Suppose that a community contains 100,000 people who are susceptible to a certai
ID: 2986597 • Letter: S
Question
Suppose that a community contains 100,000 people who are susceptible to a certain virulent disease. At time=0 the number of the people that have the disease is 1000. Three days later there are 1625 who have caught it. N(t) is the number of the people who have the disease and its rate of change is proportional to the product of the number of people who have the disease and the number that do not. Find the particular solution N(t) that satisfies the given conditions and then find how long it will take until 50000 people have been infected.
I know the answer is 28 days.
Also, i know that there are similar problems out there but i need help with this one. Please dont answer with the solution to another problem and tell me to just plug the numbers in.
Explanation / Answer
N(t = 0) =1000
dN/dt (t = 0) =1625/3 =541 per day
dN/dt = k*N*(100000 -N) "proportional to product of number of those who caught the disease(N) and those who have not (100000 - N)
To determine k
dN/dt (t = 0) = 541 per day
= k*N(t = 0) * (100000 - N(t=0)) =k*(1000)*(100000 - 1000)
k = 0.5464e-5
dN/ [N*(50000 - N)] = kdt
1/[N*(100000 - N)] = A/N + B/(100000 - N)
1 = A*(100000 - N) + B*N = (B-A)*N + A*100000
A*100000 =1 B-A = 0
A =1/100000 B = A = 1/100000
k*dt = dN/ [N*(100000 - N)] = A*dN/N + B*dN/(100000 - N) = A*[dN/N +dN/(100000 - N)]
(k/A)*dt = dN/N + dN/(100000 - N)
(k/A)*t = ln(N/N0) - ln(100000 - N)/(100000 - N0)
let N = 100000 (50000 + "another 50000")
N0 = 1000
k/A = (0.546*10^-5 *100000) = 0.546
0.546*t = ln(100000/1000) - ln[(100000 - 50000)/(100000 - 10000)]
t = 28.34ays = 28days
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