Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 37

$$y=\tan(\frac{1}{2}x^{2} ).$$

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