1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook