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Thread: Differential equation

  1. #1
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    Differential equation

    For what nonzero values of k does the function y=sin(kt) satisfy the differential equation y'' + 9y = 0?
    Last edited by mr fantastic; Oct 21st 2009 at 05:16 PM. Reason: Restored deleted question
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  2. #2
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    Quote Originally Posted by jamessmith View Post
    For what nonzero values of k does the function y=sin(kt) satisfy the differential equation y'' + 9y = 0?

    $\displaystyle y=\sin(kt)$

    $\displaystyle

    y'=k\cos(kt)
    $

    $\displaystyle

    y''= -k^2\sin(kt)
    $

    $\displaystyle y''+9y=0$

    $\displaystyle
    -k^2\sin(kt)+9\sin(kt)=0
    $

    And so $\displaystyle \sin(kt)(-k^2+9)=0$

    And you can solve that for k right?
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  3. #3
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    Quote Originally Posted by jamessmith View Post
    For what nonzero values of k does the function y=sin(kt) satisfy the differential equation y'' + 9y = 0?
    Do you know how to get started? First just find the derivatives of y.

    $\displaystyle y'=kcos(kt)$

    $\displaystyle y''=-k^2sin(kt)$

    So just substitute into the differential equation:

    $\displaystyle -k^2sin(kt)+9sin(kt)=0$

    so $\displaystyle k=3$ is one:

    $\displaystyle sin(kt)=0$ is the other possible solution. Just find the k
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