Question

In 2000, the population of a country was 4U million. I he population was growing at a rate of 5% per year and every year 30 000 people emigrated from the country. 6.1 Write down an initial value problem for the population P. (1 pt) 6.2 Solve the initial value problem. (2 pts) 6.3 Estimate the population in 2010

Answer 1

$\frac{dP}{dt}=0.05P-0.03, \ P(0)=40$

6.2)

$\frac{dP}{dt}=0.05P-0.03$

$\frac{dP}{0.05P-0.03}=dt$

Let 0.05P-0.03=Q

$\frac{dQ}{dP}=0.05$

$dP=\frac{dQ}{0.05}$

Now,

$\frac{1}{0.05Q}dQ=dt$

$\frac{dQ}{Q}=0.05dt$

Integrate both sides;

$\int \frac{dQ}{Q}=\int 0.05dt$

ln Q=0.05t+C

ln(0.05P-0.03)=0.05t+C

$0.05P-0.03=e^{0.05t+C}$

$P=\frac{e^{0.05t+C}+0.03}{0.05}$

$P=\frac{e^{0.05t}*e^{C}+0.03}{0.05}$

$P=\frac{C_1e^{0.05t}+0.03}{0.05}$ ,where $C_1=e^{C}$

$P=C_2e^{0.05t}+0.6$ ,where $C_2=\frac{C_1}{0.05}$

Using P(0)=40;

40=C₂+0.6

C₂=40-0.6=39.4

$P=39.4e^{0.05t}+0.6$

6.3)

We use the equation above,

$P=39.4e^{0.05t}+0.6$

t=10

$P=39.4e^{0.05*10}+0.6$

$=64.989618+0.6$

$=65.559618million$