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Math Help - Periodic solution

  1. #1
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    Periodic solution

    I'm trying to find a periodic solution for the following inhomogeneous equation:
    u'' − u = cos^2(t).
    I'm thinking that in order for the solution to be periodic it must be equal just with the particular solution(for the complementary we put the constants C1=C2=0)
    Also I can write cos^2(t) as cos^2(t)=(1+cos2t)/2 and then I can choose the particular solution to be of the form u_p(x)=-1/2 +Asin(t)+Bcos(t).
    After I diff and substitute into the initial equation I get the answer -1/2 +1/3*cos2t which is not correct.
    How can I solve this?Also if I can leave cos^2(t) as this and find a solution of the form
    Acos^(t)+Bsin^2(t) will it still be correct?
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  2. #2
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    Your approach is right but how did you obtain \frac{1}{3} \cos 2t ?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Because is...

    (1)

    ... imposing ...

    (2)

    ... means that the 'periodic solution' is...

    (3)

    Kind regards

    \chi \sigma
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  4. #4
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    I"m sorry but I don't see how you've obtained 1/10 cost 2t?
    Danny...I know I did wrong but I diff the particular equation and substituted...and I get B=0 and A=-1/3
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  5. #5
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    If u = A \cos 2t then u'' = - 4A \cos 2t so

    u''-u = -5A \cos 2t = \frac{1}{2} \cos 2t giving A = - \frac{1}{10}
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  6. #6
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    ahhhhh...........thanks danny
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