Draw phase diagram of $x'' + V(x) = 0$ conservative system, with $V(x) = omega^2 cos{x} - alpha x$
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I want to draw the phase diagram of $x'' + V(x) = 0$ , with $V(x) = omega^2 cos{x} - alpha x$ , with all constants greater than $0$ . I write this second order differential equation as a system in this way: $$ x' = y $$ $$ y' = omega^2 cos{x} - alpha x $$ Then I linearize the second equation finding: $$ y = omega^2 x + alpha $$ The equilibrium point is $(-frac{alpha}{omega^2},0)$ I then find the matrix A: begin{bmatrix} 0 & 1 \ omega^2 & 0 end{bmatrix} The eigenvalues are w and -w, so I find that the equilibrium point is unstable because I have an Eigen value with real part positive. Is this procedure right? From now on I am stuck, because I do not know how to draw the phase diagram.
differentia