This is related to the previous question that I asked yesterday.
Suppose the functions $a,f \in C(\mathbb{R} \rightarrow \mathbb{R})$ are both periodic with period $\omega.$ There are three parts for this question.
Show that $x'+a(t)x=f(t),$ has a unique $\omega$-periodic solution if and only if $\displaystyle{\int_{0}^{\omega} a(t)~dt \neq 0}.$ (This is DONE !)
There's one more question regarding the same equation. Show that, if $\int_{0}^{\omega}a(t)~dt=0,$ then the equation may have either no $\omega$-periodic solution or all of its solution are $\omega$-periodic. Also, under the same conditions, for which functions $f$ does the equation has $\omega$-periodic solutions ?
My approach: Let $A(\omega)=\int_{0}^{\omega}a(t)~dt.$ Suppose that $\int_{0}^{\omega}a(t)~dt=0.$ Then $e^{A(\omega)}=1,$ and thus we can't solve for $x(0),$ and hence there are no $\omega$-periodic solutions. I don't know how to the rest. Any help is much appreciated.
No, that's wrong.
The general solution of this linear differential equation can be written $$x(s) = \exp(-A(s)) \left(x(0) + \int_0^s f(t) \exp(A(t))\; dt\right)$$ where $A(s) = \int_0^s a(t)\; dt$. If $A(\omega) = 0$, that gives you $x(\omega) = x(0) + \int_0^\omega f(t) \exp(A(t))\; dt$.
Whether this solution is periodic or not depends on $\int_0^\omega f(t) \exp(A(t))\; dt$.