Let $f\in C^0(\Bbb{R},\Bbb{R})$.
Show that the equation $y''+2y'+2y=f \tag{E}$ has at most one periodic solution.
I can solve the equation $y''+2y'+2y=0$ but I am not sure it's useful here. I am tempted to say that if $y$ is a $T$-periodic solution of $(\mathrm{E})$ then $f$ is too because if I denote by $G(f)$ the set of real $T$ such that for all real $x$ we have $$f(x+T)=f(x)$$ is an additive subgroup of $\Bbb{R}$. Moreover we know that the subgroups of $\Bbb{R}$ are either of the form $\{0\}, a\Bbb{Z}$ with $a>0$ or dense on $\Bbb{R}$. Unfortunately I don't know how can I deal with the question.
Hint: the nontrivial solutions of the homogeneous equation decay as $t \to \infty$. What happens when you add a decaying function to a periodic one?
BTW, it is of course true that if $y$ is a periodic solution of (E) then $f$ is periodic: the derivative of a periodic (differentiable) function with period $T$ is periodic with period $T$, and a linear combination of functions that are periodic with period $T$ is periodic with period $T$.