Question

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents. How long will it take for this population to grow to

Answer 1

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.

How long will it take for this population to grow to a hundred rodents? To a thousand rodents?

Explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

`(dp)/(dt) =kp^2\\\\1=k(10)^2\\\\k=(1)/(100)`

Seperate the differential equation and solve for the constant C.

`(dp)/(p^2)=kdt\\\\-(1)/(p)=kt+C\\\\(1)/(p)=-kt+C\\\\p=-(1)/(kt+C) \\\\2=-(1)/(0+C)\\\\-(1)/(2)=C\\\\p(t)=-(1)/((t)/(100)-(1)/(2) )\\\\p(t)=-(1)/((2t-100)/(200) )\\\\-(200)/(2t-100)`

You have 100 rodents when:

`100=-(200)/(2t-100) \\\\2t-100=-(200)/(100) \\\\2t=98\\\\t=49\ months`

You have 1000 rodents when:

`1000=-(200)/(2t-100) \\\\2t-100=-(200)/(1000) \\\\2t=99.8\\\\t=49.9\ months`