Find the solution of the system $x'=x-2y\text{ and }y'=4x+5y$ I have to find the solution of the system $x'=x-2y\:\:\&\:\:y'=4x+5y$ using eigenvalues ​​and eigenvectors of the matrix.
If I rewrite this system I get $$\begin{bmatrix}
x'\\
y'
\end{bmatrix}=\begin{bmatrix}
1 & -2 \\
4 & 5
\end{bmatrix} \begin{bmatrix}
x\\
y
\end{bmatrix}$$
So I got eigenvalue $\lambda_1=3+2i$ for eigenvector $v_1=\begin{bmatrix}
1\\
-1-i
\end{bmatrix}$ and eigenvalue $\lambda_2=3-2i$ for eigenvector $v_1=\begin{bmatrix}
1\\
-1+i
\end{bmatrix}$
Then I got solution for this system $$\begin{bmatrix}
x(t)\\
y(t)
\end{bmatrix}=C_1e^{(3+2i)t}\begin{bmatrix}
1\\
-1-i
\end{bmatrix}+C_2e^{(3-2i)t}\begin{bmatrix}
1\\
-1+i
\end{bmatrix}=\begin{bmatrix}
e^{3t}\cos(2t) & e^{3t}\sin(2t)\\
e^{3t}(-\cos(2t)+\sin(2t)) & e^{3t}(-\cos(2t)-\sin(2t))
\end{bmatrix} \begin{bmatrix}
C_1\\
C_2
\end{bmatrix}$$
Is this correct answer?

The solution for such systems (according to Perko2001 Differential Equations And Dynamical Systems) is as follows: Let
$$A= \begin{bmatrix}
1 & -2 \\
4 & 5
\end{bmatrix}$$
We have $\lambda_{1} = 3 + 2i$, $\lambda_2 = \overline {\lambda_1}= 3 - 2i$ and
$$v_1 = \begin{bmatrix}
-1+i \\
2
\end{bmatrix} = \begin{bmatrix}
-1  \\
2
\end{bmatrix} + i\begin{bmatrix}
1 \\
0
\end{bmatrix}$$
and
$$v_2 = \begin{bmatrix}
-1-i \\
2
\end{bmatrix} = \begin{bmatrix}
-1  \\
2
\end{bmatrix} + i\begin{bmatrix}
-1 \\
0
\end{bmatrix}$$
Now we construct the matrix $P$ from the imaginary and real parts of the eigenvectors;
$$P= 
\begin{bmatrix}
1 & -1 \\
0 & 2
\end{bmatrix}
\Rightarrow P^{-1} =\frac 1 {13} \begin{bmatrix}
3 & 2 \\
-2 & 3
\end{bmatrix}$$
and
$$P^{-1} A P =\begin{bmatrix}
3 & -2 \\
2 & 3
\end{bmatrix} $$
And the final solution
$$\begin{bmatrix}
x(t) \\
y(t)
\end{bmatrix} = P e^{3t} \begin{bmatrix}
\cos 2t & -\sin 2t \\
\sin 2t & \cos 2t
\end{bmatrix} P^{-1} \begin{bmatrix}
x_0  \\
y_0
\end{bmatrix}.$$
and
$$\begin{bmatrix}
x(t) \\
y(t)
\end{bmatrix} = \frac {e^{3t}} {13} \begin{bmatrix}
1 & -1 \\
0 & 2
\end{bmatrix}  \begin{bmatrix}
\cos 2t & -\sin 2t \\
\sin 2t & \cos 2t
\end{bmatrix}  \begin{bmatrix}
3 & 2 \\
-2 & 3
\end{bmatrix} \begin{bmatrix}
x_0  \\
y_0
\end{bmatrix}.$$
