Solutions to matrix differential equation x' = Ax If A is the 2-by-2 matrix 
\begin{bmatrix}0&1\\-a_0&-a_1\end{bmatrix}
and \begin{bmatrix}x_1,_1(t)\\x_1,_2(t)\end{bmatrix} and \begin{bmatrix}x_2,_1(t)\\x_2,_2(t)\end{bmatrix} are linearly independent solutions to the matrix  differential equation $x' = Ax$, then is
$x_{1,2(t)}$ and  $x_{2,2 (t)}$
a solution to the differential equation  $y'' +a_1y' + a_0y = 0$, 
I don't know how to prove if those specific entries in the solution vectors satisfy the differential equation. I know how to prove it for $x_{1,1(t)}$ and $x_{2,1(t)}$ by plugging in the solution vector to $x'=Ax$ and solving the equations to show that they fit $y'' +a_1y' + a_0y = 0$. I'm stuck when it comes to
$x_{1,2(t)} + x_{2,2 (t)}$
 A: If $x=\left[\matrix{x_1\\x_2}\right]$ is a solution to $\dot x=Ax$ then
$$
\left[\matrix{\dot x_1\\\dot x_2}\right]=
\left[\matrix{0 & 1\\-a_0 & -a_1}\right]
\left[\matrix{x_1\\x_2}\right]\ \Leftrightarrow\ 
\left\{
\begin{array}{l}
\dot x_1=x_2,\\
\dot x_2=-a_0x_1-a_1x_2
\end{array}
\right.
\Rightarrow\ \ddot x_1+a_1\dot x_1+a_0x_1=0.
$$
So what can you say about $x_{1,1}$, $x_{2,1}$ and $x_{1,1}+x_{2,1}$?
UPDATE: If $y_1$ and $y_2$ are two solutions to $y''+a_1y'+a_0y=0$ then $z=y_1+y_2$ satisfies
$$
z''+a_1z'+a_0z=(\color{red}{y_1}+\color{blue}{y_2})''+a_1(\color{red}{y_1}+\color{blue}{y_2})'+a_0(\color{red}{y_1}+\color{blue}{y_2})=\\
=\color{red}{y_1''}+\color{blue}{y_2''}+a_1\color{red}{y_1'}+a_1\color{blue}{y_2'}+a_0\color{red}{y_1}+a_0\color{blue}{y_2}=\color{red}{y_1''+a_1y_1'+a_0y_1}+\color{blue}{{y_2''+a_1y_2'+a_0y_2}}=0,
$$
that is $y_1+y_2$ is a solution too.
P.S. $x_2=\dot x_1$ $\Rightarrow$ $\ddot x_2+a_1\dot x_2+a_0x_2=\dddot x_1+a_1\ddot x_1+a_0\dot x_1=(\ddot x_1+a_1\dot x_1+a_0x_1)'=0'=0$.
