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Math Help - Laplace Transform [LEVEL 2 MATHS]

  1. #1
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    Question Laplace Transform [LEVEL 2 MATHS]

    SOLVED Question:


    My workings so far :



    As you can see, I have a problem finding the laplace transform of the middle equation of the last image. Would you please give me tips/advise as how to obtain the answer? Thank you!
    Last edited by Salcybercat; May 1st 2011 at 08:33 AM. Reason: Question is now solved
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  2. #2
    Behold, the power of SARDINES!
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    You have an error in your Laplace transform

    x''-4x'+4x=4, \quad x(0)=0 x'(0)=-2

    s^2X-s\cdot 0-(-2)-4(sX-0)+4X=\frac{4}{s}

    (s^2-4s+4)X=-2+\frac{4}{s}=\frac{-2s+4}{s}

    X=\frac{-2s+4}{s(s-2)^2}=-\frac{-2(s-2)}{s(s-2)^2}=\frac{-2}{s(s-2)}

    Now from here you can use partial fractions or notice that

    \frac{-2}{s(s-2)}=\frac{-s+(s-2)}{s(s-2)}=\frac{-s}{s(s-2)}+\frac{s-2}{s(s-2)}=\frac{-1}{s-2}+\frac{1}{s}

    both of these are known Laplace transforms

    -e^{2t}+1
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  3. #3
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    Thank you for showing me my mistake for the Laplace Transform! I've got the answer now.
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  4. #4
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    Could you help me with another equation:
    I don't think I can solve it using partial fractions?

    EDIT: I don't know if you've noticed this post but I managed to solve it
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  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    You can avoid partial fractions by using

    \mathcal{L}\left\{t^nf(t) \right\}=(-1)^n\frac{d^n}{ds^n}F(s)

     V(s)=\frac{8}{s^2+16}+\frac{8s}{(s^2+16)^2}=2\frac  {4}{s^2+4^2}-\frac{d}{ds}\left( \frac{4}{s^2+4^2}\right)

    Now finding the inverse gives

    2\sin(4t)+t\sin(4t)
    Last edited by TheEmptySet; May 1st 2011 at 09:56 AM. Reason: missing minus sign
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