Answer to Question #261575 in Differential Equations for Khasruzzaman

Question

Suppose it is known that the population of the community in problem 1 is 10000 after 3 years.

Answer 1

Let $P_0$  be the population of a community

The population of a community is known to increase at a rate proportional to the number of people present at a time $t$

$$\dfrac{dP}{dt}=kP, k=constant$$

Then

$$\dfrac{dP}{P}=kdt$$

Integrate both sides

$$\int\dfrac{dP}{P}=\int kdt$$

$$\ln(|P)|=kt+\ln C$$

$$P(t)=Ce^{kt}$$

Using the initial condition

$$P(t)=P_0e^{kt}$$

Given that an initial population $P_0$ has doubled in 5 years

$$P(5)=P_0e^{k(5)}=2P_0$$

$$e^{5k}=2$$

$$k=0.2\ln 2$$

$$P(t)=P_0\cdot2^{0.2t}$$

Given $P(3)=10000.$

i)

$$10000=P_02^{0.2(3)}$$

$$P_0=\dfrac{10000}{2^{0.6}}$$

$$P_0=6598\ people$$

ii)

$$P(10)=P_0\cdot2^{0.2(10)}$$

$$P(10)=4P_0$$

$$P(10)=10000(2)^{3.4}$$

$$P(10)=26390\ people$$

iii)

$$\dfrac{dP}{dt}|_{t=10}=0.2\ln 2\cdot P(10)$$

$$\dfrac{dP}{dt}|_{t=10}=0.2\ln 2\cdot (10000(2)^{3.4})$$

$$\dfrac{dP}{dt}|_{t=10}=3658.45\ people/year$$