Modeling with differential equations I am struggling a bit to convert a word/modeling problem into a differential equation to then solve.
The question is:
Assume a town has a population of 100,000 citizens, within a week 10,000 people are  mysteriously ill. Assume that the rate of increase
of the number who are ill is proportional to the number of people who have not yet fallen ill. How long will it be until half the town have fallen ill?
I think it should end up being something like ds/dt=r(s-100000) where s is the number of people that have fallen sick and r is some rate for it to be proportional to.
 A: $$\frac{ds}{dt} = k(100000 - s)$$
$$\frac{ds}{(100000 - s)} = k\ dt$$
$$\ln(100000 - s) = kt + C$$
$$100000 - s = Ce^{kt}$$
$$\text{When}\ t = 0, C = 100000 - 10000 = 90 000$$
$$\text{When}\ t = -7, k = \frac{\ln\frac{10}{9}}{-7} = -.01505$$
$$s = 100000 - 90000e^{-.01505t}$$
$$50000 = 90000e^{-.01505t}$$
$$t = \frac{\ln\frac{5}{9}}{-.01505}$$
$$t = 39\  \text{days}$$
A: Assuming that you mean that the rate of infection is jointly proportional to the number of healthy and infected people (a common set-up), then the differential equations in question should be
$$\frac{ds}{dt}=rs(100000-s)$$
Where $s$ is the number of infected people, $(100000-s)$ is the number of healthy people, and $r$ is the constant of proportionality.
This can be solved by separation of variables and integration by partial fraction decomposition. The result is a pretty standard logistic curve.
$$\frac{ds}{dt}=rs(100000-s)$$
$$\frac{ds}{s(100000-s)}=r dt$$
$$(\frac{1/100000}{s}+\frac{1/100000}{100000-s})ds=r dt$$
$$\frac{1}{100000}(\ln(s)-\ln(100000-s))=rt+C$$
$$ \frac{s}{100000-s}=Ce^{100000rt}$$
Using the initial value, where when $t$ is zero, $s$ is 10000, we get that
$$C=\frac{1}{9}$$
Solving for $t$ when $s$ is equal to 50000 (half the total), we get
$$t=\frac{\ln9}{100000r}$$
It isn't enough to find an answer precisely. You need another piece of information to solve for $r$.
A: 
Assume that the rate of increase of the number who are ill is proportional to the number of people who have not yet fallen ill.



*

*If $s$ is the number of ill people, "the rate of increase of the number who are ill" is $ds/dt$. 

*If $N=100000$ is the total number of people, then $N-s$ would be "the number of people who have not yet fallen ill." 

*The fact that these two quantities are proportional means
$$
\frac{ds}{dt} = r(N-s).
$$
Assuming at time zero, no one was ill, we have the initial condition $s(0) = 0$. We can solve the equation as:
$$
\frac 1{N-s}\frac{ds}{dt} = r \implies -\frac{d}{dt}\ln(N-s) = r\\ 
\implies \ln(N-s(t))-\ln(N-s(0)) = -rt\\
\implies s(t) = N(1-e^{-rt})
$$

within a week 10,000 people are mysteriously ill. 

This mean at $t=7$ days, $s=10000$. We can use this to find $r$
$$
s(7) = 100000(1-e^{-7r}) = 10000 \implies1-e^{-7r} = 0.1\implies r = -\ln(0.9)/7
$$

How long will it be until half the town have fallen ill?

We need to find $t$ at which $s(t)$ is half of the town or $N/2$:
$$
s(t) = N(1-e^{-rt})= N/2\implies 1-e^{-rt} = 1/2\\ \implies t = -\ln(0.5)/r = 7\ln(0.5)/\ln(0.9)\approx 46\,\text{days}.
$$
Note that this is $46$ days after illness starts which is $46-7=36$  days after the first 10,000 people have fallen ill. 
