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Math Help - Question about Laplace transforms

  1. #1
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    Question about Laplace transforms

    y''-4y'+5y=sin(t), t \in [0,+\infty), y(0)=0, y'(0)=0
    s^{2}Y(s)-sy(0)-y'(0)-4(sY(s)-y(0))+5\frac{1}{s^{2}}=\frac{1}{s^{2}+1} <=>
    Y(s)(s^{2}-4s)=\frac{1}{s^{2}+1}-\frac{5}{s^{2}} <=>
    Y(s)(s^{2}-4s)=\frac{-4s^{2}-5}{s^{2}(s^{2}+1)} <=>
    Y(s)=\frac{-4s^{2}-5}{s^{2}(s^{2}+1)(s^{2}-4s)} <=>
    Y(s)=\frac{-1}{5(s^{2}+1)}+\frac{5}{4s^{2}}+\frac{21}{80(s+2)}-\frac{21}{80(s-2)} <=>
    y(t)=\frac{-1}{5}sin(t)+\frac{5}{4}t+\frac{21}{80}e^{-2t}-\frac{21}{80}e^{2t}

    Y(s) is the laplace transform of y(t) function.
    Can you please tell me what I did wrong? The answer should be y''-4y'+5y=sin(t), y(0)=0, y'(0)=0 - Wolfram|Alpha .

    EDIT:
    Y(s)=\frac{1}{29}\left ( \frac{-4}{s}+\frac{1}{s^{2}}+\frac{4}{s^{3}}+\frac{s-1}{s^2+1}-\frac{129}{s-4} \right ) which means
    y(t)=\frac{1}{29}\left ( -4+t+2t^{2}+L^{-1}\left \{ \frac{s-1}{s^2+1} \right \} -129e^{4t} \right ) <=>
    y(t)=\frac{1}{29}\left ( -4+t+2t^{2}+cos(t)-sin(t)-129e^{4t} \right )

    but the answer is still wrong...
    Last edited by cristi92; May 8th 2012 at 06:33 AM.
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  2. #2
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    Re: Question about Laplace transforms

    Mistake on the second line. The LT of 5y is 5Y(s).
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  3. #3
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    Re: Question about Laplace transforms

    Now I see, thank you!!!
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