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I am learning infectious disease modeling, and I have come across something in my textbook that has me stumped. They take the equation: $$\frac{dS}{dR} = -R_0S $$ Then, they say "upon integrating with respect to R, we obtain: $S(t) = S(0)e^{-R(t)R_0}$"

My question is, where does the e come from? I haven't taken a calculus class in a decade so I may be missing something obvious, but I'm very confused, and I'd appreciate it if someone could walk me through how they got this. If it helps, $R_0$ is a parameter while S and R are variables (people susceptible to infection and people recovered from infection). $R_0$ and R are completely different things.

here's a picture of the textbook

jgd1729
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Elizabeth
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2 Answers2

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This is an example of what we call "separation of variables" in solving ordinary differential equations. The basic idea is to treat $\frac{dS}{dR}$ like a fraction and gather all instances of S to the left and R to the right, and then integrate: \begin{align*} \frac{dS}{S} &= -R_0 dR \\ \int\frac{dS}{S} &= -R_0 \int dR \\ \ln |S| &=-R_0 R +C\\ |S| &= e^{-R_0 R +C} \\ S &= e^{-R_0 R} e^C \\ &= S(0)e^{-R_0 R} \end{align*} where we've dropped the absolute value because the right hand side cannot be negative and we've replaced $e^C$ with $S(0)$, since plugging in $R=0$ gives $S(0) = e^C$.

Technically what's going on here is a bit more complex than just treating the derivative as a fraction. The justification for why we can do this comes down to the chain rule and is really just the same reason that integration by substitution works. We typically just write it the way I did above because it's easier to read then having a couple of extra differentials hanging around.

jgd1729
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You have a differential equation stating that the derivative of $S$ with respect to $R$ is $R_0 S$. In other words the derivative of $S$ is proportional to $S$ itself. This is the property of the exponential function: $$ f(x) = e^{\lambda x} \Rightarrow f'(x) = \lambda e^{\lambda x} = \lambda f(x)$$

Moeen Nehzati
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