Question

A disease is infecting a colony of 1000 penguins living on a remote island. Let P(t) be the number of sick penguins t days after the outbreak. Suppose that 50 penguins had the disease initially, and suppose that the disease is spreading at a rate proportional to the product of the time elapsed and the number of penguins who do not have the disease.

(a) Give the mathematical model(differential equation and initial condition) for P.

(b) Find the generalsolution of the differential equation in (a).

(c) Find the particular solution that satisfies the initial condition.

Answer 1

Answer:

a. $P = 1000 - Ce^{-\frac{kt^2}{2} }$

b. $C = 950$

c. $P = 1000 - 950e^{-\frac{kt^2}{2} }$

Step-by-step explanation:

a. Let the number of penguins who have the disease t days after the outbreak be P

Initial number of penguins = 1000

Therefore, current number of penguins = 1000 - P

And the rate of spread of disease according to the statement is

$\frac{dP}{dt}\alpha t(1000-P)\\\frac{dP}{dt}=kt(1000-P)$

where k is the constant of proportionality

$\frac{dP}{1000-P}=kt.dt$

Integrating both sides

$-ln(1000-P) = \frac{kt^2}{2}+c\\\frac{1}{(1000-P)} = Ce^{\frac{kt^2}{2} }\\ (1000-P) = Ce^{-\frac{kt^2}{2} }\\P = 1000 - Ce^{-\frac{kt^2}{2} }$

b. Seeing as 50 penguins had the disease initially,

t = 0

P = 50

The general solution of the differential solution becomes

50 = 1000 - C (anything raised to the power of 0 is 1, hence e is equal to 1)

$C = 1000 - 50 = 950$

c. Therefore, the solution that satisfies the initial condition is

$P = 1000 - 950e^{-\frac{kt^2}{2} }$