Solve: $ (y^2+ 2x^2y)dx + (2x^3 - xy)dy = 0 $ Solve:
$$ (y^2+ 2x^2y)dx + (2x^3 - xy)dy = 0 $$
I tried to solve the differential equation in the following manner, but I am unable to arrive at the correct solution. The correct solution is:
$$ 4(xy)^{\frac{1}{3}} - \frac{2}{3}(x/y)^{\frac{3}{2}} = C $$
Please help me identify my mistake in the solution.


RECTIFICATION:


 A: $$(y^2+2x^2y)dx+(2x^3−xy)dy=0$$
$$(y+2x^2)ydx+(2x^2−y)xdy=0$$
Is not of the form 
$$f_1(xy)ydx+f_2(xy)xdy=0$$
So you can't use that method for this DE
Be carefull Soumee. $(y+2x^2)$ is a function $f(x,y)$ But it's not a function of the form $f(xy)$

Edit
The integrating factor $\mu(x,y)=x^{-5/2}y^{-1/2}$ proposed by @Soumee:
$$y(ydx-xdy)+2x^2dxy=0$$
Multiply by IF $x^{-5/2}y^{-1/2}$
$$x^{-5/2}y^{-1/2}y(ydx-xdy)+2x^{-5/2}y^{-1/2}x^2dxy=0$$
$$-x^{-1/2}y^{1/2}d\frac {y}{x}+2x^{-1/2}y^{-1/2}dxy=0$$
$$-\left (\frac {y}{x} \right )^{1/2}d\left (\frac {y}{x} \right )+2(xy)^{-1/2}d(xy)=0$$
Integrate
A: A very simple solution:
$$(y^2+ 2x^2y)dx + (2x^3 - xy)dy = 0$$
$$y(ydx-xdy)+ 2x^2(xdy +ydx) = 0$$
$$-y d\left ( \frac {y}x \right )+ 2d(xy) = 0$$
$$-\sqrt y d\left ( \frac {y}x \right )+ \frac 2 {\sqrt y}d(xy) = 0$$
An obvious integrating factor is $\mu(x)=1/\sqrt x$
$$-\sqrt {\frac y x} d\left ( \frac {y}x \right )+ \frac 2 {\sqrt {xy}}d(xy) = 0$$
It's integrable.
$$\frac 13 \left ( {\frac y x} \right)^{\frac 32}-2 {\sqrt {xy}} = C$$
