Periodic solutions to ODE with constant coefficients Is there a straightforward way to see that the solutions of the equation $$x''  + x = 0$$ must be periodic without actually solving the equation?
 A: You can show (by differentiating) that $x^2 + (x')^2$ is constant for any solution.  Of course $x^2 + y^2 = 0$ only for the origin, which is an equilibrium point and thus a periodic solution. Parametrize the curve $x^2 + y^2 = r^2$ for $r > 0$ as $x = r \cos(\theta)$, $y = r \sin(\theta)$, and note that $\theta' < 0$  [I'm really trying hard to avoid "solving the equation" with an explicit formula for $\theta$).  If we start at $\theta(0) = \theta_0$, then as $t$ increases either we eventually reach $\theta(T) = \theta_0 - 2 \pi$, which implies $x(T) = x(0)$ and $x'(T) = x'(0)$ and thus a periodic solution, or approach some limit: $\theta(t) \to \theta_\infty$ as $t \to \infty$.  But then $(x = r \cos(\theta_\infty), y = r \sin(\theta_\infty))$ would have to be an equilibrium point other than $(0,0)$, and that does not exist.  
A: I'm not sure what you mean by "straightforward", but from your equation you know that a solution has to satisfy $x = -x''$. So, you need a function that is its own negated second derivative. Sines and cosines satisfy this requirement and they are periodic. 
A: Laplace-transforming $\ddot{x} + x = 0$, we obtain
$$s^2 X (s) - s x_0 - v_0 + X (s) = 0$$
where $x_0$ and $v_0$ are the initial conditions. Hence,
$$X (s) = \dfrac{x_0 s + v_0}{s^2 + 1}$$
The denominator tells us that the two poles are at $s = \pm i$. Since the poles are on the imaginary axis, we can conclude that the solution to the ODE is periodic.
