Question

in an isolated town of 5000 inhabitants,160 people have a disease at the beginning of the week and 1200 have it at the end of the week. How long does it take for 80% of the population to become infected? Using differential equations.

Answer 1

What , like P= P_0 e^kt, like that?

Answer 2

f(t)=160e^(kt)
let t be measured in days
when t=7, f(x)=1200
1200=160e^(7k)
1200/160=e^(7k)
7.5=e^(7k)
ln(7.5)=7k
ln(7.5)/7= k =.2878

Now we have the equation f(x)=160e^(.2878t)
So, when 80% are infected, 4000 are infected, so solve for t
4000=160e^(.2878t)
25=e^(.2878t)
ln(25)=.2878t
ln(25)/.2878=t=11.184

Answer 3

Basically, you solve for k (which is a constant) given the initial conditions of a certain value of f(x) at a given time. then solve for t when f(x)=4000

Answer 4

oh ok finally an answer

Answer 5

cause the way its written in txtbook it uses partial fractions then integration

Answer 6

That's the textbook probably showing you how you get the general form of the equation.

Answer 7

ohhhhh ok thanks

Answer 8

im not sure how to set it up using differential equations, however you can set up a function that models this situation
\[x(0) = 160, x(7) = 1200, x(\infty) = 5000\]
\[x(t) = \frac{5000}{1+30.25e^{kt}}\]
setting x(7) = 1200 and solving for k
\[k = -.3224\]
80% of population is 4000
set x(t)=4000 and solve for t
\[t = 14.875\]
So it takes about 15 days for 80% to become infected