Answer:
(a) 128 million
(b) The limiting population is 211.6 million
Step-by-step explanation:
Ordinary Differential Equation (ODE)
It has been found a country's population P(t) (t in years from 1790) is modeled by the ODE:
[tex]\displaystyle \frac{dP(t)}{dt}=0.03142P - 0.0001485 P^2[/tex]
With an initial value of P(0)=3.9. Note this value differs from the 4.1 million originally suggested.
This is a variable separable differential equation that can be rewritten as
[tex]\displaystyle \frac{dP}{0.03142P - 0.0001485 P^2}=dt[/tex]
Integrating
[tex]\displaystyle \int \frac{dP}{0.03142P - 0.0001485 P^2}=\int dt[/tex]
The right-side integral is (for a constant A)
[tex]\int dt=t+A[/tex]
Let's compute the left-hand integral
[tex]\displaystyle I=\int \frac{dP}{0.03142P - 0.0001485 P^2}[/tex]
Factor the denominator
[tex]\displaystyle I=\int \frac{dP}{(\frac{0.03142}{P} - 0.0001485) P^2}[/tex]
Substitute
[tex]\displaystyle u=\frac{0.03142}{P} - 0.0001485[/tex]
[tex]\displaystyle du=-\frac{0.03142}{P^2}dP[/tex]
Substitute into the integral
[tex]\displaystyle I=\frac{1}{-0.03142 }\int \frac{du}{u}[/tex]
Solving
[tex]\displaystyle I=\frac{1}{-0.03142 }ln(u)+C[/tex]
Undo substitution
[tex]\displaystyle I=\frac{1}{-0.03142 }ln\left(\frac{0.03142}{P} - 0.0001485\right)+C[/tex]
Solving the ODE
[tex]\displaystyle \frac{1}{-0.03142 }ln\left(\frac{0.03142}{P} - 0.0001485\right)+C=t+A[/tex]
Taking exponentials and using a generic constant K
[tex]\displaystyle \frac{0.03142}{P} - 0.0001485=Ke^{-0.03142t}[/tex]
Solving for P
[tex]\displaystyle P =\frac{0.03142}{Ke^{-0.03142t}+0.0001485}[/tex]
Use the initial value t=0, P=3.9
[tex]\displaystyle 3.9 =\frac{0.03142}{Ke^{-0.03142(0)}+0.0001485}=\frac{0.03142}{K+0.0001485}[/tex]
We get the value of K
[tex]K=0.007908[/tex]
The final solution for P is
[tex]\displaystyle P =\frac{0.03142}{0.007908\cdot e^{-0.03142t}+0.0001485}[/tex]
(a) Compute P for the year 1930, t=1930-1790=140 years
[tex]P\approx 128\ million[/tex]
(b) For larger values of t we evaluate the limit value of P, making [tex]t=\infty[/tex]
[tex]\displaystyle P(limit) =\frac{0.03142}{0.007908\cdot e^{-0.03142(\infty)}+0.0001485}=\frac{0.03142}{0.0001485}=211.6[/tex]
The limiting population is 211.6 million
