Linearized equation of $x'(t) = f(x(t))$ in the neighbourhood of the equilibrium points found We consider the ODE of the damped pendulum :
(1) $$y''(t)+\mu y'(t)+\sin(y(t))=0$$
We note $x=\begin{pmatrix}
x_1\\ 
x_2
\end{pmatrix} = \begin{pmatrix}
y\\ 
y'\end{pmatrix}$ and we have :
(2)
$$x'(t) = \begin{pmatrix}
x_1'\\ 
x_2'
\end{pmatrix} = \begin{pmatrix}
x_2\\ 
-\mu x_2 - \sin(x_1)
\end{pmatrix} = f(x(t))$$
I have to find the equilibrium points of the equation (2). I find $(x_1,x_2) = (0,0)$ and $(x_1,x_2) = (\pi,0)$. And I have to write the linearized equation of (2) in the neighbourhood of the equilibrium points found. I don't understand the question. Soemone could help me ?
 A: When we linearise a system of ODEs, we are looking for a linear approximation of the system around the equilibrium points.
If we had a real-valued function $f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ we would approximate it around a point $(x_0,y_0)$ using a Taylor's approximation:
$$f(x,y) \approx f(x_0,y_0) + \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)$$
So if we look at the system, similar to yours:
\begin{matrix}
\frac{dx}{dt} = f(x,y)\\ 
\frac{dy}{dt} = g(x,y)
\end{matrix}
If we approximate it linearly around a point $(x_0,y_0)$ we look at the linearly approximations of $f$ and $g$ respectively:
\begin{matrix}
\frac{dx}{dt} = f(x_0,y_0) + \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)\\
\frac{dy}{dt} = g(x_0,y_0) + \frac{\partial g}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial g}{\partial y}(x_0,y_0)(y-y_0)
\end{matrix}
If $(x_0,y_0)$ is an equilibrium point, then $f(x_0,y_0) = 0$ and $g(x_0,y_0) = 0$, so our system becomes:
\begin{matrix}
\frac{dx}{dt} = \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)\\
\frac{dy}{dt} = \frac{\partial g}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial g}{\partial y}(x_0,y_0)(y-y_0)
\end{matrix}
As we can see, this is a linear system (the highest power of $x$ or $y$ is $1$).
The matrix
$$J(x_0,y_0) = \begin{pmatrix}
\frac{\partial f}{\partial x}(x_0,y_0) & \frac{\partial f}{\partial y}(x_0,y_0) \\ 
\frac{\partial g}{\partial x}(x_0,y_0) & \frac{\partial g}{\partial y}(x_0,y_0)
\end{pmatrix}
$$
is then called the Jacobian matrix of the system around the point $(x_0,y_0)$
The linearised system around the equilibrium point $(x_0,y_0)$ is then:
$$\begin{pmatrix}
\frac{dx}{dt}\\ 
\frac{dy}{dt}
\end{pmatrix} = \begin{pmatrix}
\frac{\partial f}{\partial x}(x_0,y_0) & \frac{\partial f}{\partial y}(x_0,y_0) \\ 
\frac{\partial g}{\partial x}(x_0,y_0) & \frac{\partial g}{\partial y}(x_0,y_0)
\end{pmatrix}\begin{pmatrix}
x(t)\\ 
y(t)
\end{pmatrix}$$.
A: You can write your system of equations in the form
$$
\frac{{\rm d}{\bf x}}{{\rm d}t} = {\bf F}({\bf x}) \tag{1}
$$
where the vector field ${\bf F}$ is defined by your last equation. The problem is asking you to find a version of the form
$$
\frac{{\rm d}{\bf z}}{{\rm d}t} = {\bf J}{\bf z} \tag{2}
$$
where ${\bf J}$ is a linear operator and ${\bf z}$ is your new dynamical variable. To do that, imagine that ${\bf x}_0$ is a fixed point of Eq. (1), that is, a point such that ${\bf F}({\bf x}_0) = 0$, call ${\bf z} = {\bf x} - {\bf x}_0$ and note that
$$
{\bf F}({\bf x}) \approx \underbrace{{\bf F}({\bf x}_0)}_{=0} + {\bf J}({\bf x}_0) ({\bf x} - {\bf x_0}) = {\bf J} {\bf z} \tag{3}
$$
where ${\bf J}({\bf x}_0)$ is the Jacobian of ${\bf F}$ evaluated at the fixed point ${\bf x}_0$. Note that at first order we then write Eq. (1) as 
$$
\frac{{\rm d}{\bf x}}{{\rm d}t} = \frac{{\rm d}({\bf z} + {\bf x}_0)}{{\rm d}t}= \frac{{\rm d}{\bf z}}{{\rm d}t} ={\bf F}({\bf x}) \approx {\bf J}{\bf z} ~~~~\Rightarrow~~~ \frac{{\rm d}{\bf z}}{{\rm d}t} = {\bf J}{\bf z}
$$
