Let $F:\Bbb{R} \to \Bbb{R}$ be a positive, smooth and periodic function with period $p>0$. Prove that if $\Phi(t)$ is a solution to the differential equation $x'=F(x)$ and
$$ T=\int_0^p {1\over F(y)}dy$$
then: $$\Phi(t+T)-\Phi(t)=p,\, \forall t \in \Bbb{R}$$
I already proved that if G is the antiderivate function of ${1\over F(y)}$ then $G(y+p)-G(y)=C$ where $C$ is a constant, but I don't know how to use this to solve the problem, it will help a lot a hint on how can I use this or where I can start to solve this problem. (Sorry if my english is bad)
Since $F$ (hence $1/F$) is $p$-periodic you have $$ \forall \xi \in \mathbb{R}: ~ \int_\xi^{\xi+p} \frac{1}{F(y)} dy=T. $$ By the method of separation of variables $$ \int_{\Phi(t)}^{\Phi(t+T)} \frac{1}{F(y)}dy = \int_{t}^{t+T} \frac{\Phi'(s)}{F(\Phi(s))}ds = T = \int_{\Phi(t)}^{\Phi(t)+p} \frac{1}{F(y)} dy. $$ Hence $\Phi(t+T)= \Phi(t)+p$.