Differential equations are solved by reducing them to the exact differential of an expression in $x$ and $y$ i.e., they are reduced to the form $d(f(x,y)) = 0$.

e.g.: $\frac{x\,dx + y\,dy}{\sqrt{x^2 + y^2}} = \frac{y\,dx - x\,dy}{x^2}$
$\Rightarrow \frac{1}{2}\frac{2x\,dx + 2y\,dy}{\sqrt{x^2 + y^2}} = -\frac{x\,dy - y\,dx}{x^2} \Rightarrow d\left(\sqrt{x^2 + y^2}\right) = -d\left(\frac{y}{x}\right)$
$\Rightarrow d\left(\sqrt{x^2 + y^2} + \frac{y}{x}\right) = 0$

$\therefore$ solution is $\sqrt{x^2 + y^2} + \frac{y}{x} = c$

Use the above method to answer the following question:

The general solution of $(2x^3 - xy^2)\,dx + (2y^3 - x^2y)\,dy = 0$ is