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:
$\frac{dx}{dt}=-\frac{3}{2}x-\frac{5}{2}y$,
$\frac{dy}{dt}=\frac{5}{2}x+\frac{3}{2}y$,
- 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? :-/
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.