Answer:
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?
Step-by-step explanation:
Use the initial condition when dp/dt = 1, p = 10 to get k;
[tex]\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}[/tex]
Seperate the differential equation and solve for the constant C.
[tex]\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2} }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}[/tex]
You have 100 rodents when:
[tex]100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months[/tex]
You have 1000 rodents when:
[tex]1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months[/tex]
