Answer
$x_1(t)=2e^{-t}+e^{2t}\\
x_2(t)=-e^{-t}+e^{2t}$
Work Step by Step
$x_1'=2x_2$
$x_2'=x_1+x_2$
$x_1(0)=3\\
x_2(0)=0$
We begin by rewriting the system in operator form as $x_1=x_2'-x_2 \rightarrow x_1'=x_2''-x_2'$
Since $x_1'=2x_2$
we have $2x_2=x_2''-x_2' \rightarrow 2x_2-x_2''+x_2'=0$
This constant coefficient differential equation has auxiliary polynomial
$P(r)=r^2-r-2$
Consequently, $x_2(t)=c_1e^{-t}\cos t+c_3e^{2t}$
From equation $x_1=x_2'-x_2$
$x_1(t)=-2c_1e^{-t}+c_2e^{2t}$
Substitute:
$c_1+c_2=0\\
-2c_1+c_2=3$
Solve the system of equation
$c_1=-1\\
c_2=1$
Hence,
$x_1(t)=2e^{-t}+e^{2t}\\
x_2(t)=-e^{-t}+e^{2t}$
