미분방정식 전체 문제

We know that the amount of Polonium-210 in a sample after $t$ days is given by

$$\begin{aligned} Q(t)&=Ce^{-kt} \\ \text{where $C$ is the original amount of the sample, and $k$ is some positive constant.} \\ \\ \text{The question asks us what percentage of the sample decays in a day. Since $t$ is measured in days, the amount that decays in a day is $Q(t)-Q(t+1)$. The percentage of $Q(t)$ that this represents is $100\dfrac{Q(t)-Q(t+1)}{Q(t)}$. (For example, if there were two grams at time $t$, and one gram at time $t+1$, then $100\dfrac{2-1}{1}=50$: 50\% of the sample decayed in a day.)} \\ \\ \text{In order to simplify, we should figure out a better expression for $Q(t)$. As usual, we make use of the half-life.} \\ Q(138)&=\frac{C}{2}\\ Ce^{-k\cdot138}&=\frac{C}{2}\\ \left(e^{-k}\right)^{138}&=\frac{1}{2}=2^{-1}\\ e^{-k}&=2^{-\tfrac{1}{138}} \\ \text{Now, we have a better formula for $Q(t)$:} \\ Q(t)&=C\left(e^{-k}\right)^t\\ Q(t)&=C\cdot2^{-\tfrac{t}{138}} \\ \text{Finally, we can evaluate what percentage of the sample decays in a day.} \\ 100\frac{Q(t)-Q(t+1)}{Q(t)}&=100\frac{C\cdot2^{-\tfrac{t}{138}}-C\cdot2^{-\tfrac{t+1}{138}}}{C\cdot2^{-\tfrac{t}{138}}}\left(\frac{\frac{1}{C}}{\frac{1}{C}}\right)\\ &=100\frac{2^{-\tfrac{t}{138}}-2^{-\tfrac{t+1}{138}}}{2^{-\tfrac{t}{138}}}\\ &=100\left(2^{-\tfrac{t}{138}}-2^{-\tfrac{t+1}{138}}\right)2^{\tfrac{t}{138}}\\ &=100\left(1-2^{-\tfrac{1}{138}}\right)\approx 0.5 \end{aligned}$$

About 0.5% of the sample decays in a day.

Remark: when we say that half a percent of the sample decays in a day, we don't mean half a percent of the original sample. If a day starts out with, say, 1 microgram, then what decays in the next 24 hours is about half a percent of that 1 microgram, regardless of what the “original" sample (at some time $t=0$) held.

In particular, since the sample is getting smaller and smaller, that half of a percent that decays every day represents fewer and fewer actual atoms decaying. That's why we can't say that half of the sample (50%) will decay after about 100 days, even though 0.5% decays every day and $100\times 0.5 = 50$.

The amount of Uranium-232 in the sample of ore at time $t$ will be

$$\begin{aligned} Q(t)&=Q(0)e^{-kt} \\ \text{where $6.9 \leq Q(0) \leq 7.5$. We don't exactly know $Q(0)$, and we don't exactly know the half-life, so we also won't exactly know $Q(10)$: we can only say that is it between two numbers. Our strategy is to find the highest and lowest possible values of $Q(10)$, given the information in the problem.} \\ \\ \text{In order for the most possible Uranium-232 to be in the sample after 10 years, we should start with the most and have the longest half-life (since this represents the slowest decay). So, we take $Q(0)=7.5$ and $Q(70)=\frac{1}{2}(7.5)$.} \\ Q(t)&=7.5e^{-kt}\\ \frac{1}{2}(7.5)=Q(70)&=7.5e^{-k(70)}\\ \frac{1}{2}&=\left(e^{-k}\right)^{70}\\ 2^{-\tfrac{1}{70}}&=e^{-k} \\ \text{So, in this secenario,} \\ Q(t)&=7.5\cdot 2^{-\tfrac{t}{70}} \\ \text{After ten years,} \\ Q(10)&=7.5\cdot 2^{-\tfrac{10}{70}}\approx 6.79 \\ \text{So after ten years, the sample contains \textit{at most} 6.8 $\mu$g.} \\ \\ \text{Now, let's think about the least possible amount of Uranium-232 that could be left after 10 years. We should start with as little as possible, so take $Q(0)=6.9$, and the sample should decay quickly, so take the half-life to be 68.8 years.} \\ Q(t)&=6.9e^{-kt}\\ \frac{1}{2}6.9=Q(68.8)&=6.9e^{-k(68.8)}\\ \frac{1}{2}&=\left(e^{-k}\right)^{68.8}\\ 2^{-\tfrac{1}{68.8}}&=e^{-k} \\ \text{In this scenario,} \\ Q(t)&=6.9\cdot 2^{-\tfrac{t}{68.8}} \\ \text{After ten years,} \\ Q(10)&=6.9\cdot 2^{-\tfrac{10}{68.8}}\approx 6.24 \\ \text{So after ten years, the sample contains \textit{at least} 6.2 $\mu$g.} \\ \end{aligned}$$

After ten years, the sample contains between 6.2 and 6.8 $\mu$g of Uranium-232.

What we're given is an equation relating $y$ to the integral of a function of $y$. What we know how to solve is an equation relating the derivative of $y$ to a function of $y$. We can create this by differentiating the given integral equation. By the Fundamental Theorem of Calculus, part 1:

$$y'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\int_0^x\big(y(t)^2-3y(t)+2\big)\sin t\ \mathrm{d}t\right\} =\big(y(x)^2-3y(x)+2\big)\sin x$$

So $y(x)$ satisfies the differential equation $y'=\big(y^2-3y+2\big)\sin x =(y-2)(y-1)\sin x$ and the initial equation $y(0)=3$ (just substitute $x=0$ into $(*)$). For $y\ne 1,2$:

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=(y-2)(y-1)\sin x\\ \frac{\mathrm{d}y}{(y-2)(y-1)}&=\sin x\ \mathrm{d}x \\\int \frac{\mathrm{d}y}{(y-2)(y-1)}&=\int \sin x\ \mathrm{d}x \\ \text{Using the method of partial fractions, we see $\frac{1}{(y-2)(y-1)}=\frac{1}{y-2}-\frac{1}{y-1}$.} \\ \int\Big[\frac{1}{y-2}-\frac{1}{y-1}\Big]\mathrm{d}y&=\int\sin x\ \mathrm{d}x\\ \log|y-2|-\log|y-1|&=-\cos x+c\\ \log\left|\frac{y-2}{y-1}\right|&=-\cos x+c\\ \left|\frac{y-2}{y-1}\right|&= e^{c-\cos x} \\ \text{The condition $y(0)=3$ forces $\big|\frac{3-2}{3-1}\big|= e^{c-1}$ or $e^c=\frac{1}{2} e$, hence} \\ \left|\frac{y-2}{y-1}\right|&=\frac{1}{2} e^{1-\cos x} \end{aligned}$$

Observe that, when $x=0$, $\frac{y-2}{y-1}=\frac{1}{2}>0$. Furthermore $\frac{1}{2} e^{1-\cos x}$, and hence $\big|\frac{y-2}{y-1}\big|$, can never take the value zero. As $y(x)$ varies continuously with $x$, $y(x)$ must remain larger than 2. Consquently, $\frac{y-2}{y-1}$ remains positive and we may drop the absolute value signs. Hence

$$\begin{aligned} \frac{y-2}{y-1}=\frac{1}{2} e^{1-\cos x} \\ \text{ Solving for $y$,} \\ \frac{y-2}{y-1}&=\frac{1}{2} e^{1-\cos x}\\ 2(y-2)&=e^{1-\cos x}(y-1)\\ 2y-4&=ye^{1-\cos x}-e^{1-\cos x}\\ y\big(2-e^{1-\cos x}\big)&=4-e^{1-\cos x} \\ y&=\frac{4-e^{1-\cos x}}{2-e^{1-\cos x}} \end{aligned}$$

To avoid division by zero in the last step, we need

$$\begin{aligned} e^{1-\cos x}&\neq2\\ 1-\cos x&\neq \log 2\\ \cos x&\neq 1-\log 2 \end{aligned}$$

Let $L=1-\log 2$, for brevity, and note that $L>0$. (This can be seen by observing $2<e$, so, $\log 2<\log e = 1$, hence $1-\log 2>0$.)

We know $x=0$ is in the domain of our function, but the points $x=\pm \arccos(L) = \pm\arccos(1-\log 2)$ are not.

Therefore, the largest interval for which our answer makes sense is $$-\arccos(1-\log 2))>x>\arccos(1-\log 2)$$ or approximately $- 1.259 < x < 1.259$.

(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 $y\ne 1,2$,

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

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 $x$.

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