Transform an equation of order $2$ ($y''(t) + \mu y'(t) + \sin(y(t)) = 0$) to a system of order $1$ We consider the ODE of the damped pendulum :
$y''(t) + \mu y'(t) + \sin(y(t)) = 0$. $(1)$
We note $x = (x_1,x_2) = (y, y')$, we want to show that the solutions of $(1)$ can be described by the solutions of a system of equations of order $1$ : 
$x'(t) = f(x(t))$, $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$. $(2)$ and determine explicitly $f$.
I would like to know how to transform an equation of order $2$ to a system of order $1$ ? (sorry for my bad english)
 A: The usual trick is to set $y'= z$.  Then
$$z' = y'' = -\mu y' -\sin y = \mu z - \sin y$$
That is:
$$y'=z $$
$$ z'= - \mu z - \sin y$$
a first-order system.
A: Set
$x_1(t) = y(t), \tag{1}$
and
$x_2(t) = y'(t); \tag{2}$
then the equation
$y''(t) + \mu y'(t) + \sin(y(t)) = 0 \tag{3}$
becomes
$x_2'(t) + \mu x_2(t) + \sin (x_1(t)) = 0, \tag{4}$
since from (2) we have
$x'_2(t) = y''(t).  \tag{5}$
We also see from (1) and (2) that
$x'_1(t) = y'(t) = x_2(t), \tag{6}$
so if we set
$\vec x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \tag{7}$
then we may define the function $f:\Bbb R^2 \to \Bbb R^2$ by
$f(\vec x) = f(x_1, x_2) = \begin{pmatrix} x_2 \\  -\mu x_2 - \sin x_1 \end{pmatrix}; \tag{8}$
we thus have
$\vec x'(t) = \begin{pmatrix} x'_1(t) \\ x'_2(t) \end{pmatrix} =  \begin{pmatrix} x_2(t) \\  -\mu x_2(t) - \sin (x_1(t)) \end{pmatrix} = f(\vec x(t)) . \tag{9}$
(9) is a first-order system equivalent to the given second-order system (3).  
This derivation illustrates the general policy for creating first order systems from second order systems:  introduce a new variable (in this case $x_2$) which equal to, indeed, merely "renames" the first derivative(s) of the dependent variable in the second order system (here $y'$).  Then $x'_2(t) = y''(t)$ und so weiter.
