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Solution of Initial Value Problems, Laplace Transform

  1. Apr 29, 2005 #1
    Use the Laplace transform to solve the given initial value problem.

    y"+[w^(2)]y=cos2t, w^(2) does not equal 4; y(0)=1, y'(0)=0

    I tried doing the problem, and I got up to Y(s)=[(s^(3)+5s]/[s^(2)+w^(2)],
    which hopefully is correct. Now I'm having trouble using the Laplace
    transforms to finish solving the problem.

    Thanks for your help!
  2. jcsd
  3. Apr 30, 2005 #2


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    What's the integral u have to solve to revert the Laplace transform...?

  4. May 3, 2005 #3


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    Taking the Laplace transform of both sides:

    [tex]\mathcal{L}\{y^{''}+\omega^2 y=Cos[2t]\}[/tex]





    Solving for f(s) yields:


    Now, you can reduce this using partial fractions with quadratic denominators:


    When solving for A,B,C,and D, you'll come up against a homogeneous system for B and D, and the constraint imposed on [itex]\omega^2[/itex] will force you to make a conclusion about what B and D can be. Try solving it to completion and report the results here. If you don't, I'll wrap it up tomorrow.
  5. May 4, 2005 #4


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    Just some wrap-up:

    Solving for A and C:



    The expressions for B and D are:



    You know, I'm not sure about the part below deciding about B and D: Really, I think B and D HAVE to be zero in this case. Someone can correct me if my rational is not so.

    Noting that a homogeneous equation has a non-zero solution iff determinant=0, we get:


    However, since constraint above restricts such, only the trival solution is allowed. Thus:


    Finally, taking the Laplace transform of both sides:

    [tex]y(t)=ACos[2t]+C(Cos[\omega t])[/tex]

    A plot is attached.

    Attached Files:

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