Solution of non exact differential equations with integration factor depend both $x$ and $y$ I'm not finding any general description to solve a non exact equation which's integrating factor depend both on $x$ and $y$. 
I'm on this problem $$(2x^{2}-y)dx+(x+y^2)dy=0 $$
I am trying to solve and kind of stuck now which is given below.

 A: A simple change of variables leads to a separable ODE and to the general solution on parametric form :

The arbitrary constant is included in the undefined integral.
A: $(2x^2-y)~dx+(x+y^2)~dy=0$
$(y^2+x)\dfrac{dy}{dx}=y-2x^2$
Which relates to an ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=166.
Let $u=\dfrac{1}{x}$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-\dfrac{1}{x^2}\dfrac{dy}{du}=-u^2\dfrac{dy}{du}$
$\therefore-\left(y^2+\dfrac{1}{u}\right)u^2\dfrac{dy}{du}=y-\dfrac{2}{u^2}$
$(u^2y^2+u)\dfrac{dy}{du}=\dfrac{2}{u^2}-y$
Let $t=uy$ ,
Then $y=\dfrac{t}{u}$
$\dfrac{dy}{du}=\dfrac{1}{u}\dfrac{dt}{du}-\dfrac{t}{u^2}$
$\therefore(t^2+u)\left(\dfrac{1}{u}\dfrac{dt}{du}-\dfrac{t}{u^2}\right)=\dfrac{2}{u^2}-\dfrac{t}{u}$
$\left(\dfrac{t^2}{u}+1\right)\dfrac{dt}{du}-\dfrac{t^3}{u^2}-\dfrac{t}{u}=\dfrac{2}{u^2}-\dfrac{t}{u}$
$\left(\dfrac{t^2}{u}+1\right)\dfrac{dt}{du}=\dfrac{t^3+2}{u^2}$
$(t^3+2)\dfrac{du}{dt}=t^2u+u^2$
$\dfrac{du}{dt}-\dfrac{t^2u}{t^3+2}=\dfrac{u^2}{t^3+2}$
Luckily this becomes a Bernoulli ODE.
If we go back to the original problem,
and if we let $t=\dfrac{y}{x}$ ,
Then $y=xt$
$\dfrac{dy}{dx}=x\dfrac{dt}{dx}+t$
$\therefore(x^2t^2+x)\left(x\dfrac{dt}{dx}+t\right)=xt-2x^2$
$(xt^2+1)x^2\dfrac{dt}{dx}+x^2t^3+xt=xt-2x^2$
$(xt^2+1)x^2\dfrac{dt}{dx}=-(t^3+2)x^2$
$(t^3+2)\dfrac{dx}{dt}=-t^2x-1$
$\dfrac{dx}{dt}+\dfrac{t^2x}{t^3+2}=-\dfrac{1}{t^3+2}$
Which is also luckily this becomes a linear ODE.
I.F.$=e^{\int\frac{t^2}{t^3+2}dt}=e^{\int\frac{d(t^3+2)}{3(t^3+2)}}=e^\frac{\ln(t^3+2)}{3}=\sqrt[3]{t^3+2}$
$\therefore\dfrac{d\left(x\sqrt[3]{t^3+2}\right)}{dt}=-\dfrac{1}{(t^3+2)^\frac{2}{3}}$
$x\sqrt[3]{t^3+2}=C-\int^t\dfrac{d\tau}{(\tau^3+2)^\frac{2}{3}}$
$x\sqrt[3]{\dfrac{y^3}{x^3}+2}=C-\int^\frac{y}{x}\dfrac{d\tau}{(\tau^3+2)^\frac{2}{3}}$
$\sqrt[3]{2x^3+y^3}+\int^\frac{y}{x}\dfrac{d\tau}{(\tau^3+2)^\frac{2}{3}}=C$
