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Laplace transform question

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve by Laplace transforms the following
    y'' + y = t when 0</=t<1, and = 1 if t>/=1

    2. Relevant equations

    L{y''} + L{y} = L{f(t)}

    3. The attempt at a solution

    By Laplace transforms I get
    L{f(t)} = (1 - e^-s) / s^2
    Y(s) = (1-e^-2 + s^2) / s^2 (s^2 +1)

    But I cannot simplify Y(s) in order to get y = L^-1{Y(s)}!!!
    Last edited: Nov 30, 2008
  2. jcsd
  3. Nov 30, 2008 #2
    Hi Jaejoon,

    You didn't tell us what the initial conditions are, but from your work I'll assume they were that y(0)=0 and y'(0)=1.

    To find the inverse Laplace transform of [tex]Y(s) = \frac{1-e^{-s} + s^2}{s^2(s^2 +1)}[/tex], first split it up to

    [tex]\frac{1}{s^2} - \frac{e^{-s}}{s^2(s^2 +1)}[/tex].

    To transform the second term, use the formula [tex]e^{-cs}F(s) \mapsto u(t-c)f(t-c)[/tex], where u is the Heaviside function and f is the inverse Laplace of F. Note that you will need to split F(s) into partial fractions to transform using tables.
  4. Nov 30, 2008 #3
    Last edited: Nov 30, 2008
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