The population model is simple and yet, as we shall see later in the course, it is perhaps the most fundamental of all models. Consider a petri dish of bacteria. We want to predict how many bacteria will be in the dish at a given time.

The measurable quantites in this model will be population, \(P\) and time, \(t\text{,}\) which is often measured in days in biological experiments. Our goal is to determie the right hand side of the equation: \begin{equation*} \frac{dP}{dt} = f(t,P) = ? \end{equation*} To do so, we often do thought experiments where we rely upon our intuititon to devise a reasonable function for the right hand side.

Consider two petri dishes …

\begin{equation} \boxed{\dfrac{dP}{dt} = kP} \label{eqn-population-model}\tag{2.2.1} \end{equation}

\begin{align} \frac{dP}{P} \amp = k dt\notag\\ \int \frac{dP}{P} \amp = \int k \; dt\notag\\ \ln \abs{P} \amp = kt + C\notag\\ e^{\ln \abs{P}} \amp = e^{kt+C}\notag\\ e^{\ln \abs{P}} \amp = e^{kt}\cdot e^C , \textrm{ let } P_0 = e^C\notag\\ P(t) \amp = P_0 e^{kt}\label{eqn-population-model-sol}\tag{2.2.2} \end{align}

Another extremely important separable equation comes about from modeling diffusion. Diffusion is the spreading of something from a concentrated state to a less concentrated state.

We will model the diffusion of salt across a semi--permeable membrane such as a cell wall. Imagine a cell, which contains a salt solution that is immersed in a bath of saline solution. If the salt concentration inside the cell is higher than outside the cell, then salt will on average, mostly flow out of the cell, and vice versa. Let's assume that the rate of change of salt concentration in the cell is proportional to the difference between the concentrations outside and inside the cell. Also, let's assume that the surrounding bath is so much larger in volume than the cell, that its concentration remains essentially constant because the outflow from the cell is miniscule. We must translate these ideas into a model. If we let \(y(t)\) represent the salt concentration inside the cell, and \(A\) the constant concentration of the surrounding bath, then we get the diffusion model:

\begin{equation} \boxed{ \dfrac{dy}{dt} = k(A-y)} \label{eqn-diffusion-model}\tag{2.2.3} \end{equation}

Again, \(k\) is a constant of proportionality with units, 1/time, and we assume \(k>0\text{.}\) This is a separable equation, so we know how to solve it.

\begin{align*} \int \dfrac{dy}{A-y} \amp= \int k \; dt \qquad u=A-y, \: -du=dy\\ -\int \dfrac{du}{u} \amp= \int k \; dt\\ -\ln \abs{A-y} \amp= kt + C\\ \abs{A-y} \amp= e^{-kt-C} \quad \textrm{ let } C = e^{-C}\\ \abs{A-y} \amp= C e^{-kt}\\ A-y = \amp \begin{cases} \;\;\,C e^{-kt} \amp A \gt y \\ -C e^{-kt} \amp A \lt y \end{cases}\\ y = A - \amp \begin{cases} \;\;\,C e^{-kt} \amp A \gt y \\ -C e^{-kt} \amp A \lt y \end{cases} \end{align*}

Thus we get two solutions depending on which concentration is initially higher.

\begin{equation} y(t) = \begin{cases} A - C e^{-kt} \amp y \lt A \\ A + C e^{-kt} \amp y \gt A \end{cases} \label{eqn-diffusion-model-sol}\tag{2.2.4} \end{equation}

Actually, there is a third curious solution which occurs when \(y = A\text{.}\) In this case, the right hand side of equation (2.2.3) is 0, which implies that \(y\) will never change. This in turn forces \(y(t)=A\text{,}\) i.e. a constant solution.