Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 42

$$ y= \tan(\sin^{-1}x).$$

By separation of variables, we have $$\frac{dy}{y^2+1}=\frac{dx}{\sqrt{1-x^2}} $$ then by integration, we get $$\tan^{-1}y=\sin^{-1}x+c\Longrightarrow y= \tan(\sin^{-1}x+c).$$ Now, since $y(0)=0$, then $c=0$. So the general solution is given by $$ y= \tan(\sin^{-1}x).$$