$$
det(A-{\lambda}I)=0\\
\begin{vmatrix}
2-\lambda & -3 \\
4 & -2-\lambda
\end{vmatrix}={\lambda}^2+8=0
$$
So
$$
\lambda_1=\sqrt{8}i\\
\lambda_2=-\sqrt{8}i
$$
$$
\text{when } \lambda=\sqrt{8}I\\
\begin{vmatrix}
2-\sqrt{8}i & -3 \\
4 & -2-\sqrt{8}i
\end{vmatrix} = \begin{vmatrix}
0 \\
0
\end{vmatrix}
$$
RREF:
$$
\begin{pmatrix}
2-\sqrt{8}i & -3 \\
0 & 0
\end{pmatrix}
\quad = \begin{pmatrix}
0 \\
0
\end{pmatrix}
\quad
$$
Let x_2=t
So we can get:
$$
{(2-\sqrt{8}i})x_1=3x_2=3t\\
x_1=\frac{3t}{2-\sqrt{8}i}
$$
So
$$
t*\begin{pmatrix}
\frac{3}{2-\sqrt8i} \\
1
\end{pmatrix}
\quad
$$
Therefore:
$$
e^{i\sqrt8t}\begin{pmatrix}
\frac{3}{2-\sqrt8i} \\
1
\end{pmatrix}
\quad =
(cos(\sqrt8t)+isin(\sqrt8t))\begin{pmatrix}
\frac{3}{2-\sqrt8i} \\
1
\end{pmatrix}
\quad=\begin{pmatrix}
\frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} +\frac{isin(\sqrt8t)+i\sqrt2cos(\sqrt8t)}{2}\\
cos(\sqrt8t )+ isin(\sqrt8t))
\end{pmatrix}
\quad
$$
So
$$
y=c_1\begin{pmatrix}
\frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} \\
cos(\sqrt8t)
\end{pmatrix}
\quad +c_2\begin{pmatrix}
\frac{sin(\sqrt8t)+\sqrt2cos(\sqrt8t)}{2}\\
sin(\sqrt8t))
\end{pmatrix}
\quad
$$
OK, except LaTeX sucks:
1) * IS NOT a sign of multiplication
2) "operators" should be escaped: \cos, \sin, \tan, \ln
