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Math Help - Inverse Laplace Transforms

  1. #1
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    Inverse Laplace Transforms

    Determine the inverse laplace transforms of

    s+7/s^2+4s+3

    and

    3s-8/s^2+4
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  2. #2
    Super Member Matt Westwood's Avatar
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    1: Complete the square down below and you can split it into two LTs of the form:
    \frac {s-b}{(s-b)^2 + a^2}, \frac {1}{(s-b)^2 + a^2}
    multiplied by whatever constants. Or they could be - a^2 on the bottom. Whatever, you use your inverse tables and get e^{bt} \sin a t / a or it could be cos or sinh or cosh, whatever, see what you get when you rip it apart.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    You can simply factor:

    1. (s+7)/[s^2+4s+3] = (s+7)/[(s+3)(s+1)]

    = (s+3)/[[(s+3)(s+1)] + 4/[(s+3)(s+1)]

    = 1/(s+1) + 4/[(s+3)(s+1)]


    y(t) = e^(-t ) + 4 (e^(-3t)-e^(-t))/-2

    y(t) = 3e^(-t) - 2e^(-3t)

    2. write as 3s/(s^2+4) - 8 /(s^2+4) = 3{s/(s^2+4)} - 4{2 /(s^2+4)}

    y(t) = 3cos(2t) - 4sin(2t)
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