# Find the general solution ?

Mar 20, 2018

See below.

#### Explanation:

Using the Laplace transform is handy

$X = \left(\begin{matrix}{x}_{1} \\ {x}_{2}\end{matrix}\right)$
$A = \left(\begin{matrix}1 & 1 \\ - 4 & 1\end{matrix}\right)$

$\dot{X} = A X \Rightarrow s X \left(s\right) = A X \left(s\right) + {x}_{0}$ then

$X \left(s\right) = {\left(s {I}_{2} - A\right)}^{-} 1 {x}_{0}$

here

${\left(s {I}_{2} - A\right)}^{-} 1 = \frac{1}{{s}^{2} - 2 s + 5} \left(\begin{matrix}s - 1 & 1 \\ - 4 & s - 1\end{matrix}\right)$

and ${x}_{0} = \left({x}_{10} , {x}_{20}\right)$ are initial conditions.

then

$X \left(s\right) = \frac{1}{{s}^{2} - 2 s + 5} \left(\begin{matrix}\left(s - 1\right) {x}_{10} + {x}_{20} \\ - 4 {x}_{10} + \left(s - 1\right) {x}_{20}\end{matrix}\right)$

and

inverting

$X \left(t\right) = \left(\begin{matrix}{x}_{1} \left(t\right) \\ {x}_{2} \left(t\right)\end{matrix}\right) = \left(\begin{matrix}\frac{1}{2} \left(2 \cos \left(2 t\right) {x}_{10} + \sin \left(2 t\right) {x}_{20}\right) \\ \cos \left(2 t\right) {x}_{20} - 2 \sin \left(2 t\right) {x}_{10}\end{matrix}\right) {e}^{t} u \left(t\right)$

here $u \left(t\right)$ is the unitary step function.