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Inverse Laplace involving heaviside function

  1. Dec 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Give the inverse Laplace transform of F(s) = (-3/s) + (e^-4s)/(s^2) + (3e^-4s)/s

    2. Relevant equations
    Inverse Laplace [e^(-cs) F(s)] = f(x-c)u(x-c)

    3. The attempt at a solution

    I'll break this into 3 parts.

    Part 1 - (-3/s)
    -3/s = -3(1/s) -> inverse laplace of -3(1/s) = -3

    Part 2 - (e^(-4s))/(s^2)
    By looking at the e component, I see that in this case "c" = 4.

    Rearrange to (e^(-4s)) (1/s^2)
    So the inverse laplace = u(x-4)*x [ u(x-4) being for the heaviside component and x being the inverse laplace of 1/s^2]

    Part 3 - (3e^-4s / s)

    e component still says that c = 4.
    Rearrange into 3(e^-4s) (1/s)
    Inverse laplace becomes 3 * u(x-4) * 1 = 3u(x-4)

    So MY answer would be that the inverse laplace of the complete F(s) is

    -3 + u(x-4)x + 3u(x-4)

    The CORRECT answer that wolfram spat out and that is an answer choice on the online practice test is
    f(x) = u(x-4)x - 3 - u(x-4)

    I see the similarities but that makes it even more confusing to me how they got
    -u(x-4) instead of +3u(x-4). I don't see how that works out.
    Any and all help is greatly appreciated.
     
  2. jcsd
  3. Dec 14, 2014 #2

    Ray Vickson

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    You should have -3 + u(x-4)(x-4) + 3u(x-4)
     
  4. Dec 14, 2014 #3

    Oh ok, I see it now. That's a careless mistake. So when fixing that , the u(x-4)(x-4) becomes u(x-4)x - u(x-4)4.
    Combining it all together you get -3 + u(x-4)x -1u(x-4).

    Thank you for your help!
     
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