Let $x'=f(x,t)$. Assume that $f$ is smooth, $T$-periodic (i.e. $f(x,t)=f(x,t+T)$ and $T$ is the smallest such value), and $\frac{\partial f}{\partial x}>0$ for all $x,t$. Show there is at most one $T$ periodic solution.
My attempt: Suppose $x$ and $y$ are two different $T$-periodic solutions. WLOG assume $x(t_0)\geq y(t_0)$. Then since $\frac{\partial f}{\partial x}>0$, we have $x'(t_0)=f(x(t_0),t_0)>f(y(t_0),t_0)=y'(t_0)$, so we must have that $x(t)>y(t)$ for $t_0 If anyone could point out a mistake, I would appreciate it. Thank you.