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How did my professor get from here to here? DiffEq

  1. Feb 28, 2016 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    X(x)=(Ae^kx+Be^-kx)
    Y(y)=(Csin(ky)+D(cos(ky))
    V(x,y)=(Ae^kx+Be^-kx)(Csin(ky)+D(cos(ky))


    2. Relevant equations
    separation of variables

    3. The attempt at a solution
    so our boundary condition says that as x->infinity , V=0
    this is only possible if A=0

    so A=0 and Be^-kx= 0
    so X= Be^-kx
    and
    V(x,y)=(e^-kx)(Csin(ky)+D(cos(ky))

    why is there an e^-kx on the front. Shouldn't there also be a B?
     
  2. jcsd
  3. Feb 28, 2016 #2

    TSny

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    B can be "absorbed" by the arbitrary constants C and D.
     
  4. Feb 28, 2016 #3

    grandpa2390

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    ah! you're right. distribute the B into the the parenthesis. I knew it was something so simple otherwise he wouldn't have glossed over it. lol.
    Thanks!

    just one last question. so B^e^-kx we keep because the constant is not equal to zero?
    because the next part we solve for Y, and D ends up being 0 but we keep Csin(ky) which equals 0 because sin(0)=0. so we keep that one but eliminate the part where D=0.

    so we keep the part where the constant is not equal to zero?
     
  5. Feb 28, 2016 #4

    TSny

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    Yes. If a boundary condition forces a constant factor of a term to be zero, then you drop that term. Otherwise, keep the term.
     
  6. Feb 28, 2016 #5

    LCKurtz

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    Apparently you are solving a homogeneous PDE with boundary conditions. The identically zero function is always a solution, but you want non-trivial solutions. So you hope not to have to take all the constants be zero. So you probably have something like ##Ce^{-kx}\sin(ky)## and you get your last boundary condition to be zero by choosing particular values of ##k## so the sine term can be zero without making ##C=0##. These values of ##k## determine the eigenvalues of the problem and allow nontrivial solutions.
     
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