로지스틱 방정식

(a) From the form of the given logistic equation, we can see the carrying capacity is 10000 fish. (b) In this situation, the initial condition is $P(0) = 10$, and by removing $20$% of the population per year, we subtract $0.2P$ from the rate that defines the differential equation, thus finding $$\frac{dP}{dt} = 0.1P(10 - P) - 0.2P\,.$$ (c) Expanding and rewriting the differential equation in (b), we see that $$\frac{dP}{dt} = P - 0.1P^2 - 0.2P = 0.8P - 0.1P^2 = 0.1P(8 - P)\,.$$ This, too, is a logistic equation, with carrying capacity $8000$ fish. (d) We can solve this logistic equation in the usual way (keeping in mind the initial condition $P(0) = 10$, and noting that the carrying capacity is $8$): $$P(t) = \frac{8}{\frac{8-10}{10}e^{-0.1 \cdot 8 t}+1} = \frac{8}{-0.2e^{-0.8t}+1}$$ After one year, the population of fish will thus be $$P(1) \approx 8.7899$$ thousand fish. (e) To be within 10% of the carrying capacity, we need $0.9(8) \le P(t) \le 1.1(8)$. Since the initial condition is larger than the carrying capacity, we want to know when $P(t) = 1.1(8) = 8.8$. Thus we solve the equation $$8.8 = \frac{8}{-0.2e^{-0.8t}+1}\,,$$ which implies $-0.2e^{-0.8t}+1 = 8/8.8 = 10/11$. Thus, $e^{-0.8t} = 5/11$, so $t = -1.25 \ln(5/11) \approx 0.986$ years.