Explain how second order differential equations of the form $\ddot{y}+y=0$ exhibit osciallatory dynamics I'm trying to build a skillset for research in computational neuroscience (and loving math even more as I go along) and have just jumped into the world of differential equations – very simple ones. 
One of the models I've encountered is defined by the second order differential equation: 

Which when integrated (with scipy) yields oscillatory dynamics . 
I'm looking for an explanation of why oscillatory dynamics emerge from this form of differential equation. The same way the graph of $f(x)=2x$ is so clear, the best answer will relate the equation to the dynamics in such an intuitive way. 
 A: Not sure if this is what you're asking but$\dots$

This is the equation of a spring: $$m\ddot x =-kx$$
By Hooke's law: The force ($F=m\ddot x$ by Newton's law) done by a spring is equal to a constant $k$ (hardness of the spring) times the displacement $x$ (from the equilibrium position). From this viewpoint, it's only natural that the solutions are oscillatory.
A solution could be written as $x=R\cos \left(\sqrt\frac{k}{m}t+\phi_0\right)$ where $R$ is the amplitude, $\sqrt\frac{k}{m}$ is called the natural frequency and $\phi_0$ is the initial phase.
The extra factor $mg$ in your equation, is the force of gravity: I suppose the spring is hanging from the ceilling.
A: $$\frac{d^2}{dt^2} (\sin t) = -\sin t$$
$$\frac{d^2}{dt^2} (\cos t)  = -\cos t$$
A: Let $V(y_1,y_2) = {1 \over 2} (m y_1^2 +k (y_2-{mg \over k})^2)$
and consider $\phi(t) = V(\dot{x}(t), x(t))$, where
$t \mapsto x(t)$ is a solution of the differential equation.
$\phi$ represents the kinetic & potential energy of the system.
A quick computation shows that $\dot{\phi} = 0$, hence $V$ is constant on
trajectories of the differential equation.
For $c \ge 0$, the set $V^{-1} \{c\}$ describes an ellipse centred on $(0, -{mg \over k})$.
If we let $y_1=\dot{x}, y_2 = x$, we can write the differential equation as the first order equation:
$\dot{y} = \begin{bmatrix}0 & - { k \over m} \\
1 & 0\end{bmatrix} y + \begin{bmatrix}g \\ 0 \end{bmatrix}$. A little work shows that if $V(\dot{x}(0), x(0)) \neq 0$ there there is some $\underline{v} >0$  such that $\|\dot{y}(t)\| \ge \underline{v}$ for all $t \ge 0$, hence the trajectory is continuously moving on the translated ellipse. Hence the
oscillatory nature of the solution.
Note that a one dimensional (time invariant) system cannot exhibit an
oscillatory nature.
A: A Bit of Intuition
$\ddot{x}=-k^2x$ says that when $x\gt0$, the slope of $x$ is decreasing, and when $x\lt0$, the slope of $x$ is increasing. This means that $x$ is concave above the $x$-axis and convex below the $x$-axis. This is the behavior of a periodic function, but this behavior does not guarantee periodicity since this behavior is also exhibited by $\tanh(x)$

We need to solve the equation to prove periodicity.
Full Solution
$$
D^2+k^2=(D-ik)(D+ik)\tag{1}
$$
To invert the operator $D+ik$, note that
$$
\begin{align}
(D+ik)u=f
&\implies D\left(e^{ikx}u(x)\right)=e^{ikx}f(x)\\
&\implies u(x)=e^{-ikx}\int e^{ikx}f(x)\,\mathrm{d}x\tag{2}
\end{align}
$$
we can apply $(2)$ for $k$ and $-k$ to $(1)$ to get
$$
\left(D^2+k^2\right)u=f
\implies u(x)=e^{ikx}\int e^{-2ikx}\int e^{ikx}f(x)\,\mathrm{d}x\tag{3}
$$
Setting $f(x)=0$ in $(3)$ yields
$$
\begin{align}
u(x)
&=c_1e^{ikx}+c_2e^{-ikx}\\[6pt]
&=(c_1+c_2)\cos(kx)+i(c_1-c_2)\sin(kx)\\[6pt]
&=b_1\cos(kx)+b_2\sin(kx)\tag{4}
\end{align}
$$
where $c_1,c_2$ come from the constants of integration in $(3)$.
