Finding suitable $x:[-T,T]\to\mathbb{R}^n, A\in\mathbb{R}^{n\times n}$ such that $x'(t) = Ax(t)$ when $\frac{d^2}{dt^2}B(t)=B(t)$

Question

Suppose that $\frac{d^2}{dt^2}B(t) = B(t)$ for some matrix $B$ when $t\in [-T, T], T > 0$. I am tasked to determine suitable $x:[-T,T]\to\mathbb{R}^n, A\in\mathbb{R}^{n\times n}$ such that $x'(t) = Ax(t)$.

Edit: To my knowledge I have to determine such $x$ and $A$ that $x'(t) = Ax(t)$ contains the same information as $\frac{d^2}{dt^2}B(t) = B(t)$, i.e. somehow transform the given equation to a matrix ODE problem.

I know from my linear algebra textbook that $x'(t) = Ax(t)$ is solved by $x(t) = e^{tA}x(0)$ which suggests that some substitution might be in place.

(Question) How should one solve this sort of a problem?

(Thoughts) I was initially thinking that maybe we could write $\frac{d}{dt}B(t) = A_1B(t)$ for some suitable matrix/constant $A_1$, thus yielding $B(t) = e^{tA_1}B(0)$ and $\frac{d^2}{dt^2}B(t) = A_1^2B(t) = B(t)$. But I cannot come up with any argument why the first derivative of $B$ would have this self-similarity property.

My second idea was to just integrate $B$ w.r.t. $t$ and see where that leads us. But in that case $\frac{d}{dt}B(t) = \int B(t)dt$ and it's kinda hard to see the pattern $x'(t) = Ax(t)$ from this.

Answer 1

In the original equation there is no matrix multiplication or other coupling of the components of $B$. So you get the equation $\ddot b_{ij}=b_{ij}$ independent of the other components.

For the scalar case $\ddot y=y$ you get a first-order system the usual way, set $z=\dot y$, stack $x=\pmatrix{y\\z}$ and get $$\dot x=\pmatrix{0&1\\1&0}x.$$ Now decide on some kind of component order in the matrix and stack the first-order systems in some way, component and derivative alternating or first all components then all derivatives, or ...