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.
Best Answer
$\begin{pmatrix} 2+i \\ 1\end{pmatrix}$ is proportional to $\begin{pmatrix} 5 \\ 2-i\end{pmatrix}$, so your solution is equivalent to the one given. You didn't do anything wrong.