In many population growth problems, there is an upper limit beyond which the population cannot grow. Many scientists agree that the earth will not support a population of more than 16 billion. There were 2 billion people on earth in 1925 and 4 billion in 1975. If is the population years after 1925, an appropriate model is the differential equationdy/dt=ky (16-y) Note that the growth rate approaches zero as the population approaches its maximum size. When the population is zero then we have the ordinary exponential growth described by y'=16ky. As the population grows it transits from exponential growth to stability. (a) Solve this differential equation. (b) The population in 2015 will be (c) The population will be 9 billion some time in the year

a) (y-16) / y = - 7*e∧ (-0.016946*t)

b) y = 6.34

c) t = 129.66 years in 2055

Step-by-step explanation:

a) dy/dt = ky * (16-y)

Solving the differential equation we have

dy / (y * (y-16)) = - k dt

∫ dy / (y * (y-16)) = ∫ - k dt

(-1/16) * Ln (y) + (1/16) * Ln (y-16) = - k*t + C

(1/16) Ln ((y-16) / y) = - k*t + C

Ln ((y-16) / y) = - 16*k*t + C

(y-16) / y = C*e∧ (-16*k*t)

If t = 0 and y = 2

(2-16) / 2 = C*e∧ (0)

C = - 7 then we have

(y-16) / y = - 7*e∧ (-16*k*t)

In 1975 we have t = 1975 - 1925 = 50 years and y = 4

(4-16) / 4 = - 7*e∧ (-16*k*50)

k = - Ln (3/7) / 800 = 0.001059

Finally, the differential equation will be

(y-16) / y = - 7*e∧ (-16*0.001059*t)

(y-16) / y = - 7*e∧ (-0.016946*t)

b) In 2015 we have t = 2015 - 1925 = 90 years

(y-16) / y = - 7*e∧ (-0.016946*90)

Solving the equation we get

y = 6.34

c) If y = 9

(9-16) / 9 = - 7*e∧ (-0.016946*t)

t = 129.66 years in 2055