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Quick laplace transform question

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    So I know 1/(s-a)=e^(a1), but why is say, 2/((s+4)^2) equal to 2xe^-4x? Do I just simply add an X if the numeration is a constant other than 1?
     
  2. jcsd
  3. Oct 19, 2015 #2
    Why don't you evaluate the Laplace transform of the expression and see what you get?
     
  4. Oct 19, 2015 #3
    2*e^(-4*x) is what I get, I don't know where the other x came from.

    Well I mean I know it's because it's squared, I just don't see how my laplace transform formula is giving me that
     
  5. Oct 19, 2015 #4
    Like I said, evaluate the LT of the expression with the x's and report back with what you got.
     
  6. Oct 19, 2015 #5
    L(2x*e^(-4x)) = 2x* 1/(s-4)
     
  7. Oct 19, 2015 #6
    Am I suppose to treat it as two different Laplace transforms combined? Not sure how that would work but I know the two individual formulas below
    2/((s+4)^2)

    1/s^n = t^n-1/(n-1)! = x*2
    and
    1/(s-a) = e^-4x
     
  8. Oct 20, 2015 #7
    Let's see you do the integration. How come you didn't include the leading x in the integration?

    Chet
     
  9. Oct 20, 2015 #8

    HallsofIvy

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    By the way, your original statements, that "1/(s-a)=e^(at)" and that "2/((s+4)^2) is equal to 2xe^-4x" are non-sense. You knew that when you wrote them, didn't you? What you meant was that "1/(s- a)" is the Laplace Transform" of "e^(at)" and that "2/((s+ 4)^2)" is the Laplace Transform of "2xe^(-x)". It is always better to say what you mean!
     
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