Answer
$x_1(t)=-c_1e^t +\frac{c_2}{2}e^{4t}$
$x_2(t)=c_1e^t +c_2e^{4t}$
Work Step by Step
$x_1'=2x_1+x_2$
$x_2'=2x_1+3x_2$
We begin by rewriting the system in operator form as $(D−2)x_1 - x_2=0$ (1)
$−2x_1+(D-3)x_2 =0 $ (2)
To eliminate $x_2$ between these two equations, we first operate on equation (2) with $D-2$ to obtain:
$−2(D-2)x_1+(D-3)(D - 2)x_2 =0 $
Adding twice equation (1) to this equation eliminates $x_2$ and yields
$−2x_2+(D-3)(D - 2)x_2 =0$
$(D^2-5D+4)x_2 =0$.
This constant coefficient differential equation has auxiliary polynomial
$P(r)=r^2-5r +4=(r - 4)(r − 1)$
Consequently, $x_2(t)=c_1e^t+c_2e^{4t}$
From equation $x_2'=2x_1+3x_2$ we have
$2x_1 (t)=x_2'-3x_2(t)$
$2x_1 (t) =c_1e^t+4c_2e^{4t} - 3c_1e^t - 3c_2e^{4t}$
$2x_1 (t) =-2c_1e^t+c_2e^{4t}$
$x_1(t)=-c_1e^t +\frac{c_2}{2}e^{4t}$
$x_2(t)=c_1e^t +c_2e^{4t}$
