A general solution of a partial differential equation with $f(x,y)$ I need to find a general solution to such a PDE:
$$u_x-u_y=f(x,y)$$
I am able to find a solution if $f(x,y)=0$ or $f(x,y)=u$. But I have no idea how to get the general solution. Has anybody got any ideas?
 A: Change variables:
$$
\xi=x,\quad\eta=x+y.
$$
Then
$$
u_x=u_\xi+u_\eta,\quad u_y=u_\eta.
$$
The equation becomes
$$
u_\eta=f(\xi,\eta-\xi).
$$
To solve it just integrate in the variable $\eta$.
A: Your equation is a transport equation. Usually we write it with one variable identified as "time" and the rest identified as space, like this:
$$u_t + v u_x = S(t,x,u)$$
where $v$ is a fixed velocity and $S$ is a source. In the physical situation modeled by this equation, the particles which are currently present are moving at a velocity $v$. We can use this to take the "transport" out of the problem by following the paths of the transported particles, i.e. the paths $x=x_0+vt$. If we consider this situation then $\frac{du}{dt} = \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} \frac{dx}{dt} = u_t + v u_x = S(t,x,u)$. That is:
$$\frac{d}{dt} u(t,x_0+vt) = S(t,x_0+vt,u(t,x_0+vt)).$$
This is now an ordinary differential equation along each of the parallel lines $x=x_0+vt$. In the special case where $S$ does not depend on $u$, it is easy to solve; we find
$$u(t,x_0+vt)=u(0,x_0)+\int_0^t S(s,x_0+vs) ds.$$
Usually we want a formula for $u(t,x)$; in this case $x_0=x-vt$ so
$$u(t,x)=u(0,x-vt)+\int_0^t S(s,x+v(s-t)) ds.$$
This is the simplest case of the method of characteristics.
