How to solve this initial value problem on $(-\infty, \ +\infty)$? I've managed to solve the following initial-value problem on the interval $(0, +\infty)$: $$x y^\prime - 2y = 4x^3 y^{1/2} $$ with $y = 0$ when $x = 1$. The unique solution is $y = (x^3 - x)^2$. 
How to solve this problem on the interval $(-\infty, \ +\infty)$? 

With the scaling $x \to \alpha x$, $y \to \beta y$ we can see that the equation is invariant whenever $\alpha^{1/2} = \beta^{3}$. It means that $y^{1/2}/x^{3}$ is invariant over the above mentioned scaling. It suggests the variable change $u = y/x^{6}$ or/and $y = x^{6} u$. It leads to:
$$
{1 \over 4\sqrt{u} - 6u}\,{{\rm d}u \over {\rm d}x} = {1 \over x}
$$ 
