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