Answer
See below
Work Step by Step
Given:
$x_1(t)=\begin{bmatrix}
t\sin t\\
\cos t
\end{bmatrix}$
$x_2(t)=\begin{bmatrix}
-t\cos t\\
\sin t
\end{bmatrix}$
$A=\begin{bmatrix}
\frac{1}{t} & t\\
-\frac{1}{t} & 0
\end{bmatrix}$
Differentiating the given vector functions with respect to t yields,
$x_1'(t)=\begin{bmatrix}
t\cos t+\sin t\\
-\sin t
\end{bmatrix}$
$x_2'(t)=\begin{bmatrix}
t\sin t-\cos t\\
\cos t
\end{bmatrix}$
whereas
$Ax_1(t)=\begin{bmatrix}
\frac{1}{t} & t\\
-\frac{1}{t} & 0
\end{bmatrix}\begin{bmatrix}
t\sin t\\
\cos t
\end{bmatrix}=\begin{bmatrix}
t\cos t+\sin t\\
-\sin t
\end{bmatrix}=x_1'(t)$
$Ax_2(t)=\begin{bmatrix}
\frac{1}{t} & t\\
-\frac{1}{t} & 0
\end{bmatrix}\begin{bmatrix}
-t\cos t\\
\sin t
\end{bmatrix}=\begin{bmatrix}
t\sin t-\cos t\\
\cos t
\end{bmatrix}=x_2'(t)$
The Wronskian of these solutions is
$W_{[x_1,x_2]}(t)=\begin{vmatrix}
t\sin t & -t\cos t\\
\cos t &\sin t
\end{vmatrix}=t$
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
$x(t)=\begin{bmatrix}
c_1t\sin t-c_2t\cos t \\
c_1\cos t+c_2\sin t
\end{bmatrix}$
