Explanation

(c)
Given, $y = e^{a\sin^{- 1}x}$
On differentiating w.r.t. $x,\$ we get
$y_{1} = e^{a\sin^{- 1}x}a.\frac{1}{\sqrt{1 - x^{2}}}$
$\Rightarrow \ y_{1}\sqrt{1 - x^{2}} = ay$
$\Rightarrow \ \left( 1 - x^{2} \right)y_{1}^{2} = a^{2}y^{2}$
Again, differentiating w.r.t. $x,\$we get
$\left( 1 - x^{2} \right)2y_{1}y_{2} - 2xy_{1}^{2} = a^{2}2yy_{1}$
$\Rightarrow \left( 1 - x^{2} \right)y_{2} - xy_{1} - a^{2}y = 0\ \$
Using Leibnitz’s rule,
$\left( 1 - x^{2} \right)y_{n + 2} +^{n}C_{1}y_{n + 1}( - 2x) +^{n}C_{2}y_{n}( - 2)$
$- xy_{n + 1} -^{n}c_{1}y_{n} - a^{2}y_{n} = 0$
$\Rightarrow$ $\left( 1 - x^{2} \right)y_{n + 2} + xy_{n + 1}( - 2n - 1)$
$+ y_{n}\left\lbrack - n(n - 1) - n - a^{2} \right\rbrack = 0$
$\Rightarrow \ \ \left( 1 - x^{2} \right)y_{n + 2} - (2n + 1)xy_{n + 1} = \left( n^{2} + a^{2} \right)y_{n}$