Find the general solution of the PDE $x^2u_x + y^2u_y = (x+y)u$ without any boundary conditions.

Question

Find the general solution of $x^2u_x + y^2u_y = (x+y)u$, where $u = u(x,y)$ with $x>0$ and $y>0$.

First, I solved the ODE $$\frac{dy}{dx} = \frac{y^2}{x^2},$$ which gives $$\frac{1}{y} - \frac{1}{x} = C,$$ where $C \in \mathbb{R}$ (implicit form). If we had a homogeneous PDE, I believe that the general solution would become $$u(x,y) = f\left( \frac{1}{y} - \frac{1}{x} \right),$$ where $f$ is an arbitrary differentiable function. However, as we have $(x+y)u$, I am not sure what to do next, especially since I do not have boundary conditions. I tried reading a similar problem (Oria's answer) but did not understand the motivation for the steps given, especially since the problem I have has $u$ on the RHS, versus that problem where the RHS does not have $u$. (I also wanted to know if there was a way to get an arbitrary function of one variable, rather than two as in that answer.)

Answer 1

Hint: $$\dfrac{dx}{x^2}=\dfrac{dy}{y^2}=\dfrac{dx-dy}{x^2-y^2}=\dfrac{d(x-y)}{(x-y)(x+y)}=\dfrac{du}{u(x+y)}$$

Answer 2

$$x^2u_x + y^2u_y = (x+y)u$$ A short writing of the characteristic ODEs is : $$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{(x+y)u}=ds$$ A first family of characteristic curves comes from $\quad \frac{dx}{x^2}=\frac{dy}{y^2}\quad$ leading to : $$\frac{1}{x}-\frac{1}{y}=c_1$$ A second family of characteristic curves comes from $\quad \frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dx-dy}{x^2-y^2}=\frac{dx-dy}{(x-y)(x+y)}=\frac{du}{(x+y)u}$

$\frac{dx-dy}{x-y}=\frac{du}{u}\quad$ which leads to : $$\frac{u}{x-y}=c_2$$ General solution of the PDE, from : $\quad c_2=F(c_1)=\frac{u}{x-y}=F\left( \frac{1}{x}-\frac{1}{y}\right)$

$F$ is an arbitrary function (to be determined according to some boundary condition if the condition was specified). $$u(x,y)=(x-y)F\left( \frac{1}{x}-\frac{1}{y}\right)$$