How to solve this nonlinear ODE? It is a nonlinear ODE.
$$x(4ydx+2xdy)+y^3(3ydx+5xdy)=0$$
I can not solve it by dividing $x$ and $y$ into two parts.
Also, if I let $\frac{dU}{dy}=2x^2+5xy^3$ and $\frac{dU}{dx}=4xy+3y^4$, $\frac{d^2U}{dxdy}$ will not equal to $\frac{d^2U}{dydx}$.
 A: $$(4xy+3y^4)dx+(2x^2+5xy^3)dy=0$$
This is not an exact differential. To make it exact we have to find an integrating factor $\mu(x,y)$ so that 
$$(4xy+3y^4)\mu dx+(2x^2+5xy^3)\mu dy=0$$
be an exact differential, which implies :
$$\frac{\partial}{\partial y}\bigg((4xy+3y^4)\mu\bigg) =\frac{\partial}{\partial x}\bigg((2x^2+5xy^3)\mu\bigg)$$
An integrating factor is :
$$\mu=x^2y$$
This can be found by intuition, or by trial and error or by inspection (in case of academic exercise one can try $\mu=x^ay^b$ and determine $a$ and $b$). Of course a more reliable method exists and I used it to find $\mu=x^2y$. But this involves more complicated calculus, too boring to be edited here.
So, we have to solve :
$$(4x^3y^2+3x^2y^5)dx+(2x^4y+5x^3y^4)dy=0$$
This is an exact differential :
$$d(x^4y^2+x^3y^5)=0$$
$$x^4y^2+x^3y^5=C$$
A: Substitute $$y(x)=\sqrt[3]{v(x)}$$ and we get
$$2x^2v'(x)+9v(x)^2+x(5v'(x)+12)v(x)=0$$
now let $$v(x)=xu(x)$$
then we get
$$x^2(2xu'(x)+14u(x)^2+(5xu'(x)+14u(x))u(x))=0$$
this is $$\frac{du(x)}{dx}=-\frac{14(u(x)^2+u(x))}{x(5u(x)+2)}$$
