Being constrained to move on the real line forces solutions to increase or decrease monotonically, or remain constant (fixed point). More precisely, non-constant solutions either approach a fixed point, or diverge to $\pm\infty$ monotonically. To state it more geometrically, a particle flowing on the real line according to $\dot{x}=f(x)$ never reverses direction. Hence there are no periodic solutions to such equations. Let’s prove analytically the absence of periodic solutions. Suppose on the contrary that $x(t)$ is a nontrivial periodic solution, i.e., $x(t)=x(t+T)$ for some $T>0$, and $x(t)\neq x(t+s)$ for all $0~~
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$$ \int_{t}^{t+T} f(x) \frac{\text{d}x}{\text{d}t} \ \text{d}t=\int_{x(t)}^{x(t+T)} f(x)\ \text{d}x=0. $$

The first equality follows from the chain rule, and the second from the assumption that $x(t)=x(t+T)$. On the other hand, since $\dot{x}=\text{d}x/\text{d}t=f(x)$,$$ \int_{t}^{t+T} f(x) \frac{\text{d}x}{\text{d}t} \ \text{d}t=\int_{t}^{t+T} \left(\frac{\text{d}x}{\text{d}t}\right)^2 \text{d}t\ >0 $$

by assumption that $T>0$ and $\text{d}x/\text{d}t$ does not vanish identically. This is a contradiction, thus there is no periodic solution.