TITLE: How to solve the following differential equation: $(2x^3y^2-y)dx+(2x^2y^3-x)dy=0$?
QUESTION [5 upvotes]: 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!

REPLY [4 votes]: $$(2x^3y^2-y)dx+(2x^2y^3-x)dy=0$$
$$2(x^3y^2dx+x^2y^3dy)-(xdy + ydx)=0$$
$$(x^2y^2)*2(xdx+ydy) =  d(xy)$$
$$d(x^2+y^2)=\frac{d(xy)}{(xy)^2}$$
which upon Integration gives
$$x^2+y^2+\frac{1}{xy}=c$$
Some Results:
$xdy+ydx=d(xy)$
$xdx+ydy=\frac{1}{2}d(x^2+y^2)$