two solutions to a PDE? For this question in Partial Differential Equations: An Introduction, 2nd Edition,

Solve $au_{x}+bu_{y}+cu=0$.

I defined $v$ to be $v=e^{\int \frac{c}{a}dx}=e^{\frac{c}{a}x}$ and then multiplied both sides by $v$. Now define $z(x,y)=vu(x,y)$, we'll get $az_{x}+bz_{y}=0$. $\implies \nabla z\cdot (a,b)=0$ $\implies$ z is constant on the characteristic lines defined by $\frac{dy}{dx}=\frac{b}{a}$. The characteristic lines are $y=\frac{b}{a}x+C$ for some constant $C$. $z(x,y)=z(x, \frac{b}{a}x+C)=z(0,C)=f(C)=f(y-\frac{b}{a}x)$ for some function $f$. Therefore, the final solution is $u(x,y)=f(y-\frac{b}{a}x)e^{-\frac{c}{a}x}$. I've checked my solution by taking the derivatives of $u$ and I think this is correct. However, the solution given by the book is $u(x,y)=f(ay-bx)e^{-c(ax+by)/(a^{2}+b^2)}$. I feel like both answers look right. What is going wrong here?
 A: Nothing is going wrong!
In order to compare, let me write your solution with $g$ instead of $f$:
$$
u(x,y) = g\bigl( y - \tfrac{b}{a} x \bigr) \exp\bigl(- \tfrac{c}{a} x \bigr)
.
$$
In this expression you can let
$$
g(t) = f(at) \exp\bigl(- \tfrac{bc}{a^2+b^2} t \bigr)
$$
to get the book's answer:
$$
\begin{split}
u(x,y)
&
= g\bigl( y - \tfrac{b}{a} x \bigr) \exp\bigl(- \tfrac{c}{a} x \bigr)
\\ &
= f\Bigl( a \bigl(y - \tfrac{b}{a} x \bigr) \Bigr) \exp\Bigl(- \tfrac{bc}{a^2+b^2} \bigl( y - \tfrac{b}{a} x \bigr) \Bigr) \exp\bigl(- \tfrac{c}{a} x \bigr)
\\ &
= f(ay-bx) \exp\Bigl( - \tfrac{c (ax+by)}{a^2+b^2} \Bigr)
.
\end{split}
$$
It should be clear that you can go the other way too (from $g$ to $f$), so the two expressions give the same set of solutions (when $f$, or equivalently $g$, ranges over the set of all $C^1$ functions).
Your way of writing the answer is natural if one thinks in terms of data given on the $y$-axis,
$$
u(0,y)=g(y)
,
$$
while the book's answer is obtained if one starts from data given along the line $ax+by=0$ (which is orthogonal to the characteristics),
$$
u\Big( \tfrac{-bs}{a^2+b^2}, \tfrac{as}{a^2+b^2} \Bigr) = f(s)
.
$$
