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Homework Help: Where did I go wrong? [Laplace Transform]

  1. Sep 11, 2005 #1
    [tex] L[/tex] is Laplace Transform operator.
    Question is:

    Let [tex] f(t) = t^2[/tex]. Derive [tex]L[f][/tex] from [tex]L[1][/tex]

    So I know [tex]f(1) =1[/tex] and [tex] L[1] = \frac{1}{s}[/tex]

    Carrying out the Transform...

    [tex]L[f] = \int_{0}^{\infty} e^{-st}t^2 dt[/tex]

    Integration by parts [tex] u = t^2, dv = e^{-st} dt [/tex]

    I do the integration, the uv terms go to zero. after I clean up I get

    [tex] - \int_{0}^{\infty} \frac{e^{-st}}{-s} 2t\ dt[/tex]

    Doing IBP again. [tex] u = t,\ dv = e^{-st} dt[/tex] Rinse and repeat.

    [tex] - \frac{2}{s} \int_{0}^{\infty} \frac{e^{-st}}{-s} (1) dt \\
    = - \frac{2}{s} \left( \frac{e^{-st}}{s^2}\right)_{0}^{\infty} = - \frac{2}{s^3} [/tex]

    Ok. So is that right? I don't think the negative is right at the end. But what the bigger thing that worries me is how do I use L[1] in this? Or is that a different approach?
    Last edited: Sep 11, 2005
  2. jcsd
  3. Sep 11, 2005 #2


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    Homework Helper

    It seems something like this is wanted
    [tex]L[t^a \ f(t)]=\left(-\frac{d}{ds}\right)^aL[f][/tex]
    [tex]L[t^2]=L[t^2 \ 1]=\left(-\frac{d}{ds}\right)^2L[1]=\frac{d^2}{{ds}^2} \ \frac{1}{s}=\frac{2}{s^3}[/tex]
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