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Laplace Inverse of a solution

  1. May 4, 2012 #1
    1. The problem statement, all variables and given/known data

    laplace inverse of (40/(s^2+4s+5)^2)?
    2. Relevant equations

    I completed the square in the denominator to get 40/((s+2)^2+1)^2
    I know that I will get cosines and sines from the shape of it in the laplace inverse; however I'm stuck.

    3. The attempt at a solution
  2. jcsd
  3. May 4, 2012 #2
    I get the follwing result:
    [itex] 20e^{-2t}(\sin(t)-t\cos(t))[/itex]
    But couldn't find any formula, so i did it with Maple 15
  4. May 4, 2012 #3

    Ray Vickson

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    It will be a lot easier if you first factor the expression p = s^2 + 4s + 5, then convert your expression 1/p^2 to partial fractions.

  5. May 4, 2012 #4
    dikmikkel This answer is correct; however I don't know how to get there.
    Ray Vickson, we are not used to getting complex numbers.
  6. May 4, 2012 #5

    Ray Vickson

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    You should get used to it; they are part of a standard toolkit and are used routinely in Physics, Engineering and Applied Math. However, if you don't want to use complex quantities you can use the convolution theorem instead: first get the inverse Laplace transform of 1/(s^2 + 4s + 5), then find that of 1/(s^2 + 4s + 5)^2 by convolution.

    Last edited: May 4, 2012
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