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Math Help - Second order non homogeneous diff. equation at constant coefficients

  1. #1
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    Second order non homogeneous diff. equation at constant coefficients

    Hi there. I had some trouble trying to solve this:

    y''+y=\cos x +3\sin \color{red}2x (1)

    At first I just found the solution for the homogeneous equation:
    y_h=e^{\lambda x} \rightarrow \lambda^2+1=0 \rightarrow \lambda_1,\lambda_2=\pm i

    Then y_h=C_1\cos x+ C_2 \sin x

    So I've tried to find the particular solution. I thought I should suggest an equation like:
    y_p=A\cos x +b\sin x+ C\cos 2x +D\sin 2x
    y_p'=-A\sin x +B\cos x - 2C\sin 2x + 2D\cos 2x
    y_p''=-A\cos x -B\sin x - 4C\cos 2x - 4D\sin 2x
    But then, when I've tried to find the undetermined coefficients for y_p coefficients replacing in (1):

    -A\cos x -B\sin x - 4C\cos 2x - 4D\sin 2x+A\cos x +b\sin x+ C\cos 2x +D\sin 2x=\cos x +3\sin x \rightarrow -3C\cos 2x -3D\sin 2x=cos(x)+3\sin 2x
    Then A=0,B=0,C=0 and D=-1.

    So the general solution should be
    y(x)=y_h+y_p=C_1\cos x+ C_2 \sin x-\sin 2x
    But with wolfram alpha I've corroborated my solution is wrong: y''+y=cos(x)+3 sin(2x) - Wolfram|Alpha

    So, where is the mistake and how should I do this?

    I think I've found unless one mistake, just noted it. I should use y_p=Ax \cos x +B x \sin x + C\sin 2x+ D\cos 2x instead of y_p=A\cos x +b\sin x+ C\cos 2x +D\sin 2x, because the first two terms are linearly dependent with the homogeneous solution, right?
    Last edited by Ulysses; May 3rd 2011 at 10:11 AM. Reason: Correction
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  2. #2
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    Don't use any terms like \cos(2x). You're just creating unnecessary work for yourself. Do multiply the homogeneous solution by powers of x, though, to get linearly independent solution candidates. I think you'll only need one power of x. If that doesn't work out, try two.
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  3. #3
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    I'm sorry, I've made a mistake* in (1). It was 2x instead of x. I think I need the additional terms. Am I right?

    *Just corrected.
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  4. #4
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    Try

    y_{p}=Ax\cos(x)+Bx\sin(x).
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