y$=cos(msin^{-1}x) \; \; \; \; \ldots (i) $

Diff. equation (i) w.r.t.x,

$ \therefore y_1 \;=\; -sin(msin^{-1}x) . \dfrac{m}{\sqrt{1-x^2}} \\ \; \\ \; \\ \therefore \sqrt{1-x^2} y_1 \;=\; -msin(msin^{-1}x) $

Squaring on both sides,

$ (1-x^2) y_1^2 \;=\; m^2[sin(msin^{-1}x)]^2 \\ \; \\ \; \\ = m^2 [1-cos^2(msin^{-1}x)] \\ \; \\ \; \\ \therefore (1-x^2)y_1^2 \;=\; m^2[1-y^2] \; \; \; \ldots From \; (i) $

Differentiating again w.r.t.x,

$ \therefore (1-x^2)2y_1y_2+y_1^2(-2x) \;=\; m^2(-2y)y_1 \\ \; \\ \; \\ \therefore (1-x^2)y_2 - xy_1 \;=\; -m^2y \\ \; \\ \; \\ \therefore (1-x^2)y_2 - xy_1 + m^2y \;=\; 0 $

Diff. equation (ii) n times by using Leibnitz’s theorem,

$ \therefore (1-x^2)y_{n+2} + n(-2x)y_{n+1} + \dfrac{n(n-1)}{2} .(-2) y_n - xy_{n+1} - ny_n - m^2y_n =0 \\ \; \\ \; \\ \therefore (1-x^2)y_{n+2} - x(2n+1)y_{n+1} + (-n^2+n-n+m^2) y_n =0 \\ \; \\ \; \\ \therefore (1-x^2)y_{n+2} - (2n+1)xy_{n+1} + (m^2-n^2)y_n =0 \; \; \; \; \; \ldots Hence \; Proved. $