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Taking the inverse laplace of this?

  1. May 28, 2012 #1
    How do I take the inverse laplace transform of something that looks like this? It's part of a larger piecewise-defined second order differential equation, but this is the part I'm stuck on.

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

    Completing the square doesn't work for the bottom, so I figure I need to separate the whole thing into two separate fractions. I still can't figure out how to take the inverse laplace of either resulting fraction, however.
     
    Last edited: May 28, 2012
  2. jcsd
  3. May 28, 2012 #2
    You can simplify the numerator by breaking across the difference, using the rule [itex]\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}[/itex].

    For the denominator, just looking at it I can't see why you can't complete the square, if I'm wrong, please show us why.

    Also, if you come to something in your steps that you're not sure how to take the inverse Laplace of, show us, we might be able to give an idea.
     
  4. May 28, 2012 #3
    I see.

    So, doing so would give me

    [itex]\frac{-s}{(s+1/8)^{2}+63/64}+\frac{-1/4}{(s+1/8)^{2}+63/64}[/itex]

    The second fraction would be simple to use with the [itex]e^{at} sin(bt)[/itex] identity since it's just a coeffecient on top, but what about the first?
     
  5. May 28, 2012 #4

    hunt_mat

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    This is just standard look up tables, look it up on wikipedia.
     
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