(1)
Rewrite the first equation: $x_{2} = \frac{1}{2}x_{1}'$
Then: $x_{2}'' = \frac{1}{2}x_{1}''$
Plug into the second equation: $\frac{1}{2}x_{1}'' = -2x_{1}$
Therefore, we get $x_{1}'' + 4x_{1} = 0$ which is a second order ODE of $𝑥_{1}$
(2)
\begin{align*}
Let A &=
\begin{bmatrix}
0 & 2 \\
-2 & 0
\end{bmatrix}\\
A-\lambda I &=
\begin{bmatrix}
-\lambda & 2 \\
-2 & -\lambda
\end{bmatrix}\\
det(A-\lambda I) &= \lambda^{2} + 4 = 0\\
\lambda &= \pm 2i\\
\end{align*}
For $\lambda = 2i$:
\begin{align*}
A-\lambda I &=
\begin{bmatrix}
-2i & 2 \\
-2 & -2i
\end{bmatrix}\\
null\begin{bmatrix} -2i & 2 \\ -2 & -2i \end{bmatrix}
&= span{\begin{bmatrix} 1\\ i \end{bmatrix}\\}\\
so, \ the \ eigenvector \ V &= \begin{bmatrix} 1\\ i \end{bmatrix}\\
Then, \ e^{\lambda t}V &= e^{2it}\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= (cos2t + isin2t)\begin{bmatrix} 1\\ i \end{bmatrix}\\
&= \begin{bmatrix} cos2t + isin2t\\ icos2t - sin2t \end{bmatrix}\\
Thus, \ \phi_{1}(t) &= \begin{bmatrix} cos2t\\- sin2t \end{bmatrix}\\
\phi_{2}(t) &= \begin{bmatrix} sin2t\\cos2t \end{bmatrix}\\
Therefore, \ x_{1} &= c_{1}cos2t + c_{2}sin2t\\
x_{2} &= -c_{1}sin2t + c_{2}cos2t\\
Since, \ x_{1}(0) &= 3 \ and \ x_{2}(0) = 4\\
So, \ c_{1} &= 3 \ and \ c_{2} = 4\\
\end{align*}
Therefore,
\begin{align*}
x_{1} &= 3cos2t + 4sin2t\\
x_{2} &= -3sin2t + 4cos2t\\
\end{align*}
