= Solution
The equation requires $x\ne0$. Put $v=y^2$, giving $v'=2yy'$ and
$$
v'-\frac4xv=2x.
$$
An <integrating factor> is $x^{-4}$, so $(vx^{-4})'=2x^{-3}$. Integrating gives $vx^{-4}=C-x^{-2}$, hence
$$
\boxed{y^2=Cx^4-x^2,\qquad y=\pm\sqrt{Cx^4-x^2}.}
$$
Real differentiable branches require $Cx^4-x^2>0$, with one sign on each connected solution interval. At $y=0$, the original equation would demand $0=x$, contrary to $x\ne0$; the zeros of the radicand are therefore not regular solution points. In particular, $y\equiv0$ is not an additional solution.
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