Mathematics - Application of Derivative Question with Solution | TestHub

$y=e^{a{\sin }^{-1}x}$ On differentiating w.r.t. $x$, we get $$\begin{array}{l}{y}_1=e^{a{\sin }^{-1}x}a\cdot \frac{1}{\sqrt{1-x^2}}\\ \Rightarrow y_1\sqrt{1-x^2}=ay\\ \Rightarrow \left(1-x^2\right)y_1^2=a^2{y}^2\end{array}$$ Again differentiating wr.t. $x$, we get $$\begin{array}{l}\left(1-x^2\right)2y_1{y}_2-2x{y}_1^2=a^{2}2y{y}_1\\ \Rightarrow \left(1-x^2\right)y_2-x{y}_1-a^{2}y=0\end{array}$$ Using Leibnitz's rule, $$\begin{array}{c}\left(1-x^2\right)y_{n+2}+{}^n{C}_1{y}_{n+1}(-2x)+{}^n{C}_2{y}_n(-2)\\ -x{y}_{n+1}-{}^n{C}_1{y}_n-a^2{y}_n=0\\ \Rightarrow \left(1-x^2\right)y_{n+2}+x{y}_{n+1}(-2n-1)\\ +y_n\left[-n(n-1)-n-a^2\right]=0\\ \Rightarrow \left(1-x^2\right)y_{n+2}-(2n+1)x{y}_{n+1}=\left(n^2+a^2\right)y_n\end{array}$$