Not sure if this is the right place to ask but I am trying to solve systems of ODE's using eigenvectors and eigenvalues.
$$ x'(t) = \begin{bmatrix}5 & 4\\-2 & 1\end{bmatrix} \begin{bmatrix}x_1(t)\\x_2(t)\end{bmatrix}, \ \ x_1(0)=0, x_2(0)=1 $$
I have worked out the eigenvectors and eigenvalues to be $ λ_1 = 3 +2j $ and $ V_1 = \begin{bmatrix} -1-j \\ 1\end{bmatrix}$ and $ λ_2 = 3 - 2j $ and $ V_2 = \begin{bmatrix} -1+j \\ 1\end{bmatrix}$
Subbing them into the general formula gives:
$$ x(t)= e^{3t}(d_1(\cos2t\begin{bmatrix} -1 \\ 1\end{bmatrix} -\sin2t\begin{bmatrix} -1 \\ 1\end{bmatrix})+d_2(\cos2t\begin{bmatrix} -1 \\ 1\end{bmatrix}+\sin2t\begin{bmatrix} -1 \\ 1\end{bmatrix})) $$
However when trying to work out the values of the d_1 and d_2 I always get an error.
Can anyone help me understand where i've gone wrong, thank you.
You have two eigenvectors $v_\pm = v_1\pm jv_2$ with $$ v_1 = \begin{bmatrix} -1\\ 1 \end{bmatrix} $$ and $$ v_2 = \begin{bmatrix} -1\\ 0 \end{bmatrix} $$ for eigenvalues $\lambda_\pm = a\pm jb$. Recall that the general solution to the ode is of the form $$ x(t) = e^{at}\left(\ \ d_1(v_1\cos(bt) - v_2\sin(bt)) + d_2(v_1\sin(bt) + v_2\cos(bt))\ \ \right) $$ (see more details in eq (2) here). Or equivalently $$x(t) = e^{at}\left(\ \ d_1\left(\begin{bmatrix} -1\\ 1 \end{bmatrix}\cos(bt) - \begin{bmatrix} -1\\ 0 \end{bmatrix}\sin(bt)\right) + d_2\left(\begin{bmatrix} -1\\ 1 \end{bmatrix}\sin(bt) + \begin{bmatrix} -1\\ 0 \end{bmatrix}\cos(bt)\right)\ \ \right)$$
Compare this with your result. You basically used $v_1=v_2$