Q:

Solve the partial differential equation $y\frac{\partial f}{\partial x} -x \frac{\partial f}{\partial y}=x^3y+xy^3, x> 0, y> 0$

Problem is to solve the differential equation
$$y\frac{\partial f}{\partial x} -x \frac{\partial f}{\partial y}=x^3y+xy^3,  x> 0, y> 0$$
using variables $u = x^2+y^2, v = x^2-y^2$
I have started at a solution with the chain rule, I have
$$\frac{\partial f}{\partial x} = 2x\frac{\partial f}{\partial u}+ 2x\frac{\partial f}{\partial v} $$
$$\frac{\partial f}{\partial y} = 2y\frac{\partial f}{\partial u}-2y\frac{\partial f}{\partial v}$$
Plug in that in the differential equation I get
$$y(2x\frac{\partial f}{\partial u}+2x\frac{\partial f}{\partial v})-x(2y\frac{\partial f}{\partial u}-2y\frac{\partial f}{\partial v})=x^3y+xy^3 \iff 4xy \frac{\partial f}{\partial v} = x^3y+xy^3 $$ 
But I am stuck here. 

A:

With cancellation you get 
$$\frac{\partial f}{\partial v}=\frac{1}{4}(x^2+y^2)=\frac{1}{4}u$$
The right hand side is constant in $v$, and so it can be directly integrated, yielding
$$f(u,v)=\frac{1}{4}uv+g(u)$$

