Write the ODE $\ddot{x}+k\dot{x}+V'(x)=0$ as a system 
Write the ODE $\ddot{x}+k\dot{x}+V'(x)=0$ as a system. 

Here, $\dot{x}$ is differentiation with respect to $t$. I wrote this as $$\begin{pmatrix} \dot{x}\\\dot{u}\\\dot{1}\end{pmatrix}=\begin{pmatrix}0&1&0\\0&-k&-V'(x)\\0&0&0\end{pmatrix}\begin{pmatrix} x\\u\\1\end{pmatrix}$$ where $\dot{x}=u$. 
Is this the correct representation. I'm not sure because of the extra row that I had to add to the system.
 A: The equation
$\ddot x + k \dot x + V'(x) = 0 \tag 1$
cannot in general be wrtten as a linear system (as proposed in the original question, before the present edits were in effect), for the simple reason that $V'(x)$ is not in general a linear function of the variable $x$. 
We can however write (1) as a first order system, in which only the first derivative(s) of the state variable(s) are present, by denoting 
$\dot x$ by $y$:
$y = \dot x, \tag 2$
whence
$\dot y = \ddot x; \tag 3$
then (1) may be expressed in the form
$\dot y + ky + V'(x) = 0, \tag 4$
that is,
$\dot y = -ky - V'(x); \tag 5$
this together with (2) form a first order system in the two variables $x(t)$ and $y(t)$; such systems are often writen in vector-field form; that is, we may define a vector field $\mathbf X(x, y)$ on $\Bbb R^2$ by
$\mathbf X(x, y) = \begin{pmatrix} y \\ -ky - V'(x) \end{pmatrix}; \tag 6$
then setting
$\mathbf r = \begin{pmatrix} x \\ y \end{pmatrix} \tag 7$
we have
$\dot{\mathbf r} = \begin{pmatrix} \dot x \\ \dot y \end{pmatrix} = \begin{pmatrix} y \\ -ky - V'(x) \end{pmatrix} = \mathbf X(x, y). \tag 8$
