How did the author construct the differential equation from the given information In the book of Differential Equations: A Dynamical Systems Approach: Ordinary Differential Equations by West&Hubbard, at page 85, it is given that

Example 2.5.4: Carbon dating. The carbon in living matter contains a
  minute proportion of the radioactive isotope $C^{14}$. This
  radiocarbon arises from cosmic ray bombardment in the upper atmosphere
  and enters living systems by exchange processes, reaching an
  equilibrium concentration in these organisms. This means that in
  living matter, the amount of $C^{14}$ is in constant ratio to the
  amount of the stable isotope $C^{12}$ • After the death of an
  organism, exchange stops, and the radiocarbon decreases at the rate of
  one part in 8000 per year.
Therefore carbon dating enables calculation of the moment when an
  organism died, and we set $t =$ the number of years after death. Then
  $x(t) =$ ratio of $C^{14}$ to $C^{12}$ satisfies the differential
  equation
$$x' = \frac{ -1}{8000 } x$$ [...]

However, I couldn't understand what exactly what does the authors mean by "decreases at the rate of one part in 8000 per year", and how did they construct the given differential equation from that information. Can someone explain, or give some hints about it ?
 A: I think perhaps the author is not quite correct, strictly speaking. I can get to the author's answer, assuming we know something about radioactive decay. Getting there will show what the problem is.
Radioactive decay is an instantaneous process, by which at any given time, the rate of change, $dx(t)/dt$, of the amount of material, $x(t)$, is proportional to the amount of material present. This means $x(t)$ it should satisfy a differential equation of the type
$$\frac{dx(t)}{dt} = a x(t),$$
where $a$ is the ''rate'', which is negative for decay. The unique solution to this differential equation is
$$x(t) = y e^{a t},$$ 
where $y$ is an initial condition, interpreted as the amount of material at time zero.
The author tells us in his story (the last sentence of the first paragraph) that $x(t+1) - x(t) = -1/8000\,\, x(t)$. Putting in our general solution, we can solve for $a$, and find that $a= \log(1-1/8000)$. This means that $x(t)$ satisfies the differential equation
$$\frac{dx(t)}{dt} = \log(1-1/8000) x(t),$$
unlike what the author claims.
The author is more or less correct however, in the sense that $\log(1-x) = -x + \mathcal{O}(x^2)$. So since $1/8000$ is small, we can approximate the logarithm and get exactly the differential equation the author writes.
I realize this does not explain why radioactive decay should work according to the differential equation I wrote, but I wanted to see if this clarifies things -- I hope it shows that it is impossible to follow the author's reasoning (when limited to this excerpt) without knowing the general answer to begin with.
Addendum: 
Why does radioactive decay result in a differential equation of the type
$$\frac{dx(t)}{dt} = a x(t) \, \, ?$$
This is ultimately because of the laws of quantum mechanics, which I will not get in to. The main point is that radioactive decay is a process by which there is some chance that a particle decays, and this chance is the same for every particle of its kind, at any time. In other words, if a particle has a 10% chance of decaying in 1 second, and if it does not decay in this first second, it will have a 10% chance to decay in the next second, and so on. Now, if we are interested in seconds, this is all good: the chance the particle is still around after $n$ seconds is $(0.9)^n$ in decimal notation. What about smaller timescales? Say we look at timescales of 1/5 of a second. What could be the chance $c$ of decay? We know that $(1-c)^5 = 0.9$, so you can compute $c$, and it will be slightly larger than $0.1/5 = 0.02$. Now you can keep playing this game, dividing into ever smaller intervals, and then you get some infinitesimal chance per infinitesimal time interval, call it $a$, some constant (in our example $a = - \log(0.9)$, the solution for $x$ in $\lim_{n\to\infty} (1-x/n)^n = 0.9$).
How does this relate to the differential equation we want to get? Well, if one particle has a certain chance of decaying, if you take a very large number of particles, that chance effectively becomes a rate of decay (flip a coin once and you may or may not get heads, but flip it a million times and the ratio of heads/(total number of tosses) will be pretty close to 0.5). So then the total number of particles will decay at a rate proportional to the their number, with proportionality the (infinitesimal) chance of decay per time $a$, giving us
$$\frac{dx(t)}{dt} = a x(t)$$.
Finally, when the author says "decreases at the rate of one part in eight thousand per year", he means that in the course of one year, 1/8000 of the number of particles (i.e. amount of the material) will decay.
A: The rate is the change of $x$ in time, that is
$$x'=\frac{dx}{dt}$$
since, for a given amount at time t, it is decreasing with the rate of $$\frac1{8000}x$$
we obtain
$$x'=\frac{dx}{dt}=-\frac1{8000}x$$
We need the minus sign because the rate is negative and $x$ decreases with time.
