# Thread: PDEs solvable as ODEs

1. ## PDEs solvable as ODEs

I have one remaining problem in this category that is tripping me up. The PDE is:

u sub y + 2yu = 0

solve for u = u(x,y)

Seems to me that this problem needs to be put into the form u'' - u = 0, but since my answer doesn't quite make sense, is this the right approach? Help or guidance will be appreciated.

2. Originally Posted by dsprice
I have one remaining problem in this category that is tripping me up. The PDE is:

u sub y + 2yu = 0

solve for u = u(x,y)

Seems to me that this problem needs to be put into the form u'' - u = 0, but since my answer doesn't quite make sense, is this the right approach? Help or guidance will be appreciated.
Since your PDE is literally independent of x, we can separate. So

$u_y + 2 y u = 0\;\; \Rightarrow \frac{du}{u} = - 2y \,dy$ so $\ln u = -y^2 + \ln f(x),$ then solve for $u$.

3. Thanks! I had to work thru this a few times, but I think I know what to do. Will I know under a timed exam...that's another story! :-)