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Laplace transforms and partial fractions

  1. Feb 3, 2006 #1
    Hey guys, Im supposed to find the Laplace transform of a set of ODEs.

    Ive broken it down a bit and Im left with finding the Laplace transform of:

    (-2e^-s)/(s(s+4)(s+1))

    Is this something I have to use partial fractions for? Or is there another way? Im a bit confused.
     
  2. jcsd
  3. Feb 3, 2006 #2

    benorin

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    Do you mean the inverse Laplace transform of (-2e^-s)/(s(s+4)(s+1))?
     
  4. Feb 3, 2006 #3
    Ahh, yeah that would be correct. oops
     
  5. Feb 3, 2006 #4

    benorin

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    If so, factor out the exponential and just deal with the rational part:

    [tex]-\frac{2e^{-s}}{s(s+4)(s+1)}=-2e^{-s}\frac{1}{s(s+4)(s+1)}[/tex]

    So [tex]\frac{1}{s(s+4)(s+1)}=\frac{A}{s}+\frac{B}{s+4}+\frac{C}{s+1}[/tex]
     
  6. Feb 3, 2006 #5
    Ok, so after finding values for A,B,C, do I just use the tables in order to work out the individual bits?
    ie ((-2e^-s)*A)/s etc?
     
  7. Feb 3, 2006 #6

    benorin

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    Cross-multiply to get:

    [tex]1=A(s+4)(s+1)+Bs(s+1)+Cs(s+4)[/tex]

    then plug, say, s=0,-4,-1 into the above to solve for A,B, and C
     
  8. Feb 3, 2006 #7

    benorin

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    yep, use tables if ya got 'em.
     
  9. Feb 3, 2006 #8

    benorin

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    s=0 gives 1=4A or [tex]A=\frac{1}{4}[/tex]

    s=-4 gives 1=12B or [tex]B=\frac{1}{12}[/tex]

    and s=-1 gives 1=-3C or [tex]C=-\frac{1}{3}[/tex]
     
  10. Feb 3, 2006 #9
    Cheers Benorin, partial fractions are no worries, just that exponential that was confusing me, seems pretty simple now.
     
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