Question

If $y=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$ then $y_n=\left(0\right)$

• A. ${(n+1)}^2{y}_{n-2}\left(0\right)$
• B. $n^2{y}_{n-2}\left(0\right)$
• C. ${(n-1)}^2{y}_{n-2}\left(0\right)$
• D. ${(n-1)}^2{y}_{n-1}\left(0\right)$

Answer 1

Answer: (C)

${(n-1)}^2{y}_{n-2}\left(0\right)$

$(1-x^2)y^2={\left({\sin }^{-1}x\right)}^2\Rightarrow -2x{y}^2+(1-x^2)2y{y}_1$
$=\frac{2{\sin }^{-1}x}{\sqrt{1-x^2}}=2y$
$\Rightarrow -xy+(1-x^2)y_1=2\Rightarrow (1-x^2)y_2-2x{y}_1-x{y}_1-y=0\Rightarrow (1-x^2)y_2-3x{y}_1-y=0$
Differentiating n times , we obtain
$(1-x^2)y_{n+2}-(2n+3)x{y}_{n+1}-{(n+1)}^2{y}_n=0$
Putting $x=0$, we get $y_{n+2}\left(0\right)={(n+1)}^2{y}_n\left(0\right)$