Answer
See below
Work Step by Step
Given:
$x_1(t)=\begin{bmatrix}
2t\\
e^t
\end{bmatrix}$
$x_2(t)=\begin{bmatrix}
0\\
3e^{t}
\end{bmatrix}$
$A=\begin{bmatrix}
\frac{1}{t} & 0\\
0 & 1
\end{bmatrix}$
Differentiating the given vector functions with respect to t yields,
$x_1'(t)=\begin{bmatrix}
2\\
e^t
\end{bmatrix}$
$x_2'(t)=\begin{bmatrix}
0\\
3e^{t}
\end{bmatrix}$
whereas
$Ax_1(t)=\begin{bmatrix}
\frac{1}{t} & 0\\
0 & 1 \end{bmatrix}\begin{bmatrix}
2t\\
e^t
\end{bmatrix}=\begin{bmatrix}
2\\
e^t
\end{bmatrix}=x_1'(t)$
$Ax_2(t)=\begin{bmatrix}
\frac{1}{t} & 0\\
0 & 1 \end{bmatrix}\begin{bmatrix}
0\\
3e^{t}
\end{bmatrix}=\begin{bmatrix}
0\\
3e^{t}
\end{bmatrix}=x_2'(t)$
The Wronskian of these solutions is
$W_{[x_1,x_2]}(t)=\begin{vmatrix}
2t & 0\\
e^{t} & 3e^{t}
\end{vmatrix}=6te^{t}$
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
$x(t)=\begin{bmatrix}
2c_1t \\
c_1e^{t}+3c_2e^{t}
\end{bmatrix}$
