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Math Help - Don't know how to even start with this one.

  1. #1
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    Don't know how to even start with this one.

    If y_1 and y_2 satisfy

    y_1 '' + \lambda_1 y_1 = 0  \ \ \ y_1 ' (0) = 0, \ \ y_1(L) = 0

    y_2 '' + \lambda_2 y_2 = 0 \ \ \ y_2 ' (0) = 0, \ \ y_2(L) = 0

    Without determining the explicit form of both y_1 and y_2, show that

    \int_0^L y_1 y_2 \ dx = 0 if \lambda_1 \neq \lambda_2 .
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  2. #2
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    Multiply the first by y_2 and the second by  y_1 and substract so

    y_2 y_1'' - y_1 y_2'' + (\lambda_1-\lambda_2) y_1 y_2 = 0.

    Now integrate [0,L]

    \displaystyle \int_0^L y_2 y_1'' - y_1 y_2''dx + (\lambda_1-\lambda_2) \int_0^L y_1 y_2dx = 0,

    \displaystyle y_2 y_1' - y_1 y_2'|_0^L + (\lambda_1-\lambda_2) \int_0^L y_1 y_2dx = 0,

    then sub in your BC's.
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  3. #3
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    To start off, when this question asks without determining the explicit form of both y_1 and y_2, what does that mean exactly? In other words, what is the explicit form?

    Anyway we have....

    y_2(L)y_1'(L) - y_1(L)y_2'(L) - y_2(0)y_1'(0) + y_1(0)y_2'(0) + (\lambda_1 - \lambda_2) \int_0^L y_1y_2 \ dx = 0

    Since all these BC's equal zero we therefore have

    \int_0^L y_1 y_2 \ dx = 0 if \lambda_1 \neq \lambda_2

    This is right, right?
    Thanks very much for your help.
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  4. #4
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    Exactly!
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