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.
On
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:
$$ \begin{cases} \text{X}_1\left(\text{s}\right)=\frac{\text{k}\cdot\text{X}_2\left(\text{s}\right)+3\cdot\text{s}}{\text{s}^2+2\cdot\text{k}}\\ \\ \text{X}_2\left(\text{s}\right)=\frac{\text{k}\cdot\left(\text{X}_1\left(\text{s}\right)+\text{X}_3\left(\text{s}\right)\right)}{\text{s}^2+2\cdot\text{k}}\\ \\ \text{X}_3\left(\text{s}\right)=\frac{\text{k}\cdot\text{X}_2\left(\text{s}\right)+\text{s}}{\text{s}^2+2\cdot\text{k}} \end{cases}\tag4 $$
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 $X_3= \mathcal{L}\{x_3\}$.