How can I determine $P(t)$ and $B$? Please consider the following theorem:
Theorem (Floquet): Consider the equation $\dot{x} = A(t)x$ with $A(t)$ a continuous $T$-periodic $n\times n$-matrix. Each fundamental matrix $\Phi(t)$ of the equation $\dot{x} = A(t)x$ can be written as the product of two $n\times n$-matrices
$$
\Phi(t) = P(t)e^{Bt}
$$ 
with $P(t)$ $T$-periodic and $B$ a constant $n\times n$-matrix.
I want to solve the following exercise:
Consider the equation $\dot{x} = A(t)x$ with $A(t)$ a smooth $T$-periodic $n\times n$-matrix, $x\in\mathbb{R}^n$, $f(t)$ a smooth scalar $T$-periodic function. Consider the case when $n = 1, A(t) = f(t)$. Determine $P(t)$ and $B$ in the Floquet theorem. 
What I've tried: If $n = 1, A(t) = f(t)$ the fundamental matrix $\Phi(t) = x(t)$ is a scalar function and the solution to the equation $\dot{x} = f(t)x$. According to the Floquet theoren we can write $\Phi(t) = x(t) = P(t)e^{Bt} = p(t)e^{bt}$ where $p(t)$ is a $T$-periodic scalar function and $b$ a scalar. If I substitute this into the original equation I get the following
$$
\dot{x} = f(t)p(t)e^{bt}
$$
Unfortunately I don't really know how to proceed from here..
Question: How should I solve this exercise?
 A: Given that $x(t)=p(t)e^{bt}$ we get by substituting into the original equation $\frac{dx}{dt}=f(t)x$ that $p(t)$ must satisfy:
$$\dot{p}+(b-f(t))p=0\Rightarrow e^{\int^{t} (b-f(t'))dt'}p(t)=C\Rightarrow p(t)=Ce^{-bt}e^{\int^{t}f(t')dt'}$$
and thus the most general solution to the problem is:
$$x(t)=Ce^{\int^t f(t')dt'}=x(t_0)\exp\int_{t_0}^{t}f(t')dt'$$
for some arbitrary time $t_0$.
Note that there is some arbitrariness in the way we can define b in this simple problem. To finish we should choose it's value such that $p(t)$ is periodic. We have that
$$\frac{d}{dt}\ln\frac{p(t+T)}{p(t)}=\frac{d}{dt}\int_t^{t+T}(f(t')-b)dt'=f(t+T)-f(t)=0$$
and therefore for every t we get that:
$$p(t+T)=p(t)\exp\int_{t_0}^{t_0+T}(f(t')-b)dt', A\in\mathbb{R}^+$$
However from mean value theorem we are guaranteed that there exists a $\xi\in(0,T)$ (the mean value of f in that interval) s.t:
$$\frac{1}{T}\int_{t_0}^{t_0+T}f(t')dt'=f(\xi)$$
If we pick $b=f(\xi)$ we get a periodic $p(t)$ and the Floquet theorem requirements are fulfilled.
