differential equation of periodic function $f: \mathbb{R} \to \mathbb{R}$ positive and periodic function with period $p>0$,
$y(x)$ is the solution to $y'(x) = f(y(x))$ ,let $T = \int \limits_{0}^{p} \frac{dt}{f(t)}$ prove that $y(x+T)-y(x) = p$ ?
 A: Because $f$ is periodic
$$T = \int \limits_{0}^{p} \frac{dt}{f(t)}=\int \limits_{s}^{p+s} \frac{dt}{f(t)}$$
Now, the variable change $t=y(z)$ simplifies the integral 
$$T =\int \limits_{s}^{p+s} \frac{dt}{f(t)}=\int \limits_{y^{-1}(s)}^{y^{-1}(p+s)} \frac{y'(z)dz}{f(y(z))}$$
because, by hypothesis,  $y'(z) = f(y(z))$
$$T=\int \limits_{y^{-1}(s)}^{y^{-1}(p+s)} \frac{f(y(z))dz}{f(y(z))}=\int \limits_{y^{-1}(s)}^{y^{-1}(p+s)}dz=y^{-1}(p+s)-y^{-1}(s))$$
Now
$T+y^{-1}(s)=y^{-1}(p+s)$ and $y(T+y^{-1}(s))=p+s$
Finally, define $y^{-1}(s)=x$  yielding  $s=y(x)$ and
$$y(T+x)-y(x)=p$$
A: $f(t)$ is periodic with period $p$.
So $T = \int \limits_{0}^{p} \frac{dt}{f(t)} = \int \limits_{x}^{x + p} \frac{dt}{f(t)} \,\, \, \forall \, x \, \epsilon \, \mathbb{R}$.
Let $G: \mathbb{R} \to \mathbb{R}$ such that $\frac{\mathrm{d}G(x)}{\mathrm{d}x}=\frac{1}{f(x)}$
$G$ is the antiderivative of $\frac{1}{f(x)}$.
Then
$G(x+p)-G(x) \tag{1}\label{eq1} = T$
By definition, $y'(t)=f(y(t))$ .
Let $H(t)=G(y(t))$
Then $H'(t) = G'(y(t)) \, y'(t) = \frac{1}{f(y(t))} \, f(y(t)) = 1$
So $H(t) = t + C$ for some constant $C$. And
$H(t+T) - H(t) = T$
Then
$G(y(t+T))-G(y(t)) = T \tag{2}\label{eq2}$
Replacing $y(t)$ with $x$ in the equation $\eqref{eq2}$, we get
$G(y(t+T))-G(x) = T \tag{3}\label{eq3}$
$G'(x) = \frac{1}{f(x)} > 0 \Rightarrow G(t)$ is injective.
Comparing $\eqref{eq1}$ with $\eqref{eq3}$ and using the injective property of $G$, we can conclude that
$y(t+T)=x+p=y(t)+p$.
$y(t+T) - y(t) = p$
