Given, $y=e^{a \sin^{-1}} x$
On differentiating w.r.t $x$, we get
$ y _{1}= e ^{ a \sin^{-1}} x a \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) 2 y_{1} y_{2}-2 x y_{1}^{2}=a^{2} 2 y 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}(-2 x)+{ }^{n} c_{2} y_{2}(-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}(-2 n-1)+y_{n}\left[-n(n-1)-n-a^{2}\right]=0 $
$\Rightarrow \left(1-x^{2}\right) y_{n+2}-(2 n+1) x y_{n+1}=\left(n^{2}+a^{2}\right) y_{n}$