변수분리형 미분방정식

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.

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

(a) Setting $x=0$ gives

$$\begin{aligned} f(0)=3+\int_0^0\big(f(t)-1\big)\big(f(t)-2\big)\ \mathrm{d}t= 3 \end{aligned}$$

(b) By the Fundamental Theorem of Calculus part 1,

$$\begin{aligned} f'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\int_0^x\big(f(t)-1\big)\big(f(t)-2\big)\ \mathrm{d}t =\big(f(x)-1\big)\big(f(x)-2\big) \end{aligned}$$

Thus $y=f(x)$ obeys the differential equation $y'=(y-1)(y-2)$.

(c) If $y\ne 1,2$,

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=(y-1)(y-2) \\ \frac{\mathrm{d}y}{(y-1)(y-2)}&=\mathrm{d}x\\ \int\frac{\mathrm{d}y}{(y-1)(y-2)}&=\int\mathrm{d}x \\ \text{Using the method of partial fractions,} \\ \int\left(\frac{1}{y-2}-\frac{1}{y-1}\right)\mathrm{d}y&=\int \mathrm{d}x \\ \log|y-2|-\log|y-1|&=x+C\\ \log\left|\frac{y-2}{y-1}\right|&=x+C \\ \text{ Observe that $\frac{\mathrm{d}y}{\mathrm{d}x}=(y-1)(y-2)>0$ for all $y\ge 2$. That is, $f(x)$ is increasing at all $x$ for which $f(x)>2$. As $f(0)=3$, $f(x)$ increases for all $x\ge 0$, and $f(x)\ge 3$ for all $x\ge 0$. So we may drop the absolute value signs.} \\ \log\frac{f(x)-2}{f(x)-1}&=x+C\\ \frac{f(x)-2}{f(x)-1}&=e^Ce^x \\ \text{ At $x=0$, $\frac{f(x)-2}{f(x)-1}=\frac{1}{2}$ so $e^C=\frac{1}{2}$.} \\ \frac{f(x)-2}{f(x)-1}&=\frac{1}{2} e^x\\ 2f(x)-4&=[f(x)-1]e^x\\ [2-e^x]f(x)&=4-e^x \\ f(x)&=\frac{4-e^x}{2-e^x} \end{aligned}$$

Suppose that at time $t$ (measured in hours starting at, say, noon), the water in the tank has height $y$, which is between 0 and 2 metres. At that time, the top surface of the water forms a circular disk of radius $r=y^p$ and area $A(y)=\pi y^{2p}$. Thus, by Torricelli's law,

$$\begin{aligned} \pi y^{2p}\frac{\mathrm{d}y}{\mathrm{d}t}&=-c\sqrt{y}\\ -\frac{\pi}{c}\cdot y^{2p-{1\over 2}}\mathrm{d}y&=\,\mathrm{d}t\\ \int-\frac{\pi}{c}\cdot y^{2p-{1\over 2}}\mathrm{d}y&=\,\int \mathrm{d}t\\ -\frac{\pi}{c}\cdot\frac{y^{2p+{1\over 2}}}{2p+{1\over 2}}+d &=t \\ \text{for some constant $d$. At time $t=0$, the height is $y=2$, so $d=\displaystyle\frac{\pi}{c}\cdot\frac{2^{2p+{1\over 2}}}{2p+{1\over 2}}$ \ .} \\ t&=\frac{\pi}{c}\bigg(\frac{2^{2p+{1\over 2}}}{2p+{1\over 2}} -\frac{y^{2p+{1\over 2}}}{2p+{1\over 2}}\bigg)\\ &=\frac{\pi}{c(2p+\frac12)}\left(2^{2p+\frac12}-y^{2p+\frac12}\right) \end{aligned}$$

The time at which the height is $1$ is obtained by subbing $y=1$ into this formula. The time at which the height is $0$ is obtained by subbing $y=0$ into this formula. Thus the condition that the top half ($y=2$ to $y=1$) takes exactly the same amount of time to drain as the bottom half ($y=1$ to $y=0$) is:

$$\begin{aligned} t(2)-t(1)&=t(1)-t(0)\\ 0-t(1)&=t(1)-t(0)\\ t(0)&=2t(1)\\ \frac{\pi}{c(2p+\frac12)}\left(2^{2p+\frac12}-0^{2p+\frac12}\right)&= 2\frac{\pi}{c(2p+\frac12)}\left(2^{2p+\frac12}-1^{2p+\frac12}\right)\\ 2^{2p+\frac12}&= 2\left(2^{2p+\frac12}-1\right)\\ 2^{2p+\frac12}&= 2\cdot 2^{2p+\frac12}-2\\ 2&=2^{2p+\frac12}\\ 1&=2p+\frac12\\ p&=\frac14 \end{aligned}$$

We start by antidifferentiating both sides with respect to $x$.

$$\begin{aligned} \int \left(\frac{\mathrm{d}^2y}{\mathrm{d}{x}^2}\right)\,\mathrm{d}x&=\int\left(\frac{2}{y^3}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}\right)\,\mathrm{d}x \\ \text{The right integral is in exactly the form we would use for a change of variables (substitution) to $y$.} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\int\left(\frac{2}{y^3}\right)\,\mathrm{d}y = -\frac{1}{y^2}+C \\ \text{When $y=1$, $\frac{\mathrm{d}y}{\mathrm{d}x}=3$.} \\ 3& =-\frac{1}{1}+C\\ C&=4 \\ \text{So,} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{1}{y^2}+4 \\ \text{ This is a separable differential equation. } \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{4y^2-1}{y^2} \\ \frac{y^2}{4y^2-1}\,\mathrm{d}y&=\mathrm{d}x\\ \int \frac{y^2}{4y^2-1}\,\mathrm{d}y&=\int \mathrm{d}x\tag{$*$} \end{aligned}$$

We can evaluate the left integral with partial fractions, but because the numerator has the same degree as the denominator, we have to simplify first. We do this by inspection, but you can also use long division.

$$\begin{aligned} \frac{y^2}{4y^2-1}&=\frac{\frac{1}{4}(4y^2-1)+\frac{1}{4}}{4y^2-1}\\ &=\frac{1}{4}\left(1+\frac{1}{4y^2-1}\right)\\ &=\frac{1}{4}\left(1+\frac{1}{(2y-1)(2y+1)}\right)\\ &=\frac{1}{4}\left(1+\frac{1/2}{2y-1}-\frac{1/2}{2y+1}\right) \end{aligned}$$

Now, we return to ($*$).

$$\begin{aligned} \int \mathrm{d}x&=\int \frac{y^2}{4y^2-1}\,\mathrm{d}y\\ &= \int \frac{1}{4}\left(1+\frac{1/2}{2y-1}-\frac{1/2}{2y+1}\right)\,\mathrm{d}y\\ &= \frac{1}{4}\left(y+\frac{1}{4}\log|2y-1| - \frac{1}{4}\log|2y+1|\right)\\ &= \frac{1}{4}\left(y+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\\ x+C&=\frac{1}{4}\left(y+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right) \\ \text{When $x=-\frac{1}{16}\log 3$, $y=1$.} \\ -\frac{1}{16}\log 3 +C &=\frac{1}{4}\left(1+\frac14\log\left| \frac{2-1}{2+1}\right|\right) = \frac{1}{4}+\frac{1}{16}\log\frac{1}{3}\\ C&=\frac{1}{4} \\ \text{So,} \\ x+\frac{1}{4}&=\frac{1}{4}\left(y+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\\ x&=\frac{1}{4}\left(y-1+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right) \end{aligned}$$

We can check our answer by differentiating with respect to $x$.

$$\begin{aligned} x&=\frac{1}{4}\left(y-1+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\\ 4x&=y-1+\frac{1}{4}\log|2y-1| -\frac{1}{4}\log|2y+1|\\ \frac{\mathrm{d}}{\mathrm{d}x}\{4x\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{y-1+\frac{1}{4}\log|2y-1| -\frac{1}{4}\log|2y+1|\right\}\\ 4&=\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{4}\cdot\frac{2\frac{\mathrm{d}y}{\mathrm{d}x}}{2y-1} - \frac{1}{4}\cdot\frac{2\frac{\mathrm{d}y}{\mathrm{d}x}}{2y+1}\\ 4&=\frac{\mathrm{d}y}{\mathrm{d}x}\left(1+\frac{1/2}{2y-1} - \frac{1/2}{2y+1}\right) = \frac{\mathrm{d}y}{\mathrm{d}x}\left(\frac{4y^2}{4y^2-1}\right)\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{4y^2-1}{y^2}=4-\frac{1}{y^2} \tag{$**$} \\ \text{Differentiating with respect to $x$ again, using the chain rule,} \\ \frac{\mathrm{d}^2y}{\mathrm{d}{x}^2}&=\frac{2}{y^3}\cdot\frac{\mathrm{d}y}{\mathrm{d}x} \end{aligned}$$