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

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

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

Squaring on both sides,

$ (1-x^2)y_1^2 \;=\; 4(sin^{-1}x)^2 \\ \; \\ \; \\ (1-x^2)y_1^2 \;=\; 4y^2 \; \; \; \ldots (ii) \; From \; (i) $

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

$ \therefore (1-x^2)2y_1y_2 + y_1^2(-2x) \;=\; 4y_1 \\ \; \\ \; \\ \therefore (1-x^2)(2y_2) -2xy_1 \;=\; 4 \\ \; \\ \; \\ (1-x^2)y_2 -xy_1 - 4 \;=\; 0 $

Differentiating ‘n’ times by using Leibnitz’s theorem the above equation,

$ \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 \;=\; 0 \\ \; \\ \; \\ (1-x^2)y_{n+2}-(2n+1)xy_{n+1}-n^2y_n \;=\; 0 \; \; \; \; \ldots (iii) \\ \; \\ \; \\ When \; x=0 , y(0)=0 \\ \; \\ y_1(0) = 0 , \; y_2(0)=2 \\ \; \\ y_{n+2}(0)=n^2y_n(0) \; \; \; \ldots (iv) $

If n is odd: Put n=1,3,5 in equation (iv),

$ \therefore y_1=y_3=0 \\ \ y_5=(3)^2 y_3 \;=\; 0 $

Hence in general, $ (y_n)_0=0 $ when n is odd

If n is even:

Put n=2,4,6,… in equation (iv),

$ \therefore y_4(0)=(2)^2y_2(0) \;=\; 4(2) \;=\; 8 \;=\; 2^2 . 2 \\ \therefore y_6(0)=(4)^2y_4(0) \;=\; 4^2(8) \;=\; 4^2 . 2^2 . 2 $

$\\ \; \\ \; \\ $

Hence in general, $ y_n=2.2^2.4^2.6^2\ldots(n-2)^2$ if‘n’ is even

$ y_n =0$ For n is odd