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Transform of a piecewise continuous function

  1. Jun 1, 2012 #1
    We know that the [itex]\mathcal L\{f(t)\} = \int^{\infty}_0 e^{-st}f(t) dt[/itex].

    Say we want to, for example, solve the following IVP: [tex]y'' + y = f(t)[/tex] where [itex]f(t) = \begin{cases}
    0 & 0 \leq t < \pi \\
    1 & \pi \leq t < 2\pi\\
    0 & 2\pi \leq t
    \end{cases}[/itex]

    and [itex]y(0) = 0 , y'(0) = 0[/itex]

    We apply Laplace on both side of the DE, and we get [itex](s^2 + 1)Y(s) = \mathcal L\{f(t)\}[/itex]. Using the cases above, do we divide the integral from 0 to [itex]\infty[/itex] into three integrals?

    I did that and [itex]\mathcal L\{f(t)\} = \int^{\pi}_0 e^{-st}(0) dt + \int^{2\pi}_{\pi} e^{-st}(1) dt + \int^{\infty}_{2\pi} e^{-st}(0) dt[/itex]. The first and third integrals are zeros so we need to integrate the second one. We get [itex]-(1/s)[e^{-2\pi s} - e^{-\pi s}][/itex]. Right?

    Back to the DE, [itex]Y(s) = -[e^{-2\pi s} - e^{-\pi s}] / s(s^2 + 1)[/itex]. How exactly do we find [itex]\mathcal L^{-1}\{Y(s)\}[/itex]?
     
    Last edited: Jun 1, 2012
  2. jcsd
  3. Jun 1, 2012 #2
    Okay, so I tried solving it. [itex]1/s(s^2 + 1) = 1/s - s/(s^2 + 1)[/itex], and after a litle work, [itex]Y(s) = e^{-\pi s}/s - e^{-\pi s}/(s^2 + 1) - e^{-2\pi s}/s + e^{-2\pi s}/(s^2 + 1)[/itex]. Is this correct? Now, the Laplace inverse of [itex]Y(s)[/itex] is [itex]\delta(t - \pi) - \delta(t - \pi)sin(t - \pi) - \delta(t - 2\pi) + \delta(t - 2\pi)cos(t - 2\pi)[/itex]. I'm not sure about this, but I think it's fine?
     
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