y$= \dfrac{sin^{-1}x}{\sqrt{1-x^2}} \; \; \; \ldots (i)$

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

$ \therefore y_1 \;=\; \dfrac{ \sqrt{1-x^2} . \dfrac{1}{\sqrt{1-x^2}} -sin^{-1}x . \dfrac{1}{2\sqrt{1-x^2}} (-2x) }{1-x^2} \\ \; \\ \; \\ \therefore y_1 \;=\; \dfrac{1+x \dfrac{sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2} \\ \; \\ \; \\ \therefore y_1 \;=\; \dfrac{1+xy}{1-x^2} \; \; \; \ldots (ii) \\ \; \\ \; \\ \therefore (1-x^2)y_1 = 1+xy \\ \; \\ \; \\ \therefore (1-x^2)y_1 - 1 - xy = 0 $

Differentitating above equation n times and using Leibnitz's theorem,

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