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Math Help - Variation of Parameters

  1. #1
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    Variation of Parameters

    Hello everyone. I have tried this problem a couple times and seem to get it wrong.

    Find a particular solution to
    Any help would be appreciated
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  2. #2
    Member Mauritzvdworm's Avatar
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    its a bit tedeous...

    firstly we have to consider the homogenous equation

    y^{\prime\prime}-4y^{\prime}+4=0

    which has solution

    y_{c}(t)=c_1 e^{2t}+c_2 t e^{2t}

    using the variation of parameters theorem we see that

    y_{p}(t)=-e^{2t}\int\frac{te^{2t}f(t)}{W(t)}dt+te^{2t}\int\f  rac{e^{2t}f(t)}{W(t)}dt

    with W(t)=e^{4x} calculted using the Wronskian determinant

    so we can then see that

    y_{p}(t)=-e^{2x}\int\frac{te^{2t}f(t)}{e^{4t}}dt+te^{2t}\int  \frac{e^{2t}f(t)}{e^{4t}}dt

    and after substituting f(t)=\frac{e^{2t}}{t^{2}+1} and turning the crank we end up with

    y_{p}(t)=-e^{2t}\int\frac{t}{t^{2}+1}dt+te^{2t}\int\frac{1}{  t^{2}+1}dt

    which is simple to integrate, we then find the final answer of

    y_{p}(t)=\frac{1}{2}e^{2t}\ln \left(t^{2}+1\right)+te^{2t}\tan^{-1}\left(x\right)+C

    The final anwer will then just be the superposition of the particular and complimentary solutions
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  3. #3
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    Thanks for the help!
    I assume that the arctan(x) is supposed to be arctan(t) eh? I still get the wrong answer with that, but it does give me something to build off of. Thanks again!
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