로지스틱 방정식

$\alpha =0.0081$

(a) The equilibrium solutions for this logistic equation are $p = 0$ and $p = 1$. Since $p = 1$ is the carrying capacity and thus a stable equilibrium, $p(t) \to 1$ as $t \to \infty$ provided $p(0) > 0$. (b) We know that the general solution to the logistic equation $\frac{dP}{dt} = kP(N-P), \ P(0) = P_0$ is $$P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right) e^{-kNt} + 1}\,.$$ In the context of this problem, we know that $k = 0.2$, $N = 1$, and $p(0) = 0.1$ and therefore the solution is $$p(t) = \frac{1}{\left(\frac{1-0.1}{0.1}\right) e^{-0.2t} + 1} = \frac{1}{9e^{-0.2t} + 1}\,.$$ (c) We know that $p$ is changing most rapidly when its derivative, $p'$, is largest. Recalling the original differential equation, we know that $$p' = 0.2p(1-p)$$ which shows that $p'$ is a quadratic function of $p$. That quadratic function has zeros at $p = 0$ and $p = 1$, so the vertex of this concave down quadratic function lies at $p = \frac{1}{2}$, and at this value of $p$ the rate of change of $p$ is largest. Thus, we solve the equation $p(t) = 0.5$ to find the time at which the population is increasing fastest. We see that $$0.5 = \frac{1}{9e^{-0.2t} + 1}$$ so $$9e^{-0.2t} + 1 = 2$$ and $e^{-0.2t} = 1/9$. It follows that $t = -5 \ln(1/9) \approx 10.986$ days. (d) Since $p(t)$ tells us the percentage of people who've heard the rumor, we need to solve the equation $p(t) = 0.8$. Thus we have $$0.8 = \frac{1}{9e^{-0.2t} + 1}$$ so $$9e^{-0.2t} + 1 = 1.25$$ and $e^{-0.2t} = 0.25/9$. It follows that $t = -5 \ln(0.25/9) \approx 17.19$ days.