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Laplace Transformation

  1. Sep 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Obtain the Laplace transformation of the fucntion defined by

    f(t) = 0 t<0

    = t2e-at t>=0


    2. Relevant equations



    3. The attempt at a solution

    I'm a little unsure of what I'm doing here, so bear with me.

    L {t2e-at} = ∫inf0 t2e-at dt

    = ∫0inf t2e-(a+s)tdt

    How do I integrate this? I tried using my TI-89 but it told me it is undefined. Any help would be greatly appreciated.

    EDIT: The complete solution is required, which I think means I can't just take it out of the Laplace tables
     
  2. jcsd
  3. Sep 4, 2013 #2

    Ray Vickson

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    First of all: never, never write what you did above, which was
    [tex] \int_0^{\infty} t^2 e^{-at} \, dt = \int_0^{\infty} t^2 e^{-(a+s)t} \, dt. [/tex]
    This is only true if s= 0.

    Anyway, for your final (correct) integral, change variables from t to x = (a+s)t (assuming s > -a).

    I strongly recommend that you put away the TI-89 until after you have learned how to do these problems, or at least restrict its use to numerical computation.
     
  4. Sep 4, 2013 #3
    Sorry, I missed out a part.

    There should have been an e-st

    But I don't understand your point about changing the variable from t to x. Could you explain further please?
     
  5. Sep 4, 2013 #4

    Ray Vickson

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    I assume you have taken integration already. If so, you already studied change-of-variable methods (or so I hope).

    If you have not done integration yet you would need a longer explanation than I am prepared to give here: we would need to go over material that takes several weeks to present in coursework! However, such information is widely available in books and on-line.
     
  6. Sep 4, 2013 #5

    Dick

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    Yes, it's just integration. But the trick you really need to handle the t^2 factor is integration by parts. By all means, review integration.
     
  7. Sep 5, 2013 #6

    vanhees71

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    It's much simpler to evaluate
    [tex]\tilde{g}(s)=\int_0^{\infty} \mathrm{d} t \exp[-(a+s) t][/tex]
    and then taking the 2nd derivative wrt. [itex]s[/itex]. It's very easy to see that
    [tex]\tilde{f}(x)=\tilde{g}''(s).[/tex]
     
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