Answer
See below
Work Step by Step
The given system can be written as $x′ = Ax$,
where $A=\begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}$
The transformed system is $y′ = (S−1AS)y$ where $x = Sy$
To determine S, we need the eigenvalues and eigenvectors of A. The characteristic polynomial of A is
$p(λ) = \begin{bmatrix}
-\lambda & 1\\
-1 & -\lambda
\end{bmatrix}=(-\lambda)(-\lambda)+1=(\lambda-i)(\lambda+i)$
Hence, A is nondefective by Corollary 7.2.10. The eigenvectors are easily computed:
$\lambda_1=i: v=r(1,i)\\
\lambda_2=-i: v=s(1,-i)$
Set $S = \begin{bmatrix}
1 & 1\\
i & -i
\end{bmatrix}$
then from Theorem 7.3.4, $S−1AS = diag(i, -i)$
so that the system becomes:
$\begin{bmatrix}
y_1'\\
y_2'
\end{bmatrix}=\begin{bmatrix}
i & 0\\
0 & -i
\end{bmatrix}\begin{bmatrix}
y_1\\
y_2
\end{bmatrix}$
Hence, $y_1'=iy_1\\
y_2'=-iy_2$
Both of these equations can be integrated to obtain $y_1(t)=C_1e^{it}\\
y_2(t)=C_2e^{-it}$
Return to the original variables, we have:
$x=Sy=\begin{bmatrix}
1 & 1\\
i & -i
\end{bmatrix}\begin{bmatrix}
C_1e^{it}\\
C_2e^{-it}
\end{bmatrix} \\
\rightarrow x_1(t)=C_1e^{it}+C_2e^{-it}\\
x_2(t)=iC_1e^{it}-iC_2e^{-it}$
