Question

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

Answer 1

Answer:

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

Answer 2

The solution for (a)$a) (y-16)/y = -7*e^{(-0.016946*t)}$b) y = 6.34 and (c) the value of t is  129.66 years  in 2055

We have given that,

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

By using variable separable form we have,

What is the variable separable form?

A variable separable differential equation is any differential equation in which variables can be separated

Therefore by solving the differential equation we have

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

integrating both side with respect to t

$\int dy / (y*(y-16)) = \int -k dt$

Solve the integration of the above

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

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

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

$(y-16)/y = C*e^{(-16*k*t)}$

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

$(9-16)/9 = -7*e^{(-0.016946*t)}$

Therefore we get the value of t is  129.66 years  in 2055

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