I had a ODE of second order, then I built the following system of differential equations:
$$x' = y$$ $$y' = -\lambda sin(x) -2sin(x) -sin(x)cos(x)$$
With $\lambda = \frac{w^2}{\Omega^2}$ . I want to find the bifurcation points of the system. First of all I find the critical points, which are all of the form $(\pi k,0)$ with $k \in \mathbb{Z}$. Let's focus on k = 0 for simplicity.
I tried to linearize it, and I got:
$$x' = y$$ $$y' = -\lambda x -2x $$
I hope it is right. Now, I find that the matrix of the system is
\begin{bmatrix} 0 & 1 \\ -\lambda & -2 \end{bmatrix}
The corresponding eigenvalues are: $x_{1,2} = -1 \pm \sqrt{1-\lambda}$
Is the procedure correct until now? What can I say about the stability?