Why is a solution to $y'' + y = 0$ periodic

Question

Solutions to $y'' + y = 0$ are 2$\pi$ periodic. Is this accidental or does this ODE have some symmetries associated with it that force the solutions to be periodic?

Answer 1

The characteristic polynomial is $$\lambda ^2+1=0, \lambda = \pm i$$ which implies the solutions are $$y=c_1 \cos(t)+c_2 \sin(t)$$

Thus the solutions are periodic with period $2\pi$

You may also check the phase plane of the system $$ y'=u,u'=-y$$ which implies $$y^2+u^2=C$$ hence the solutions are periodic.

Answer 2

Well, the equation $y'' + y = 0$ it conserves the quantity $y'^2/2+y^2/2$, so the trajectories are confined to this circle in phase space. Moreover, the equation is autonomous and has unique solutions, so the system can't wobble back and forth around the circle; it can only spiral around it. This is almost sufficient to show that any solution must be periodic; I think it only remains to show that the system continues moving at a rate which is bounded below by a positive number, rather than approaching a steady state.

(Answer based on a comment by Ian.)