Unique $T$-periodic solution to $x'=a(t)x+b(t)$ I think that conceptually I understand this question but I'm struggling to formally it in writing. The questions states "Let $a,b : \mathbb{R} → \mathbb{R}$ be $T$-periodic continuous functions. Suppose the equation $x′ = a(t)x$ has no nonzero $T$-periodic solutions, prove that the equation $x′ = a(t)x + b(t)$ has a unique $T$-periodic solution".
I can see that a periodic solution would exist because we simply would have the period of $b(t)$, but I am unsure how to prove uniqueness.
 A: Let $x=x(t)$ be a particular solution of $x'(t) = a(t)x(t)+b(t)$. Note that since $a(t+T) = a(t)$ and $b(t+T)=b(t)$, then
$$x'(t+T) = a(t)x(t+T)+b(t).$$
This means that $y(t) := x(t+T)$ is a solution. But then $y$ must be of the form
$$y(t) = x(t+T) = x(t)+Ce^{\int_0^t a(s)\,ds}, \quad C \in \mathbb R.$$
To see this, note that $x$ and $y$ solutions means
\begin{align*}
y'(t) & = a(t)y(t)+b(t), \\
x'(t) & = a(t)x(t)+b(t),
\end{align*}
so that $y-x$ is a solution of the homogeneous equation $z' = a(t)z$, whose general form is $Ce^{\int_0^t a(s)\,ds}$.
Let us look for $T$-periodic solutions now. If $y(t+T) = y(t)$, then necessarily $y(0) = y(T)$, which by the above implies
$$x(T) = y(0) = y(T) = x(T)+Ce^{\int_0^T a(s)\,ds},$$
so $C=0$. Then $y(t) = x(t+T) = x(t)$, namely the particular solution we took is indeed $T$-periodic. This shows that a $T$-periodic solution exists.
As for uniqueness, take $x_1 = x_1(t)$ and $x_2 = x_2(t)$ two $T$-periodic solutions. Then $x_1-x_2$ is a $T$-periodic solution of $z' = a(t)z$, so it must be zero by your assumption, and hence $x_1 = x_2$
