Answer to Question #165063 in Differential Equations for baloch

Question

The population of a town grows at a rate proportional to the

population present at time t. The initial population of 700 increases by 20% in 15 years. What

will be the population in 35 years? How fast is the population growing at t=35?

Answer 1

The population $P$ after $t$ years obeys the differential equation

$\frac{dP}{t}=kP;$

where $k$ is a positive castant; the initial condition is $P(0)=700$.

To solve this, we use separation of variables:

$\int\frac{1}{P}dP=\int kdt$

$ln|P|=kt+C$

$|P|=e^Ce^{kt}$

$P=Ae^{kt}.$

Using $P(0)=700$ gives $700=Ae^0\Rarr A=700$.

Thus, $P=700e^{kt}.$ Furthermore, $P(15)=700\times120 \%=840$ so

$840=700e^{15k}$

$e^{15k}=1.2$

$15k=ln(1.2)$

$k=\frac{ln(1.2)}{15}\approx 0.012.$

Thus, $P=700e^{0.012t}.$ The population after 35 years is therefore

$P=700e^{0.012(35)}=1065.37\approx 1065$ people.

Rate of population growth is

$\frac{dP}{t}=\frac{d(700e^{0.012t})}{dt}=(700*0.012)e^{0.012t}$

putting $t=35$, we get

$\frac{dP}{t}=(700*0.012)e^{0.012(35)}\approx 13$ people/year.