Answer
$x_1 (t) = -c_1e^{2t} - 2c_2e^{3t}$
$x_2(t)=c_1e^{2t}+c_2e^{3t}$
Work Step by Step
$x_1'=4x_1 + 2x_2$
$x_2'=-x_1 + x_2$
We begin by rewriting the system in operator form as $(D−4)x_1 - 2x_2=0$ (1)
$x_1+(D-1)x_2 =0 $ (2)
To eliminate $x_2$ between these two equations, we first operate on equation (2) with $-(D-4)$ to obtain:
$-(D-4)x_1-(D-4)(D-1)x_2 =0 $
Adding equation (1) to this equation eliminates $x_2$ and yields
$-2x_2-(D-1)(D-4)x_2 =0 $
$-(D^2-5D+6)x_2 =0$.
This constant coefficient differential equation has auxiliary polynomial
$P(r)=-(r^2-5r+6)=r^2-5r+6=(r-2)(r-3)$
Consequently, $x_2(t)=c_1e^{2t}+c_2e^{3t}$
Thus:
$x_1 (t) = -c_1e^{2t} - 2c_2e^{3t}$
