What is “first order" and “second order" in time? What is the meaning of the text quoted below?

In the physical world, if a system is described by an equation that is
  first order in time, the system is general dissipative (has energy
  loss). If the equation is second order in time, the system may be non
  dissipative. Such a system has time-reversal symmetry.

Can somebody explain what it really means to be first order and second order in plain English?
 A: Orders for differential equations refer to the largest number of derivatives that are taken with respect to the independent variable.  In partial differential equations, there can be multiple independent variables, so you have to specify which one you're talking about -- hence the "in time" part.
Thus, first-order in time means just one derivative with respect to time.  The classic example is the basic one-dimensional heat equation:
\begin{equation}
\frac{\partial \phi}{\partial t} = \frac{\partial^2 \phi}{\partial x^2}~.
\end{equation}
You can see that this is second-order in space (because $\partial^2 / \partial x^2$ is two spatial derivatives), but there's just one time derivative, so it's first-order in time.  The heat equation is the classic example of dissipation because heat flows out irreversibly.  This is related to the fact that this equation does not have time-reversal symmetry.  That is, if you replace $t$ with $-t$, this equation becomes
\begin{equation}
\frac{\partial \phi}{\partial t} = -\frac{\partial^2 \phi}{\partial x^2}~,
\end{equation}
which is really a different equation because of that negative sign.
Second-order in time means two derivatives with respect to time.  The classic example is the basic one-dimensional wave equation:
\begin{equation}
\frac{\partial^2 \phi}{\partial t^2} = \frac{\partial^2 \phi}{\partial x^2}~.
\end{equation}
This equation does have time-reversal symmetry, because if you replace $t$ with $-t$, you get the exact same equation back.  And in particular, this equation is non-dissipative; as the wave travels along, it doesn't lose any energy, and -- with the right boundary conditions -- it could even return to exactly its initial data.
But the author said that second-order equations may be non-dissipative.  An example of a second-order equation that is dissipative is the damped wave equation:
\begin{equation}
\frac{\partial^2 \phi}{\partial t^2} = -\frac{\partial \phi}{\partial t} + \frac{\partial^2 \phi}{\partial x^2}~.
\end{equation}
Note that this has both a single and a double derivative with respect to time, but it's still called second-order in time because order is defined as the highest number of derivatives.  Solutions to this equation are wave-like, except that the waves gradually lose energy, so there is dissipation.  And you can see that this is not time-reversible because the equation changes if you flip the sign of $t$, and becomes the "anti-damped" wave equation (the waves grow in time).
A: I think the author refers to the order of the differential equations describing the dynamical system that means the order of time derivative in the equation. How is it related to dissipation? To get answer and understand it one has to be able to solve following differential equations:
$$\frac{dx}{dt}=-\alpha x$$
$$\frac{d^2x}{dt^2}=-\alpha x$$
The answer is in solutions of these simple equations
A: "First Order" is short for "First-Order Ordinary Differential Equation"
And the same goes for "Second order".
Lets break that:
The definition of an "Ordinary Differential Equation" is an equation containing a function of one independent variable and its derivatives. They have many forms.
The order of a differential equation is determined by the power of the differential (aka if you are doing a derivative once or twice, or n times).
