Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation
$$y'' + qy = 0. \tag{1}$$
Let $y_1$ and $y_2$ be the solutions of (1) with initial conditions $y_1(0)=1, y_1'(0)=0$ and $y_2(0)=0, y_2'(0)=1$.
Show that:
$$\forall t \in \mathbb{R}, \quad y_1(t) y_2'(t) - y_1'(t) y_2(t) = 1.$$