## Transform of a piecewise continuous function

We know that the $\mathcal L\{f(t)\} = \int^{\infty}_0 e^{-st}f(t) dt$.

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

and $y(0) = 0 , y'(0) = 0$

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

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

Back to the DE, $Y(s) = -[e^{-2\pi s} - e^{-\pi s}] / s(s^2 + 1)$. How exactly do we find $\mathcal L^{-1}\{Y(s)\}$?
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Okay, so I tried solving it. $1/s(s^2 + 1) = 1/s - s/(s^2 + 1)$, and after a litle work, $Y(s) = e^{-\pi s}/s - e^{-\pi s}/(s^2 + 1) - e^{-2\pi s}/s + e^{-2\pi s}/(s^2 + 1)$. Is this correct? Now, the Laplace inverse of $Y(s)$ is $\delta(t - \pi) - \delta(t - \pi)sin(t - \pi) - \delta(t - 2\pi) + \delta(t - 2\pi)cos(t - 2\pi)$. I'm not sure about this, but I think it's fine?

 Similar discussions for: Transform of a piecewise continuous function Thread Forum Replies Calculus 5 Calculus 1 Calculus 0 Precalculus Mathematics Homework 5 Calculus & Beyond Homework 0