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Inital value problem, Laplace transform

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve inital value problem: x''+2x'+x=g(t)

    g(t)=
    t 0<t<1
    2-t 1<t<2
    0 t>2


    2. Relevant equations

    Second shift theorem, Heaviside function and Laplace transforms. I denote Heaviside,functuon H(t-a), and Laplace transform with L

    3. The attempt at a solution

    I rewrote the problem with Heaviside functions:

    g(t)=t+(2-t)H(t-1)
    L(t+(2-t)H(t-1))= [itex]\frac{1}{s^{2}}[/itex]+[itex]\frac{2e^{-s}}{s^{2}}[/itex]

    After this I transformed the original diff. equation and plugged in my transform of g(t). After some simplification I got:


    G(s)=(1-[itex]2e^{-s}[/itex])([itex]\frac{1}{s^{2}(s+1)^{2}}[/itex])
    And after partial fraction decomp. I got:

    G(s)=(1-[itex]2e^{-s}[/itex])([itex]\frac{2}{s}[/itex]+[itex]\frac{1}{s^{2}}[/itex]+[itex]\frac{1}{(s+1)^{2}}[/itex])

    If I use inverse Laplace transform I get

    x(t)=f(t)-2f(t-1)H(t-1)

    where

    f(t)=(t+2)e[itex]^{-t}[/itex]+t-2

    According to my book, this is only about half of the answer. I am missing some step and I cant figure out what is it! All help is appreciated

    Thanks in advance!
     
    Last edited: Sep 5, 2011
  2. jcsd
  3. Sep 5, 2011 #2

    LCKurtz

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    I didn't check your work, but I notice that you haven't mentioned the initial conditions. Unless you were given x(0) = 0 and x'(0) = 0 you have left something out, which might explain why you don't have the whole answer.
     
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