Coordinate method for solving first order linear PDE Task: Solving the PDE $au_x+bu_y+cu=0$.
(Source: PDE, 2ndE by Walter A. Strauss, Exercise 1.2.19. Lots of books have it, though.)
Solution 1
The PDE can be transformed by the coordinate method via
$$\begin{cases}x'=ax+by\\y'=bx-ay\\\end{cases}$$
and we shall obtain $(a^2+b^2)u_{x'}+cu=0$, which gives $u=f(bx-ay)\exp(-\frac{c}{a^2+b^2}x)$.
Solution 2
When I was googling I came across this second solution here (links to PDF) which states that we may do another transformation by letting
$$\begin{cases}x'=-\frac{b}{a}x+y\\y'=x\\\end{cases}$$
and getting $u=f(-\frac{b}{a}x+y)\exp(-\frac{c}{a}x)$
By the link, I sense that by choosing $f(bx-ay)$ smartly, between the two generic solutions will shine equivalence.
(I'm aware of a third solution in which we divide the left side of the PDE by $u$ and then set $v=\ln(u)$ which will yield the same solution as solution II. My God, so many solutions??)
Thoughts and Questions
I think that the way we solve PDEs in Solution I is nice and clean because we chose another orthogonal coordinate system, and that the Solution II is harder to envisage because the coordinates are not orthogonal. I'm not so confident, so I would like to ask the following questions:

*

*How to show the equivalence of the two generic solutions?


*What's the goal of Solution II? And how is it achieving its desired goal (why does it transform like that by going non-orthogonal)?


*Are there any advantages in using one transform over another? If so, why?
(I see for sure that the link's author likes this Solution II)
I'd like some cool suggestions.


*(Thank you! And something extra on the other side of the fence.)
I can't solve PDEs like:
$au_x+bu_y+cu=g(x,y)=\exp(dx+ey)$
What should I do about it?
Thanks!
 A: The equation involves the directional derivative of $u$. So we want to solve it by considering the lines with direction vector $(a,b)$, since on each such line the equation becomes an ODE. Specifically, it becomes a linear homogeneous ODE with constant coefficients: $u_t+ku=0$ and we know that the standard approach to such ODE is to look for solutions of the form $u=C\exp(\lambda t)$. The difference between solutions II and III is the order of operation: we could recognize the constant-coefficient structure first (hence take an exponential ansatz) or the directional structure first (hence introduce new coordinate axes). 
Whether to use I or II depends on how initial values are given. Since we can solve the equation separately on each line in the direction $(a,b)$ , the solution is uniquely determined by its values on any line going in any other direction. If we know the values of $u$ on a line perpendicular to $(a,b)$, the orthogonal transformation of I fits the task perfectly. But if we are given $u(0,y)$ or $u(x,0)$ instead, then we better keep that axis as one of our coordinates, orthogonality be damned.
As for IV, it's a nonhomogeneous equation which reduces to a nonhomogeneous ODE of the form $u_t+cu=\alpha \exp(\beta t)$. The solution is obtained as a sum of the general solution of the homogeneous ODE and a particular solution, which you will find in the form of the right hand side (or, if you are unlucky to hit resonance $\beta=-c$, RHS times a linear function). This is standard ODE material, http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients
