1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Laplace transforms
  1. Laplace Transformation (Replies: 5)

  2. Laplace Transform (Replies: 1)

  3. Laplace Transforms (Replies: 4)

  4. Laplace transformation (Replies: 2)

  5. Laplace transform (Replies: 3)