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Math Help - Method of Undetermined Coefficients

  1. #1
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    Method of Undetermined Coefficients

    For the equation y''+4y=t^2+3e^t: I have split up the particular solution into two parts, so that I have

    Y_p=At^2+Bt+C=t^2 and Y_{p2}=Ae^t=3e^t

    which gives me

    Y_p=\frac{1}{4}t^2 and Y_{p2}=\frac{1}{5}e^t

    The solution of the corresponding homogeneous equation is of course y=c_1cos(2t)+c_2sin(2t). So for my answer I get \frac{1}{4}t^2+\frac{1}{5}e^t+\frac{1}{5}cos(2t)+\  frac{9}{10}sin(2t)

    Where did I go wrong?
    Last edited by petition Edgecombe; November 13th 2008 at 06:50 PM.
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  2. #2
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    Quote Originally Posted by petition Edgecombe View Post
    For the equation y''+4y=t^2+3e^t: I have split up the particular solution into two parts, so that I have

    Y_p=At^2+Bt+C=t^2 and Y_{p2}=Ae^t=3e^t Mr F says: It's {\color{red}Y_{p2} = \frac{3}{5} e^t}. I realise you've got it almost correct down below, but it's still careless here.

    which gives me

    Y_p=\frac{1}{4}t^2 Mr F says: This is wrong. Show all your working on how you got it (when you do this you will probably realise for yourself why it's wrong).

    and Y_{p2}=\frac{1}{5}e^t Mr F says: Wrong. Show your working.

    The solution of the corresponding homogeneous equation is of course y=c_1cos(2t)+c_2sin(2t). So for my answer I get \frac{1}{4}t^2+\frac{1}{5}e^t+\frac{1}{5}cos(2t)+\  frac{9}{10}sin(2t)

    Where did I go wrong?
    ..
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    and Mr F says: It's . I realise you've got it almost correct down below, but it's still careless here.
    I did have \frac{3}{5} e^t, but I forgot how I got it, so I changed it (that's why I edited my post).

    I see what I did there. I included the 3 in the initial guess so that it looked like
    y_{p2}=3Ae^t instead of y_{p2}=Ae^t

    Mr F says: This is wrong. Show all your working on how you got it (when you do this you will probably realise for yourself why it's wrong).
    I never really learned about equating coefficients. I briefly used it in calculus for partial fraction decomposition, that's it.

    y_{p}=At^2+Bt+C

    y_{p}'=2At+B

    y_{p}''=2A

    2A+4At^2+4Bt+4C=t^2+0t+0c

    4At^2=t^2

    A=1/4


    4Bt=0t

    B=0


    C=0c

    C=0
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  4. #4
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    Quote Originally Posted by petition Edgecombe View Post
    I did have \frac{3}{5} e^t, but I forgot how I got it, so I changed it (that's why I edited my post).

    I see what I did there. I included the 3 in the initial guess so that it looked like
    y_{p2}=3Ae^t instead of y_{p2}=Ae^t

    I never really learned about equating coefficients. I briefly used it in calculus for partial fraction decomposition, that's it.

    y_{p}=At^2+Bt+C

    y_{p}'=2At+B

    y_{p}''=2A

    2A+4At^2+4Bt+4C=t^2+0t+0c Mr F says: This can be re-arranged as 4At^2 + 4Bt + (2A + 4C) = t^2 + 0t + 0. The constant term on the left hand side is what you've failed to see properly ..... See below.

    4At^2=t^2

    A=1/4


    4Bt=0t

    B=0


    C=0c Mr F says: Wrong.

    This is where I knew the mistake would be. I was hoping you would have realised it while typing your work out. The correct equation is 2A + 4C = 0. Substitute A = 1/4 and solve for C.

    C = 0 Mr F says: No. C = ....
    ..
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  5. #5
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    c=-\frac{1}{8}

    And c_1=-\frac{19}{40}

    So my problem was with the method of equating coefficients, as I had thought.
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  6. #6
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    y=c_1cos(2t)+c_2sin(2t) + \frac{1}{4}t^2 - \frac{1}{8} + \frac{3}{5}e^t
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