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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 $x_{1,2(t)} + x_{2,2 (t)}$

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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. $x_2=\dot x_1$ $\Rightarrow$ $\ddot x_2+a_1\dot x_2+a_0x_2=\dddot x_1+a_1\ddot x_1+a_0\dot x_1=(\ddot x_1+a_1\dot x_1+a_0x_1)'=0'=0$.

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