Question

A population, initially consisting of M₀ mice, has a per-capita birth rate of
8 week
and a per-capita death rate of $2\frac{1}{week}$. Also, 20 mouse traps are set each fortnight and they are always filled.
(a) Write down the word equation for the mice population M(t).
(b) Write the differential rate equation for the number of mice.
(c) Solve the differential rate equation to obtain the formula for the mice population M(t) at any time t in terms of the initial population M₀.
(d) Find the equilibrium solution M_e.
(e) Find the long-term solution with dependence on M₀. What happens when M₀ = M_e?
(f) Find the time interval on which M(t) ≥ 0 with dependence on M₀.

UPDATE: The terms and numbers may not appear clear, I admit, but that is the question asked. The questioning has been queried and is correct. Please attempt an answer based on what is there, thank you so much.

For reference and to start off part (a) this is the answer I have gotten:

(Rate of change of number of mice)=(Rate of Births)−(Normal rate of Reaths)−(Rate of deaths by mousetraps)(Rate of change of number of mice)=(Rate of Births)−(Normal rate of Reaths)−(Rate of deaths by mousetraps)

Answer 1

Initial population = M_o Per capita birth rate = 8 per week for M mice birth rate = 8 M per week per capita death rate = 2 per week. For M mice death rate = 2 M per week death due to mouse trap = 20 per fortnight = 20 per two weeks = 10 per week. Word equation (Rate of change of number of mice) = (Rate of Normal Birth) - (Rate of Normal Deaths) - (Rate of deaths by Mouse traps) From the world equation dM/dt = 8 M - 2 M - 10 dM/dt = 6 M - 10 dM/dt = 6 (m - 10/6 dM/M - 5/3 = 6 dt Rightarrow int dM/M - 5/3 = 6 dt ln (M - 5/3 =