The Hill's equation $(H)$ is defined as $y'' + p(t)y=0$ where $p(t+T)=p(t) \forall t$. Let $x_1=y$ and $x_2=y'$, and let $x=\begin{bmatrix}{x_1}\\{x_2}\end{bmatrix}$, so $x'=\begin{bmatrix}{0}&{1}\\{-p(t)}&{0}\end{bmatrix}x$.
Now we consider the solutions of $(H)$ with the following initial conditions: $\varphi_1(0) = 1$, $\varphi_2(0) = 0$, $\varphi_1'(0) = 0$ and $\varphi_2'(0) = 1$.
My Floquet's Theory notes says that $\phi(t) = \begin{bmatrix}{\varphi_1}&{\varphi_2}\\{\varphi_1'}&{\varphi_2'}\end{bmatrix}$ is a fundamental matrix of $(H)$, but I don't know why. Could anybody explain that? Thanks
The condition that $p(t)$ is periodic is not necessary. You also have an error in the definition of $\varphi_1, \varphi_2$.
Consider a linear system of ODEs, $x'(t)=A(t)x(t)$, where $x(t)$ is an $\mathbb{R}^n$-valued unknown function and $A(t)$ is a given $n\times n$ matrix-valued function. This system defines for each $t\in\mathbb{R}$ a linear transformation $F(t):\mathbb{R}^n\to\mathbb{R}^n$, sending $v\in\mathbb{R}^n$ to the value at $t$ of the solution to the system whose value at $t=0$ is $v$. The matrix of $F(t)$ with respect to the standard basis of $\mathbb{R}^n$, as a function of $t$, is called the fundamental matrix of solutions of the system. Its columns form a basis of the space of solutions of the system. The $j$-th column is the solution whose value at $t=0$ is the $j$-th standard vector (1 at the $j$-th place, the rest 0). If you apply this to the Hill's equation you get that, in your notation, $\varphi_1(0)=1,\varphi'_1(0)=0, \varphi_2(0)=0,\varphi'_2(0)=1$.