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This is a problem from MIT's publicly available course on differential equations.

We are given the initial value problem \begin{align} y'' + \omega^2y & = h(t)\sin t - h(t-c)\sin t \end{align} with $c > 0$ and $y(0) = y'(0) = 0$. (Note also that $h$ is the Heaviside step function.)

We are to solve the equation and then show that $y(0) = y'(0) = y''(0) = 0$. I have a solution to the equation, but it's clearly wrong, in that it does not satisfy the initial conditions. I have been staring at it for a couple of days; I'm hoping someone here can point out what I'm doing wrong. (Hopefully it's not a dumb algebra mistake.)

Taking Laplace transforms, we have \begin{align} Y(s)(s^2 + \omega^2) & = \frac{1}{s^2 + 1} - e^{-sc}\left(\frac{\cos c + s\sin c}{s^2 + 1}\right)\\ \end{align}

Note that I obtained the second term by observing that \begin{align} \int_0^{\infty}e^{-st}h(t-c)\sin tdt &= \int_0^{c}e^{-st}h(t-c)\sin tdt +\int_c^{\infty}e^{-st}h(t-c)\sin tdt\\ & = \int_c^{\infty}e^{-st}\sin tdt \end{align} and then computing the integral.

Thus we have to find the inverse Laplace transform of \begin{align} \frac{1}{(s^2 + 1)(s^2 + \omega^2)} - e^{-sc}\left(\frac{\cos c + s\sin c}{(s^2 + 1)(s^2 + \omega^2)}\right) \end{align}
which, I believe, is \begin{align} \frac{1}{\omega^2 - 1}\left[\sin t - \frac{1}{\omega}\sin \omega t - \cos c\left(\sin(t-c) - \frac{1}{\omega}\sin(\omega (t - c))\right)\\ - \sin c[\cos(t - c) - \cos(\omega (t - c))]\right] \end{align}

The difficulty, of course, is that \begin{align} y(0) & = \frac{1}{\omega^2 - 1}\left[-\cos c\left(\sin(-c) - \frac{1}{\omega}\sin(-\omega c)\right) - \sin c(\cos(-c) - \cos(-\omega c))\right]\\ & = \frac{1}{\omega^2 - 1}\left(\sin c\cos(\omega c)-\frac{\cos c}{\omega}\sin (\omega c)\right)\\ & \neq 0 \end{align}

unless $c$ is an integer multiple of $\pi$. Any thoughts would be appreciated.

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  • $\begingroup$ No (I believe) to the first, but a resounding (and embarrassing) yes to the second. Would you like to write up an answer so that I can accept it and you can get reputation? $\endgroup$ – solitaireartist Aug 11 '16 at 11:38
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As the user Moo was kind enough to point out (before mysteriously removing all traces of her/himself from the comment thread), the inverse Laplace I seek should be

\begin{align} \frac{1}{\omega^2 - 1}\left[\sin t - \frac{1}{\omega}\sin \omega t - h(t-c)\left(\cos c\left(\sin(t-c) - \frac{1}{\omega}\sin(\omega (t - c))\right)\\ - \sin c[\cos(t - c) - \cos(\omega (t - c))]\right)\right] \end{align}

which neatly eliminates the issue I was having.

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