Question

Suppose a population p of rodents satisfies the differential equation dpdt=kp2. dpdt=kp2. initially, there are p(0)=3p(0)=3 rodents, and their number is increasing at the rate of dpdt=1 rodent per month when there are p=12 rodents. how long will it take for this population to grow to 101 rodents? months

Answer 1

This is a separable differential equation, so can be solved as follows.

$\displaystyle\frac{dp}{dt}=kp^2=\frac{p^2}{144}\qquad\text{using dp/dt=1 at p=12}\\\\\frac{144\,dp}{p^2}=dt\qquad\text{separate the variables}\\\\144\int{p^{-2}}\,dp=\int{}\,dt\qquad\text{integrate}\\\\\frac{-144}{p}=t+C\\\\p=\frac{144}{48-t}\qquad\text{using p=3 at t=0 gives C=-48}$

The question asks when the population will reach 101 rodents. That will be the solution to

$101=\dfrac{144}{48-t}\\\\t=48-\dfrac{144}{101}\approx 46.574$

It will take about 46.6 months for the rodent population to grow to 101.