So as I was reading this chapter I came across this example:
SPREAD OF A DISEASE A contagious disease—for example, a flu virus—is
spread throughout a community by people coming into contact with other people. Let $x(t)$ denote the number of people who have contracted the disease and $y(t)$ denote the number of people who have not yet been exposed. It seems reasonable to assume that the rate $$\frac{{dx}}{{dt}}
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$$ at which the disease spreads is proportional to the number of encounters,
or interactions, between these two groups of people. If we assume that the number
of interactions is jointly proportional to $x(t)$ and $y(t)$ —that is, proportional to the
product $xy$—then,
$$\frac{{dx}}{{dt}} = kxy
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$$
where $k$ is the usual constant of proportionality. Suppose a small community has a
fixed population of $n$ people. If one infected person is introduced into this community,
then it could be argued that $x(t)$ and $y(t)$ are related by $$x + y = n + 1
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$$ .
Using this last equation to eliminate y in above equation gives us the model:
$$\frac{{dx}}{{dt}} = kx(n + 1 – x)
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$$
What I don't understand specifically is their remark…
An obvious initial condition accompanying equation is
$$x(0) = 1
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$$
I substituted in dx/dt into the other side of $$\frac{{dx}}{{dt}} = kx(n + 1 – x)
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$$
but how would you do that without the independent variable $t$ in the equation? Also…when substitution $0$ into $x$, wouldn't the whole thing zero out? Thanks in advance.
Best Answer
The remark of starting at $x(0) = 1$ means that when $t = 0$, i.e. at the start of our model, there is a single infected person. That makes sense, since for most infections there is a single starting point from which the disease spreads. By comparison, if $x(0) = 0$ then you'll find that the solution is the constant $x(t) = 0$, i.e. if no-one is infected at the start, no-one magically becomes infected later on too (if there's no source, there's no spread).
Also, for the differential equation involving $\frac{dx}{dt}$, you can take the reciprocal of everything to get $\frac{dt}{dx} = \frac{1}{kx(n + 1 - x)}$, then when you solve that to get $t$ as a function of $x$, you then need to rearrange to get $x$ as a function of $t$.