Stability of $y = \cos(2t)$ for the ode $y'' + 4y = 0$ I want to examine the stability of the solution $\phi = \cos(2t)$ of the ode $y'' + 4y = 0$.
I know the general solution of this ode is $y = c_1 \cos(2t) + c_2 \sin(2t)$. So to examine the stability of my solution, I need to see if other solutions stay close to it starting at $t = 0$. It'll be asymptotically stable if other solutions converge to it.
So, when $t = 0$, $\phi(0) = \cos(0) = 1$. and $\phi'(0) = -\sin(0) =0$. But.. how do I pick $y(t)$ so that $y(0)$ starts out close to $1$ and $y'(0)$ is close to $0$? Then what?
Well, $y(t) = 1$ when $t = 0$ and when $t = 0$, $y'(0) = 1$. I'm not sure how to proceed. 
 A: Since all solutions are periodic, this is stable but not asymptotically stable: solutions that start close stay close, but do not converge to your solution.
A: Note that :
$$-1 \leq \sin(2t) \leq 1 \quad \text{and} \quad -1 \leq \cos(2t) \leq 1$$
That means that your solution will be bounded, by let's say, some $m \in \mathbb R$ and $M \in \mathbb R$, such that :
$$m \leq y(t) \leq M$$
This automatically tells us that the system is stable, since its solutions do not diverge for any values of $c_1,c_2 \in \mathbb R$ arbitrary constants.
But, note that the limits $\lim_{t\to  \infty} \sin(2t)$ and $\lim_{t\to  \infty} \cos(2t)$ do not exists, which means that your solutions will not get arbitrary close to each other.
Thus, the given ODE solution is simply stable.
For a more formal definition (which proves our conclusions above) recall that :

Definition : A solution of an ODE is stable if for every $\varepsilon >0$ there exists $\delta >0 $ when if $\hat{y}(t)$ satisfies the ODE, it is :
  $$\|\hat{y}(t) - y(t_0) \| \leq \delta \implies  \|\hat{y}(t) - y(t)\| \leq \varepsilon \; \forall t \geq t_0$$

