln{\left(\frac{a-x}{a+x}\right)}\)
উত্তরঃ \((n-1)!\left[\frac{(-1)^n}{(a+x)^{n}}-\frac{1}{(a-x)^{n}}\right]\)
\(Q.3.(xix)\) \(y=\ln{\left(\frac{a-x}{a+x}\right)}\)
উত্তরঃ \((n-1)!\left[\frac{(-1)^n}{(a+x)^{n}}-\frac{1}{(a-x)^{n}}\right]\)
সমাধানঃ
দেওয়া আছে,
\(y=\ln{\left(\frac{a-x}{a+x}\right)}\)
\(\Rightarrow y=\ln{(a-x)}-\ln{(a+x)}\) ➜ \(\because \ln\left(\frac{M}{N}\right)=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\ln{a-x})-\frac{d}{dx}(\ln{a+x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{a-x}\frac{d}{dx}(a-x)-\frac{1}{a+x}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{1}=\frac{1}{a-x}.(-1)-\frac{1}{a+x}.1\)
\(\Rightarrow y_{1}=-(a-x)^{-1}+(-1)^1.(a+x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{(-1).(a-x)^{-1}+(-1)^1.(a+x)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-(-1)(a-x)^{-1-1}\frac{d}{dx}(a-x)+(-1)^1.(-1)(a+x)^{-1-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{2}=-(-1)(a-x)^{-2}(-1)+(-1)^2.1(a+x)^{-2}.1\)
\(\Rightarrow y_{2}=-1.(a-x)^{-2}+(-1)^2.1(a+x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=-1\frac{d}{dx}\{(a-x)^{-2}\}+(-1)^2.1\frac{d}{dx}\{(a+x)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-1(-2)(a-x)^{-2-1}\frac{d}{dx}(a-x)+(-1)^2.1.(-2)(a+x)^{-2-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{3}=-1(-2)(a-x)^{-3}(-1)+(-1)^2.1.(-2)(a+x)^{-3}.1\)
\(\Rightarrow y_{3}=-1.2(a-x)^{-3}+(-1)^3.1.2(a+x)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=-1.2\frac{d}{dx}\{(a-x)^{-3}\}+(-1)^3.1.2\frac{d}{dx}\{(a+x)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=-1.2.(-3)(a-x)^{-3-1}\frac{d}{dx}(a-x)+(-1)^3.1.2.(-3)(a+x)^{-3-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{4}=-1.2.(-3)(a-x)^{-4}(-1)+(-1)^3.1.2.(-3)(a+x)^{-4}.1\)
\(\Rightarrow y_{4}=-1.2.3(a-x)^{-4}+(-1)^4.1.2.3(a+x)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=-1.2.3 .........(n-1)(a-x)^{-n}+(-1)^n.1.2.3 ....(n-1)(a+x)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=-(n-1)!\frac{1}{(a-x)^{n}}+(-1)^n(n-1)!\frac{1}{(a+x)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\Rightarrow y_{n}=(-1)^n(n-1)!\frac{1}{(a+x)^{n}}-(n-1)!\frac{1}{(a-x)^{n}}\)
\(\therefore y_{n}=(n-1)!\left[\frac{(-1)^n}{(a+x)^{n}}-\frac{1}{(a-x)^{n}}\right]\)
\(y=\ln{\left(\frac{a-x}{a+x}\right)}\)
\(\Rightarrow y=\ln{(a-x)}-\ln{(a+x)}\) ➜ \(\because \ln\left(\frac{M}{N}\right)=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\ln{a-x})-\frac{d}{dx}(\ln{a+x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{a-x}\frac{d}{dx}(a-x)-\frac{1}{a+x}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{1}=\frac{1}{a-x}.(-1)-\frac{1}{a+x}.1\)
\(\Rightarrow y_{1}=-(a-x)^{-1}+(-1)^1.(a+x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{(-1).(a-x)^{-1}+(-1)^1.(a+x)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-(-1)(a-x)^{-1-1}\frac{d}{dx}(a-x)+(-1)^1.(-1)(a+x)^{-1-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{2}=-(-1)(a-x)^{-2}(-1)+(-1)^2.1(a+x)^{-2}.1\)
\(\Rightarrow y_{2}=-1.(a-x)^{-2}+(-1)^2.1(a+x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=-1\frac{d}{dx}\{(a-x)^{-2}\}+(-1)^2.1\frac{d}{dx}\{(a+x)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-1(-2)(a-x)^{-2-1}\frac{d}{dx}(a-x)+(-1)^2.1.(-2)(a+x)^{-2-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{3}=-1(-2)(a-x)^{-3}(-1)+(-1)^2.1.(-2)(a+x)^{-3}.1\)
\(\Rightarrow y_{3}=-1.2(a-x)^{-3}+(-1)^3.1.2(a+x)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=-1.2\frac{d}{dx}\{(a-x)^{-3}\}+(-1)^3.1.2\frac{d}{dx}\{(a+x)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=-1.2.(-3)(a-x)^{-3-1}\frac{d}{dx}(a-x)+(-1)^3.1.2.(-3)(a+x)^{-3-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{4}=-1.2.(-3)(a-x)^{-4}(-1)+(-1)^3.1.2.(-3)(a+x)^{-4}.1\)
\(\Rightarrow y_{4}=-1.2.3(a-x)^{-4}+(-1)^4.1.2.3(a+x)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=-1.2.3 .........(n-1)(a-x)^{-n}+(-1)^n.1.2.3 ....(n-1)(a+x)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=-(n-1)!\frac{1}{(a-x)^{n}}+(-1)^n(n-1)!\frac{1}{(a+x)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\Rightarrow y_{n}=(-1)^n(n-1)!\frac{1}{(a+x)^{n}}-(n-1)!\frac{1}{(a-x)^{n}}\)
\(\therefore y_{n}=(n-1)!\left[\frac{(-1)^n}{(a+x)^{n}}-\frac{1}{(a-x)^{n}}\right]\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xx)\) \(y=\ln{(ax+x^2)}\)
উত্তরঃ \((-1)^{n-1}(n-1)!\left[\frac{1}{(x)^{n}}+\frac{1}{(a+x)^{n}}\right]\)
\(Q.3.(xx)\) \(y=\ln{(ax+x^2)}\)
উত্তরঃ \((-1)^{n-1}(n-1)!\left[\frac{1}{(x)^{n}}+\frac{1}{(a+x)^{n}}\right]\)
সমাধানঃ
দেওয়া আছে,
\(y=\ln{(ax+x^2)}\)
\(\Rightarrow y=\ln{\{x(a+x)\}}\)
\(\Rightarrow y=\ln{x}+\ln{(a+x)}\) ➜ \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\ln{x})+\frac{d}{dx}\{\ln{(a+x)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x}-\frac{1}{a+x}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{1}=\frac{1}{x}+\frac{1}{a+x}\)
\(\Rightarrow y_{1}=x^{-1}+(a+x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{x^{-1}+(a+x)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)x^{-1-1}+(-1)(a+x)^{-1-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{2}=(-1)x^{-1-1}+(-1)(a+x)^{-1-1}.1\)
\(\Rightarrow y_{2}=(-1)^{1}x^{-2}+(-1)^{1}(a+x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^{1}\frac{d}{dx}(x^{-2})+(-1)^{1}\frac{d}{dx}\{(a+x)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^{1}(-2)x^{-2-1}+(-1)^{1}(-2)(a+x)^{-2-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{3}=(-1)^{2}.1.2x^{-3}+(-1)^{2}.1.2(a+x)^{-3}.1\)
\(\Rightarrow y_{3}=(-1)^{2}.1.2x^{-3}+(-1)^{2}.1.2(a+x)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^{2}.1.2\frac{d}{dx}(x^{-3})+(-1)^{2}.1.2\frac{d}{dx}\{(a+x)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=(-1)^{2}.1.2.(-3)x^{-3-1}+(-1)^2.1.2.(-3)(a+x)^{-3-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{4}=(-1)^{3}.1.2.3x^{-4}+(-1)^3.1.2.3(a+x)^{-4}.1\)
\(\Rightarrow y_{4}=(-1)^{3}.1.2.3x^{-4}+(-1)^3.1.2.3(a+x)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n-1}.1.2.3 .....(n-1)x^{-n}+(-1)^{n-1}.1.2.3 .......(n-1)(a+x)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!x^{-n}+(-1)^{n-1}(n-1)!(a+x)^{-n}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!\frac{1}{x^{n}}+(-1)^{n-1}(n-1)!\frac{1}{(a+x)^{n}}\)
\(\therefore y_{n}=(-1)^{n-1}(n-1)!\left[\frac{1}{x^{n}}+\frac{1}{(a+x)^{n}}\right]\)
\(y=\ln{(ax+x^2)}\)
\(\Rightarrow y=\ln{\{x(a+x)\}}\)
\(\Rightarrow y=\ln{x}+\ln{(a+x)}\) ➜ \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\ln{x})+\frac{d}{dx}\{\ln{(a+x)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x}-\frac{1}{a+x}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{1}=\frac{1}{x}+\frac{1}{a+x}\)
\(\Rightarrow y_{1}=x^{-1}+(a+x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{x^{-1}+(a+x)^{-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)x^{-1-1}+(-1)(a+x)^{-1-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{2}=(-1)x^{-1-1}+(-1)(a+x)^{-1-1}.1\)
\(\Rightarrow y_{2}=(-1)^{1}x^{-2}+(-1)^{1}(a+x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^{1}\frac{d}{dx}(x^{-2})+(-1)^{1}\frac{d}{dx}\{(a+x)^{-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^{1}(-2)x^{-2-1}+(-1)^{1}(-2)(a+x)^{-2-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{3}=(-1)^{2}.1.2x^{-3}+(-1)^{2}.1.2(a+x)^{-3}.1\)
\(\Rightarrow y_{3}=(-1)^{2}.1.2x^{-3}+(-1)^{2}.1.2(a+x)^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^{2}.1.2\frac{d}{dx}(x^{-3})+(-1)^{2}.1.2\frac{d}{dx}\{(a+x)^{-3}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=(-1)^{2}.1.2.(-3)x^{-3-1}+(-1)^2.1.2.(-3)(a+x)^{-3-1}\frac{d}{dx}(a+x)\)
\(\Rightarrow y_{4}=(-1)^{3}.1.2.3x^{-4}+(-1)^3.1.2.3(a+x)^{-4}.1\)
\(\Rightarrow y_{4}=(-1)^{3}.1.2.3x^{-4}+(-1)^3.1.2.3(a+x)^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n-1}.1.2.3 .....(n-1)x^{-n}+(-1)^{n-1}.1.2.3 .......(n-1)(a+x)^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!x^{-n}+(-1)^{n-1}(n-1)!(a+x)^{-n}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!\frac{1}{x^{n}}+(-1)^{n-1}(n-1)!\frac{1}{(a+x)^{n}}\)
\(\therefore y_{n}=(-1)^{n-1}(n-1)!\left[\frac{1}{x^{n}}+\frac{1}{(a+x)^{n}}\right]\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxi)\) \(y=\cos{2x}\cos{x}\)
উত্তরঃ \(\frac{1}{2}\left[3^n\cos{\left(\frac{n\pi}{2}+3x\right)}+\cos{\left(\frac{n\pi}{2}+x\right)}\right]\)
\(Q.3.(xxi)\) \(y=\cos{2x}\cos{x}\)
উত্তরঃ \(\frac{1}{2}\left[3^n\cos{\left(\frac{n\pi}{2}+3x\right)}+\cos{\left(\frac{n\pi}{2}+x\right)}\right]\)
সমাধানঃ
দেওয়া আছে,
\(y=\cos{2x}\cos{x}\)
\(\Rightarrow 2y=2\cos{2x}\cos{x}\) ➜ উভয় পার্শে \(2\) গুণ করে।
\(\Rightarrow 2y=\cos{(2x-x)}+\cos{(2x+x)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow 2y=\cos{x}+\cos{3x}\)
\(\Rightarrow 2\frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}(\cos{x})+\frac{d^n}{dx^n}(\cos{3x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow 2y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\cos{ax})=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\), \(\frac{d^n}{dx^n}(\cos{x})=\cos{\left(\frac{n\pi}{2}+x\right)}\)
\(\therefore y_{n}=\frac{1}{2}\left[\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\right]\)
\(y=\cos{2x}\cos{x}\)
\(\Rightarrow 2y=2\cos{2x}\cos{x}\) ➜ উভয় পার্শে \(2\) গুণ করে।
\(\Rightarrow 2y=\cos{(2x-x)}+\cos{(2x+x)}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(\Rightarrow 2y=\cos{x}+\cos{3x}\)
\(\Rightarrow 2\frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}(\cos{x})+\frac{d^n}{dx^n}(\cos{3x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow 2y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\cos{ax})=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\), \(\frac{d^n}{dx^n}(\cos{x})=\cos{\left(\frac{n\pi}{2}+x\right)}\)
\(\therefore y_{n}=\frac{1}{2}\left[\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\right]\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxii)\) \(y=\cos^3{x}\)
উত্তরঃ \(\frac{1}{4}\left[3\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\right]\)
\(Q.3.(xxii)\) \(y=\cos^3{x}\)
উত্তরঃ \(\frac{1}{4}\left[3\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\right]\)
সমাধানঃ
দেওয়া আছে,
\(y=\cos^3{x}\)
\(\Rightarrow 4y=4\cos^3{x}\) ➜ উভয় পার্শে \(4\) গুণ করে।
\(\Rightarrow 4y=3\cos{x}+\cos{3x}\) ➜ \(\because 4\cos^3{A}=3\cos{A}+\cos{3A}\)
\(\Rightarrow 4\frac{d^n}{dx^n}(y)=3\frac{d^n}{dx^n}(\cos{x})+\frac{d^n}{dx^n}(\cos{3x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow 4y_{n}=3\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\cos{ax})=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\), \(\frac{d^n}{dx^n}(\cos{x})=\cos{\left(\frac{n\pi}{2}+x\right)}\)
\(\therefore y_{n}=\frac{1}{4}\left[3\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\right]\)
\(y=\cos^3{x}\)
\(\Rightarrow 4y=4\cos^3{x}\) ➜ উভয় পার্শে \(4\) গুণ করে।
\(\Rightarrow 4y=3\cos{x}+\cos{3x}\) ➜ \(\because 4\cos^3{A}=3\cos{A}+\cos{3A}\)
\(\Rightarrow 4\frac{d^n}{dx^n}(y)=3\frac{d^n}{dx^n}(\cos{x})+\frac{d^n}{dx^n}(\cos{3x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow 4y_{n}=3\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\cos{ax})=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\), \(\frac{d^n}{dx^n}(\cos{x})=\cos{\left(\frac{n\pi}{2}+x\right)}\)
\(\therefore y_{n}=\frac{1}{4}\left[3\cos{\left(\frac{n\pi}{2}+x\right)}+3^n\cos{\left(\frac{n\pi}{2}+3x\right)}\right]\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxiii)\) \(y=\sin{x}\sin{3x}\)
উত্তরঃ \(2^{n-1}\cos{\left(2x+\frac{n\pi}{2}\right)}-2^{2n-1}\cos{\left(4x+\frac{n\pi}{2}\right)}\)
\(Q.3.(xxiii)\) \(y=\sin{x}\sin{3x}\)
উত্তরঃ \(2^{n-1}\cos{\left(2x+\frac{n\pi}{2}\right)}-2^{2n-1}\cos{\left(4x+\frac{n\pi}{2}\right)}\)
সমাধানঃ
দেওয়া আছে,
\(y=\sin{x}\sin{3x}\)
\(\Rightarrow 2y=2\sin{3x}\sin{x}\) ➜ উভয় পার্শে \(2\) গুণ করে।
\(\Rightarrow 2y=\cos{(3x-x)}-\cos{(3x+x)}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow 2y=\cos{2x}-\cos{4x}\)
\(\Rightarrow 2\frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}(\cos{2x})-\frac{d^n}{dx^n}(\cos{4x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow 2y_{n}=2^n\cos{\left(\frac{n\pi}{2}+2x\right)}-4^n\cos{\left(\frac{n\pi}{2}+4x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\cos{ax})=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\(\Rightarrow y_{n}=\frac{1}{2}\left[2^n\cos{\left(\frac{n\pi}{2}+2x\right)}-2^{2n}\cos{\left(\frac{n\pi}{2}+4x\right)}\right]\)
\(\therefore y_{n}=2^{n-1}\cos{\left(2x+\frac{n\pi}{2}\right)}-2^{2n-1}\cos{\left(4x+\frac{n\pi}{2}\right)}\)
\(y=\sin{x}\sin{3x}\)
\(\Rightarrow 2y=2\sin{3x}\sin{x}\) ➜ উভয় পার্শে \(2\) গুণ করে।
\(\Rightarrow 2y=\cos{(3x-x)}-\cos{(3x+x)}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(\Rightarrow 2y=\cos{2x}-\cos{4x}\)
\(\Rightarrow 2\frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}(\cos{2x})-\frac{d^n}{dx^n}(\cos{4x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow 2y_{n}=2^n\cos{\left(\frac{n\pi}{2}+2x\right)}-4^n\cos{\left(\frac{n\pi}{2}+4x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\cos{ax})=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\(\Rightarrow y_{n}=\frac{1}{2}\left[2^n\cos{\left(\frac{n\pi}{2}+2x\right)}-2^{2n}\cos{\left(\frac{n\pi}{2}+4x\right)}\right]\)
\(\therefore y_{n}=2^{n-1}\cos{\left(2x+\frac{n\pi}{2}\right)}-2^{2n-1}\cos{\left(4x+\frac{n\pi}{2}\right)}\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxiv)\) \(y=\ln{x}\)
উত্তরঃ \(\frac{(-1){n-1}(n-1)!}{x^n}\)
\(Q.3.(xxiv)\) \(y=\ln{x}\)
উত্তরঃ \(\frac{(-1){n-1}(n-1)!}{x^n}\)
সমাধানঃ
দেওয়া আছে,
\(y=\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x}\)
\(\Rightarrow y_{1}=(x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(x^{-1})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-1(x)^{-1-1}\)
\(\Rightarrow y_{2}=(-1)^1.1(x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^1.1\frac{d}{dx}(x^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^1.1.(-2)x^{-2-1}\)
\(\Rightarrow y_{3}=(-1)^2.1.2.x^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^2.1.2\frac{d}{dx}(x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=(-1)^2.1.2.(-3)x^{-3-1}\)
\(\Rightarrow y_{4}=(-1)^3.1.2.3.x^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n-1}.1.2.3 .........(n-1)x^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!\frac{1}{(x)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!}{(x)^{n}}\)
\(y=\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\ln{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=\frac{1}{x}\)
\(\Rightarrow y_{1}=(x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}(x^{-1})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-1(x)^{-1-1}\)
\(\Rightarrow y_{2}=(-1)^1.1(x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^1.1\frac{d}{dx}(x^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^1.1.(-2)x^{-2-1}\)
\(\Rightarrow y_{3}=(-1)^2.1.2.x^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{3})=(-1)^2.1.2\frac{d}{dx}(x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{4}=(-1)^2.1.2.(-3)x^{-3-1}\)
\(\Rightarrow y_{4}=(-1)^3.1.2.3.x^{-4}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n-1}.1.2.3 .........(n-1)x^{-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n-1}(n-1)!\frac{1}{(x)^{n}}\) ➜\(\because 1.2.3 .........n-1=(n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n-1}(n-1)!}{(x)^{n}}\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxv)\) \(y=\frac{1}{x}\)
উত্তরঃ \(\frac{(-1){n}(n)!}{x^{n+1}}\)
\(Q.3.(xxv)\) \(y=\frac{1}{x}\)
উত্তরঃ \(\frac{(-1){n}(n)!}{x^{n+1}}\)
সমাধানঃ
দেওয়া আছে,
\(y=\frac{1}{x}\)
\(\Rightarrow y=(x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{-1})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(x)^{-1-1}\)
\(\Rightarrow y_{1}=(-1)^1.1(x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=(-1)^1.1\frac{d}{dx}(x^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)^1.1.(-2)x^{-2-1}\)
\(\Rightarrow y_{2}=(-1)^2.1.2.x^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^2.1.2\frac{d}{dx}(x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^2.1.2.(-3)x^{-3-1}\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.x^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n}.1.2.3 .........nx^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n}n!\frac{1}{(x)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{(-1)^{n}n!}{(x)^{n+1}}\)
\(y=\frac{1}{x}\)
\(\Rightarrow y=(x)^{-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{-1})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-1(x)^{-1-1}\)
\(\Rightarrow y_{1}=(-1)^1.1(x)^{-2}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=(-1)^1.1\frac{d}{dx}(x^{-2})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=(-1)^1.1.(-2)x^{-2-1}\)
\(\Rightarrow y_{2}=(-1)^2.1.2.x^{-3}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=(-1)^2.1.2\frac{d}{dx}(x^{-3})\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=(-1)^2.1.2.(-3)x^{-3-1}\)
\(\Rightarrow y_{3}=(-1)^3.1.2.3.x^{-(3+1)}\)
\(...............\)
\(\Rightarrow y_{n}=(-1)^{n}.1.2.3 .........nx^{-(n+1)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=(-1)^{n}n!\frac{1}{(x)^{n+1}}\) ➜\(\because 1.2.3 .........n=n!\)
\(\therefore y_{n}=\frac{(-1)^{n}n!}{(x)^{n+1}}\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxvi)\) \(y=\frac{x^2+1}{(x-1)(x-2)(x-3)}\)
উত্তরঃ \((-1)^nn!\left[\frac{1}{(x-1)^{n+1}}-\frac{5}{(x-2)^{n+1}}+\frac{5}{(x-3)^{n+1}}\right]\)
\(Q.3.(xxvi)\) \(y=\frac{x^2+1}{(x-1)(x-2)(x-3)}\)
উত্তরঃ \((-1)^nn!\left[\frac{1}{(x-1)^{n+1}}-\frac{5}{(x-2)^{n+1}}+\frac{5}{(x-3)^{n+1}}\right]\)
সমাধানঃ
দেওয়া আছে,
\(y=\frac{x^2+1}{(x-1)(x-2)(x-3)}\)
\(\Rightarrow y=\frac{1^2+1}{(x-1)(1-2)(1-3)}+\frac{2^2+1}{(2-1)(x-2)(2-3)}+\frac{3^2+1}{(3-1)(3-2)(x-3)}\)| আংশিক ভগ্নাংশে পরিণত করে।
\(\Rightarrow y=\frac{1+1}{(x-1)(-1)(-2)}+\frac{4+1}{1.(x-2)(-1)}+\frac{9+1}{(2)(1)(x-3)}\)
\(\Rightarrow y=\frac{2}{2(x-1)}+\frac{5}{-(x-2)}+\frac{10}{2(x-3)}\)
\(\Rightarrow y=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{5}{(x-3)}\)
\(\Rightarrow \frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}\left[\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{5}{(x-3)}\right]\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow y_{n}=\frac{d^n}{dx^n}\{\frac{1}{(x-1)}\}-5\frac{d^n}{dx^n}\{\frac{1}{(x-2)}\}+5\frac{d^n}{dx^n}\{\frac{1}{(x-3)}\}\) ➜ \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow y_{n}=\frac{(-1)^{n}n!}{(x-1)^{n+1}}-5\frac{(-1)^{n}n!}{(x-2)^{n+1}}+5\frac{(-1)^{n}n!}{(x-3)^{n+1}}\) ➜ \(\because \frac{d^n}{dx^n}\left(\frac{1}{x-a}\right)=\frac{(-1)^{n}n!}{(x-a)^{n+1}}\)
\(\therefore y_{n}=(-1)^{n}n!\left[\frac{1}{(x-1)^{n+1}}-\frac{5}{(x-2)^{n+1}}+\frac{5}{(x-3)^{n+1}}\right]\)
\(y=\frac{x^2+1}{(x-1)(x-2)(x-3)}\)
\(\Rightarrow y=\frac{1^2+1}{(x-1)(1-2)(1-3)}+\frac{2^2+1}{(2-1)(x-2)(2-3)}+\frac{3^2+1}{(3-1)(3-2)(x-3)}\)| আংশিক ভগ্নাংশে পরিণত করে।
\(\Rightarrow y=\frac{1+1}{(x-1)(-1)(-2)}+\frac{4+1}{1.(x-2)(-1)}+\frac{9+1}{(2)(1)(x-3)}\)
\(\Rightarrow y=\frac{2}{2(x-1)}+\frac{5}{-(x-2)}+\frac{10}{2(x-3)}\)
\(\Rightarrow y=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{5}{(x-3)}\)
\(\Rightarrow \frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}\left[\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{5}{(x-3)}\right]\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow y_{n}=\frac{d^n}{dx^n}\{\frac{1}{(x-1)}\}-5\frac{d^n}{dx^n}\{\frac{1}{(x-2)}\}+5\frac{d^n}{dx^n}\{\frac{1}{(x-3)}\}\) ➜ \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow y_{n}=\frac{(-1)^{n}n!}{(x-1)^{n+1}}-5\frac{(-1)^{n}n!}{(x-2)^{n+1}}+5\frac{(-1)^{n}n!}{(x-3)^{n+1}}\) ➜ \(\because \frac{d^n}{dx^n}\left(\frac{1}{x-a}\right)=\frac{(-1)^{n}n!}{(x-a)^{n+1}}\)
\(\therefore y_{n}=(-1)^{n}n!\left[\frac{1}{(x-1)^{n+1}}-\frac{5}{(x-2)^{n+1}}+\frac{5}{(x-3)^{n+1}}\right]\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxvii)\) \(y=\cos{(ax)}\)
উত্তরঃ \(a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\(Q.3.(xxvii)\) \(y=\cos{(ax)}\)
উত্তরঃ \(a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
সমাধানঃ
দেওয়া আছে,
\(y=\cos{(ax)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(ax)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(ax)}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{1}=-\sin{(ax)}.a\)
\(\Rightarrow y_{1}=a\cos{\left(\frac{\pi}{2}+ax\right)}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\cos{\left(\frac{\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-a\sin{\left(\frac{\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{\pi}{2}+ax\right)\)
\(\Rightarrow y_{2}=-a\sin{\left(\frac{\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{2}=a^2\cos{\left(\frac{2\pi}{2}+ax\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\cos{\left(\frac{2\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-a^2\sin{\left(\frac{2\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+ax\right)\)
\(\Rightarrow y_{3}=-a^2\sin{\left(\frac{2\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{3}=a^3\cos{\left(\frac{3\pi}{2}+ax\right)}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
\(y=\cos{(ax)}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{\cos{(ax)}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{(ax)}\frac{d}{dx}(ax)\)
\(\Rightarrow y_{1}=-\sin{(ax)}.a\)
\(\Rightarrow y_{1}=a\cos{\left(\frac{\pi}{2}+ax\right)}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a\frac{d}{dx}\{\cos{\left(\frac{\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-a\sin{\left(\frac{\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{\pi}{2}+ax\right)\)
\(\Rightarrow y_{2}=-a\sin{\left(\frac{\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{2}=a^2\cos{\left(\frac{2\pi}{2}+ax\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2\frac{d}{dx}\{\cos{\left(\frac{2\pi}{2}+ax\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-a^2\sin{\left(\frac{2\pi}{2}+ax\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+ax\right)\)
\(\Rightarrow y_{3}=-a^2\sin{\left(\frac{2\pi}{2}+ax\right)}.a\)
\(\Rightarrow y_{3}=a^3\cos{\left(\frac{3\pi}{2}+ax\right)}\)
\(...............\)
\(\Rightarrow y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=a^n\cos{\left(\frac{n\pi}{2}+ax\right)}\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxviii)\) \(y=2\sin{3x}\cos{2x}\)
উত্তরঃ \(5^n\sin{\left(\frac{n\pi}{2}+5x\right)}+\sin{\left(\frac{n\pi}{2}+x\right)}\)
\(Q.3.(xxviii)\) \(y=2\sin{3x}\cos{2x}\)
উত্তরঃ \(5^n\sin{\left(\frac{n\pi}{2}+5x\right)}+\sin{\left(\frac{n\pi}{2}+x\right)}\)
সমাধানঃ
দেওয়া আছে,
\(y=2\sin{3x}\cos{2x}\)
\(\Rightarrow y=\sin{(3x+2x)}+\sin{(3x-2x)}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(\Rightarrow y=\sin{(5x)}+\sin{(x)}\)
\(\Rightarrow \frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}(\sin{5x})+\frac{d^n}{dx^n}(\sin{x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow y_{n}=5^n\sin{\left(\frac{n\pi}{2}+5x\right)}+\sin{\left(\frac{n\pi}{2}+x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\sin{ax})=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
\(\therefore y_{n}=5^n\sin{\left(\frac{n\pi}{2}+5x\right)}+\sin{\left(\frac{n\pi}{2}+x\right)}\)
\(y=2\sin{3x}\cos{2x}\)
\(\Rightarrow y=\sin{(3x+2x)}+\sin{(3x-2x)}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(\Rightarrow y=\sin{(5x)}+\sin{(x)}\)
\(\Rightarrow \frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}(\sin{5x})+\frac{d^n}{dx^n}(\sin{x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow y_{n}=5^n\sin{\left(\frac{n\pi}{2}+5x\right)}+\sin{\left(\frac{n\pi}{2}+x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\sin{ax})=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
\(\therefore y_{n}=5^n\sin{\left(\frac{n\pi}{2}+5x\right)}+\sin{\left(\frac{n\pi}{2}+x\right)}\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxix)\) \(y=4\sin^3{x}\)
উত্তরঃ \(3\sin{\left(\frac{n\pi}{2}+x\right)}-3^n\sin{\left(\frac{n\pi}{2}+3x\right)}\)
\(Q.3.(xxix)\) \(y=4\sin^3{x}\)
উত্তরঃ \(3\sin{\left(\frac{n\pi}{2}+x\right)}-3^n\sin{\left(\frac{n\pi}{2}+3x\right)}\)
সমাধানঃ
দেওয়া আছে,
\(y=4\sin^3{x}\)
\(\Rightarrow y=3\sin{x}-\sin{3x}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(\Rightarrow \frac{d^n}{dx^n}(y)=3\frac{d^n}{dx^n}(\sin{x})-\frac{d^n}{dx^n}(\sin{3x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow y_{n}=3\sin{\left(\frac{n\pi}{2}+x\right)}-3^n\sin{\left(\frac{n\pi}{2}+3x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\sin{x})=\sin{\left(\frac{n\pi}{2}+x\right)}\), \(\frac{d^n}{dx^n}(\sin{ax})=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
\(\therefore y_{n}=3\sin{\left(\frac{n\pi}{2}+x\right)}-3^n\sin{\left(\frac{n\pi}{2}+3x\right)}\)
\(y=4\sin^3{x}\)
\(\Rightarrow y=3\sin{x}-\sin{3x}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(\Rightarrow \frac{d^n}{dx^n}(y)=3\frac{d^n}{dx^n}(\sin{x})-\frac{d^n}{dx^n}(\sin{3x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow y_{n}=3\sin{\left(\frac{n\pi}{2}+x\right)}-3^n\sin{\left(\frac{n\pi}{2}+3x\right)}\) ➜ \(\because \frac{d^n}{dx^n}(\sin{x})=\sin{\left(\frac{n\pi}{2}+x\right)}\), \(\frac{d^n}{dx^n}(\sin{ax})=a^n\sin{\left(\frac{n\pi}{2}+ax\right)}\)
\(\therefore y_{n}=3\sin{\left(\frac{n\pi}{2}+x\right)}-3^n\sin{\left(\frac{n\pi}{2}+3x\right)}\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxx)\) \(y=(ax+b)^{-m}\)
উত্তরঃ \(\frac{(-1)^n(m+n-1)!a^n}{(m-1)!(ax+b)^{m+n}}\)
\(Q.3.(xxx)\) \(y=(ax+b)^{-m}\)
উত্তরঃ \(\frac{(-1)^n(m+n-1)!a^n}{(m-1)!(ax+b)^{m+n}}\)
সমাধানঃ
দেওয়া আছে,
\(y=(ax+b)^{-m}\)
\(\Rightarrow \frac{d}{dx}(y)=a\frac{d}{dx}\{(ax+b)^{-m}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(-m)(ax+b)^{-m-1}\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{1}=(-1)^1.m(ax+b)^{-m-1}.a\)
\(\Rightarrow y_{1}=a(-1)^1.m(ax+b)^{-m-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a(-1)^1.m\frac{d}{dx}\{(ax+b)^{-m-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a(-1)^1.m.(-m-1)(ax+b)^{-m-2}.\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{2}=a(-1)^2.m.(m+1).(ax+b)^{-m-2}.a\)
\(\Rightarrow y_{2}=a^2(-1)^2.m.(m+1).(ax+b)^{-m-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2(-1)^2.m.(m+1).\frac{d}{dx}\{(ax+b)^{-m-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2(-1)^2.m.(m+1).(-m-2)(ax+b)^{-m-3}.\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{3}=a^2(-1)^3.m.(m+1).(m+2).(ax+b)^{-m-3}.a\)
\(\Rightarrow y_{3}=a^3(-1)^3.m.(m+1).(m+2).(ax+b)^{-m-3}\)
\(...............\)
\(\Rightarrow y_{n}=a^n(-1)^{n}.m.(m+1).(m+2).... (m+n-1).(ax+b)^{-m-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=a^n(-1)^{n}.m.(m+1).(m+2).... (m+n-1).\frac{1}{(ax+b)^{m+n}}\)
\(\Rightarrow y_{n}=a^n(-1)^{n}.1.2.3....(m-1).m.(m+1).(m+2).... (m+n-1).\frac{1}{1.2.3....(m-1)(ax+b)^{m+n}}\)
\(\Rightarrow y_{n}=a^n(-1)^{n}.1.2.3....(m-1).m.(m+1).(m+2).... (m+n-1).\frac{1}{(m-1)!(ax+b)^{m+n}}\) ➜\(\because 1.2.3. ....(m-1)=(m-1)!\)
\(\Rightarrow y_{n}=a^n(-1)^{n}(m+n-1)!\frac{1}{(m-1)!(ax+b)^{m+n}}\) ➜\(\because 1.2.3....(m-1).m.(m+1).(m+2).... (m+n-1)=(m+n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n}(m+n-1)!a^n}{(m-1)!(ax+b)^{m+n}}\)
\(y=(ax+b)^{-m}\)
\(\Rightarrow \frac{d}{dx}(y)=a\frac{d}{dx}\{(ax+b)^{-m}\}\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=(-m)(ax+b)^{-m-1}\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{1}=(-1)^1.m(ax+b)^{-m-1}.a\)
\(\Rightarrow y_{1}=a(-1)^1.m(ax+b)^{-m-1}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=a(-1)^1.m\frac{d}{dx}\{(ax+b)^{-m-1}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=a(-1)^1.m.(-m-1)(ax+b)^{-m-2}.\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{2}=a(-1)^2.m.(m+1).(ax+b)^{-m-2}.a\)
\(\Rightarrow y_{2}=a^2(-1)^2.m.(m+1).(ax+b)^{-m-2}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=a^2(-1)^2.m.(m+1).\frac{d}{dx}\{(ax+b)^{-m-2}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=a^2(-1)^2.m.(m+1).(-m-2)(ax+b)^{-m-3}.\frac{d}{dx}(ax+b)\)
\(\Rightarrow y_{3}=a^2(-1)^3.m.(m+1).(m+2).(ax+b)^{-m-3}.a\)
\(\Rightarrow y_{3}=a^3(-1)^3.m.(m+1).(m+2).(ax+b)^{-m-3}\)
\(...............\)
\(\Rightarrow y_{n}=a^n(-1)^{n}.m.(m+1).(m+2).... (m+n-1).(ax+b)^{-m-n}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\Rightarrow y_{n}=a^n(-1)^{n}.m.(m+1).(m+2).... (m+n-1).\frac{1}{(ax+b)^{m+n}}\)
\(\Rightarrow y_{n}=a^n(-1)^{n}.1.2.3....(m-1).m.(m+1).(m+2).... (m+n-1).\frac{1}{1.2.3....(m-1)(ax+b)^{m+n}}\)
\(\Rightarrow y_{n}=a^n(-1)^{n}.1.2.3....(m-1).m.(m+1).(m+2).... (m+n-1).\frac{1}{(m-1)!(ax+b)^{m+n}}\) ➜\(\because 1.2.3. ....(m-1)=(m-1)!\)
\(\Rightarrow y_{n}=a^n(-1)^{n}(m+n-1)!\frac{1}{(m-1)!(ax+b)^{m+n}}\) ➜\(\because 1.2.3....(m-1).m.(m+1).(m+2).... (m+n-1)=(m+n-1)!\)
\(\therefore y_{n}=\frac{(-1)^{n}(m+n-1)!a^n}{(m-1)!(ax+b)^{m+n}}\)
নিচের ফাংশনটির \(n\)তম অন্তরজ \(y_{n}\) নির্ণয় কর।
\(Q.3.(xxxi)\) \(y=e^{3x}\sin^2{x}\)
উত্তরঃ \(\frac{1}{2}e^{3x}\left[3^{n}-(\sqrt{13})^n\cos{(2x+n\tan^{-1}{\frac{2}{3}})}\right]\)
\(Q.3.(xxxi)\) \(y=e^{3x}\sin^2{x}\)
উত্তরঃ \(\frac{1}{2}e^{3x}\left[3^{n}-(\sqrt{13})^n\cos{(2x+n\tan^{-1}{\frac{2}{3}})}\right]\)
সমাধানঃ
দেওয়া আছে,
\(y=e^{3x}\sin^2{x}\)
\(\Rightarrow 2y=e^{3x}2\sin^2{x}\) ➜ উভয় পার্শে \(2\) গুণ করে।
\(\Rightarrow 2y=e^{3x}(1-\cos{2x})\)
\(\Rightarrow 2y=e^{3x}-e^{3x}\cos{2x}\)
\(\Rightarrow 2\frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}(e^{3x}-e^{3x}\cos{2x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow 2y_{n}=\frac{d^n}{dx^n}(e^{3x})-\frac{d^n}{dx^n}(e^{3x}\cos{2x})\)
\(\Rightarrow 2y_{n}=3^ne^{3x}-(3^2+2^2)^{\frac{n}{2}}e^{3x}\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\) ➜ \(\because \frac{d^n}{dx^n}(e^{ax})=a^ne^{ax}\), \(\frac{d^n}{dx^n}(e^{ax}\cos{bx})=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\(\Rightarrow 2y_{n}=3^ne^{3x}-(9+4)^{\frac{n}{2}}e^{3x}\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\)
\(\Rightarrow 2y_{n}=3^ne^{3x}-(13)^{\frac{n}{2}}e^{3x}\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\)
\(\Rightarrow y_{n}=\frac{1}{2}\left[3^ne^{3x}-(\sqrt{13})^ne^{3x}\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\right]\)
\(\therefore y_{n}=\frac{1}{2}e^{3x}\left[3^n-(\sqrt{13})^n\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\right]\)
\(y=e^{3x}\sin^2{x}\)
\(\Rightarrow 2y=e^{3x}2\sin^2{x}\) ➜ উভয় পার্শে \(2\) গুণ করে।
\(\Rightarrow 2y=e^{3x}(1-\cos{2x})\)
\(\Rightarrow 2y=e^{3x}-e^{3x}\cos{2x}\)
\(\Rightarrow 2\frac{d^n}{dx^n}(y)=\frac{d^n}{dx^n}(e^{3x}-e^{3x}\cos{2x})\) ➜ \(x\)-এর সাপেক্ষে \(n\) সংখ্যকবার অন্তরীকরণ করে।
\(\Rightarrow 2y_{n}=\frac{d^n}{dx^n}(e^{3x})-\frac{d^n}{dx^n}(e^{3x}\cos{2x})\)
\(\Rightarrow 2y_{n}=3^ne^{3x}-(3^2+2^2)^{\frac{n}{2}}e^{3x}\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\) ➜ \(\because \frac{d^n}{dx^n}(e^{ax})=a^ne^{ax}\), \(\frac{d^n}{dx^n}(e^{ax}\cos{bx})=(a^2+b^2)^{\frac{n}{2}}e^{ax}\cos{\{bx+n\tan^{-1}{\left(\frac{b}{a}\right)}\}}\)
\(\Rightarrow 2y_{n}=3^ne^{3x}-(9+4)^{\frac{n}{2}}e^{3x}\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\)
\(\Rightarrow 2y_{n}=3^ne^{3x}-(13)^{\frac{n}{2}}e^{3x}\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\)
\(\Rightarrow y_{n}=\frac{1}{2}\left[3^ne^{3x}-(\sqrt{13})^ne^{3x}\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\right]\)
\(\therefore y_{n}=\frac{1}{2}e^{3x}\left[3^n-(\sqrt{13})^n\cos{\left(2x+n\tan^{-1}{\frac{2}{3}}\right)}\right]\)
\(Q.3.(xxxii)\) \(y=\cos{x}\) হলে, \(y_{n}\) নির্ণয় কর এবং \((y_{n})_{0}\) এর মাণ নির্ণয় কর।
উত্তরঃ \(\cos{\left(\frac{n\pi}{2}+x\right)}\); \(n\) জোড় হলে, \(-1\) এবং \(n\) বিজোড় হলে, \(0\)
উত্তরঃ \(\cos{\left(\frac{n\pi}{2}+x\right)}\); \(n\) জোড় হলে, \(-1\) এবং \(n\) বিজোড় হলে, \(0\)
সমাধানঃ
দেওয়া আছে,
\(y=\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{x}\)
\(\Rightarrow y_{1}=\cos{\left(\frac{\pi}{2}+x\right)}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{\cos{\left(\frac{\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-\sin{\left(\frac{\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{\pi}{2}+x\right)\)
\(\Rightarrow y_{2}=-\sin{\left(\frac{\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{2}=\cos{\left(\frac{2\pi}{2}+x\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}\{\cos{\left(\frac{2\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-\sin{\left(\frac{2\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+x\right)\)
\(\Rightarrow y_{3}=-\sin{\left(\frac{2\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{3}=\cos{\left(\frac{3\pi}{2}+x\right)}\)
\(...............\)
\(\Rightarrow y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\)
এখন,
\(x=0\) বসিয়ে,
\((y_{n})_{0}=\cos{\left(\frac{n\pi}{2}+0\right)}\)
\(=\cos{\left(\frac{2m\pi}{2}\right)}\) যখন \(n=2m\) জোড় সংখ্যা ।
\(=\cos{m\pi}\)
\(=-1\) যখন \(m\) স্বাভাবিক সংখ্যা।
আবার,
\(=\cos{\left(\frac{(2m-1)\pi}{2}\right)}\) যখন \(n=2m-1\) বিজোড় সংখ্যা ।
\(=\cos{m\pi-\frac{\pi}{2}}\)
\(=-\cos{\frac{\pi}{2}}\) যখন \(m\) স্বাভাবিক সংখ্যা।
\(=-0\)
\(=0\)
\(y=\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\cos{x})\) ➜ \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{1}=-\sin{x}\)
\(\Rightarrow y_{1}=\cos{\left(\frac{\pi}{2}+x\right)}\) ➜ \(\because -\sin{A}=\cos{\left(\frac{\pi}{2}+A\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{1})=\frac{d}{dx}\{\cos{\left(\frac{\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{2}=-\sin{\left(\frac{\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{\pi}{2}+x\right)\)
\(\Rightarrow y_{2}=-\sin{\left(\frac{\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{2}=\cos{\left(\frac{2\pi}{2}+x\right)}\)
\(\Rightarrow \frac{d}{dx}(y_{2})=\frac{d}{dx}\{\cos{\left(\frac{2\pi}{2}+x\right)}\}\) ➜ আবার, \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y_{3}=-\sin{\left(\frac{2\pi}{2}+x\right)}\frac{d}{dx}\left(\frac{2\pi}{2}+x\right)\)
\(\Rightarrow y_{3}=-\sin{\left(\frac{2\pi}{2}+x\right)}.1\)
\(\Rightarrow y_{3}=\cos{\left(\frac{3\pi}{2}+x\right)}\)
\(...............\)
\(\Rightarrow y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\) যেখানে \(n\) একটি ধনাত্মক পূর্ণ সংখ্যা।
\(\therefore y_{n}=\cos{\left(\frac{n\pi}{2}+x\right)}\)
এখন,
\(x=0\) বসিয়ে,
\((y_{n})_{0}=\cos{\left(\frac{n\pi}{2}+0\right)}\)
\(=\cos{\left(\frac{2m\pi}{2}\right)}\) যখন \(n=2m\) জোড় সংখ্যা ।
\(=\cos{m\pi}\)
\(=-1\) যখন \(m\) স্বাভাবিক সংখ্যা।
আবার,
\(=\cos{\left(\frac{(2m-1)\pi}{2}\right)}\) যখন \(n=2m-1\) বিজোড় সংখ্যা ।
\(=\cos{m\pi-\frac{\pi}{2}}\)
\(=-\cos{\frac{\pi}{2}}\) যখন \(m\) স্বাভাবিক সংখ্যা।
\(=-0\)
\(=0\)
- About
- Author
-
Contact
Call me at +880-1715-651163Email: Golzarrahman1966@gmail.com
Visitors online: 000010