Find the general solution to this PDE I'm asked to find the general solution of
$$au_{xx}+bu_{xy}=0$$
With $u=u(x,y)$ and $a$, $b$ real constants.
I'm just starting with PDE's, haven't seen any resolution technique except for basic changes of variables and factorizations, and sometimes assuming the solutions might be of a certain form and work with that.
First thing I tried was writting:
$$au_{xx}+bu_{xy}=(au_x+bu_y)_x=0$$
Which implies
$$au_x+bu_y=f(y)$$
It should be simpler to solve this PDE, but due to my amateur state I haven't been able to. I've tried a linear change of independent variables to see if I could get some cancellation by choosing proper coeficients
$$w=Ax+By$$
$$v=Cx+Dy$$
Substituting back in the equation
$$Aau_w+Cau_v+Bbu_w+Dbu_v=f(y)$$
It doesn't look like I could simplify that, plus I'm not sure how to deal with that arbitrary $f(y)$. I tried this change directly in the original second order equation to no avail. Any help?
 A: You can successfully apply the method of characteristic curves. In your case, you should consider straight lines in the direction of $(a,b)$. So let $\phi$ the straight line through say $(x_0,0)$ with direction $(a,b)$. Then
\begin{equation}
\phi(t) = (x_0,0) + t(a,b).
\end{equation}
Let $U(t)=u(\phi(t))$. Then by the chain rule, $U$ is differentiable and
\begin{equation}
U'(t) =  u_x(\phi(t)) a + u_y(\phi(t)) b = f(bt)
\end{equation}
You can now recover $U(t)$ through integration. There is still a small issue to consider when $b=0$.
EDIT: More information added in response to OP's comments:
Now, let $(x,y) \in \mathbb{R}^2$ be given. We which to recover $u(x,y)$. Assuming $b \not = 0$, we pick $t = y/b$ and set $x_0 = x - at$, such that
\begin{equation}
(x,y) = (x_0 + at, 0 + tb) = (x_0,0) + t(a,b) = \phi(t).
\end{equation}
Then by the above reasoning we have
\begin{equation}
u(x,y) - u(x_0,0) = U(t) - U(0) = \int_0^t f(bs) ds
\end{equation}
It follows, that
\begin{multline}
u(x,y) = u(x_0,0) + \int_0^t f(bs)dx = u(x - ay/b,0) + \int_0^{y/b} f(bs)ds \\ = u(x - ay/b,0) + \frac{1}{b} \int_0^y f(\tau) d\tau.
\end{multline} 
We notice that the constant $a$ and $b$ determine the point were we intersect with the $x$ axis and that they figure prominently in the final expression. It is necessary to specify the solution along a curve which crosses the characteristic curves. In this case it was convenient to use the $x$ axis.
In general, the characteristic curves are not straight lines, but the principle is the same. You find suitable curves so that you exploit the chain rule.
