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 $$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$ Let $\mu_1, \mu_2$ be the complex roots of the equation with unknown $x$: $$x^2 - \left(y_1(T) + y_2'(T)\right) x + 1 = 0$$ Suppose that $\mu_1 = \mu_2$. Show that $\mu_1 = \mu_2 = \pm 1$ and that equation (1) admits a periodic solution in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$.
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
$$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$
Let $\mu_1, \mu_2$ be the complex roots of the equation with unknown $x$:
$$x^2 - \left(y_1(T) + y_2'(T)\right) x + 1 = 0$$
Suppose that $\mu_1 = \mu_2$. Show that $\mu_1 = \mu_2 = \pm 1$ and that equation (1) admits a periodic solution in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$.