I've spent a lot of time trying to figure out where the error is, but it has not been possible to find where the error is, unfortunately. This process of "debugging" is really tiresome, and this is probably the first time that I can't figure out what's wrong. I would certainly appreciate some help!

The system of ODEs given:



  • Rewrite in matrix form:

$\frac{d\vec{x}}{dt}=\begin{bmatrix} -3/2 & -5/2 \\ 5/2 & 3/2 \end{bmatrix}\vec{x}$

  • Find eigenvalues and eigenvectors:

$\lambda=\pm 2\rm{i}$

$\vec{v}=\begin{bmatrix} -3+4i\\ 5 \end{bmatrix}$ (multiplied by 5 to remove the fraction).

Now, $\vec{a}=\begin{bmatrix} -3\\ 5 \end{bmatrix}$, $\vec{b}=\begin{bmatrix} 4\\ 0 \end{bmatrix}$.

  • Solution

We can write the solution as $\vec{x}=c_1 (\vec{a}\cos(2t)-\vec{b}\sin(2t))+c_2(\vec{a}\sin(2t)+\vec{b}\cos(2t))$, thus

$\vec{x}(t) = c_1 \begin{bmatrix} -3\cos(2t) -4\sin(2t) \\ 5\cos(2t) \end{bmatrix} + c_2\begin{bmatrix} -3\sin(2t) +4\cos(2t) \\ 5\sin(2t) \end{bmatrix}$.

However, this solution is wrong. What is/are my errors? :-/

  • $\begingroup$ What is the given solution? $\endgroup$
    – Ian
    Dec 4, 2015 at 2:56
  • $\begingroup$ $\vec{x}(t) = \begin{bmatrix} \frac{1}{4} c_1 (4\cos(2t) -3\sin(2t))-\frac{5}{4} c_2 \sin(2t)\\ \frac{5}{4} c_1 \sin(2t)+\frac{1}{4}c_2 (3\sin(2t)+4\cos(2t)) \end{bmatrix}$. $\endgroup$
    – sequence
    Dec 4, 2015 at 3:02

1 Answer 1


Your result is :

$\vec{x}(t) = C_1 \begin{bmatrix} -3\cos(2t) -4\sin(2t) \\ 5\cos(2t) \end{bmatrix} + C_2\begin{bmatrix} -3\sin(2t) +4\cos(2t) \\ 5\sin(2t) \end{bmatrix}$.

$\vec{x}(t) = \begin{bmatrix} -3C_1+4C_2 \\ 5C_1 \end{bmatrix}\cos(2t) + \begin{bmatrix} -4C_1-3C_2 \\ 5C_2 \end{bmatrix}\sin(2t)$.

The textbook result is :

$\vec{x}(t) = \begin{bmatrix} \frac{1}{4} c_1 (4\cos(2t) -3\sin(2t))-\frac{5}{4} c_2 \sin(2t)\\ \frac{5}{4} c_1 \sin(2t)+\frac{1}{4}c_2 (3\sin(2t)+4\cos(2t)) \end{bmatrix}$

$\vec{x}(t) = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}\cos(2t)+ \begin{bmatrix} -\frac{3}{4} c_1 -\frac{5}{4} c_2 \\ \frac{5}{4} c_1 +\frac{3}{4}c_2 \end{bmatrix}\sin(2t)$

Let $c_2=5C_1$ and $c_1=-3C_1+4C_2$

$ \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}\cos(2t)+ \begin{bmatrix} -\frac{3}{4} c_1 -\frac{5}{4} c_2 \\ \frac{5}{4} c_1 +\frac{3}{4}c_2 \end{bmatrix}\sin(2t) = \begin{bmatrix} -3C_1+4C_2 \\ 5C_1 \end{bmatrix}\cos(2t) + \begin{bmatrix} -4C_1-3C_2 \\ 5C_2 \end{bmatrix}\sin(2t)$.

So, your result is correct. It is the same as the expected result, but with a different writting of the arbitrary constants.

  • $\begingroup$ But if we substitute my result into the equation, we'll get different LHS and RHS. $\endgroup$
    – sequence
    Dec 4, 2015 at 8:06
  • 1
    $\begingroup$ Your result is the same as the textbook result. Do you think that the textbook be mistaken ? No. Don't worry. If you make no mistake in substituing your result into the equation, you'll get equal LHS and RHS $\endgroup$
    – JJacquelin
    Dec 4, 2015 at 8:19

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