Suppose both $a$ and $f$ are real-valued, periodic with period $\omega.$
(a)Show that $x'+a(t)x=f(t)$ has a unique solution iff $\int_{0}^{\omega}a(t) dt \neq0.$
(b) Show that, if $\int_{0}^{\omega}a(t) dt =0,$ then the equation in (a) may have either no $\omega$-periodic solution or all of its solutions are $\omega$-periodic.
My approach:From the periodicity, we can write $f(t+\omega)=f(t)$ and $a(t+\omega)=a(t)$ for all $t \in \mathbb{R}.$ For part (a), assuming there is a unique $\omega$-periodic solution, by using the method of integrating factor, we get $$x(t)=e^{-a(t)} \int_{0}^{t} f(s) e^{a(s)}~ds + K.$$ That's all I can get. Can anyone help me to solve this problem. Thank you for your time.
Some hints: Let $A(t)=\int a(t) dt$ be an indefinite integral of $a$ and show that the equation is the same as $$ \frac{d}{dt} \left( x(t) e^{A(t)} \right) = f(t) e^{A(t)}$$
Suppose $x_1$ and $x_2$ are two w-periodic solutions and set $z=x_1-x_2$.
[I presume that in the statement (a) it should read: ... unique w-periodic solution iff ...]
Taking differences of the odes for $x_1$ and $x_2$ we get $\frac{d}{dt}(z(t)e^{A(t)})'=0$ so $z(t)e^{A(t)}$ is constant in time. Thus $z(w)e^{A(w)}=z(0)e^{A(0)}$. Now $A(w)-A(0)=\int_0^w a(t)dt$. If the integral is non-zero then $A(w)$ and $A(0)$ are different. So either $z(w)=z(0)=0$ (and the solutions are the same) or $z(w)\neq z(0)$ and at least one of the solutions is not w-periodic. If $x_1$ is an arbitrary solution and $A(w)\neq A(0)$ then you construct $x_2(t)=x_1(t)+ c e^{A(t)}$ and find the value of $c$ (it's a linear equation) so that $x_2$ becomes w-periodic.
When $A(w)=A(0)$ then $z(w)=z(0)$ no matter how. So if $x_1$ is any solution, then either it is w-periodic and any other solution $x_2=x_1+ce^A$ is also w-periodic or no solution is w-periodic.