Solve initial-value differentialsystem corresponding to matrix $A$ of degree $2$. I have proved that the eigenvalues of the following matrix are $-4,-1,2$ and $3$ by calculation of $p_A(\lambda) = (x+1)(x-2)(x-3)(x+4)$.

How can I solve the differentialsystem $$x_1^{''}=2x_2^{'}+x_1+2x_2, \\ x_2^{''}=-x_1^{'}-x_4+16x_2$$
with initial values $$x_1(0)=0, x_2(0)=-4 \\ x_1^{'}(0)=4, x_2^{'}(0)=26 $$
Up until now I have only solved systems of the form $x^{'} = Ax$ ?
 A: Introduce two new variables $y_1$ and $y_2$ such that
$y_1 = x_1^{'}, \; y_2 = x_2^{'}; \tag{1}$
then set up (i.e., define) a vector of variables $\mathbf r$, thusly:
$\mathbf r = \begin{pmatrix} x_1 \\ x_2 \\ y_1 \\ y_2 \end{pmatrix}. \tag{2}$
Noting that the equations
$x_1^{''}=2x_2^{'}+x_1+2x_2, \tag{3}$
$ x_2^{''}=-x_1^{'}-4x_1+16x_2, \tag{4}$
can now be written
$y_1^{'} =2y_2+x_1+2x_2, \tag{5}$
$ y_2^{'}=-y_1-4x_1+16x_2, \tag{6}$
we see that (1), (5) and (6) may be combined into a single matrix-vector linear ODE
$\mathbf r^{'} = A\mathbf r, \tag{7}$
where $A$ is the matrix
$A = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 2 & 0 & 2 \\ -4 & 16 & -1 & 0 \end{bmatrix}. \tag{8}$
(7) is of the familiar form $z' = Bz$, no explicit second derivatives here!  We take as intitial data for (7) $\mathbf r(0)$ with entries
$x_1(0)=0, x_2(0)=-4, y_1(0)=4, y_2(0)=26; \tag{9}$
the solution is then
$\mathbf r(t) = e^{At}\mathbf r(0), \tag{10}$
where $e^{At}$ may be evaluated using typical diagonalization methods; since the eigenvalues of $A$, which we already have at our disposal, are distinct, such techniques are relatively easy to apply.  I leave the details to you, my friends.
The key idea is to introduce new variables which are themselves first derivatives of the original variables, so that the resulting equation is of first order.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
