Question

PDE's: Show that v(x,y) = xF(2x+y) is a general solution of xv_x - 2xv_y = v, where F is an arbitrary differentiable function.

Answer 1

Attached is my solution, somehow I did it wrong, please point my mistake out.

Answer 2

I'm pretty sure there's an error in your PDE. It should be
\[ xv_x - 2x^2v_y = v \]
This is a change of -2x in front of v_y to -2x^2

Now,
\[ v(x,y) = xF(2x+y) \]
\[ v_x = F(2x + y) + 2xF'(2x + y) \]
\[ v_y = F'(2x + y) \]

Write F for F(2x + y), F' for F'(2x + y)
Hence
\[ xv_x - 2x^2v_y \]
\[ = xF + 2x^2F' - 2x^2F' \]
\[= xF\] \[ = v\]

Answer 3

Hm, thanks, I'll check if there's an errata list on this book, I have a very old copy (though same ISBN as the newer copies) of Schaum's Outline Series for Fourier Analysis with applications to BVPs.

Answer 4

But really, I copied the exact equation in the book, except for the notation.

Answer 5

I believe you. But as it is, that form of v does not satisfy the equation.

Answer 6

Haha thank you, I'll tell my prof.

Answer 7

Wait, wait. I made a mistake.

\[ v_y = xV'(2x+y) \]

Answer 8

Now, substitute that back in and it DOES satisfy the equation. Apologies.

Answer 9

For example, let F(t) = t, and thus
\[ v = xF(2x+y) = x(2x+y) = 2x^2 + xy \]
Hence
\[ v_x = 4x + y, v_y = x \]
Therefore
\[ xv_x - 2xv_y = x(4x + y) - 2x^2 \]
\[ = 2x^2 + xy \]
\[ = x(2x + y) = xF(2x+y) \]

Answer 10

Haha thanks, my prof also pointed out that the way I was doing it was making the problem much more difficult than it really is. Now my problem is finding a particular solution if v(1,y)=y^2.