Find the particular solution for the following differential equation: , given that

Rewrite as: $\frac{dx}{dy} + \frac{1}{\sqrt{1 - y^2}} x = \frac{\sin^{-1} y}{\sqrt{1 - y^2}}$. This is a linear differential equation in $x$ with integrating factor: $I.F. = e^{\int \frac{1}{\sqrt{1 - y^2}} dy} = e^{\sin^{-1} y}$. The general solution is: $x e^{\sin^{-1} y} = \int e^{\sin^{-1} y} \frac{\sin^{-1} y}{\sqrt{1 - y^2}} dy + C = (\sin^{-1} y - 1)e^{\sin^{-1} y} + C$. $x = \sin^{-1} y - 1 + C e^{-\sin^{-1} y}$. Given $y(0) = 0$ (i.e. $y = 0$ when $x = 0$): $0 = \sin^{-1}(0) - 1 + C e^0 \implies C = 1$. Hence, the particular solution is $x = \sin^{-1} y - 1 + e^{-\sin^{-1} y}$.