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Math Help - Help with inverse laplace

  1. #1
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    Help with inverse laplace

    Hi, i have attempted to solve the following:

    L^-1: 3/(4s^2-9)

    and got

    (1/8)sinh(3t/2).

    I first moved a (1/4) outside of the laplace in order to get the bottom closer to the laplace of sinh(kt) = k/(s^2-k^2). I then pulled a (1/2) out to get the 3 on the top to be 3/2 so that it would fit the equation. Is this correct? Thanks for any help!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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     \frac{3}{4s^2-9}=\frac14\cdot\frac{3}{s^2-9/4} = \frac{3}{4}\cdot\frac23\cdot\frac{3/2}{s^2-(3/2)^2} = \frac12\cdot\frac{3/2}{s^2-(3/2)^2}

    Thus  \mathcal{L}^{-1}\left(\frac{3}{4s^2-9}\right) = \frac12\cdot\sinh\left(\frac{3t}{2}\right)

    I believe your mistake is your last step where you pulled out a  \frac12 when you should've pulled out a  2 .
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  3. #3
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    ok, that makes sense. Thanks for the help!
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