Respuesta :
Answer:
The country's population for the year 2030 is 368.8 million.
Step-by-step explanation:
The differential equation is:
[tex]\frac{dP}{dt}=kP(600 - P)\\\frac{dP}{P(600 - P)} =kdt[/tex]
Integrate the differential equation to determine the equation of P in terms of t as follows:
[tex]\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}[/tex]
At t = 0 the value of P is 300 million.
Determine the value of constant C as follows:
[tex]\frac{P}{600-P} = Ce^{600kt}\\\frac{300}{600-300}=Ce^{600\times0\times k}\\\frac{1}{300} =C\times1\\C=\frac{1}{300}[/tex]
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 [tex]\frac{dP}{dt}=1[/tex].
Determine k as follows:
[tex]\frac{dP}{dt}=kP(600 - P)\\1=k\times300(600-300)\\k=\frac{1}{90000}[/tex]
For the year 2030, P (2030) = P (70).
Determine the value of P (70) as follows:
[tex]\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[/tex]
Thus, the country's population for the year 2030 is 368.8 million.
The expected population of the country for the year 2030 is about 866 million.
How to analyse an ordinary differential equation
Let be the following ordinary differential equation:
[tex]\frac{dP}{dt} = \frac{k\cdot P}{600-P}[/tex] (1)
Where [tex]k[/tex] is constant, no unit.
We separate variables and integrate the resulting expression:
[tex]\int {\frac{600-P}{P} } \, dP = k \int {dt}[/tex]
[tex]600\int {\frac{dP}{P} } - \int {dP} = k \int {dt} + C[/tex]
[tex]600\cdot \ln P - P = k\cdot t + C[/tex] (2)
Where [tex]C[/tex] is the integration constant, in millions of people, no unit.
Now we proceed to determine the value of the constant [tex]k[/tex] by (1) and the integration constant by (2): ([tex]\frac{dP}{dt} = 1[/tex], [tex]t = 1960[/tex], [tex]P = 300[/tex])
[tex]k = \left(\frac{600-P}{P}\right)\cdot \frac{dP}{dt}[/tex]
[tex]k = \left(\frac{600-300}{300} \right)\cdot 1[/tex]
[tex]k = 1[/tex]
[tex]600\cdot \ln 300-300 = 1\cdot 1960 + C[/tex]
[tex]C = 1162.269[/tex]
Now we proceed to find the expected population for the year 2030: ([tex]t = 2030[/tex])
[tex]600\cdot \ln P - P = 2030 +1162.269[/tex]
[tex]600\cdot \ln P - P = 3192.269[/tex]
Since [tex]\frac{dP}{dt} > 0[/tex] for the year 1960, then it is expected a greater population for the year 2030. The only solution for this expression is:
[tex]P \approx 866.202[/tex]
The expected population of the country for the year 2030 is about 866 million. [tex]\blacksquare[/tex]
To learn more on ordinary differential equations, we kindly invite to check this verified question: https://brainly.com/question/25731911
