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Y''+4y'+4y = 6.5e^(-2t), I guessed y = At^2*e^(-2t) but still wrong! any ideas?

  1. Feb 24, 2006 #1
    Hello everyone!
    I'm stuck on this problem! as usual.

    Find a particular solution to
    y'' + 4 y' + 4 y = 6.5 e^{-2 t}.

    Here is my work:
    y''+4y'+4y = 6.5e^(-2t)
    r^2+4r+4 = 0;
    r = -2, repeated root
    So i made the guess:

    y = A*t^2*e^(-2t);
    y' = A(2t*e^(-2t) + t^2(-2)*e^(-2t))
    y'' = A(2*e^(-2t)+t^2*4*e^(-2t))


    So i equated co-efficents of e^(-2t) and got:
    2A +4A = 6.5
    6A = 6.5
    A = 13/12

    y = c1*e^(-2t)+c2*t*e^(-2t)+(13/12)*t^2*e^(-2t);
    So as the answer i submitted was:
    [​IMG]
    but it was wrong any ideas why?
    Thanks!
     
    Last edited: Feb 24, 2006
  2. jcsd
  3. Feb 24, 2006 #2

    arildno

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    Your second derivative is wrong; besides, try an expansion y=(At^2+Bt)e^-2t
     
  4. Feb 24, 2006 #3
    Thanks for the help arildno, but i'm still stuck now. I don't seem to know what to do about equating co-efficents, here is my work:
    [​IMG]
     
  5. Feb 24, 2006 #4
    N/m, i got it right now, seems like you only equate e^(-2t) for A's because verything else must go to Zero! i got
    [​IMG]
    Thanks for the help. Quick question though, how did you know to guess:
    y=(At^2+Bt)e^-2t ? and not just y = At^2*e^(-2t)?
     
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