It is widely known that the course Difference- and Differential Equations is one of the most hated courses among students from the Bachelor EOR. This is probably due to the fact that only a small fraction of the students is able to pass the course. However, this should not take away from the fact that difference- and differential equations are very useful in real life. One of the fields in which differential equations are used is pharmacokinetics, the study of how substances (e.g drugs) are processed within the body. This article will shortly introduce you to modeling the change in concentration of drugs in the body over time.
Since the retake exam of Difference- and Differential Equations was not too long ago, it is reasonable to assume that most of you have some kind of knowledge of differential equations. Nonetheless, for those who did not follow the course yet or for those who have developed such fear for this topic, I will shortly repeat what differential equations are. A differential equation is an equation that relates one or more unknown functions and their derivatives. Here, we will only use a first order differential equation which is a function of the form $F(t, x(t), x’(t)) = 0$ where t denotes the time. When applied in real life, it describes the relation between certain quantities and their rate of change.
For modeling the concentration of a certain drug in someone’s body, we focus on a single compartmental model. In this model, we take a look at how a drug is absorbed and eliminated by a single compartment (see graphical representation below) of the body through which the drug travels (e.g blood stream, organs, etcetera).
In order to represent this mathematically, we need to define some variables.
Let
- $C(t)$ = drug concentration in the compartment at time t
- $r$ = the growth rate (decrease rate if negative)
Then the change in drug concentration is given by
$$C’ = dC\dt = C * r , (1)$$
which is a first order differential equation.
The problem here is that we can only measure the initial drug concentration in the body at time $t=0$ and the drug concentration at some time t. Consequently, we do not know the growth rate and the rate of change in drug concentration $C’(t)$, we only know $C(t)$ thus we need to rewrite the equation.
Since (1) is separable, rewriting en integration yields the following:
$$\frac{dC}{dt} = C * r \Longleftrightarrow \frac{1}{C}dC = rdt$$
$$\int \frac{1}{C}dC = \int r dt$$
$$ln|C| = rt + Z, Z \in \mathbb{R}$$
$$ C = e^{rt + Z} = e^{rt}e^c = C_0 e^{rt}$$
Thus, we see that the drug concentration at time t is the original drug concentration at time $t= 0$ multiplied by some factor depending on $t$ and the growth rate $r$. Now we are able to calculate the growth rate for a certain drug. In order to calculate $r$ for a certain drug, we measure the drug concentration in the body $C$ at time $t=0$ and at some time $t$. Solving for $r$ will give us the growth rate.
Although you might still hate the course Difference- and Differential equations because you have to take the resit for the 5th time, I hope that after reading this article you might be a bit more motivated to study even harder for that 5th resit and finally have some understanding of these widely used equations.