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Homework Help: Inverse laplace transform (easy one)

  1. Feb 3, 2010 #1
    I am asked to find inverse laplace of (3s+5)/(s^2-6s+25)

    I do partial frac and get (3s)/((s-3)^2+16) + (5/(s-3)^2+16)

    I do the translation and get [(3s)/s^2+16 + 5/(s^2+16)]*e^3t

    My final answer is e^3t[3cos(4t) + 5/4sin(4t)]

    but the book says the sin term should be 7/2sin(4t)? I say 5/4 because a 4 needs to be in numerator to make use of the table for k/(s^2+k^2) so if you multiply 5 by (5/4) the 4 can be in numerator as 4/(s^2+16) because it will cancel and be the original 5.

    I just need to know how the 7/2 gets in the sin term and why its not (5/4)?
     
  2. jcsd
  3. Feb 3, 2010 #2

    vela

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    You didn't handle the shift correctly. The bottom has (s-3) in it, but the top doesn't for the cosine term. Fixing that will probably fix the sine term.
     
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