solve the given system of differential equations. (dx_1)/(dt)=2x_1-3x_2 (dx_2)/(dt)=x_1-2x_2

Step 1
Consider the following system of differential equations:
${\displaystyle \frac{{dx}_1}{dt}=2x_1-3x_2}$...(1)
${\displaystyle \frac{{dx}_2}{dt}=x_1-2x_2}$
Substitute ${\displaystyle x1=\frac{{dx}_2}{dt}+2x_2}$ in the equation (1):
${\displaystyle \frac{d}{dt}(\frac{{dx}_2}{dt}+2x_2)=2(\frac{{dx}_2}{dt}+2x_2)-3x_2}$
${\displaystyle \frac{d^2{x}_2}{{dt}^2}+2\frac{{dx}_2}{dt}=2\frac{{dx}_2}{dt}+4x_2-3x_2}$
${\displaystyle \frac{d^2{x}_2}{{dt}^2}-x_2=0}$
Step 2
The auxiliary equation is ${\displaystyle m^2-1=0}$.
The roots of auxiliary equation are 1,−1.
Hence, the solution of homogeneous equation is ${\displaystyle x_2=A{e}^t+B{e}^{-t}}$.
A and B are constants.
Substitute value of ${\displaystyle x_2}$ in the equation ${\displaystyle x_1=\frac{{dx}_2}{dt}+2x_2}$:
${\displaystyle x_1=\frac{d}{dt}(A{e}^t+B{e}^{-t}+2(A{e}^t+B{e}^{-t})}$
${\displaystyle =A{e}^t-B{e}^{-t}+2A{e}^t+2B{e}^{-t}}$
${\displaystyle =3A{e}^t+B{e}^t}$
Step 3
Hence, the solution is
${\displaystyle x_1=3A{e}^t+B{e}^t}$
${\displaystyle x_2=A{e}^t+B{e}^{-t}}$.