Answer
See below
Work Step by Step
$x_1'=2x_1 $
$x_2'=x_2-x_3$
$x_3'=x_2+x_3$
We begin by rewriting the system in operator form as $(D−1)x_2 + x_3=0$ (1)
$-x_2+(D -1)x_3 =0 $ (2)
To eliminate $x_2$ between these two equations, we first operate on equation (2) with $(D-1)$ to obtain:
$-(D-1)x_2+(D-1)^2x_3=0$
Adding equation (1) to this equation eliminates $x_2$ and yields
$(D^2-2D+2)x_3 =0$.
This constant coefficient differential equation has auxiliary polynomial
$P(r)=r^2-2r+2$
Consequently, $x_3(t)=c_2e^{t}\cos t+c_3e^{t}\sin t$
From equation $-x_2+(D -1)x_3 =0 $
$x_2=(D -1)x_3$
$x_2=c_3e^t\cos t-c_2e^t\sin t$
