Question

Suppose that the population​ P(t) of a country satisfies the differential equation dP/dt = kP (600 - P) with k constant. Its population in 1960 was 300 million and was then growing at the rate of 1 million per year. Predict this​ country's population for the year 2030.

Answer 1

Answer:

The country's population for the year 2030 is 368.8 million.

Step-by-step explanation:

The differential equation is:

$\frac{dP}{dt}=kP(600 - P)\\\frac{dP}{P(600 - P)} =kdt$

Integrate the differential equation to determine the equation of P in terms of t as follows:

$\int\limits {\frac{1}{P(600-P)} } \, dP =k\int\limits {1} \, dt \\(\frac{1}{600} )[(\int\limits {\frac{1}{P} } \, dP) - (\int\limits {\frac{}{600-P} } \, dP)]=k\int\limits {1} \, dt\\\ln P-\ln (600-P)=600kt+C\\\ln (\frac{P}{600-P} )=600kt+C\\\frac{P}{600-P} = Ce^{600kt}$

At t = 0 the value of P is 300 million.

Determine the value of constant C as follows:

$\frac{P}{600-P} = Ce^{600kt}\\\frac{300}{600-300}=Ce^{600\times0\times k}\\\frac{1}{300} =C\times1\\C=\frac{1}{300}$

It is provided that the population growth rate is 1 million per year.

Then for the year 1961, the population is: P (1) = 301

Then $\frac{dP}{dt}=1$.

Determine k as follows:

$\frac{dP}{dt}=kP(600 - P)\\1=k\times300(600-300)\\k=\frac{1}{90000}$

For the year 2030, P (2030) = P (70).

Determine the value of P (70) as follows:

$\frac{P(70)}{600-P(70)} = \frac{1}{300} e^{\frac{600\times 70}{90000}}\\\frac{P(70)}{600-P(70)} =1.595\\P(70)=957-1.595P(70)\\2.595P(70)=957\\P(70)=368.786$

Thus, the country's population for the year 2030 is 368.8 million.