I am working through some practice problems, and I am getting a different answer from what the back of my book says:
Express the general solution of the given system of equations in terms of real valued functions:
$$\begin{equation*} x' = \begin{pmatrix} 2 & -5 \\ 1 & -2 \\ \end{pmatrix}x \end{equation*}$$
I get eigenvalues of $$\pm i$$ and eigenvectors
$$\begin{equation*} v_1 = \begin{pmatrix} 2+i \\ 1 \\ \end{pmatrix} \end{equation*}$$
$$\begin{equation*} v_2 = \begin{pmatrix} 2-i \\ 1 \\ \end{pmatrix} \end{equation*}$$
The general solution with complex values is
$$\begin{equation*} x = e^{it} \begin{pmatrix} 2+i \\ 1 \\ \end{pmatrix} \end{equation*}$$
which becomes
$$\begin{equation*} x = (\cos(t)+i\sin(t)) \begin{pmatrix} 2+i \\ 1 \\ \end{pmatrix} \end{equation*}$$
Expanding this and simplifying, I get
$$\begin{equation*} x = \begin{pmatrix} 2\cos(t)-\sin(t) \\ \cos(t) \\ \end{pmatrix} + i \begin{pmatrix} \cos(t)+2\sin(t) \\ \sin(t) \\ \end{pmatrix} \end{equation*}$$
and then adding the constants, I get
$$\begin{equation*} x = c_1 \begin{pmatrix} 2\cos(t)-\sin(t) \\ \cos(t) \\ \end{pmatrix} + c_2 \begin{pmatrix} \cos(t)+2\sin(t) \\ \sin(t) \\ \end{pmatrix} \end{equation*}$$
which is my final answer.
Looking at the solution at the back of the book, it says the correct answer is
$$\begin{equation*} x = c_1 \begin{pmatrix} 5\cos(t) \\ 2\cos(t)+\sin(t) \\ \end{pmatrix} + c_2 \begin{pmatrix} 5\sin(t) \\ -\cos(t)+2\sin(t) \\ \end{pmatrix} \end{equation*}$$
Where did I go wrong? Thank you.
