Answer
See below
Work Step by Step
Given:
$x_1(t)=\begin{bmatrix}
e^{2t}\\
-e^{2t}
\end{bmatrix}$
$x_2(t)=\begin{bmatrix}
e^{2t}(1+t)\\
-te^{2t}
\end{bmatrix}$
$A=\begin{bmatrix}
3& 1\\
-1 & 1
\end{bmatrix}$
Differentiating the given vector functions with respect to t yields,
$x_1'(t)=\begin{bmatrix}
2e^{2t}\\
-2e^{2t}
\end{bmatrix}$
$x_2'(t)=\begin{bmatrix}
2te^{2t}+3e^{2t}\\
-2te^{2t}-e^{2t}
\end{bmatrix}$
whereas
$Ax_1(t)=\begin{bmatrix}
3 & 1\\
-1 & 1
\end{bmatrix}\begin{bmatrix}
e^{2t}\\
-e^{2t}
\end{bmatrix}=\begin{bmatrix}
2e^{2t}\\
-2e^{2t}
\end{bmatrix}=x_1'(t)$
$Ax_2(t)=\begin{bmatrix}
3 & 1\\
-1 & 1
\end{bmatrix}\begin{bmatrix}
te^{2t}+e^{2t}\\
-te^{2t}
\end{bmatrix}=\begin{bmatrix}
2te^{2t}+3e^{2t}\\
-2te^{2t}-e^{2t}
\end{bmatrix}=x_2'(t)$
The Wronskian of these solutions is
$W_{[x_1,x_2]}(t)=\begin{vmatrix}
e^{2t} & te^{2t}+e^{2t}\\
-e^{2t} & te^{2t}
\end{vmatrix}=e^{4t}$
Since the Wronskian is never zero, it follows that $\{x_1, x_2\}$ is linearly independent on any interval and so is a fundamental set of solutions for the given vector differential equation. Therefore, the general solution to the system is
$x(t)=\begin{bmatrix}
c_1e^{2t} +c_2e^{2t}(t+1) \\
-c_1e^{2t}- c_2te^{2t}
\end{bmatrix}$
Since $x(0)=\begin{bmatrix}
-2 \\
1
\end{bmatrix} \rightarrow C_1=1,C_2=-1$
we have
$x(t)=\begin{bmatrix}
e^{-t}(-3+e^{5t}) \\
e^{-t} (-1+2e^{5t})
\end{bmatrix}$
