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Inverse Laplace Transform

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data
    L-1{(2s2+3)/(s2+3s-4)2}

    3. The attempt at a solution
    I factored the denominator

    f(t)=(2s2+3)/((s-1)(s+4))2

    now I've tried partial fractions to get

    (2s2+3)/((s-1)(s+4))2 = A/(s-1)2 + B(s+4)2

    (2s2+3)=A(s+4)2 + B(s-1)2

    by substitution, s=1 and s=-4

    5=A(25)
    A=1/5

    35=B(25)
    B=7/5

    (1/5) 1/(s-1)2 + (7/5) 1/(s+4)2

    At this point I'm not sure if I am on the right track, but I did start to see some identities that may help.

    1/5 L-1{1/(s-1)2} +7/5 L-1{1/(s+4)2}

    I'm starting to see a pattern for n!\sn+1 and eat

    Am I on the right track, or did I go on a tangent?

    Any help would be appreciated!!!!!!!!!!
     
  2. jcsd
  3. Nov 19, 2015 #2
    You need to take another look at your partial fraction expansion. How do you expand when you have double roots in the denominator?
     
  4. Nov 19, 2015 #3
    I'm not sure if splitting the roots, or can I?
     
  5. Nov 19, 2015 #4
    sorry, I'm not sure if I can split the roots to two rationals.
     
  6. Nov 19, 2015 #5
    Note, when I say "double roots", I mean, how do you expand into separate terms when, for instance, one of the zeroes of the denominator is squared (or cubed, etc.)?
     
  7. Nov 19, 2015 #6
    A polynomial would then be formed, right?
     
  8. Nov 19, 2015 #7
    You may want to look back at your old notes regarding how to expand it using partial fractions. You're missing a couple of terms in your expansion.
     
  9. Nov 19, 2015 #8
    Okay. Thank you, I'll reply as soon as I see my mistakes.
     
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