# Thread: help with integrals needed

1. ## help with integrals needed

I have to solve the following:

$
dy=\frac{du}{x-u}$
, where $u=u(x, y)$.

How do I solve this?
Thank you so much for all your help!

2. Originally Posted by georgel
I have to solve the following:

$
dy=\frac{du}{x-u}$
, where $u=u(x, y)$.

How do I solve this?
Thank you so much for all your help!
Is this exactly as the original question is stated?

3. Originally Posted by mr fantastic
Is this exactly as the original question is stated?
No. I have to solve a quasilinear equation:

$x u_x + y u_y= xy-yu$

I'm trying to find the general soultion so I do the standard procedure from my textbook:

$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{y(x-u)}$

$\frac{dx}{x}=\frac{dy}{y}$ gives me
$ln x=ln y + ln c$, $\phi (x, y)=c=\frac{x}{y}$

And now I try to do the same with $\frac{dy}{y}=\frac{du}{y(x-u)}$ to get $\psi(x, y, u)$, but don't know how.

4. I've had a similar problem:

$\frac{dy}{y+u}=\frac{du}{y-u}$, and this is what I did:

$ydy-udy=ydu+udu$
$(y-u)dy-(y-u)du=2udu$
Then I substituted $z=y-u$ and got $zdz=2udu$.

But I don't know what to do with the problem I stated above..
Please, could anyone help?

I'm stuck and unfortunately, I'm running out of time to solve it.