TITLE: Differential Equation (Application)
QUESTION [0 upvotes]: In an experiment to study the spread of a soil disease, an area of $10m^2$ of soil was exposed to infection. In a simple model, it's assumed that the infected area grows at a rate which is proportional to the product of the infected area and the uninfected area. Initially, $5m^2$ was infected and the rate of growth of the infected area was $0.1m^2$ per day. At time $t$ days after the start of the experiment, an area $am^2$ is infected and an area $(10-a)m^2$ is uninfected.
How to show that $\frac{da}{dt}=0.004a(10-a)$

REPLY [1 votes]: From the information in the problem formulation, you have $$\frac{da}{dt} \propto [10-a(t)]a(t)$$
and you know that $a(0)=5$ and $\frac{da}{dt}|_{t=0}=0.1$. So to get the constant of proportionality $k$, simply input these values:
$$0.1=k(10-5)5 \leftrightarrow k=\frac{0.1}{25}=0.004$$
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