Help Reducing PDE Uxx + 3*Uyy - 2*Ux + 24*Uy +5*U = 0 to Vxx + Vyy + C*V=0

[DarthBane]

Homework Statement

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.

Homework Equations

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

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!

 

[mighty2000]

Hello,

Thank you for your forum post. It seems like you are on the right track with your solution. The gamma term in this equation represents the coefficient of the second derivative with respect to y. In other words, it is the coefficient of the Vyy term.

In order to reduce the equation to the form Vxx + Vyy + C*V = 0, you need to factor out the exponential term and divide both sides by it. This will leave you with the following equation:

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

Now, you can compare this to the desired form and see that the coefficient of the Vyy term is 3 instead of 1. This is where the gamma term comes in. In order to make the coefficient of Vyy equal to 1, you need to multiply both sides by 1/gamma^2, which will give you:

(Vxx + 3*Vyy)/gamma^2 + (alpha^2 + Beta^2 + 24*Beta - 2*alpha + 5)*V = 0

Finally, you can see that this is equivalent to the desired form of Vxx + Vyy + C*V = 0, with C = (alpha^2 + Beta^2 + 24*Beta - 2*alpha + 5)/gamma^2.

I hope this explanation helps you understand how the gamma term works in this problem. Let me know if you have any further questions. Good luck with your studies!