I'm having trouble on this problem from Strogatz. Given a $T$-periodic, smooth function $f(t)$, is true that $\dot x +x=f(t)$ necessarily has a stable, $T$-periodic solution $x(t)$? I must either prove this or provide a counter example.
In my attempt to prove it, I have that the solution must be of the form
$$x(t)=e^{-t} \int e^t f(t)dt$$
Can I generalize a $T$-periodic function in any way that would allow me to solve this integral in a general sense?
Another approach would be to take advantage of the periodicity condition $f(t+T)=f(t)$ for all $t$ and the solution of the ODE. The general solution of the ODE is given by $$x(t)=e^{-(t-t_0)}x_0+\int_{t_0}^t{e^{-(t-s)}f(s)ds}\qquad\qquad (1)$$ Considering time $t+T$ we have $$x(t+T)=e^{-(t+T-t_0)}x_0+\int_{t_0}^{t+T}{e^{-(t+T-s)}f(s)ds}\qquad\qquad (2)$$ Changing the variable of integration to $w=s-T$ and using the periodicity of $f$ and (1) we obtain $$x(t+T)=e^{-(t+T-t_0)}x_0+\int_{t_0-T}^{t}{e^{-(t-w)}f(w+T)dw}\\ =e^{-(t+T-t_0)}x_0+\int_{t_0-T}^{t}{e^{-(t-w)}f(w)dw}\\ =e^{-(t+T-t_0)}x_0+e^{-t}\int_{t_0-T}^{t_0}{e^{w}f(w)dw}+\int_{t_0}^{t}{e^{-(t-w)}f(w)dw}\\ =x(t)+\left[e^{-(t+T-t_0)}-e^{-(t-t_0)}\right]x_0+e^{-t}\int_{t_0-T}^{t_0}{e^{w}f(w)dw}\\ =x(t)+e^{-t}\bigg\{\left[e^{t_0-T}-e^{t_0}\right]x_0+\int_{t_0-T}^{t_0}{e^{w}f(w)dw}\bigg\}\qquad\qquad (3)$$ From (3), for all intial times $t=t_0$, there exists an initial point $x_0$ given by $$x_0=\frac{1}{e^{t_0}-e^{t_0-T}}\int_{t_0-T}^{t_0}{e^w}f(w)dw$$ such that the trajectory which starts at $t_0$ from $x_0$ is $T$-periodic since $$x(t+T)=x(t)\qquad \forall t\geq t_0$$ i.e. a periodic solution always exists.