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 in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$ to (1) satisfying $$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$ Show that if $y \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ is a solution of (1), then the function $t \mapsto y(t+T)$ is also one. Deduce that for all $t \in \mathbb{R}$: $$y(t+T) = y(T) y_1(t) + y'(T) y_2(t)$$
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 in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$ to (1) satisfying
$$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$
Show that if $y \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ is a solution of (1), then the function $t \mapsto y(t+T)$ is also one. Deduce that for all $t \in \mathbb{R}$:
$$y(t+T) = y(T) y_1(t) + y'(T) y_2(t)$$