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Inverse Laplace transform

  1. May 17, 2009 #1

    tony873004

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    L-1(1/(s+2)3)
    I don't see this one in the table. How do I solve the inverse Laplace transform?

    I know from class notes that the answer is (1/2) t2e-2t
    But I don't know to get it.
    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 17, 2009 #2

    gabbagabbahey

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    I assume that [tex]\mathcal{L}^{-1}\left[ \frac{1}{s^3} \right][/tex] is in your table?

    If so, just begin by applying the frequency shift rule:

    [tex]\mathcal{L}^{-1}\left[ f(s+a) \right]=e^{-at}\mathcal{L}^{-1}\left[ f(s) \right][/tex]
     
  4. May 17, 2009 #3

    tony873004

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    ok, thanks. That gives me
    [tex]\frac{1}{(e^{-2t}L^{-1}(s))^3}[/tex]

    And the table gives for [tex]L^{-1}\frac{1}{s^3}[/tex] as [tex]\frac{t^2}{2}[/tex]

    What do I do from here?
     
  5. May 17, 2009 #4

    gabbagabbahey

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    How do you get that ugly expression?

    You should get

    [tex]\mathcal{L}^{-1}\left[ \frac{1}{(s+2)^3} \right]=e^{-2t}\mathcal{L}^{-1}\left[ \frac{1}{s^3} \right]
    [/tex]
     
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