$y =\cos (m \sin^{-1}x) \cdots (A)$ Differentiating w.r.t x, and using chain rule, $y_1=-\sin (m \sin^{-1}x) \times (m \sin^{-1}x)' = -\sin (m \sin^{-1}x)\times \dfrac {m}{\sqrt{(1-x^2)}} \ y_1 = -m \dfrac {\sin ( m \sin^{-1}x)}{\sqrt{(1-x^2)}} \cdots (B)$ Differentiating w.r.t y again, using chain and quatient rule, $y_2 = \dfrac {-m \cos (m \sin^{-1} x)(m \sin^{-1}x)' (\sqrt{1-x^2})+ m \sin (m \sin^{-1}x) \left ( \sqrt{1-x^2} \right )'} { \left [\sqrt{(1-x^2)} \right ]}$ $y_2 = \dfrac {-my \frac {m}{\sqrt{(1-x^2)}}\left ( \sqrt{(1-x^2)} \right )+ \frac {\left ( -y_1 \sqrt{(1-x^2)}\right )}{2 \sqrt{(1-x^2)}} (-2x)}{1-x^2}\ \cdots { from \ (A) \ and \ (B)} $ $y_2 = \dfrac {-m^2y+xy_1}{1-x^2} \ (1-x^2)y^2 -xy_1 +m^2 y=0 $ Differentiating n' times using Lebnitz' Theorem, $(1-x^2)y_{n+2}+ (-2x)(n)y_{n+1}+ (-2) \left ( \dfrac {n (n-1)}{2} \right ) y_n-xy_{n+1}-(1)(n)y_n + m^2y_n=0$ $(1-x^2)y_{n+2}- (2n+1)xy_{n+1}+ (-n^2 +n-n+m^2)=0 \ (1-x^2)y_{n+2} - (2n+1)xy_{n+1}+ (m^2 -n^2)y_n=0 $