Question

$y=e^{a\sin -1}x\Rightarrow \left(1-x^2\right)y_n+2-\left(2n+1\right)x{y}_n+1$ is equal to

• A. $-(n^2+a^2)y_n$
• B. $(n^2-a^2)y_n$
• C. $(n^2+a^2)y_n$
• D. $-(n^2-a^2)y_n$

Answer 1

Answer: (C)

Given, $y=e^{a{\sin }^{-1}}x$
On differentiating w.r.t $x$, we get
$y_1=e^{a{\sin }^{-1}}xa\cdot \frac{1}{1-x^2}$
$\Rightarrow y_1\overline{1-x^2}=ay$
$\Rightarrow 1-x^2{y}_1^2=a^2{y}^2$
Again, differentiating w.r.t $x$, we get
$\left(1-x^2\right)2y_1{y}_2-2x{y}_1^2=a^{2}2y{y}_1$
$\Rightarrow \left(1-x^2\right)y_2-x{y}_1-a^{2}y=0$
Using Leibntiz's rule,
$\left(1-x^2\right)y_{n+2}+{}^n{c}_1{y}_{n+1}(-2x)+{}^n{c}_2{y}_2(-2)-x{y}_{n+1}-{}^n{c}_{1}yn-a^{2}yn=0$
$\Rightarrow \left(1-x^2\right)y_{n+2}+x{y}_{n+1}(-2n-1)+y_n\left[-n(n-1)-n-a^2\right]=0$
$\Rightarrow \left(1-x^2\right)y_{n+2}-(2n+1)x{y}_{n+1}=\left(n^2+a^2\right)y_n$