Solve $(y^2+2x^2y)dx+(2x^3-xy)dy=0$ Solve $(y^2+2x^2y)dx+(2x^3-xy)dy=0$
My attempt
$Mdx + Ndy$ form
$M = (y^2+2x^2y)$, $N =(2x^3-xy) , M_y=2y+2x^2, N_x=6x^2-y$
I am not getting in of standard forms here...
Another attempt:
$2x^2 d(xy)+y(ydx-xdy)=0$, 
$2d(xy)=yd(y/x)$ Any clue pls???
 A: $$y(y+2x^2)dx+x(2x^2-y)dy=0$$
$$y(y+2x^2)(2xdx)+2x^2(2x^2-y)dy=0$$
An obvious change of variable appears : $\quad t=x^2$
$dt=2xdx \quad\implies\quad y(y+2t)dt+2t(2t-y)dy=0$
$$\frac{dy}{dt}=\frac{y}{2t}\frac{y+2t}{y-2t}$$
This is an ODE of the homogeneous kind. The usual change of function is : $\quad y=tu$
$\frac{dy}{dt}=u+t\frac{du}{dt} \quad\implies\quad u+t\frac{du}{dt}=\frac{tu}{2t}\frac{tu+2t}{tu-2t}=\frac{u}{2}\frac{u+2}{u-2}$
$$t\frac{du}{dt}=\frac{u}{2}\frac{u+2}{u-2}-u$$
This is a separable ODE. I suppose that you can take it from here.
HINT: 
First, solve for $t(u)$, not for $u(t)$.
Second, $\quad \begin{cases} t=x^2\\ u=\frac{y}{x^2} \end{cases}\quad$ to come to the relationship between $y$ and $x$.
A: Here is another way
$$y(y+2x^2)dx+x(2x^2-y)dy=0$$
Substitute $y=tx$ 
$$t(t+2x)=(t-2x)y'$$
Note that $y'=t'x+t$
$$t(t+2x)=(t-2x)(t'x+t)$$
After some simplifications you get:
$$t'(t-2x)=4t$$
Consider now $x'=\frac {dx}{dt}$
$$x'+\frac x {2t}=\frac 1 4$$
Which is easy to solve.....
A: I thought it might be useful to demonstrate Jacquelin's suggestion from the comments. We're given the ODE
$$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$$
Distributing an integrating factor $\mu(x,y)$ across both sides we get
$$\mu M\,\mathrm dx+\mu N\,\mathrm dy=0$$
To meet the conditions of exactness, we require
$$(\mu M)_y=(\mu N)_x\implies\mu_y M+\mu M_y=\mu_xN+\mu N_x$$
$$\implies \mu_y(y^2+2x^2y)-\mu_x(2x^3-xy)=\mu(4x^2-3y)$$
Let's attempt the traditional guess of $\mu(x,y)=x^ay^b$. Substituting into the PDE above, we have
$$\begin{align*}
bx^ay^{b-1}(y^2+2x^2y)-ax^{a-1}y^b(2x^3-xy)&=x^ay^b(4x^2-3y)\\[1ex]
(a+b)x^ay^{b+1}-(2a-2b)x^{a+2}y^b&=-3x^ay^{b+1}+4x^{a+2}y^b
\end{align*}$$
$$\implies\begin{cases}a+b=-3\\2a-2b=-4\end{cases}\implies a=-\frac52,b=-\frac12$$
(You can verify that $(\mu M)_y=(\mu N)_x$.) Now we seek a solution $\Psi(x,y)=C$ for which $\Psi_x=\mu M$ and $\Psi_y=\mu N$. We have
$$\Psi_x=x^{-5/2}y^{3/2}+2x^{-1/2}y^{1/2}\implies \Psi=-\frac23x^{-3/2}y^{3/2}+4x^{1/2}y^{1/2}+f(y)$$
$$\Psi_y=2x^{1/2}y^{-1/2}-x^{-3/2}y^{1/2}=-x^{-3/2}y^{1/2}+2x^{1/2}y^{-1/2}+\frac{\mathrm df}{\mathrm dy}$$
$$\implies\frac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$$
where $C$ is an arbitrary constant. We thus arrive at a solution
$$\Psi(x,y)=\frac23x^{-3/2}y^{1/2}\left(6x^2-y\right)=C$$
A: $$(y^2+2x^2y)dx+(2x^3-xy)dy=0....(i)\\
\implies y^2dx+2x^2ydx+2x^3dy-xydy=0\\
\implies y(ydx-xdy)+x^2(2ydx+2xdy)=0$$
Which is the form of :
$$x^ay^b(mydx+nxdy)+x^{a'}y^{b'}(m'ydx+n'xdy)=0$$
So, comparing both equations we get:
$a=0, b=1, m=1, n=-1, a'=2, b'=0, m'=2, n'=2$
Now,
$$\frac{a+h+1}{m}=\frac{b+k+1}{n}\\
\implies \frac{h+1}{1}=\frac{1+k+1}{-1}\\
\implies h+k+3=0.....(ii)$$
and $$\frac{a'+h+1}{m'}=\frac{b'+k+1}{n'}$$
$$\implies \frac{2+h+1}{2}=\frac{k+1}{2}\\
\implies h-k+2=0.....(iii)$$
Solving equation (ii) & (iii) we get, $h=-\frac{5}{2}, k=-\frac{1}{2}.$
So,I.F.= $x^{-\frac{5}{2}}y^{-\frac{1}{2}}$
Multiplying equation (i) with the I.F. we get,
$$(x^{-\frac{5}{2}}y^\frac{3}{2}dx+2x^{-\frac{1}{2}}y^\frac{1}{2})dx+(2x^\frac{1}{2}y^{-\frac{1}{2}}-x^{-\frac{3}{2}}y^\frac{1}{2})dy=0$$
Therefore the solution is:
$$\int{Mdx}+\int{(\text{terms of N not containing x})dy}=c\\
\implies \int{x^{-\frac{5}{2}}y^\frac{3}{2}dx}+2\int{x^{-\frac{1}{2}}y^\frac{1}{2}dx}=c\\
\implies -\frac{2}{3}x^{-\frac{3}{2}}y^\frac{3}{2}+4(xy)^\frac{1}{2}=c\\
\implies 4(xy)^\frac{1}{2}-\frac{2}{3}\left(\frac{y}{x}\right)^\frac{3}{2}=c$$
