Answer
SEE BELOW
Work Step by Step
Given:
$x_1(t)=\begin{bmatrix}
e^{4t}\\
2e^{4t}
\end{bmatrix}$
$x_2(t)=\begin{bmatrix}
3e^{-t}\\
e^{-t}
\end{bmatrix}$
$A=\begin{bmatrix}
-2 & 3\\
-2 & 5
\end{bmatrix}$
$x(0)=\begin{bmatrix}
-2 \\
1
\end{bmatrix}$
Differentiating the given vector functions with respect to t yields,
$x_1'(t)=\begin{bmatrix}
4e^{4t}\\
8e^{4t}
\end{bmatrix}$
$x_2'(t)=\begin{bmatrix}
-3e^{-t}\\
-e^{-t}
\end{bmatrix}$
whereas
$Ax_1(t)=\begin{bmatrix}
-2 & 3\\
-2 & 5
\end{bmatrix}\begin{bmatrix}
e^{4t}\\
2e^{4t}
\end{bmatrix}=\begin{bmatrix}
4e^{4t}\\
8e^{4t}
\end{bmatrix}$
$Ax_2(t)=\begin{bmatrix}
-2 & 3\\
-2 & 5
\end{bmatrix}\begin{bmatrix}
3e^{-t}\\
e^{-t}
\end{bmatrix}=\begin{bmatrix}
-3e^{-t}\\
-e^{-t}
\end{bmatrix}$
The Wronskian of these solutions is
$W_{[x_1,x_2]}(t)=\begin{vmatrix}
e^{4t} & 3e^{-t}\\
2e^{4t} & e^{-t}
\end{vmatrix}=-5e^{3t}$
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
$x(t)=\begin{bmatrix}
c_1e^{4t} +3c_2e^{-t} \\
2c_1e^{4t} + c_2e^{-t}
\end{bmatrix}$
Since $x(0)=\begin{bmatrix}
-2 \\
1
\end{bmatrix} \rightarrow C_1=1,C_2=-1$
we have
$x(t)=\begin{bmatrix}
e^{-t}(-3+e^{5t}) \\
e^{-t} (-1+2e^{5t})
\end{bmatrix}$
