A rumor spreads through a population of 5000 people at a rate proportional to the product of the number of people who have heard it and the number who have not. Suppose that 100 people initiate the rumor and that a total of 500 people know the rumor after two days.
Let y(t) denotes the number of people who know the rumor at time t. Take log 9/log49 = 129/229
(i) The time required for half the population to hear rumor is:
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Ans.
(i)
Sol. Let y(t) denote the number of people who know the rumor at time t. Maximum value of y(t) is 5000, y(0) = 100 and y(2) = 500. Also,
α y (5000 –y) ⇒ 
= ky (5000 –y)
Separating the variables and integrating we have
⇒ 
= kt + C
⇒ 
= kt + C
⇒ 
log = kt + C
⇒ 
= const. e 5000kt = Ke 5000kt ⇒ y(1 + Ke
5000kt ) = 5000 Ke 5000kt ... (i)
Put y (0) = 100, we have 100 (1 + K) = 5000 K
⇒ K = 1/49. Using y(2) = 500, we have
500 (1 + Ke 10,000k ) = 5000 Ke 10,000k ⇒ 1 = 9 Ke
10,000k ⇒ e 10,000k = 
⇒ k = 
log 
To determine how long it will take for half the population to hear rumor, we solve
2500 = 
⇒ 1 + 49 e –5000 kt = 2 ⇒ e –5000 kt = 
5000 kt = log 49 ⇒ t = 
⇒ t = 
= 
= 
= 4. 58 days
(ii)
Sol. Let y(t) denote the number of people who know the rumor at time t. Maximum value of y(t) is 5000, y(0) = 100 and y(2) = 500. Also,
α y (5000 –y) ⇒ 
= ky (5000 –y)
Separating the variables and integrating we have
⇒ 
= kt + C
⇒ 
= kt + C
⇒ 
log = kt + C
⇒ 
= const. e 5000kt = Ke 5000kt ⇒ y(1 + Ke
5000kt ) = 5000 Ke 5000kt ... (i)
Put y (0) = 100, we have 100 (1 + K) = 5000 K
⇒ K = 1/49. Using y(2) = 500, we have
500 (1 + Ke 10,000k ) = 5000 Ke 10,000k ⇒ 1 = 9 Ke
10,000k ⇒ e 10,000k = 
⇒ k = 
log 
= 5000 
= 5000 log 
(iii)
Sol. Let y(t) denote the number of people who know the rumor at time t. Maximum value of y(t) is 5000, y(0) = 100 and y(2) = 500. Also,
α y (5000 –y) ⇒ 
= ky (5000 –y)
Separating the variables and integrating we have
⇒ 
= kt + C
⇒ 
= kt + C
⇒ 
log = kt + C
⇒ 
= const. e 5000kt = Ke 5000kt ⇒ y(1 + Ke
5000kt ) = 5000 Ke 5000kt ... (i)
Put y (0) = 100, we have 100 (1 + K) = 5000 K
⇒ K = 1/49. Using y(2) = 500, we have
500 (1 + Ke 10,000k ) = 5000 Ke 10,000k ⇒ 1 = 9 Ke
10,000k ⇒ e 10,000k = 
⇒ k = 
log 
Therefore, (i)
⇒ y = 
= 
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