미분방정식 전체 문제

$\begin{array}{l | l} x_i & y_i \\ \hline 0.00 & 1.0000 \\ 0.25 & 1.5000 \\ 0.50 & 2.3125 \\ 0.75 & 3.5938 \\ 1.00 & 5.5781 \end{array}$

$\begin{array}{l | l} x_i & y_i \\ \hline 0.0 & 1.0000 \\ 0.1 & 1.0000 \\ 0.2 & 1.0037 \\ 0.3 & 1.0110 \\ 0.4 & 1.0219 \\ 0.5 & 1.0363 \end{array}$

$\begin{array}{l | l} x_i & y_i \\ \hline 0.0 & 2.0000 \\ 0.2 & 2.4000 \\ 0.4 & 2.9197 \\ 0.6 & 3.5816 \\ 0.8 & 4.4108 \\ 1.0 & 5.4364 \end{array}$

$\begin{array}{l | l} x_i & y_i \\ \hline 0.0 & 0.0000 \\ 0.5 & 0.5000 \\ 1.0 & 1.8591 \\ 1.5 & 10.5824 \\ 2.0 & 88378.1190 \\ \end{array}$

$\begin{array}{l | l | l | l} x_i & y(x) & h = 0.2 & h = 0.1 \\ \hline 0.0 & 0.5000 & 0.5000 & 0.5000 \\ 0.2 & 0.5412 & 0.5000 & 0.5201 \\ 0.4 & 0.6806 & 0.5816 & 0.6282 \\ 0.6 & 0.9747 & 0.7686 & 0.8622 \\ 0.8 & 1.5551 & 1.1250 & 1.3132 \\ 1.0 & 2.7183 & 1.7885 & 2.1788 \end{array}$

(a) $P$ increases for $1 < P < 3$ since $\frac{dP}{dt} > 0$ for these $P$-values. (b) $P$ decreases for $P < 1$ and $3 < P < 4$ since $\frac{dP}{dt} < 0$ for these $P$-values. (c) If $P(0) = 3$, $\frac{dP}{dt}\vert_{P=3} = 0$, so $P$ will not change at all with time since its derivative is $0$. (d) If the initial population satisfies $0< P(0)< 1$, we expect that the population will decrease toward $P = 0$ with $P$ always being positive. Since $\frac{dP}{dt}$ is always negative for $0< P < 1$, $P$ is always decreasing on this interval. The population can't drop below $P = 0$ because at that value of $P$, $\frac{dP}{dt} = 0$ and at that value the population wouldn't change. (e) If the initial population satisfies $1< P(0)< 3$, we expect that the population will increase toward $P = 3$ with $P$ always being between $1$ and $3$. Since $\frac{dP}{dt}$ is always positive for $1 < P < 3$, $P$ is always increasing on this interval. The population can't increase above $P = 3$ because at that value of $P$, $\frac{dP}{dt} = 0$ and at that value the population wouldn't change. (f) If the initial population satisfies $3< P(0)$, we expect that the population will decrease toward $P = 3$ with $P$ always being above $3$. Since $\frac{dP}{dt}$ is always negative for $P > 3$, $P$ is always decreasing on this interval. The population can't drop below $P = 3$ because at that value of $P$, $\frac{dP}{dt} = 0$ and at that value the population wouldn't change. (g) It looks like there are two thresholds for the population: there's a minimum threshold of $P = 1$, since below that population, the population decreases to $0$, and there's a maximum threshold of $P = 3$, where if we are above the minimum but below the maximum, the population works to increase toward the maximum threshold, while if the population is above the maximum, its value will fall back toward it.

(a) A graph of $f$ against $P$ is given in blue in the figure below. This graph and the differential equation show that the equilibrium solutions occur when $f(P)=0$, or when $P=0$ or $P=6$. Since $P' < 0$ for $P$ slightly less than 0, the solutions with $P < 0$ decrease away from the equilibrium $P=0$. Also, $P'>0$ for $P$ slightly larger than 0 means that the solutions for $P >$ increase away from the equilibrium $P=0$. So $P=0$ is an unstable equilibrium. If $P$ is slightly less than $6$ we see that $P > 0$, so these solutions increase toward the equilibrium $P=0$. When $P$ is slightly larger than $6$, we have $P' < 0$ and so these solutions decrease toward the equilibrium $P = 6$. We conclude that the equilibrium solution $P=6$ is stable. (b) Since $P$ will decrease by $1$ thousand fish every month through harvesting, the new differential equation is $$\frac{dP}{dt} = g(P) = P(6-P)-1\,.$$ A sketch of $g$ against $P$ is given in green in the figure in (a). The equilibrium solutions now occur when $g(P) = P(6-P)-1 = 0$. This happens when $P^2-6P+1 = 0$. The quadratic formula tells us that the solutions to this quadratic equation are $$P = \frac{6+\sqrt{32}}{2} \approx 5.83 \ \ \ \text{ and } \ \ \ P = \frac{6-\sqrt{32}}{2} \approx 0.172\,.$$ The same argument as in (a) shows that the equilibrium at $P\approx 0.172$ is unstable and the equilibrium at $P \approx 5.83$ is stable. (c) Assuming that there are fish in the lake, so that $P > 0$, there are two cases to consider. If $P < \frac{6-\sqrt{32}}{2}$, then $P' < 0$ and the fish population will die out. If $\frac{6-\sqrt{32}}{2} < P < \frac{6+\sqrt{32}}{2}$, then $P' > 0$ and the fish population will increase and approach a stable equilibrium population of $\frac{6+\sqrt{32}}{2}$ thousand fish. If $P > \frac{6+\sqrt{32}}{2}$, then $P' < 0$ and the fish population will decrease and approach a stable equilibrium population of $\frac{6+\sqrt{32}}{2}$ thousand fish. (d) In this case the differential equation becomes $$\frac{dP}{dt} = g(P) = P(6-P)-h\,.$$ We will have equilibrium solutions when $P^2 - 6P + h = 0$ or when $$P = \frac{6+\sqrt{36-4h}}{2} \ \ \ \text{ and } \ \ \ P = \frac{6-\sqrt{36-4h}}{2}\,.$$ (e) We need a positive stable equilibrium. This will will happen as long as $36-4h \ge 0$ or when $h \le 9$. So the maximum harvest is $9000$ fish. Note that $P'$ is negative, so we need to start with more than $3000$ fish. Since $P(6-P)$ has a maximum value of $9$, the fish will reproduce quickly enough so that harvesting $9000$ fish will maintain a stable population.

(a) If we let $y(t)$ be the population of the mice at year $t$ in thousands and it grows at a yearly rate that is twenty times the number of mice, the governing differential equation is $\frac{dy}{dt} = 20y$. (b) On their own, the mice add $20y$ thousand mice per year to their total population. With $C$ cats present, the cats now remove $M(y) = C\frac{y}{2+y}$ thousand mice per year. By taking both the rates at which mice are added and removed from the population into account, we find the updated differential equation $$\frac{dy}{dt} = 20y - C\frac{y}{2+y}$$ (c) For positive $y$ near $0$, $M(y) = \frac{y}{2+y} \approx 0$; for large values of $y$, $M(y) = \frac{y}{2+y} \approx 1$. This means that a single cat can remove about $1$ thousand mice per year from the population when the mice population is large. (d) Letting $C=1$, we are working with the differential equation $$\frac{dy}{dt} = 20y - \frac{y}{2+y}\,.$$ To find the equilibrium solutions, we first set $\frac{dy}{dt} = 0$. Doing so, we find that $20y = \frac{y}{2+y}$, so $40y + 20y^2 = y$, or $39y + 20y^2 = 0$. Factoring, we see $y(39y + 20) = 0$ and therefore $y = 0$ or $y = -\frac{20}{39}$. Since a negative equilibrium solution doesn't make sense in the context of the problem, the only equilibrium solution to study is $y = 0$. If we plot $dy/dt$ versus $y$ for $y > 0$, we see that $dy/dt$ is always positive, and thus the population will grow away from $y = 0$, making this equilibrium unstable. And this makes sense: the population of mice will grow without bound, because one cat is insufficient to keep up with a mice population growing at a rate of $20$ times its size per year. (e) Letting $C=60$, we are working with the differential equation $$\frac{dy}{dt} = 20y - \frac{60y}{2+y}\,.$$ Similar work as in (d) shows that the equilibrium solutions are $y = 0$ or $y = 1$, and if we plot $dy/dt$ versus $y$, we see that for $0 < y < 1$, $dy/dt < 0$, while for $y > 1$, $dt/dt > 0$. Thus, with $60$ cats present, as long as there are initially fewer than $1000$ mice, the cats will drive the mice population to $0$. If there are more than $1000$ mice initially present, the mice population will grow without bound. And if there are $1000$ mice initially present, this will be an (unstable) equilibrium. (f) In order to keep the mice population from growing arbitrarily large, we need to bring enough cats that for certain initial mice populations, we can make $dy/dt \le 0$. In other words, we need enough cats that there's a positive equilibrium solution. Considering the general differential equation, $$\frac{dy}{dt} = 20y - C\frac{y}{2+y}\,,$$ we thus set $dy/dt = 0$ and solve for $C$. Doing so, we find that $0 = 20y - C\frac{y}{2+y}$, so $20y(y+2) = Cy$ or $20y^2 + 40y - Cy = 0$. Factoring, we find that $$y(20y + (40-C)) = 0\,.$$ To have a positive equilibrium solution, we require that $40-C < 0$, and therefore the farmer must bring at least $41$ cats to the farm. Even in that circumstance, whether the cats can keep the number of mice from growing without bound will depend on the initial number of mice present.

(a) The differential equation for Alice's coffee is $$\frac{dT_A}{dt} = -0.5(T-70)\,,$$ so at the instant $t = 0$, $T = 100$, and thus $\frac{dT_A}{dt} \vert_{T = 100} = -0.5(30) = -15$ degrees per minute. For Bob's coffee, $$\frac{dT_B}{dt} = -0.1(T-70)\,,$$ so $\frac{dT_B}{dt} \vert_{T = 100} = -0.1(30) = -3$ degrees per minute. These are the initial rates of change of the respective cups of coffee. (b) The difference between the cups of coffee could be explained by how insulated the containers are: Bob's coffee loses temperature less quickly, so his is likely better insulated. (c) With this varying temperature, the governing differential equation for Alice's coffee is $$\frac{dT_A}{dt} = -0.5(T_A-(70+10\sin t))\,,$$ with the inital condition $T_A(0) = 100$. Implementing Euler's Method with $\Delta t = 0.1$, we get the following (abbreviated) table of data.

$t$	$T_A(t)$
$0.0$	$100$
$0.1$	$98.5$
$0.2$	$97.12492$
$0.3$	$95.86801$
$0.4$	$94.72237$
$\vdots$	$\vdots$
$49.6$	$65.56715$
$49.7$	$65.48008$
$49.8$	$65.43816$
$49.9$	$65.44183$
$50$	$65.49103$
Plotting the data points that result from Euler's Method along with the room temperature function, we see how the room temperature drives the temperature of the coffee up and down as time goes on, as pictured in the figure above, where the data points for the coffee's approximate temperature are shown in dark blue, while the room temperature function, $T_R(t) = 70 + 10\sin(t)$ is shown in green.
(d) Similar work with the differential equation for Bob's coffee yields the following abbreviated table of approximate function values along with the plot of the data and the temperature function.
$t$	$T_B(t)$
---	---
$0.0$	$100$
$0.1$	$99.7$
$0.2$	$99.41298$
$0.3$	$99.13872$
$0.4$	$98.87689$
$\vdots$	$\vdots$
$49.6$	$69.39515$
$49.7$	$69.33946$
$49.8$	$69.29248$
$49.9$	$69.25467$
$50$	$69.22638$
(e) The behaviors of both cups of coffee are similar in that each decreases toward the room temperature initially and then is driven in an oscillating way by the room temperature function as it rises and falls. We see that Alice's coffee is more susceptible to being influenced away from the average room temperature, which is consistent with her mug being less well insulated, while in Bob's coffee, we see much less radical oscillations (and a slower overall rate of decrease to get near room temperature), which is consistent with being better insulated.