Suppose we have a monic differential equation defined by $$x'' + x = \sum_{n=0}^\infty \delta(t - 2n\pi)$$ When we take the Laplace transform of this differential equation, we get $$s^2X(s) + X(s) = \sum_{n=0}^\infty e^{-2n\pi s}$$ Solving for $X(s)$ we have $$X(s) = \frac{1}{s^2 + 1} \sum_{n=0}^\infty e^{-2n\pi s}$$ Expanding the sum will yield the following: $$X(s) = \frac{1}{s^2 + 1} + \frac{e^{-2\pi s}}{s^2 + 1} + \frac{e^{-4\pi s}}{s^2 + 1} + ...$$ From here, we can take the Inverse Laplace transforms of all terms in the sequence and sum them together to yield another sum. This operation yields: $$x(t) = sin(t) + u(t-2\pi)sin(t-2\pi) + u(t-4\pi)sin(t - 4\pi) + ... = sin(t) + \sum_{n=1}^\infty u(t - 2n\pi)sin(t - 2n\pi)$$ This is where my confusion begins. I assume that I've done everything correct up to this point (please correct me if I am wrong). My textbook states "... show that if $2n\pi \lt t \lt 2(n+1)\pi$, then $x(t) = (n + 1) sin(t)$".
How do I do this? What is the significant of the bounds set for $t$? Thanks in advance m8s.
Note: I have noticed that $x(t) = sin(t) + \sum_{n=1}^\infty u(t - 2n\pi)sin(t - 2n\pi) = sin(t) + \sum_{n=1}^\infty u(t - 2n\pi)sin(t)$. I just can't seem to grasp the intuition between the unit-step function and the bounds set on $t$.
Note that we have
$$x(t)=\sin(t)\sum_{k=0}^\infty u(t-2k\pi) \tag 1$$
where $u(t)$ is the unit step function defined as
$$u(t)=\begin{cases}1&,t\ge 0\\\\0&,t\le 0 \end{cases},$$
Now, note that if $2n\pi < t<2(n+1)\pi$, then all of the unit step functions in $(1)$ are $0$ when $k>n+1$ and all of the unit step functions in $(1)$ are $1$ when $k\le n$.
That is to say, that for $t\in (2n\pi,2(n+1)\pi)$
$$u(t-2k\pi)=\begin{cases}1&, k\le n\\\\0&,k>n\end{cases}$$
Therefore, for $t\in (2n\pi,2(n+1)\pi)$
$$\begin{align} x(t)&=\sin(t)\sum_{k=0}^\infty u(t-2k\pi)\\\\ &=\sin(t)\left(\sum_{k=0}^n (1)+\sum_{k=n+1}^\infty(0)\right)\\\\ &=(n+1)\sin(t) \end{align}$$
as was to be shown!