I am reading Partial Differential Equations: An Introduction by Walter Strauss, and there's an exercise that comes with its solution at the end of the book. I also found that problem solved here: http://www1.maths.leeds.ac.uk/~kersale/Teach/M3414/Notes/m3414_1.pdf
The problem is that both solutions are different.
The book states that the solution is $$u(x,y) = f(bx-ay)e^{-\frac{c}{a^2 + b^2}(ax+by)}$$
Whereas the source states that it is
$$u(x,y) = f(bx-ay)e^{-cx}$$
How come both are correct answers?
First thing: your second solution is wrong, it should be $$u(x,y) = f(bx-ay)e^{-cx/a}\ .$$ Next: remember that $f$ is an arbitrary function (subject to some restrictions on differentiability). So in this solution we can take $$g(t)=f(t)e^{-mt}\quad\hbox{where}\quad m=\frac{bc}{a(a^2+b^2)}\ ;$$ then the solution is $$\eqalign{u(x,y) &=g(bx-ay)\exp\Bigl(\frac{bc}{a(a^2+b^2)}(bx-ay)\Bigr)\exp\Bigl(-\frac{cx}a\Bigr)\cr &=g(bx-ay)\exp\Bigl(\frac{bc}{a(a^2+b^2)}(bx-ay)-\frac{cx}a\Bigr)\cr &=g(bx-ay)\exp\Bigl(-\frac{c}{a(a^2+b^2)}\bigl((a^2+b^2)x-b^2x+aby\bigr)\Bigr)\cr &=g(bx-ay)\exp\Bigl(-\frac{c}{a^2+b^2}\bigl(ax+by\bigr)\Bigr)\cr}$$ which is your first solution.