Differential equation with start condition and maximum capacity The house with 1000 people have got 1 infected with an unknown virus.
The number of students who are newly infected is proportional to the number of healthy product of current and the current number of infected students. After 2 days is 5 people infected. How many of them will be infected in a week?
I have calculated this with diferencial equation and i have got 1000 people. Is that OK??
Thanks for answers
My solution:
MAX people: 1000


*

*day: 1 infected

*day: 5 infected
$P (7) = (1000* (\frac{2}{998} e^{10})*e^{5*7}) / ( 1 + (\frac{2}{998} e^{10}) * e^{5*7}) = 999,99 = 1000$
 A: Hint: The logistic equations sounds like a good model.
A: Let $y$ be the number infected at time $t$, where $y(0)=1$. The model we are asked to use is 
$$y'=ky(1000-y).$$
This is a separable differential equation, which can be rewritten as 
$$\frac{dy}{y(1000-y)}=k\,dt,$$
and then, using partial fractions, as 
$$\frac{1}{1000}\left(\frac{1}{y}+\frac{1}{1000-y}\right)\,dy=k\,dt$$
and then as
$$\left(\frac{1}{y}+\frac{1}{1000-y}\right)\,dy=1000k\,dt.$$
Integrate. We get, since $y$ will stay between $1$ and $1000$,
$$\ln\left(\frac{y}{1000-y}\right)=1000kt+C.\tag{1}$$
Since $y(0)=1$, we get $C=\ln(1/999)=-\ln 999$. 
We have $y(2)=5$. Substituting in our equation we get 
$$\ln(5/995)=2000k -\ln 999,$$
and therefore 
$$k=\frac{1}{2000}\ln(999/199).$$
Finally, take the exponential of both sides of (1). We get
$$\frac{y}{1000-y}=\frac{e^{kt}}{999},$$
where $k$ is known. Set $t=7$. With some manipulation, we get a linear equation for $y(7)$. Solve.  
Remark: The model should not be taken entirely seriously. One can expect at best a rough fit with reality. 
