How to solve the following differential equation: $(2x^3y^2-y)dx+(2x^2y^3-x)dy=0$? I'd love your help with solving this following differential equation: $$(2x^3y^2-y)dx+(2x^2y^3-x)dy=0.$$
I tried to use check if this is an exact equation and find a integration, but it didn't work. Then I tried to divide the equation by some factor and to use some kind of assignment such as $z=\frac{y}{x}$, but it didn't work for me either.
Any suggestions?
($y$ is a function of $x$)
Even though Julian answer is great,  I wonder if there's a solution without an integration depends both on x and y.
Thanks a lot!
 A: Look for an integrating factor of the form $\mu(x,y)=\mu(x\, y)$. Taking into account that
$$
\frac{\partial\mu(x\,y)}{\partial y}=x\,\mu'(x\,y)\text{ and }\frac{\partial\mu(x\,y)}{\partial x}=y\,\mu'(x\,y)
$$
the equation
$$
\frac{\partial}{\partial y}\Bigl(\mu\,(2\,x^3y^2-y)\Bigr)=\frac{\partial}{\partial x}\Bigl(\mu\,(2\,x^2y^3-x)\Bigr)
$$
leads to
$$
\frac{\mu'(x\,y)}{\mu(x\,y)}=-\frac{2}{x\,y}.
$$
Since the right hand side depends only on $x\,y$, such an integrating factor exists. Calling $t=x\,y$, we have $\mu'(t)/\mu(t)=-2/t$. Form this $\mu(t)=1/t^2$. Then $\mu(x,y)=1/(x\,y)^2$ is an integrating factor. Multiply the original equation by $1/(x\,y)^2$ to get
$$
\Bigl(2\,x-\frac{1}{x^2y}\Bigr)dx+\Bigl(2\,y-\frac{1}{x\,y^2}\Bigr)dy=0.
$$
I leave it to you to check that it is exact and solve it.
A: $$(2x^3y^2-y)dx+(2x^2y^3-x)dy=0$$
$$2(x^3y^2dx+x^2y^3dy)-(xdy + ydx)=0$$
$$(x^2y^2)*2(xdx+ydy) =  d(xy)$$
$$d(x^2+y^2)=\frac{d(xy)}{(xy)^2}$$
which upon Integration gives
$$x^2+y^2+\frac{1}{xy}=c$$
Some Results:
$xdy+ydx=d(xy)$
$xdx+ydy=\frac{1}{2}d(x^2+y^2)$
A: $\mu(x,y)=\frac{1}{x^2 y^2}$  is an integrating factor. The solution is
$x^2+y^2+ \frac{1}{xy}=c$.
I found the integrating factor using the Prelle-Singer algorithm.
I write 
$\dot x = x-2x^2 y^3 =f_1  $
$\dot y=2x^3 y^2-y=f_2$
It is clear that $x$ and $y$ are Darboux polynomials with cofactor $\Lambda_1=2x y^3-1$ and $\Lambda_2=2 y x^3-1$.  However there no linear combination of $\Lambda_1$, $\Lambda_2$ which gives $0$.  Therefore, we consider second degree polynomials.  The obvious choice is $J_1=x^2$, $J_2=y^2$, $J_3=xy$ with cofactors $\Lambda_1=2 -4 xy^3$, $\Lambda_2=4x^3 y-2$, $\Lambda_3=2x^3 y- 2 xy^3$ respectively.  There is  a linear combination which gives zero but it produces the trivial integral $1$.  However, if we solve $\sum_i \mu_i \Lambda_i=-( \partial_x f_1+\partial_y f_2) $ we find the solution  $\mu_1=0$, $\mu_2=0$, $\mu_3=-2$. According to the Prelle-Singer algorithm  $\prod_i J_i^{\mu_i}$ is an integrating factor. This gives $(xy)^{-2}$. 
I wrote a paper on this subject but unfortunately it is in Greek!
