Answer
$x_1(t)=3e^{3t}-te^{3t}\\
x_2(t)=2e^{3t}-te^{3t}$
Work Step by Step
$x_1'=2x_1+x_2+5e^{4t}$
$x_2'=-x_1+4x_2$
$x_1(0)=1\\
x_2(0)=3$
We begin by rewriting the system in operator form as $(D−2)x_1 - x_2=0$ (1)
$x_1+(D -4)x_2 =0 $ (2)
To eliminate $x_2$ between these two equations, we first operate on equation (2) with $-(D-2)$ to obtain:
$-(D−2)x_1 -(D-2)(D-4)x_2=0$
Adding equation (1) to this equation eliminates $x_2$ and yields
$-(D^2-6D+9)x_2 =0$.
This constant coefficient differential equation has auxiliary polynomial
$P(r)=r^2-6r+9$
Consequently, $x_2(t)=c_1e^{3t}+c_2te^{3t}$
and $x_1(t)=(c_1-c_2)e^{3t}+c_2te^{3t}$
Since $x_1(0)=1\\
x_2(0)=3$
Substitute:
$c_1+c_2=1\\
c_1-c_2=3$
Solve the system of equation
$c_1=2\\
c_2=-1$
Hence,
$x_1(t)=3e^{3t}-te^{3t}\\
x_2(t)=2e^{3t}-te^{3t}$
