If every century the human population doubles. What is the annual growth rate? If every century the human population doubles. What is the annual growth rate?
Use differential equations. 
My try: 
Let $P$ the human population.
We know that $\frac{dP}{dt} = 2P\quad$ then $\quad\frac{dP}{2P} = dt\quad$
If we integrate from both sides:
$\quad\int\frac{dP}{2P} = \int dt\quad$
$\quad\frac{In|P|}{2} = t + c$
But since the population is always positive we got: 
$\quad\frac{In(P)}{2} = t + c$
$\quad In(P) = 2(t + c)$
$\quad P = e^{2(t + c)}$
We want to know the change rate of $P$ with respect to $t$ so let's differentiate from both sides
$\quad \frac{dP}{dt} = e^{2(t + c)}dt$ 
I don't know if I'm on the right track. Any hints? 
 A: The population growth follows exponential model, $$P(t)=P(0)e^{rt}$$
You know that $$2P(0) = p(0)e^{100r}$$ that is $$e^{100r}=2$$ which implies $$r=ln(2)/100\approx 0.0069$$
A: Let $r$ be the annual growth rate. Thus, the growth factor after $n$ years is $r^n$, so we have $r^{100} = 2$. Thus, $r = 2^{\frac{1}{100}} \approx 1.00696$. That is, the population increases by approximately $0.696\%$ every year.

With differential equations: lonza leggiera has a good answer!
Going through the motions of your working using the premise $\frac{dP}{dt} = 2P$, you indeed get $P(t) = e^{2(t+c)}$. Since you are using a timescale of "hundreds of years", this means that if $P(t)$ is the population now, then $P\big(t + \frac{1}{100}\big)$ is the population 1 year from now. Thus, we want to determine $\frac{P(t+1/100)}{P(t)}$, which simplifies to $e^{2/100} ≈ 1.0202$. However, "2.02% per year" isn't the correct answer, and so the premise itself must be false.
If you work from the premise $\frac{dP}{dt} = \mathrm{ln}(2) \cdot P$, then you get the right answer, but forming this premise requires you to already know that the growth is exponential a priori, and that if the effective growth rate is 2, then the continuous growth rate is $\mathrm{ln}(2)$. You essentially already know the solution $P(t) = 2^{t+c}$ (where $c$ is an arbitrary constant whose value depends on the population at time $t=0$) before you even write down the differential equation!
Separately, you might observe that the population grows proportionally to its current size, leading to the equation $\frac{\mathrm{d}P}{\mathrm{d}t} = kP$ as a premise, like lonza leggiera uses in their answer, but it is not immediately obvious that this is equivalent to the population having a "doubling life" (always doubling after some fixed amount of time) and so the statement "the population doubles after every $x$ years" cannot be used to directly infer $\frac{\mathrm{d}P}{\mathrm{d}t} = kP$ unless we are already mathematicians who know about exponentials and that they have these unique properties.
As such, your attempt to solve the problem using differential equations will not work based solely on the statement "the population doubles after every 100 years." This is because nothing is directly said about the rate at which the population changes — only the result after 100 years — and so a differential equation cannot be formed from this info alone. It does, however, imply exponential growth, but philosophically, you should derive this in a different way, such as by exclusively using the fact that $P(t+1) = 2 P(t)$ for all $t$,* which is formulated directly from the info available in the question.

The distinction between effective and continuous growth rates is of quite important note, particularly in finance, where it tends to trip up consumers and those who are studying formal mathematical finance for the first time. The terms "gross interest rate" and "equivalent interest rate over a period", e.g. "annual equivalent rate" or "AER", are used to describe these exact quantities. For example, if a bank lets you save money with them and they award an amount of interest such that if you deposit £200 with them, then after a year you will have £300, then the AER is 50%; your account balance grew by a factor of 1.5. However, the bank will actually compound your interest on a daily basis (in case you want to withdraw before a year has elapsed) using a daily interest rate $r$ such that $r^{365} = 1.5$. In this case, $r ≈ 1.00111$, a percentage increase of 0.111% per day. Since $0.111...\% \times 365 ≈ 40.57\%$, they will stipulate "Interest awarded at 40.57% gross per annum, calculated daily" in their terms, but this is equivalent to an AER of 50%. Similarly, in finance terms, the AER of your population growth is 100%, but the gross growth rate is 69.38% per year when calculated daily.
In the limit, as the calculation period lessens from daily, to hourly, to every minute, to every second, to eventually continuous compounding, this gross rate approaches $\mathrm{ln}(r)$ (by definition!), e.g. $\mathrm{ln}(1.5) \approx 0.4055 = 40.55\%$ for our bank example,** and $\mathrm{ln}(2) \approx 0.6931 = 69.31\%$ for our population example. You are almost certainly familiar with this definition of $e$, which arises from this exact financial situation of continuously compounding interest:
$$e := \lim_{n \to \infty} {\bigg(1+\frac{1}{n}\bigg)^{n}}$$


*This is very similar to how raising a number to a non-integer power is defined; we want the property $a^x a^y = a^{x+y}$, which we know holds for integers $x$ and $y$, to hold for non-integers, too. It turns out that a "pure" exponential function $f(t) := a^t$ is the only kind of function which has the general property $f(x)f(y) = f(x+y)$, and so your $P(t)$ is implicitly an exponential function, since it warrants a similar property.
In particular, your $P(t)$ is a general exponential of the form $P(t) = a^{t+c}$ for some additional constant $c$ — or equivalently, $P(t) = ka^t$ for the positive constant $k = a^c$ — so it is not necessarily a "pure" exponential, since $P(0)$ is not necessarily equal to 1. However, there is still a shifted/scaled version of $P(t)$ which satisfies the defining property of "pure" exponentials, namely $Q(t) := P(t-c) = \frac{P(t)}{k}$ has the property $Q(x)Q(y) = Q(x+y)$.
Without referencing that property, we can make the observation that $k = P(0)$, and thus instead say that $P(t)$ warrants the property $\frac{P(x)}{P(y)} = \frac{P(x-y)}{P(0)}$. This more general property is possessed by "pure" exponentials $f(t) = a^t$ as well; but since $f(0) = 1$ for them, it reduces to $\frac{f(x)}{f(y)} = f(x-y)$, i.e. $f(x) = f(x-y)f(y)$, which is equivalent to saying $f(x)f(y) = f(x+y)$.

**Since these instantaneous/continuous growth rates differ negligibly in practice from the daily rates, banks won't bother to calculate your interest more often than daily.
A: You're on the right track, in that the differential equation satisfied by the population $\ P\ $ has the form
$$
\frac{dP}{dt}=kP\ ,
$$
which has the solution $\ P=Ae^{kt}\ $ for some positive constant $\ A\ $. However, $\ k\ $ is the instantaneous proportional rate of increase in $\ P\ $ and won't necessarily be $2$ if the population takes $100$ years to double.
If you take the unit of time to be centuries, then the population after one century is $\ Ae^{k\cdot 1}=Ae^k\ $, whose ratio to the initial population (i.e. the population at $\ t=0\ $, namely $\ Ae^0=A\ $) is $\ e^k\ $. Thus, if the population doubles in $100$ years, then we must have $\ e^k=2\ $, or $\ k=\ln 2\ $.
Since $1$ year is $\ \frac{1}{100}^\text{th}\ $ of a century, the population grows by a factor of  $\ e^\frac{\ln 2}{100}=2^\frac{1}{100}\approx1.007\ $ in a year, or, in other words, by about $0.7\%\ $ per year.
