I have the equation $$(x+y)uu_x + (x-y)uu_y = x^2 + y^2$$ with the initial condition $u(x, 2x) = 0$. I managed to write down the characteristic $$\frac{dx}{(x+y)u} = \frac{dy}{(x-y)u} = \frac{du}{x^2 + y^2}.$$ After some manipulations on the equation $$\frac{dx}{(x+y)u} = \frac{dy}{(x-y)u},$$ I got $$x^2-2xy-y^2 = c_1$$ for some constant $c_1$.
However, I couldn't manage to derive anything about $c_2$, the other constant which will show up in the solution.
Any help would be highly appreciated.
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For the sake of completeness i want to add the general solution of the PDE:
$$u\! \left(x,y\right)\mapsto-\sqrt{2 x y+f_{1}\! \left(-\frac{1}{\sqrt{-x^{2}+2 x y+y^{2}}}\right)}$$
For the initial condition $u(x,2x)=0$ we get the equation $4x^{2}+f_{1}\left(-\frac{\sqrt{7}}{7 \sqrt{x^{2}}}\right)=0$ with the solution
$$f_{1} (x)\mapsto -\frac{4}{7\cdot x^{2}}$$
and hence we get the final solution
$$u(x,y)\mapsto-\frac{\sqrt{28 \left(x+2 y\right) \left(x-\frac{y}{2}\right)}}{7}$$
Hint: Can you think about multipliers?
$\frac{xdx-ydy}{x^2u+xyu-xyu+y^2u} = \frac{du}{x^2 + y^2}$
$\implies xdx-ydy=udu$
$\implies \frac{1}{2}d(x^2-y^2-u^2)=0$
$\implies x^2-y^2-u^2=C$