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 $y_2(0)=0, y_2'(0)=1$. 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 of (1) with initial conditions $y_1(0)=1, y_1'(0)=0$ and $y_2(0)=0, y_2'(0)=1$.
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 $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$.