How do I solve the differential equation for the motion of pendulum? In order to determine the equation for displacement, we need to solve the following second order differential equation: 
$$\frac{d^2x}{dt^2}=-g\sin \left(\frac{x}{l}\right)$$
where $\theta = x/l$. I am not sure how to get started on this. 
 A: Going off of what paulinho said in the comments, we first approximate that $\sin(\frac{x}{l})\approx\frac{x}{l}$, meaning our equation reduces to: 
$$ x'' +\frac{gx}{l}=0$$
This is a homogeneous second order linear differential equation with constant coefficients and auxiliary polynomial $t^2+\frac{g}{l}$. This can be rewritten as $(t-\frac{i\sqrt{g}}{\sqrt{l}})(t+\frac{i\sqrt{g}}{\sqrt{l}})$. It follows that the following set should be a basis for the solution space:
$$ \{e^{it\sqrt{\frac{g}{l}}},e^{-it\sqrt{\frac{g}{l}}}\}\longrightarrow \{\cos t\sqrt{\frac{g}{l}},\sin t\sqrt{\frac{g}{l}}\}$$
Since 
$$ \cos t\sqrt{\frac{g}{l}} = \frac{1}{2}\left(e^{it\sqrt{\frac{g}{l}}}+e^{-it\sqrt{\frac{g}{l}}}\right)\quad\textrm{and}\quad \sin t\sqrt{\frac{g}{l}}=\frac{1}{2i}\left(e^{it\sqrt{\frac{g}{l}}}-e^{-it\sqrt{\frac{g}{l}}}\right)$$
it follows that $\{\cos t\sqrt{\frac{g}{l}},\sin t\sqrt{\frac{g}{l}}\}$ should be a basis for the solution space since if $\{x,y\}$ is a basis for a vector space over $\mathbb{C}$, then so is $\{\frac{1}{2}(x+y),\frac{1}{2i}(x+y)\}$ and we would prefer this basis since we are familiar with sine and cosine. Thus we know that the solution should be of the form:
$$x(t)=c_1 \cos t\sqrt{\frac{g}{l}}+c_2\sin t\sqrt{\frac{g}{l}}\quad c_2,c_2\in\mathbb{R}$$ 
