Consider the equation $\dot{y} = (acos(t) + b)y - y^3$ $a > 0, b>0$. I know that I need to recast the equation as a first order system $\dot{y} = (acos(x) +b)y - y^3, \dot{x} = 1$. Also, we are given that $\alpha \in R$ let $(x(t, \alpha), y(t, \alpha))$ denote the solution of the system with initial values $x(0, \alpha) = 0$, $y(0, \alpha) = \alpha$. Also, we know the line $\sum = \{(x, y)| x = 0\}$ with each line $ \{(x, y)| x = 2\pi*L\}$. L is an integer. We also have $T$ x $R$ where $T$ is the torus. $\alpha \in R$ is the coordinates of the point $\sum$, the Poincare' map is given by $P(\alpha) = y(2\pi, \alpha)$
I know that the fixed point of the Poincare' map corresponds to a periodic orbit by definition where 0 is a trivial solution.
I want to show $ P'(0) > 1)$ I know for this I need to look into the equation that satisfies the partial $y_{\alpha}$ at $\alpha = 0$.
I also need to use $P'(0) > 1$ , and continuity of $P'(\alpha)$ to show that there exists $ 0 < c < a + b + 1$ such that $P$ maps $[ c, a+b+1]$ into itself.
and finally I need to Prove P has a fixed point in [c, a+b+1].
I also know that I will take the derivative and plug in the critical value it's got to be greater than or less than some number for it to be stable. Any help will be greatly appreciated. Thanks.