We know that the [itex]\mathcal L\{f(t)\} = \int^{\infty}_0 e^{-st}f(t) dt[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

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]?

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# Transform of a piecewise continuous function

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