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Laplace transforms

  1. May 9, 2009 #1
    Im trying to find the laplace transform of t^1 x e^(3t)
    but looking it at the table, it looks like there's two different possible solutions for it.
    one is for t^(n) x f(t)
    and the other is for e^(at) x f(t)
    which one do i choose?
     
  2. jcsd
  3. May 9, 2009 #2

    Defennder

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    Do you get different answers for both? In this case, the table I'm using would point to the one for e^(at)f(t).
     
  4. May 9, 2009 #3
    i do get two different answers.
    doesnt ur table have t^(n) x f(t)? wouldnt that also satisfy it?
     
  5. May 10, 2009 #4

    Defennder

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    My table doesn't have that one. What are your answers for each?
     
  6. May 10, 2009 #5
    this is what the table has.
    http://users.on.net/~rohanlal/2222.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  7. May 10, 2009 #6

    HallsofIvy

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    Okay, in your case n= 1 so it is just -F(s). What is F(s), the Laplace transform of e3t?

    Of course, you could use the basic formula for Laplace transform:
    [tex]L(s)= \int_0^\infty t e^{3t}e^{-st}dt[/tex]
    using integration by parts.
     
  8. May 10, 2009 #7
    so i choose the other one for t^1 x e^(3t), the one that defender mentioned, because n = 1?
    so if i used the one in my previous post, that would that be incorrect?
     
  9. May 10, 2009 #8
    You could also use the other property, namely that

    [tex]\displaystyle \mathcal{L}[e^{at}f(t)] = F(s-a),[/tex]​

    where [tex]F(s) = \mathcal{L}[f(t)].[/tex] Using this, you only need to get the Laplace transform of [tex]t[/tex], and evaluate it at [tex]s-3[/tex]. You should get the same result with both properties.

    Good luck.
     
  10. May 10, 2009 #9

    HallsofIvy

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    The point is that all three methods:
    a)[tex]\displaystyle \mathcal{L}[e^{at}f(t)] = F(s-a),[/tex]
    where F(s) is the Laplace transform of t.

    b)[tex]\displaystyle \mathcal{L}[tf(t)]= -F'(s)[/tex]
    where F(s) is the Laplace transform of [itex]e^{3t}[/itex].

    c)[tex]\displaystyle \mathcal{L}[te^{at}]= \int_0^\infty te^{(-s+3)t}dt[/tex]

    will give the same result.

    It would be a good exercise to try each method and see.
     
  11. May 10, 2009 #10
    for the laplace transform of t^2 x e^(3t) (n is greater than 1)
    would http://users.on.net/~rohanlal/2222.jpg [Broken] be the correct one to use?
     
    Last edited by a moderator: May 4, 2017
  12. May 10, 2009 #11
    All three ways are correct, but I personally think the exponential property is the quickest, if you already know the Laplace transforms of polynomials.
     
  13. May 10, 2009 #12
    can you reread my previous post, i put in the wrong url for the image.
     
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