Results 1 to 1 of 1

Math Help - Simple Inverse Laplace Transform

  1. #1
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19

    Simple Inverse Laplace Transform

    I can see that L^{-1}\{\frac{1}{s(s^2+5)}\}=\frac{1}{5}-\frac{1}{5}cos(\sqrt{5}t),

    because,

    L\{\int_0^tf(\tau)d\tau\}=\frac{F(s)}{s}

    But when I try to use partial fractions L^{-1}\{\frac{1}{s(s^2+5)}\}=L^{-1}\{\frac{A}{s}+\frac{B}{s^2+5}\}

    A(s^2+5)+Bs=1

    s=0\rightarrow A=\frac{1}{5}

    As^2+Bs+5A=1

    This doesn't seem to work at all. Basically this gives inconsistent values of A and a value of zero for B. Am I approaching the method of partial fractions wrong? Since there isn't a squared term on the right, A must be zero right?

    EDIT: LOL, I know where I'm going wrong. I need to use As+B in the denominator because the one factor is quadratic. Sorry for wasting time.
    Last edited by adkinsjr; March 16th 2011 at 02:33 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. inverse laplace transform help
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 16th 2010, 11:24 AM
  2. Help with Inverse Laplace Transform
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: July 11th 2009, 01:25 PM
  3. Inverse Laplace Transform
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: June 9th 2009, 04:48 AM
  4. inverse laplace transform
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 30th 2008, 12:41 AM
  5. Inverse Laplace transform
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 22nd 2008, 06:12 AM

/mathhelpforum @mathhelpforum