In many population growth problems there is an upper limit b

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 equation dy/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

Solution

a) Given equation is dy ----- = ky(16 - y) dt dy ----------- = k dt y(16-y) dy ----------- = -k dt y(y-16) -dy dy ----- ----- 16 16 ------ + -------- = -k dt y y-16 -1/16 ln(y) + 1/16 ln(y-16) = -k dt + C y-16 ln(--------) = -16 k t + C y y - 16 -16 k t --------- = C e y 2 billion people in 1925 4 billion people in 1975. when t = 0 y = 2 2-16 ------- = C So C = -7 2 y-16 -16 k t ---- = -7 e y when t = 50 y = 4 4-16 -16 k 50 ------ = -7 e 4 -16 k 50 -3 = - 7 e -800 k 3/7 = e ln(3/7) = -800 k -ln(3/7) k = -------- 800 16 ln(3/7) ------------ y - 16 800 --------- = -7 e y ln(3/7) -------- t y - 16 50 --------- = -7 e y t/50 y - 16 | 3 | --------- = -7 | - | y | 7 | t = 0 y - 16 = -7 y y = 2 t = 50 y-16 = -3 y y = 4 (b) Find the population in 2015 2015 -1925 ======== 90 years (y-16)y +7 (3/7)^(90/50) == 0 {{y -> 0.0957695}, {y -> 15.9042}} (c) When will the population be 9 billion. Solve[(9-16)/9 +7 (3/7)^(t/50)==0] {{t -> 129.661}} ln(3/7) -------- t y - 16 50 --------- = -7 e y y - 16 = -7 e^(ln(3/7)/50 t) y = 0 16 y = ---------------------------- 1+7 e^(ln(3/7)/50 t)