Answer
$x_2(t)=c_1e^{2t}+c_2e^{-2t}$
Work Step by Step
$x_1'=2x_1 + 4x_2$
$x_2'=-4x_1 - 6x_2$
We begin by rewriting the system in operator form as $(D−2)x_1 - 4x_2=0$ (1)
$4x_1+(D + 6)x_2 =0 $ (2)
To eliminate $x_2$ between these two equations, we first operate on equation (2) with $-\frac{1}{4}(D-2)$ to obtain:
$-(D−2)x_1 -\frac{1}{4}(D-2)(D+6)x_2=0$
Adding equation (1) to this equation eliminates $x_2$ and yields
$-4x_2- \frac{1}{4}(D-2)(D+6)x_2 =0 $
$-\frac{1}{4}(D^2+4D+4)x_2 =0$.
This constant coefficient differential equation has auxiliary polynomial
$P(r)=-(r^2+4r+4)=r^2+4r+4=(r+2)^2$
Consequently, $x_2(t)=c_1e^{2t}+c_2e^{-2t}$
From equation $4x_1+(D + 6)x_2 =0 $
$x_1=\frac{-1}{4}(D + 6)x_2$
$x_1=(-c_1 - \frac{1}{4}c_2)e^{-2t}-c_2e^{-2t}$
