Does ther exist a relation between fundamental solutions of system of ODE and its equivalent second order DE? Let's say I have a second-order differential equation of the form
$$
y'' + p(t)y' + q(t)y=0
$$
We can transform this second-order differential equation into a system of ODEs as follows
$$
\left\{\begin{matrix}x_1'=x_2 \\ x_2'=-p(t)x_2 -q(t)x_1 \end{matrix}\right.
$$
or alternatively written as
$$
\mathbf{x'}= \begin{pmatrix} x_1' \\ x_2'\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -p & -q\end{pmatrix}\begin{pmatrix} x_1 \\ x_2\end{pmatrix} = A\mathbf{x}
$$
It is my understanding that if $y_1, \ y_2$ are fundamental solutions of the second-order differential equation, then you can say $ y = c_1 y_1 + c_2 y_2$. On the oher hand, if $\mathbf{x_1}, \ \mathbf{x_2}$ are fundamental solutions to the system of equations it's because $\mathbf{x_1'}= A\mathbf{x_1}$ and $\mathbf{x_2'}= A\mathbf{x_2}$. 
Given this understanding, my question is

Does there exist a way to relate the fundamental solutions of the system of equations $\mathbf{x_1}, \ \mathbf{x_2}$ with the fundamental solutions of the original differential equation $y_1, \ y_2$?

Thank you!
 A: With
$\mathbf x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}, \tag 1$
we take
$x_1= y, \tag 2$
and
$x_2 = \dot y, \tag 3$
whence
$\dot x_1= \dot y, \tag 4$
and
$\dot x_2 = \ddot y; \tag 5$
then the system
$\begin{pmatrix} \dot x_1 \\ \dot x_2 \end{pmatrix} = \dot {\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}} = \dot{\mathbf x} = A\mathbf x = \begin{pmatrix} 0 & 1 \\ -q & -p \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} x_2 \\ -px_2 - qx_1 \end{pmatrix}  \tag 6$
translates to
$\begin{pmatrix} \dot y \\ \ddot y \end{pmatrix} =  \begin{pmatrix} \dot y \\ -p \dot y - q y \end{pmatrix},  \tag 7$
from which
$\ddot y = -p \dot y -qy, \tag 8$
or
$\ddot y + p \dot y + qy = 0. \tag 9$
In light of (2)-(5), this argument shows the equivalence of the first-order system (6) and the second order equation (9).
It is well-known that (9) is possessed of two linearly independent solutions $y_1$ and $y_2$; it is clear from the above that these correspond to two column vector solutions $\mathbf x_1$ and $\mathbf x_2$ of (6); we may this define a matrix $\mathbf X$ in columnar form
$\mathbf X = [\mathbf x_1 \; \mathbf x_2] \tag{10}$
which satisfies
$\dot{\mathbf X} = [\dot{\mathbf x}_1 \; \dot{\mathbf x}_2] = [A\mathbf x_2 \; A\mathbf x_2] = A[\mathbf x_1 \; \mathbf x_2] = A\mathbf X; \tag{11}$
that the column vectors $\mathbf x_1$, $\mathbf x_2$ are linearly independent follows from that of the $y_1$ and $y_2$, since with
$\mathbf x_1 = \begin{pmatrix} y_1 \\ \dot y_1 \end{pmatrix}, \tag{12}$
$\mathbf x_2 = \begin{pmatrix} y_2 \\ \dot y_2 \end{pmatrix}, \tag{13}$
we have
$a\mathbf x_1 + b\mathbf x_2 = \begin{pmatrix} ay_1 + by_2 \\ a \dot y_1 + b\dot y_2 \end{pmatrix}; \tag{14}$
if there existed $a$ and $b$ such that
$a\mathbf x_1 + b\mathbf x_2 = 0 \tag{15}$
with at least one of $a, b \ne 0$, then the same of course would hold for $y_1$, $y_2$ by virtue of (14); thus we see the linear independence of $\mathbf x_1$ and $\mathbf x_2$.
From the linear independence of $\mathbf x_1$ and $\mathbf x_2$ we see that $\mathbf X$ is in fact a fundamental matrix solution of (11); the preceding discussion show how a linearly independent pair of solutions $y_1$, $y_2$ to (9) give rise of a funamental solution fo (11).
As a final note, the term "fundamantal solutions" is not often used in the context of (9), where the term "linearly independent solutions" is generally preferred.
