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Im trying to solve this ODE and find a simplified expression for $x(t)$.

$$\ddot x+4x=-2\sum_{n=1}^{4} e^{in\pi}u(t-n\pi);\space x(0)=0=\dot x(0),i=\sqrt{-1}$$

First i found the the laplace transform of the left hand side:

$$\mathcal{L}\{\ddot x+4x\}=s^2X(s)+4X(s)$$

But im not sure where to even start, to find the laplace transform of the right hand side? Cant find any similar examples, only general forms like this: $(t-\tau)^n e^{-\alpha (t-\tau)} \cdot u(t-\tau)$, that i cant tell whether or not they apply here or not?

Update #1 @ 6/3/14, 20:54) I now have obtained the simplified form:

$$x(t)=-\frac{1}{2}\sum_{n=1}^{4} (-1)^{n}u(t-n\pi)(1-\cos(2t)$$

But im really stumped on how to graph this? I would know how to graph it, if it was just $x(t)=u(t-n\pi)(1-\cos(2t)$ , but i really dont know what to do with this summation thing, in terms of plotting?

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Note that in a system with step delays at different times, the solution of the system at the time that a new step function "activates" is equivalent to a new set of initial conditions at that time.

So if you have a system with some solution trajectory, and at time $T$ you have an input multiplied by a step, then when $t < T_1$, that input has zero effect. When $t = T_1$, you use the solution up to that point as your initial conditions. Do the same at $T_2, T_3,$ etc.

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  • $\begingroup$ What are the $T$ values? Do you mean i should evaluate the sum, somehow or ? Not sure i understand really? $\endgroup$ Jun 3 '14 at 4:13
  • $\begingroup$ You should plot the right-hand side of your ODE and explore what it looks like when $t < \pi$, $t < 2\pi$, $t < 3\pi$ and $t < 4\pi$. $\endgroup$
    – Emily
    Jun 3 '14 at 18:04
  • $\begingroup$ I don't know how to do that either. The summation thing is throwing me off, cant find any examples like it. $\endgroup$ Jun 3 '14 at 18:35
  • $\begingroup$ Don't think of it as a thing that you need examples of. Just expand it: $-2 \sum e^{in\pi}u(t-n\pi) = -2 e^{i\pi} u(t-\pi) - 2e^{2i\pi}u(t-2\pi)+\cdots$. $\endgroup$
    – Emily
    Jun 3 '14 at 18:39
  • $\begingroup$ I got $x(t)$, just stuck even more now on how to graph this thing? I updated my results so far. $\endgroup$ Jun 4 '14 at 4:04

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