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Laplace transform

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the Laplace transform of the following function
    t2 - 2t

    3. The attempt at a solution
    [itex]\[\begin{gathered}
    f\left( t \right) = {t^2} - 2t \hfill \\
    l\left( {f\left( t \right)} \right) = F\left( s \right) = \int_0^\infty {{e^{ - st}} \cdot \left( {{t^2} - 2t} \right)ds} \hfill \\
    \Leftrightarrow \hfill \\
    = \int_0^\infty {{e^{ - st}}{t^2} - 2t{e^{ - st}}ds} \hfill \\
    = \int_0^\infty {{e^{ - st}}{t^2}} ds - \int_0^\infty {2t{e^{ - st}}} ds \hfill \\
    = \int_0^\infty {{e^{ - st}}{t^2}} ds - \int_0^\infty {2t{e^{ - st}}} ds \hfill \\
    {\text{Integration by parts}} \hfill \\
    \int_0^\infty {{e^{ - st}}{t^2}} ds \hfill \\
    \int_{}^{} {uv' = uv - \int_{}^{} {u'v} } \hfill \\
    v'\left( t \right) = {e^{ - st}} \Rightarrow v = \frac{{ - 1}}{s}{e^{ - st}} \hfill \\
    u\left( t \right) = {t^2} \Rightarrow u'\left( t \right) = 2t \hfill \\
    \int_{}^{} {uv' = uv - \int_{}^{} {u'v} } \hfill \\
    \int_0^\infty {{e^{ - st}}{t^2}} ds = \frac{{ - 1}}{s}{e^{ - st}} \cdot {t^2} - \int_{}^{} {2t \cdot } \frac{{ - 1}}{s}{e^{ - st}}ds = \frac{{ - 1}}{s}{e^{ - st}} \cdot {t^2} - \frac{{ - 2}}{s}\int_{}^{} {t \cdot } {e^{ - st}}ds \hfill \\
    \end{gathered} \][/itex]


    am i doing it correctly?
     
    Last edited: Feb 11, 2013
  2. jcsd
  3. Feb 11, 2013 #2
    You're supposed to integrate with respect to ##t##, not ##s##! Switch the ##ds## with ##dt## and start over.
     
  4. Feb 11, 2013 #3
    Ahh.. It makes sense now.
     
  5. Feb 11, 2013 #4
    Could you tell me what i am doing wrong with this one..

    Think i got it..
     
    Last edited: Feb 11, 2013
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