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Second order system DE $$\tag 1 x_1''=k(x_2-2x_1) $$ $$\tag 2 x_2''=k(x_1-2x_2+x_3)$$ $$\tag 3 x_3''=k(x_2-2x_3)$$ With $\vec x(0)=(x_1,x_2,x_3)=(3,0,1) \;\;\; \vec x'(0)=(x_1',x_2',x_3')=(0,0,0)$

I understand that one version is to make use of $p_i=x_i'$ making a new vector $\vec y =(x_1,x_2,x_3,x_1',x_2',x_3')$ $$\vec y'=A\vec y$$

But a $6×6$ matrix feels incredibly unpalletable for an exam. Any tips would be welcome.

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    $\begingroup$ I suggest asking your professor about what is he looking for in a solution. The matrix $A$ might be sparse actually ? $\endgroup$ – Chee Han Feb 13 '17 at 2:53
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You could try using Laplace transform methods. You'll have $3$ equations to solve for $X_1= \mathcal{L}\{x_1\}$, $X_2= \mathcal{L}\{x_2\}$, and $X_3= \mathcal{L}\{x_3\}$.

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Working out the answer given by @BobaFret:

$$ \begin{cases} \text{x}_1''\left(t\right)=\text{k}\cdot\left(\text{x}_2\left(t\right)-2\cdot\text{x}_1\left(t\right)\right)\\ \\ \text{x}_2''\left(t\right)=\text{k}\cdot\left(\text{x}_1\left(t\right)-2\cdot\text{x}_2\left(t\right)+\text{x}_3\left(t\right)\right)\\ \\ \text{x}_3''\left(t\right)=\text{k}\cdot\left(\text{x}_2\left(t\right)-2\cdot\text{x}_3\left(t\right)\right) \end{cases}\tag1 $$

Now, for the Laplace transform of this system we get:

$$ \begin{cases} \text{s}^2\cdot\text{X}_1\left(\text{s}\right)-\text{s}\cdot\text{x}_1\left(0\right)-\text{x}_1'\left(0\right)=\text{k}\cdot\left(\text{X}_2\left(\text{s}\right)-2\cdot\text{X}_1\left(\text{s}\right)\right)\\ \\ \text{s}^2\cdot\text{X}_2\left(\text{s}\right)-\text{s}\cdot\text{x}_2\left(0\right)-\text{x}_2'\left(0\right)=\text{k}\cdot\left(\text{X}_1\left(\text{s}\right)-2\cdot\text{X}_2\left(\text{s}\right)+\text{X}_3\left(\text{s}\right)\right)\\ \\ \text{s}^2\cdot\text{X}_3\left(\text{s}\right)-\text{s}\cdot\text{x}_3\left(0\right)-\text{x}_3'\left(0\right)=\text{k}\cdot\left(\text{X}_2\left(\text{s}\right)-2\cdot\text{X}_3\left(\text{s}\right)\right) \end{cases}\tag2 $$

Now, using the initial conditions:

$$ \begin{cases} \text{s}^2\cdot\text{X}_1\left(\text{s}\right)-3\cdot\text{s}=\text{k}\cdot\left(\text{X}_2\left(\text{s}\right)-2\cdot\text{X}_1\left(\text{s}\right)\right)\\ \\ \text{s}^2\cdot\text{X}_2\left(\text{s}\right)=\text{k}\cdot\left(\text{X}_1\left(\text{s}\right)-2\cdot\text{X}_2\left(\text{s}\right)+\text{X}_3\left(\text{s}\right)\right)\\ \\ \text{s}^2\cdot\text{X}_3\left(\text{s}\right)-\text{s}=\text{k}\cdot\left(\text{X}_2\left(\text{s}\right)-2\cdot\text{X}_3\left(\text{s}\right)\right) \end{cases}\tag3 $$

Solving each function out of its own equation:

$$ \begin{cases} \text{X}_1\left(\text{s}\right)=\frac{\text{k}\cdot\text{X}_2\left(\text{s}\right)+3\cdot\text{s}}{\text{s}^2+2\cdot\text{k}}\\ \\ \text{X}_2\left(\text{s}\right)=\frac{\text{k}\cdot\left(\text{X}_1\left(\text{s}\right)+\text{X}_3\left(\text{s}\right)\right)}{\text{s}^2+2\cdot\text{k}}\\ \\ \text{X}_3\left(\text{s}\right)=\frac{\text{k}\cdot\text{X}_2\left(\text{s}\right)+\text{s}}{\text{s}^2+2\cdot\text{k}} \end{cases}\tag4 $$

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  • $\begingroup$ I'm split between giving you or Boba the answer :( he was first after all but yours is way more detailed. $\endgroup$ – Katpton Liamfuppinshire Feb 17 '17 at 0:51

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