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

Higher order systems of differential equations can be complicated and difficult to solve. We can often simplify however to obtain a first order linear system which can be solved, and this solution to the first order system can be used to solve the original problem.,br /> Example: Express the system
$\ddot{y}-3 \dot{y}+2y+3 \dot{x}-x=0$

$\dot{y}=2y+\dot{x}+x=0$

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