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Using Laplace transform to solve ODE

  1. Oct 22, 2006 #1
    Here's the problem:

    Solve [tex]ty'' + 2y' + ty = 0[/tex]
    with [tex]lim_{t\rightarrow0+} y = 0[/tex]
    and
    [tex] y(\pi) = 0[/tex]

    So here's my working:

    [tex]-\frac{d}{ds}[s^2Y - sy(0) - y'(0)] + 2[sY - y(0)] - Y' = 0[/tex]

    [tex]-[2sY + s^2Y' - s] + 2[sY - y(0)] - Y' = 0[/tex]

    Arranging terms, I get:

    [tex] Y' = \frac{s-2y(0)}{s^2+1}[/tex]
    and
    [tex]-Y' = \frac{2y(0)-s}{s^2+1}[/tex]

    [tex]L^{-1}(-Y')= L^{-1}(\frac{2y(0)}{s^2+1}) - L^{-1}(\frac{s}{s^2+1})[/tex]

    [tex]ty = 2y(0)sint - cost[/tex]

    [tex] y = \frac{2y(0)sint}{t} - \frac{cost}{t}[/tex]



    The answer is [tex]y = \frac{sint}{t}[/tex]

    What's wrong?

    Thanks..
     
    Last edited: Oct 22, 2006
  2. jcsd
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