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!