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 $

Diff. above equation w.r.t. x,

$ \therefore (1-x^2)y_2 + y_1(-2x)-xy_1-y=0 \\ \; \\ \; \\ \therefore (1-x^2)y_2-3xy_1-y=0 \; \;\; \; \ldots (iii) $

Diff. equation (iii) 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 - 3xy_{n+1} - 3ny_n - y_n =0 \\ \; \\ \; \\ \therefore (1-x^2)y_{n+2} - x(2n+3)y_{n+1} + (-n^2+n-3n-1) y_n =0 \\ \; \\ \; \\ \therefore (1-x^2)y_{n+2} - (2n+3)xy_{n+1} - (n^2+2n+1)y_n =0 \; \; \; \; \; \ldots Hence \; Proved. $

To find $ y_n(0)$ , put x=0

$ \therefore y_{n+2} - 0 - (n+1)^2y_n=0 \\ \; \\ \therefore y_{n+2} = (n+1)^2 y_n $

Put x=0 in equation (i) $ \therefore y(0)=0$

Put x=0 in equation (ii) $ \therefore y_1(0)=1$

If n is odd :

Put n = 1 $ \therefore y_3=4y_1 \;=\; 4(1)=2^2.1$

Put n = 3 $ \therefore y_5=(3+1)^2y_3 \;=\; 16(4)=64 \;=\; 4^2.2^2.1$

Put n = 5 $ \therefore y_7=(7+1)^2y_5 \;=\; 6^2.4^2.2^2.1$

$ \therefore y_n \;=\; (n+1)^2(n-1)^2 \ldots 4^2. 2^2 . 1 $

If n is even

Put n=0 $ \therefore y_4 \;=\; (2+1)^2y_2 \;=\; 0 $

$ \\ \; \\ \; \\$ $ y_n \;=\; (n+1)^2(n-1)^2 \ldots 4^2. 2^2 . 1 \; \; \; \; \; \; \; if \; n \; is \; odd \ \; \ y_n \;=\; 0 \; \; \; \; \;\;\; \; \; \; \; \; \; if \; n \; is \; even. $