Solve $au_x+bu_y+cu=0$ I am asked to solve
$$au_x+bu_y+cu=0$$
I am tempted to first solve $au_x+au_y=0$ which has characteristic lines $C=ay-bx$ and thus a solution to this is given by
$$u(x,y)=f(ay-bx)$$
where $f$ is an arbitrary function. Then substituting back into the original equation yields
$$au_x+bu_y+cu=0+cu=cf(ay-bx)=0$$
implying that I have merely found the trivial solution $u(x,y)=f(ay-bx)=0$. 
So far the book I am using has only explained the method of characteristic equations and I have solved various difficult ones like $\sqrt{1-x^2}u_x+u_y=0$. So I am guessing that I should be able to solve $au_x+bu_y+cu=0$ using this method combined with maybe some clever thinking. I might be able to use the fact that the directional derivative of $u$ along the lines $C=ay-bx$ is $-cu$ and so maybe along these lines $u=e^{-cf(ay-bx)}$ or something. If anyone has any suggestions I would be thankful.
 A: Make a change of variables
$$
t = bx + ay \\
p = bx - ay
$$
So 
$$
u_x = b(u_t + u_p) \\
u_y = a(u_t - u_p)
$$
After substituting in PDE
$$
ab(u_t + u_p) + ab(u_t - u_p) + cu = 2ab\ u_t + cu = 0
$$
It can be easily integrated
$$
\frac {u_t}u = -\frac c{2ab} \\
\ln u = -\frac c{2ab}t+f(p) \\
u = F(p)e^{-\frac c{2ab}t}
$$
or, in initial variables
$$
u = F(bx-ay)e^{-\frac c{2ab}(bx+ay)}
$$
where $F(x) = e^{f(x)}$
Update
If you use 
$$
t = ax +  by \\
p = bx - ay \\
$$
so equation is
$$
(a^2+b^2)u_t + cu = 0 \\
\ln u = -\frac c{a^2+b^2}t \\
u = F(p)e^{-\frac {ct}{a^2+b^2}} \\
u = F(bx - ay)e^{-\frac c{a^2+b^2} (ax + by)}
$$
which is also a solution.
A: let $v(x,y)=e^{cx}u(x,y)$ and compute $v_x,v_y$  then we have$$v_x+v_y=0$$ by geometric method we have$$ v(x,y)=e^{\frac{-c}{a}}f(y-x)$$ easily conclude $$u(x,y)=e^{\frac{-c}{a}}f(\frac{y}{b}-\frac{x}{a})$$
A: Generally, your idea is correct. However, some implication is need.
We need to parametrize the characteristic curve with parameter r. Change the original equation to $\frac{du}{dr}+cu=0$ with
$$y=br$$
$$x=ar+c_0$$
where $c_0$ denote which characteristic line we are focus on. Second, use separation of variable to solve it.
$$\frac{du}{u}=-cdr$$
$$ln|u|=-cr+c_1$$
$$u=\pm e^{c_1}e^{-cr}$$
Actually, $c_1$ depends on which characteristic line we are studying. Therefore, $e^{c_1}$ can be seen as a constant depending on $c_0=by-ax$, written as $F(by-ax)$. Substitute $r$ with $r=\frac{y}{b}$ we got the solution $u=e^{-\frac{c}{b}y}F(bx-ay)$. PS:It may differ with other solutions by some function of $bx-ay$. Say $u=e^{-\frac{c}{b}y}f(bx-ay)\cdot e^{-\frac{bx-ay}{2ab}}=e^{-\frac{c}{2ab}(bx+ay)}$
