How do I set up this differential equation?? I am trying to tutor some students in differential equations and I am a little rusty. They presented me with this problem a TA did and I'm trying to help them understand it. It says "A population of bacteria grows at a rate proportional to its size. Write and solve a differential equation that expresses this. If there are 1000 bacteria after one hour and 2000 bacteria after two hours, how many bacteria are there after 3 hours?" Any help with this I'm really lost on this one.
 A: Let $P$ be the population size. This grows with time and so can be thought of as a function of time. If the growth of the population is proportional to the size of the population then
$$ \frac{\operatorname{d}\!P}{\operatorname{d}\!t} \propto P$$
What does this mean? Well, it means that there exists some fixed number, say $k$ for which
$$\frac{\operatorname{d}\!P}{\operatorname{d}\!t} = kP$$
We can solve this differential equation by "seperation of variables". Dividing by $P$ and then multiplying throught by $\operatorname{d}\!t$ gives
$$\frac{1}{P}\operatorname{d}\!P = k\operatorname{d}\!t$$
Does this remind you of anything? It reminds me of an integral:
$$\int \frac{1}{P}\operatorname{d}\!P = \int k\operatorname{d}\!t $$
We can perform the integrations on both sides to give
$$\ln \left|P\right| = kt+C$$
It follows that $\left|P\right| = \operatorname{e}^{kt+C} \equiv \operatorname{e}^{kt}\times \operatorname{e}^C$. Dropping the modulus gives $P = \pm\operatorname{e}^{kt}\times \operatorname{e}^C$. Since $C$ was a random constant, positive or negative, we can relabel $\pm\operatorname{e}^C$ as a random constant, say $\rho$. Hence
$$P = \rho\operatorname{e}^{kt}$$
Use your initial conditions to set up two equation to solve for $\rho$ and $k$.
A: \begin{align}
\frac{d P}{d t} = \alpha P
\end{align}
where $P$ is the population, $t$ is the time and $\alpha$ is a proportionality constant.
Hence, $P(t) = Ce^{\alpha t}$. Initial conditions will give you $C$.
EDIT: 
\begin{align}
P(t=1h)=1000=Ce^\alpha\\
P(t=2h)=2000=Ce^{2\alpha}\\
\implies \alpha=\ln 2
\end{align}
Now shift the initial time and choose 'after one hour' to mean time 0. Then you have $P(t=0)=1000$ which implies $C=1000$, thus
\begin{align}
  P(t)=1000 e^{\ln(2)t}
\end{align}
Evaluate at time $t=2$, which is equivalent to 'after 3 hours', and you get the answer, which is 4000. Hence we came up with a very convoluted way of concluding that the population doubles every hour.
