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

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