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Math Help - Heat Equation with Weird Boundary Conditions

  1. #1
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    Heat Equation with Weird Boundary Conditions

    After a few trivial steps I reached:

     X'' + \lambda^2 X = 0
    X(x) = Asin(\lambda x) + Bcos(\lambda x)

    But my boundary conditions are:

    X(-2) = 0, X(2) = 0

    How am I supposed to solve for A and B and \lambda?
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  2. #2
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    Re: Heat Equation with Weird Boundary Conditions

    If you sub. in your BC's you get

    -A \sin 2 \lambda + B \cos 2\lambda = 0,

    A \sin 2 \lambda + B \cos 2\lambda = 0,

    or

    A \sin 2\lambda = 0 and B \cos 2 \lambda = 0.

    This gives rise to two cases

    (1)\;\;A = 0,\;\; \cos 2 \lambda = 0,

    (2)\;\;  B = 0, \;\; \sin 2  \lambda = 0.
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