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PDE concept Help

  1. Sep 9, 2010 #1
    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
  2. jcsd
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