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Laplace transform

  1. Nov 22, 2006 #1
    My understanding of the laplace trasnform isnt so great so i would appreciate some help with this question please:

    find the laplace transform of (t+2)sinh2t

    now i know the laplace transform of sinh2t is 2/(s^2 -4) as this is a standard rule......

    looking through textbooks they show the multiplication by t^n rule is needed and i found that the laplace transform of t (sin kt) = 2ks/(s^2 + k^2) ^2

    how do i apply this to my equation..... :confused:
     
  2. jcsd
  3. Nov 22, 2006 #2

    arildno

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    Dearly Missed

    Do not double post! :grumpy:
     
  4. Nov 22, 2006 #3
    the Laplace transform of (t+2) is [tex] 1/s^{2}+2/s [/tex]

    If you multiply f(t) by exp(-at) then there's a shift so F(s+a) and

    [tex] 2sinh(ax)=e^{xa}+e^{-ax} [/tex]

    then next is just hand-work...
     
  5. Nov 22, 2006 #4
    thanks for the tips,

    can u find the LT of 2sinh2t and the tsinh2t and add them together which gives 4/(s^2 - 4) + 4s/(s^2 + 4) ^2

    is this correct?
     
  6. Nov 23, 2006 #5

    HallsofIvy

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    Yes. The definition of the Laplace tranform is:
    [tex]L(f(t))= \int_0^\infty f(s)e^{-st}dt[/tex]
    Since
    [tex]\int (f(x)+ g(x))dx= \int f(x)dx+ \int g(x)dx[/tex]
    It follows that you can add Laplace transforms.

    It should be easy to integrate
    [tex]\int_0^\infty (t+2)sinh t dt[/tex]
    (Break it into two integrals and use integration by parts)
    as an exercise.
     
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