## Transforming a System of Coupled Constant Coefficient Linear Differential Equations to First Order

\[\ddot{y}-3 \dot{y}+2y+3 \dot{x}-x=0\]

\[\dot{y}=2y+\dot{x}+x=0\]

this is a constant coefficient, linear second order system. Let

\[x=u_1, \: y=u_2, \: \dot{y}=u_3\]

then \[\dot{x}=\dot{u}_1, \: \dot{y}=\dot{u}_2, \: \: \ddot{y}=\dot{u}_3\]

the system becomes\[\dot{u}_3-3u_3+2u_2+\dot{u}_1-u_1 =0 \rightarrow \dot{u}_3=3u_3-2u_2-\dot{u}_1+u_1\]

\[u_3-2u_2+\dot{u}_1+u_1=0 \rightarrow \dot{u}_1=-u_1+2u_2-u_3\]

Substitute the second of these into the first, and the result, together with the second equation, becomes

\[\dot{u}_3=3u_3-2u_2-(u_1+2u_2-u_3)+u_1 =-4u_2+4u_3\]

\[\dot{u}_1=-u_1+2u_2-u_3\]

\[\dot[U]_2=u_3\]

This is a coupled first order system.

We can write this in matrix form as

\[\begin{pmatrix}\dot{u}_1\\ \dot{u}_2\\ \dot{u}_3\end{pmatrix}= \left( \begin{array}{ccc} -1 & 2 & -1 \\ 0 & 0 & 1 \\ 0 & -4 & 4 \end{array} \right) \begin{pmatrix}u_1\\u_2\\u_3 \end{pmatrix}\]