General Solution for a PDE I would like to know if someone could provide me the solutions for the next PDE:
\begin{align*}
f_{x}(x,y) - f_{y}(x,y) = \frac{y-x}{x^{2}+y^{2}}f(x,y)
\end{align*}
I empirically found the following particular solution given by $f(x,y) = \displaystyle\frac{x+y}{\sqrt{x^{2}+y^{2}}}$. Any contribution is appreciated. Thank you in advance. 
 A: One can use method of characteristics $\gamma = \left(x(t),y(t)\right)$:
$$f_x + f_y = \dfrac{y-x}{x^2+y^2}\,f
$$
$$
x' = y'=1, \qquad \frac{df}{dt} = \dfrac{y-x}{x^2+y^2}\,f
$$
Solving system of ODEs above get
$$
x = t+a,\quad y = t+b = x + c, \implies c = y-x \quad - \quad \operatorname{const}\\
 f' = \frac{f}{2t^2 + 2(a+b)t + a^2 + b^2 }
\\
\ln f = \int \frac{dt}{2t^2 + 2(a+b)t + a^2 + b^2 }
\implies f = \exp \int \frac{dt}{2t^2 + 2(a+b)t + a^2 + b^2 }
$$
Hope you can proceed from here with computing the integral.
A: Hint: notice that if you multiply your particular solution by any constant the result is also a solution, i.e.,
$$f(x,y) = C \frac{x+y}{\sqrt{x^2 + y^2}}$$
On the other hand, the method of characteristics tells us that $\mathrm{d}x + \mathrm{d}y = 0$ (can you see this?) and therefore $x + y = c$, for some constant $c$. Can you relate this constant with the aforementioned factor?
A: $$f_{x}(x,y) - f_{y}(x,y) = \frac{y-x}{x^{2}+y^{2}}f(x,y)$$
The set of characteristic ODEs is:
$$\frac{dx}{1}=\frac{dy}{-1}=\frac{df}{\frac{y-x}{x^{2}+y^{2}}f}$$
The equation of a first family of characteristic curves comes from :
$$\frac{dx}{1}=\frac{dy}{-1} \quad\to\quad x+y=c_1$$
The equation of a second family of characteristic curves comes from :
$$\frac{dx}{1}=\frac{dy}{-1}=\frac{xdx+ydy}{x-y}=\frac{df}{\frac{y-x}{x^{2}+y^{2}}f}$$
$$\frac{df}{f}=-\frac{d(x^2+y^2)}{2(x^2+y^2)}\quad\to\quad \sqrt{x^2+y^2}f=c_2$$
The general solution of the PDE can be expressed on the form of an implicit equation :
$$\Phi\left(x+y)\:,\:\sqrt{x^2+y^2}f\right)=0$$
where $\Phi$ is any differentiable function of two variables.
Or, on the equivalent explicit form : $\quad \sqrt{x^2+y^2}f=F(x+y)$
$$f(x,y)=\frac{F(x+y)}{\sqrt{x^2+y^2}}$$
where $F$ is any differentiable function.
The function $F$ has to be determined according to the boundary condition (not specified in the wording of the question).
