$y=(\sin^{-1} x)^2 \ \cdots \ \cdots (1)$ $\therefore y_1 = 2 \sin^{-1} \ x \dfrac {1}{\sqrt{1-x^2}} \ \cdots \ \cdots (2)$ $\therefore \sqrt{1-x^2} y_1 = 2 \sin ^{-1} x$ Differentiating again we get, $(1-x^2)y_2 - xy_1 = 2 \ \cdots \ \cdots (3)$ By Leibnitz's theorem $(1-x^2) y_{n+2}+ n (-2x)y_{n+1}+ \dfrac {n(n-1)(-2)y_n}{2!} - \big [ xy_{n+1} + n(1) y_n \big ] = 0$ $\therefore (1-x^2) y_{n+2} - (2n+1)xy_{n+1} - n^2 y_n = 0 \ \cdots \ \cdots (4)$ Putting $x=0, \ y_{n+2}(0) = n^2 y_n (0) \ \cdots \ \cdots (5)$ Putting x=0 in (1), (2) and (3), we get, $\therefore y(0)=0$ $y_1(0)=0$ $y_2(0)=2$ Putting n=1,2,3,... in (5), we get, $y_3(0)=1^2 y_1(0)=0$ $y_5(0)=3^2 y_3 (0)=0$ $y_7(0)=5^2 y_5(0) = 0$ $y_n(0)=0$ if n is odd. Putting n=2,4,6,.. in (5), we get, $y_4(0) = 2^2 y_2 (0) = 2^2 2$ $y_6(0)=4^2 y_4 (0) = 4^2 2^2 2$ $y_n(0)=2 \ 2^2\cdots (n-2)^2$

if n is even and n≠2