변수분리형 미분방정식

$P(t)=\frac{1600000{e}^{0.02t}}{9840+160{e}^{0.02t}}$

Let $Q(t)$ be the quantity of morphine in a patient's bloodstream at time $t$, where $t$ is measured in minutes.

Using the definition of a derivative, $$\frac{\mathrm{d}Q}{\mathrm{d}t}=\lim_{h \to 0}\frac{Q(t+h)-Q(t)}{h}\approx \frac{Q(t+1)-Q(t)}{1}$$ So, $\frac{\mathrm{d}Q}{\mathrm{d}t}$ is roughly the change in the amount of morphine in one minute, from $t$ to $t+1$.

The sentence tells us that the change in the amount of morphine in one minute is about $-0.003Q$, where $Q$ is the quantity in the bloodstream. That is: $$\frac{\mathrm{d}Q}{\mathrm{d}t}=-0.003Q(t)$$

If $p(t)$ is the proportion of times speakers use the new form, measured between 0 and 1, then $1-p(t)$ is the proportion of times speakers use the old form.

The law, then, states that $\frac{\mathrm{d}p}{\mathrm{d}t}$ is proportional to $p(t)\times \big(1-p(t)\big)$. When we say two quantities are proportional, we mean that one is a constant multiple of the other. So, the law says $$\frac{\mathrm{d}p}{\mathrm{d}t}=\alpha p(t)\big(1-p(t)\big)$$ for some constant $\alpha$.

Remark: it follows from this model that, when a new form is either very rare or entirely ubiquitous, the rate of change of its adoption is small. This makes sense: if the new form is used all the time ($p(t)\approx 1$), there's nobody left to convert; if the new form is almost never used ($p(t)\approx 0$) then people don't know about it, so they won't pick it up.

The given differential equation is separable and we solve it accordingly.

$$\begin{aligned} \frac{\mathrm{d}P}{\mathrm{d}t}&=-k\sqrt{P}\\ \frac{\mathrm{d}P}{\sqrt{P}}&=-k\,\mathrm{d}t\\ \int\frac{\mathrm{d}P}{\sqrt{P}}&=\int-k\,\mathrm{d}t\\ 2\sqrt{P}&=-kt+C \\ \text{At $t=0$, $P=90,000$ so } \\ 2\sqrt{90,000}&=-k\times 0+C\\ C&=2\times 300=600 \\ \text{Therefore,} \\ 2\sqrt{P}&=-kt+600\tag{$*$} \\ \text{Now, we find $k$. Let $t$ be measured in weeks. Then when $t=6$, $P=40,000$.} \\ 2\sqrt{40,000}&=-6k+600\\ 2\cdot 200&=-6k+600\\ k&=\frac{200}{6} = \frac{100}{3} \\ \text{Substituting our value of $k$ into ($*$):} \\ 2\sqrt{P}&=-\frac{100}{3}t+600 \\ \text{To find when the population will be 10,000, we set $P=10,000$ and solve for $t$.} \\ 2\sqrt{10,000}&=-\frac{100}{3}t+600\\ 2\cdot 100&=-\frac{100}{3}t+600\\ \frac{100}{3}t&=400\\ t&=12 \end{aligned}$$

Since we measured $t$ in weeks when we found $k$, we see that in 12 weeks the population will decrease to 10,000 individuals.

The given differential equation is separable and we solve it accordingly.

$$\begin{aligned} m\frac{\mathrm{d}v}{\mathrm{d}t}&=-(mg+kv^2)\\ \frac{m}{mg+kv^2}\,\mathrm{d}v&=-\mathrm{d}t\\ \int\frac{m}{mg+kv^2}\,\mathrm{d}v&=\int-\mathrm{d}t \\ \text{The left integral looks something like the antiderivative of arctangent. Let's factor out that $mg$ from the denominator.} \\ \frac{1}{mg}\int\frac{m}{1+\frac{k}{mg}v^2}\,\mathrm{d}v&=-t+C\\ \frac{1}{g}\int\frac{1}{1+\left(\sqrt{\frac{k}{mg}}v\right)^2}\,\mathrm{d}v&=-t+C \\ \text{Now it looks even more like the derivative of arctangent. We can guess the antiderivative from here, or use the substitution $u=\sqrt{\frac{k}{mg}}v$, $\mathrm{d}u=\sqrt{\frac{k}{mg}}\,\mathrm{d}v$.} \\ \frac{1}{g}\sqrt{\frac{mg}{k}}\arctan\left(\sqrt{\frac{k}{mg}}v\right)&=-t+C\\ \sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v\right)&=-t+C\tag{$*$} \\ \text{At $t=0$, $v=v_0$, so: } \\ \sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)&=C \\ \text{Plug $C$ into $(*)$.} \\ \sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v\right)&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)-t \\ \text{ At its highest point, the object has velocity $v=0$. This happens when $t$ obeys:} \\ \sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}0\right)&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)-t\\ 0&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)-t\\ t&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right) \end{aligned}$$

(a) The given differential equation is separable and we solve it accordingly.

$$\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}t}&=-k\,v^2\\ -\frac{\mathrm{d}v}{v^2}&=k\,\mathrm{d}t\\ \int- \frac{\mathrm{d}v}{v^2}&=\int k\,\mathrm{d}t\\ \frac{1}{v}&=kt+C \\ \text{At $t=0$, $v=40$ so } \\ \frac{1}{40}&=k\times 0+ C\\ C&=\frac{1}{40} \\ \text{Therefore,} \\ v(t)&=\frac{1}{kt+C}=\frac{1}{kt+1/40}=\frac{40}{40kt+1}\tag{$*$} \\ \text{The constant of proportionality $k$ is determined by} \\ v(10)&=20\\ 20&=\frac{40}{40k\times 10+1}\\ \frac{1}{2}&=\frac{1}{400k+1}\\ 400k+1&=2 \\ k&=\frac{1}{400} \end{aligned}$$

(b) Subbing in the value of $k$ to $(*)$,

$$\begin{aligned} v(t)&=\frac{40}{40kt+1}=\frac{40}{t/10+1}\\ \\ \text{We want to know the value of $t$ that gives $v(t)=5$.} \\ 5&=\frac{40}{t/10+1}\\ \frac{t}{10}+1&=8 \\ t&=70\text{ sec} \end{aligned}$$

(a) The rate of change of speed at time $t$ is $-kv(t)^2$ for some constant of proportionality $k$ (to be determined–but we assume it is positive, since the speed is decreasing). So $v(t)$ obeys the differential equation $\frac{\mathrm{d}v}{\mathrm{d}t}=-kv^2$ . (b) The equation $\frac{\mathrm{d}v}{\mathrm{d}t}=-kv^2$ is a separable differential equation, which we can solve in the usual way.

$$\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}t}&=-kv^2\\ \frac{\mathrm{d}v}{-v^2}&=k\mathrm{d}t\\ \int -\frac{\mathrm{d}v}{v^2} &=\int k\mathrm{d}t\\ \frac{1}{v}&=kt+C \\ \text{At time $t=0$, $v=400$, so $C=\frac{1}{400}$. Then:} \\ \frac{1}{v}&=kt+\frac{1}{400}\tag{$*$} \\ \text{At time $t=1$, $v=200$, so } \\ \frac{1}{200}&=k+\frac{1}{400}\\ k&=\frac{1}{400} \\ \text{Therefore, from ($*$),} \\ \frac{1}{v}&=\frac{t}{400}+\frac{1}{400} = \frac{t+1}{400}\\ v&=\frac{400}{t+1} \end{aligned}$$ (c) To find when the speed is 50, we set $v=50$ in the equation from (b) and solve for $t$.

$$\begin{aligned} 50&=\frac{400}{t+1}\\ 50(t+1)&=400\\ t+1&=8\\ t&=7 \end{aligned}$$

Suppose that in a very short time interval $\mathrm{d}t$, the height of water in the tank changes by $\mathrm{d}h$ (which is negative). Then in this time interval the amount of the water in the tank decreases by $\mathrm{d}V=-\pi(3)^2\mathrm{d}h$. This must be the same as the amount of water that flows through the hole in this time interval. The water flowing through the hole makes a cylinder of radius 1 cm (that is, 0.01 m) with length $v(t)\mathrm{d}t$, the distance the water moves out of the hole in $\mathrm{d}t$ seconds. So, the amount of water leaving the hole over the time interval $\mathrm{d}t$ is $\pi(0.01)^2 v(t)\,\mathrm{d}t =\pi(0.01)^2 \sqrt{2gh(t)}\,\mathrm{d}t$.

This gives us a separable differential equation. Recall $g$ is a constant.

$$\begin{aligned} -\pi(3)^2\mathrm{d}h&=\pi(0.01)^2 \sqrt{2gh(t)}\,\mathrm{d}t\\ \frac{\mathrm{d}h}{\sqrt{h}}&=-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,\mathrm{d}t\\ \int\frac{\mathrm{d}h}{\sqrt{h}}&=\int-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,\mathrm{d}t\\ 2\sqrt{h}&=-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,t+C \\ \text{At time $0$, the height is $6$, so $C=2\sqrt{6}$ and} \\ 2\sqrt{h}&=-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,t+2\sqrt{6} \\ \text{We want to know when the height of the water in the tank is 0.} \\ 0&=-\Big(\frac{0.01}{3}\Big)^{\!2} \sqrt{2g}\,t+2\sqrt{6}\\ \Big(\frac{0.01}{3}\Big)^{\!2} \sqrt{2g}\,t&=2\sqrt{6}\\ t&=\frac{2\sqrt6}{\Big(\frac{0.01}{3}\Big)^{\!2} \sqrt{2g}}\\ &=2{\Big(\frac{3}{0.01}\Big)}^{\!2} \sqrt{\frac{3}{g}}\\ &=180,000 \sqrt{\frac{3}{g}}\approx 99,591\text{ sec} \approx 27.66\text{ hr} \end{aligned}$$

Suppose that at time $t$, the mercury in the tank has height $h$, which is between 0 and 12 feet.

At that time, the top surface of the mercury forms a circular disk of radius $\sqrt{6^2-(h-6)^2}$. (We found this by applying the Pythagorean Theorem to the triangle in the diagram above. In the diagram, $h$ is shown as being larger than 6, but the same equation holds for all $h$ in $[0,12]$.) Now suppose that in a very short time interval $\mathrm{d}t$, the height of mercury in the tank changes by $\mathrm{d}h$ (which is negative). Then in this time interval the amount of the mercury in the tank decreases by $-\pi\big(\sqrt{6^2-(h-6)^2}\ \big)^2\mathrm{d}h$. (That's the volume of the red disk in the figure above.) This must be the same as the amount of mercury that flows through the hole in this time interval. The mercury comes out of the hole as a cylinder. Its radius is the radius of the hole, $\frac{1}{12}$ foot, and its length is the distance the mercury travels in $\mathrm{d}t$ seconds, $v(t)\mathrm{d}t$ feet. So, the volume of escaped mercury is $\pi\big(\frac{1}{12}\big)^2 v\,\mathrm{d}t =\pi\big(\frac{1}{12}\big)^2 \sqrt{2gh}\,\mathrm{d}t$. This gives us a separable differential equation.

$$\begin{aligned} -\pi{\big(\sqrt{6^2-(h-6)^2}\ \big)}^2\mathrm{d}h &=\pi\Big(\frac{1}{12}\Big)^2 \sqrt{2gh}\,\mathrm{d}t\\ -{\big({36-(h^2-12h+36)}\ \big)}\mathrm{d}h &=\Big(\frac{1}{12}\Big)^2 \sqrt{2gh}\,\mathrm{d}t\\ \big(h^2-12h\big)\mathrm{d}h &=\frac{1}{144} \sqrt{2g}\sqrt{h}\,\mathrm{d}t\\ \big(h^{3/2}-12h^{1/2}\big)\mathrm{d}h&=\frac{1}{144} \sqrt{2g}\,\mathrm{d}t\\ \int \big(h^{3/2}-12h^{1/2}\big)\mathrm{d}h&=\int\frac{1}{144} \sqrt{2g}\,\mathrm{d}t\\ \frac{h^{5/2}}{5/2}-12\frac{h^{3/2}}{3/2}&=\frac{1}{144} \sqrt{2g}\,t+C \\ \text{At time $0$, the height is $12$, so $C=\frac{12^{5/2}}{5/2}-12\frac{12^{3/2}}{3/2} =12^{5/2}\big(\frac{2}{5}-\frac{2}{3}\big) =-\frac{4}{15}12^{5/2}$, which yields} \\ \frac{h^{5/2}}{5/2}-12\frac{h^{3/2}}{3/2}&=\frac{1}{144} \sqrt{2g}\,t -\frac{4}{15}12^{5/2} \\ \text{We want to find the time $t$ when the height is $h=0$.} \\ 0&=\frac{1}{144} \sqrt{2g}\,t-\frac{4}{15}12^{5/2}\\ \frac{1}{144} \sqrt{2g}\,t&=\frac{4}{15}12^{5/2}\\ t&=\frac{4\times 144}{15} \sqrt{\frac{12^5}{2g}}\\ &=38.4 \sqrt{\frac{124416}{g}}\approx 2,394\,\text{sec }\approx 0.665\, \text{hr} \end{aligned}$$