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Suppose we have the following possibly nonlinear ODE:

$\dot{x} = f(x) + \cos\omega t$

Suppose that the equation has a periodic solution, i.e., $\exists$ T > 0 such that $x(t) = x(t+T)$.
Is it true that $T = k\frac{2\pi}{\omega}$ for some $k \in \mathbb{Z}^+$? If so, how would you show this?

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We have the ODE $$\begin{aligned} \frac{dx}{dt} &= f(x) + \cos{\omega t} \\ x &= \int f(x)dt + \int \cos{\omega t}\ dt \\ x(t) &= \int f(x(t)) \ dt + \frac{\sin {\omega t}}{\omega} + C \\ x(t+T) &= \int f(x(t+T)) \ d(t+T) + \frac{\sin {\omega (t+T)}}{\omega} + C \\ &= \int f(x(t+T)) \ dt + \frac{\sin {\omega (t+T)}}{\omega} + C \\ \text{if} \ T = \frac{2 \pi k}{\omega} \\ x(t+T) &= \int f(x(t+T)) \ dt + \frac{\sin {\omega t}}{\omega} + C \\ \because x(t+T) = x(t) \\ x(t+T) &= \int f(x(t)) \ dt + \frac{\sin {\omega t}}{\omega} + C \end{aligned}$$

Therefore, $x(t)$ is periodic with period $T =\frac{2 \pi k}{\omega}$

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$x$ and thus $\dot x$ are $T$-periodic. Thus $$\dot x(t)-f(x(t))$$ has $T$ as one of its period. If there is a smaller positive minimal period, it has the form $T/k$ for some integer $k$. On the other side is a function with frequency $ω$. Equality is only possible if $T/k=2\pi/ω$, or as claimed $$ T=\frac{2k\pi}{ω}. $$

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