Results 1 to 6 of 6

Math Help - Laplace Transform

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    96

    Laplace Transform

    x'' + 4x' -8x = 0
    x(0) = 0 x'(0) = 1

    So I set up the transform, and got it to (s^2)L(x) +4sL(x)-8L(x) = 1
    From here, x(t) = L^(-1) of 1/(s^2+4s-8)
    I did a sort of completing the square, to get 1/((s+2)^2-12), which almost fits the equation for a laplace transform of e^(at)cos(bt), but now this b value would be the square root of a negative number...where did I go wrong? Any clarification would be great, thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    You have constant coefficients. I would substitute x(t)=e^{mt} instead.

    Then you only need to solve m^2+4m-8=0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mistykz View Post
    x'' + 4x' -8x = 0
    x(0) = 0 x'(0) = 1

    So I set up the transform, and got it to (s^2)L(x) +4sL(x)-8L(x) = 1
    From here, x(t) = L^(-1) of 1/(s^2+4s-8)
    I did a sort of completing the square, to get 1/((s+2)^2-12), which almost fits the equation for a laplace transform of e^(at)cos(bt), but now this b value would be the square root of a negative number...where did I go wrong? Any clarification would be great, thanks!
    Factorise the denominator and then express in partial fraction form.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2007
    Posts
    96
    How would I factor the bottom? I don't see anything that works without getting complex numbers..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mistykz View Post
    How would I factor the bottom? I don't see anything that works without getting complex numbers..
    Last time I checked, (s+2)^2-12 could be factorised using the difference of two squares formula.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2007
    Posts
    96
    Ah, I didn't think of using that...got it! Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Laplace transform and Fourier transform what is the different?
    Posted in the Advanced Applied Math Forum
    Replies: 8
    Last Post: December 29th 2010, 10:51 PM
  2. Laplace transform
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: December 27th 2010, 10:55 AM
  3. Laplace transform
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 4th 2009, 03:21 AM
  4. Laplace Transform Help
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 9th 2009, 10:50 PM
  5. LaPlace Transform
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 12th 2008, 10:35 PM

Search Tags


/mathhelpforum @mathhelpforum