Finding the solutions of the PDE $x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}=(x+y)u$ How to find all solutions of the PDE
$x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}=(x+y)u$?
My way was:
The Lagrange-Charpit equations are: $\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{(x+y)u}$.
There are two linear independent solutions. For the first one I got:
$\frac{du}{(x+y)u}=\frac{dx-dy}{x^2-y^2}=\frac{dx-dy}{(x+y)(x-y)} \Leftrightarrow \frac{1}{u}du-\frac{1}{x-y}d(x-y)=0 \Leftrightarrow d\mathrm{ln}(|\frac{u}{x-y}|)=0$.
So the first solution is $f(x,y,u)=\frac{u}{x-y}$.
Now I don't know how to find the second solution. I tried to manipulate the terms $\frac{dx}{x^2}, \frac{dy}{y^2}$ and $\frac{du}{u(x+y)}$ in a similar way as above but I don't get any simplification .
Can it be done in the same way or how can the second solution be found?

$\frac{\frac{1}{x}dx}{x}=\frac{\frac{1}{y}dy}{y}=\frac{\frac{1}{u}du}{(x+y)}$
$\implies\frac{\frac{1}{x}dx+\frac{1}{y}dy}{x+y}=\frac{\frac{1}{u}du}{x+y} $
$\implies \frac{1}{x}dx+\frac{1}{y}dy=\frac{1}{u}du$
Solve and get second integral curve to form the solution surface.
