Differential equation with infinitely many solutions The problem is to solve for $-1<x<1$ $$y'(x)=\frac{4x^3y(x)}{x^2+y(x)^2}$$ with $y(0)=0$.
I need to show that this equation has infinitely many solutions. Note that $\frac{4y(x)x^3}{x^2+y(x)^2}$ is undefined for $y(0)=0$, but note that $\frac{4x^3y(x)}{x^2+y(x)^2}=2x^2(\frac{2xy(x)}{x^2+y(x)^2})=O(x^2)$ since $|\frac{2xy(x)}{x^2+y(x)^2}|\le 1$.
I do not know how solve the problem with an explicit formula. Can anyone point out how to prove that this problem has infinitely many solutions?

As you observed, the function 
$$F(x,y)=\frac{4x^3y}{x^2+y^2},$$
extended by $F(0,0)$, is continuous. Also, its partial derivative with respect to $y$ is bounded near the origin: 
$$\left|\frac{\partial F}{\partial y}\right| = 
\left|\frac{4x^3(x^2-y^2)}{(x^2+y^2)^2}\right| = 4|x| \frac{x^2|x^2-y^2|}{(x^2+y^2)^2}\le 4|x|$$
By the Picard–Lindelöf theorem the initial value problem with $y(0)=0$ has a unique solution. Of course, this solution is $y(x)\equiv 0$.
Maybe you misunderstood the problem?
