# PDE concept Help

1. Sep 9, 2010

### DarthBane

1. The problem statement, all variables and given/known data
Uxx + 3*Uyy - 2*Ux + 24*Uy +5*U = 0

Reduce this to the form Vxx + Vyy + C*V = 0

U = V*e^(alpha*x + Beta*y)

y' = gamma*y

Ok, the problem I am having is I don't know what to do with the gamma, however I am off by a factor of 3 in my answer for Vyy, so I know gamma must be 1/sqrt(3). I just don't understand how the gamma works here.

2. Relevant equations
I am not quite sure what to put in this space, first time posting, and have everything up to I believe

3. The attempt at a solution
Ok, I have simply taken partial derivatives of the equation U = V*e^(alpha*x + Beta*y)

This has churned out the following equation when plugging the partial derivatives back into the PDE above:

Vxx*e^(alpha*x + Beta*y) + 3*Vyy*e^(alpha*x + Beta*y) + (alpha -1)*Vx*e^(alpha*x + Beta*y) + (6*Beta + 24)*Vy*e^(alpha*x + Beta*y) + ((alpha)^2 + (Beta)^2 + 24*Beta - 2*alpha + 5)*V*e^(alpha*x + Beta*y) = 0

as seen, there is a factor of 3 I can't account for because it is easily found that alpha = 1 and Beta = -4. I know the 3 is killed by gamma = 1/sqrt(3), however IDK how it works.

The equation is then reduced to (I am leaving the e^(alpha*x + Beta*y) in although it could be divided out easily):

Vxx*e^(alpha*x + Beta*y) + 3*Vyy*e^(alpha*x + Beta*y) + ((alpha)^2 + (Beta)^2 + 24*Beta - 2*alpha + 5)*V*e^(alpha*x + Beta*y) = 0

I would really appreciate an explanation for why and how this gamma works.

Thanks!

Last edited: Sep 9, 2010