Мое понимание решения этой PDE.

[drzewo]

(Оффтоп)

and that is miserable view indeed

[Knight2023]

I guess I have made it.

$$ \begin{align} u(x,y) = u \left( x(x',y'), y(x',y') \right) \\ x(x',y') = ax' \\ y(x',y') = bx' + y'\\ \partial u = \frac{\partial u}{\partial x} \delta x + \frac{\partial u}{ \partial y} \delta y \\ u_{x'}= u_x \frac{\partial x}{\partial x'} + u_y \frac{\partial y}{\partial x'} \\ u_{x'} = a ~u_x + b~u_y = 0\\ \implies u = f(y') = f( y-b/ax) \end{align} $$

Still, the mystery is how did you come up with such a nice substitution for variables.

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As His words are already mentioned in this thread, I too must not keep myself from remembering

Resist not evil

[пианист]

Knight2023 в сообщении #1651298 писал(а):

the mystery is how did you come up with such a nice substitution for variables.

Some practice with variable substitution will help you.

[vpb]

1)

Knight2023 в сообщении #1651178 писал(а):

(Возникновение этой идеи связано с книгой Вальтера Штрауса)

Note that

W.Strauss писал(а):

The main prerequisite is a solid knowledge of calculus, especially multivariate.

At the moment, your knowledge is not sufficient, as the colleague already said to you.

2) By the way, the changing of variables proposed by пианист may not work (namely, in the case $a=0$).

[vpb]

P.S.

Knight2023 в сообщении #1651298 писал(а):

I guess I have made it.

Your calculation is correct, but not accurately enough written. In particular, write $y-(b/a)x$, not $y-b/ax$, because $y-b/ax$ may mean either $y-(b/a)x$, or $y-b/(ax)$ (I usually think in such cases that it means $y-b/(ax)$).