Answer to Question #122450 in Differential Equations for JSE

Question

The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increases by 15% in 10 years. What will be the population in 50 years? (Round your answer to the nearest person.)

____ persons

How fast is the population growing at t = 50? (Round your answer to two decimal places.)

______ persons / year

Answer 1

According to the question,

$\frac{dP}{dt} = kP$ where P is population at any time

solving equation,

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

$lnP = kt + C$ (1)

Applying the conditions,

at t=0, P = 500

$ln500 = C$ (2)

at t=10, P= 575

$ln(575) = k(10) + ln(500)$

$k = \frac{1}{10} ln (\frac{575}{500}) = 0.0139762$

So Equation of the Population growth is given by

$P = 500 e^{0.0139762t}$ (3)

Population in 50 years,

$P = 500 e^{0.0139762*50} = 1005.6 \approx 1006$ people

Rate of population growth is

$\frac {dP}{dt} = \frac {d (500 e^{0.0139762t})}{dt} = (500*0.0139762)e^{0.0139762t}$

putting t = 50, we get

$\frac {dP}{dt} =(500*0.0139762) e^{0.0139762*50} = 14$ people/year (approx)