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.

  • $\begingroup$ What happened to $\pmatrix{2-i\\1}$? $\endgroup$ May 5, 2021 at 23:52
  • 1
    $\begingroup$ Why do you think that your answer is different from the book? $\endgroup$ May 5, 2021 at 23:52
  • $\begingroup$ change $c_1 \to 2a - b$ and $c_2 \to a + 2b$ in your solution (this is allowed since $c_1$ and $c_2$ are arbitrary) $\endgroup$
    – guest
    May 5, 2021 at 23:56
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    $\begingroup$ @J.W.Tanner , there is a theorem saying in the system $x'=Ax$, where each value of A is real and continuous, and if $x=u(t)+iv(t)$ is a complex solution of $x'=Ax$, then its real part $u(t)$ and imaginary part $v(t)$ are also solutions to the equation. $\endgroup$ May 5, 2021 at 23:57

1 Answer 1


$\begin{pmatrix} 2+i \\ 1\end{pmatrix}$ 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.


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