Start with the "exponential ansatz":
\begin{align*}
\binom{x(t)}{y(t)} = \binom{C}{D} e^{\lambda t} .
\end{align*}
Substituting this into
\begin{align*}
\binom{x'}{y'} =
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{x}{y}
\end{align*}
we arrive at the eigenvector problem
\begin{align*}
\lambda \binom{C}{D} =
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{C}{D}
\end{align*}
Let's find the eigenvalues:
\begin{align*}
\det
\begin{pmatrix}
2 - \lambda & 3 \\
-3 & 1 - \lambda
\end{pmatrix}
= 0
\end{align*}
or
\begin{align*}
\lambda^2 - 3 \lambda + 11 = 0
\end{align*}
so that
\begin{align*}
\lambda_1 = \frac{3}{2} + i \frac{\sqrt{35}}{2} , \qquad \lambda_2 = \frac{3}{2} - i \frac{\sqrt{35}}{2}
\end{align*}
As the eigenvalues are the complex conjugate of each other one of the eigenvectors will be proportional to the complex conjugate of the other. The eigenvector equation is
\begin{align*}
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{C}{D} =
(\frac{3}{2} + i \frac{\sqrt{35}}{2}) \binom{C}{D}
\end{align*}
or
\begin{align*}
\begin{pmatrix}
4 & 6 \\
-6 & 2
\end{pmatrix}
\binom{C}{D} =
(3 + i \sqrt{35}) \binom{C}{D}
\end{align*}
or
\begin{align*}
\begin{pmatrix}
1 - i \sqrt{35} & 6 \\
-6 & -1 - i \sqrt{35}
\end{pmatrix}
\binom{C}{D} = 0
\end{align*}
or
\begin{align*}
(1 - i \sqrt{35}) C + 6 D = 0
\nonumber \\
-6 C - (1 + i \sqrt{35}) D = 0
\end{align*}
Using the first condition, the two eigenvectors are
\begin{align*}
\vec{v}_1 = \binom{1}{\dfrac{-1 + i \sqrt{35}}{6}} , \qquad (\lambda_1 = \frac{3}{2} + i \frac{\sqrt{35}}{2})
\nonumber \\
\vec{v}_2 = \binom{1}{\dfrac{-1 - i \sqrt{35}}{6}} , \qquad (\lambda_2 = \frac{3}{2} - i \frac{\sqrt{35}}{2})
\end{align*}
The (real) general solution of the homogeneous differential equation is
\begin{align*}
\binom{x}{y} = C \binom{1}{\dfrac{- 1 + i \sqrt{35}}{6}} e^{(\frac{3}{2} + i \frac{\sqrt{35}}{2}) t} + C^* \binom{1}{\dfrac{-1 - i \sqrt{35}}{6}} e^{(\frac{3}{2} - i \frac{\sqrt{35}}{2}) t}
\end{align*}
Writing ##C = C_1 + i C_2##, we have
\begin{align*}
\binom{x}{y} = (C_1 + i C_2) \binom{1}{\dfrac{-1 + i \sqrt{35}}{6}} e^{(\frac{3}{2} + i \frac{\sqrt{35}}{2}) t} + c.c.
\end{align*}
so that
\begin{align*}
\binom{x}{y} = (C_1 + i C_2) \binom{1}{\dfrac{-1 + i \sqrt{35}}{6}} (\cos (\frac{\sqrt{35}}{2} t) + i \sin (\frac{\sqrt{35}}{2} t)) e^{3t/2} + c.c.
\end{align*}
or
\begin{align*}
\binom{x}{y} = \binom{C_1 + i C_2}{\dfrac{(-C_1 - C_2 \sqrt{35}) + i (-C_2 + C_1 \sqrt{35})}{6} } (\cos (\frac{\sqrt{35}}{2} t) + i \sin (\frac{\sqrt{35}}{2} t)) e^{3t/2} + c.c.
\end{align*}
or
\begin{align*}
\binom{x}{y} = 2 \binom{C_1 \cos (\frac{\sqrt{35}}{2} t) - C_2 \sin (\frac{\sqrt{35}}{2} t)}
{\dfrac{-C_1 - C_2 \sqrt{35}}{6} \cos (\frac{\sqrt{35}}{2} t) - \dfrac{-C_2 + C_1 \sqrt{35}}{6} \sin (\frac{\sqrt{35}}{2} t)} e^{3t/2}
\end{align*}
Using ##x(0)=1## and ##y(0)=2##,
\begin{align*}
\binom{1}{2} = 2 \binom{ C_1 }{ - \dfrac{ C_1 + C_2 \sqrt{35} }{6} }
\end{align*}
So ##C_1 = \frac{1}{2}##, ##1 = - \dfrac{ C_1 + C_2 \sqrt{35} }{6}##, ##C_2 = - \dfrac{13}{2 \sqrt{35}}##, and ##\dfrac{4}{\sqrt{35}} = \dfrac{ -C_2 + C_1 \sqrt{35} }{6}##.
So finally we have,
\begin{align*}
\binom{x}{y} = e^{2 t/3} \cos (\frac{ \sqrt{35} }{2}t) \binom{1}{2}
+ e^{2 t/3} \dfrac{\sin (\frac{ \sqrt{35} }{2}t)}{\sqrt{35}} \binom{13}{-8} .
\end{align*}