# How do I solve this linear partial differential equation?

I haven't learned how to do this yet, but a friend gave me this question to do. He said it was on an exam he did and it was a fun puzzle.

$xu_x-yu_y=2u$

That's pretty much all he gave me. Is this even a complete question? I assume I read it like

$x\frac{\partial u}{\partial x} - y\frac{\partial u}{\partial y} =2u$

and that $u$ is a function of $x$ and $y$. After that, I'm not even sure what answering this means.

But if I think about it, $x\frac{\partial u}{\partial x}$ looks like an approximation of the function $u$ (when y is constant) at $(x,y)$ if the gradient is that of the function when $x=0$. Likewise, $y\frac{\partial u}{\partial y}$ is an approximation of $u$ (when $x$ is constant) at $(x,y)$. I don't know what I'd get subtracting one from the other, but the result somehow gives me twice the height of $u$ at every point.

How do I go about solving this?

I saw someone posted an answer and it's tempting to scroll down and look at it, but I'm refusing to until later. I'm going to try to figure this out for myself. :)

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This is a linear PDE also. –  Daryl Oct 13 '12 at 4:31
Woops, so it is. Fixed, thanks :) –  Korgan Rivera Oct 13 '12 at 12:13
$$u \left( x,y \right) = x^2F\left( x\,y \right)\,.$$