Answer to Question #147113 in Differential Equations for Divine Grace Jaravata

Question

The population of a certain country has grown at a rate proportional to the number of people in the country. At present, the country has 80 million inhabitants. Ten years ago it had 70 million. Assuming that this trend continues:

a. Find an expression for the approximate number of people living in the country at any time t (taking t=0 to be the present time)
b. Find the approximate number of people who will inhabit the country at the end of the next ten-year period.

Answer 1

a.

Let $P$ denotes the population at time $t$ years and $P_0$ be the population at time $t=0$. Then,

$\frac{dP}{dt}\varpropto P \implies \frac{dP}{dt}=kP\implies \frac{dP}{P}=kdt.$ where $k$ is a constant.

Integrating both sides, we have;

$\int\frac{dP}{P}= \int kdt\implies \ln P= kt+c\implies P= e^{kt+c}\implies P= Ae^{kt}$ .

At time $t=0, P=P_0$

$\implies P_0=Ae^{k.0}\implies A=P_0\implies P=P_0e^{kt}$

At time $t=-10, P= 70,000,000$

$\implies 7\times10^7=8 \times 10^7e^{-10k}\implies \frac{7}{8}=e^{-10k}\\\implies k= \frac{\ln \frac{7}{8}}{-10}$

At any time $t,$ the population of the country will be $P=P_0e^{kt}$ where $P_0= 8 \times 10^7$ and $k= \frac{\ln \frac{7}{8}}{-10}$ .

b.

At the end of the next 10 years, $t=10,$ the population will be approximately;

$P=P_0e^{10k}\implies P= 8 \times 10^7 \times \frac{8}{7}\\\implies P=91,428,571.43 \approx9.143 \times 10^7.$