미분방정식 전체 문제

The two pieces of information give us

$$\begin{aligned} f(0) &= A = 5 & f(7) &= A e^{7k}=\pi \end{aligned}$$

Thus we know that $A=5$ and so $\pi = f(7) = 5e^{7k}$. Hence

$$\begin{aligned} e^{7k} &= \frac{\pi}{5} \\ 7k &= \log(\pi/5) \\ k &= \frac{1}{7} \cdot \log(\pi/5). \end{aligned}$$

where we use $\log$ to mean natural logarithm, $\log_e$.

The amount of Carbon-14 in the sample $t$ years after the animal died will be $$Q(t)=5e^{-kt}$$ for some constant $k$ (where 5 is the amount of Carbon-14 in the sample at time $t=0$). So, the answer we're looking for is $Q(10000)$. We need to replace $k$ with an actual number to evaluate $Q(10000)$, and the key to doing this is the half-life. The text tells us that the half-life of Carbon-14 is 5730 years, so we know:

$$\begin{aligned} Q(5730)&=\frac{5}{2}\\ 5e^{-k\cdot5730}&=\frac{5}{2}\\ \left(e^{-k}\right)^{5730}&=\frac{1}{2}\\ e^{-k}&=\sqrt[5730]{\frac{1}{2}}=2^{-\tfrac{1}{5730}} \\ \text{So:} \\ Q(t)&=5\left(e^{-k}\right)^t\\ &=5\cdot2^{-\tfrac{t}{5730}} \\ \text{Now, we can evaluate:} \\ Q(10000)&=5\cdot 2^{-\tfrac{10000}{5730}}\approx 1.5~\mu g \end{aligned}$$

Remark: after $2(5730)=11,460$ years, the sample will have been sitting for two half-lives, so its remaining Carbon-14 will be a quarter of its original amount, or $1.25$ $\mu$g. It makes sense that at 10,000 years, the sample will contain slightly more Carbon-14 than at 11,460 years. Indeed, 1.5 is slightly larger than 1.25, so our answer seems plausible.

It's a good habit to look for ways to quickly check whether your answer seems plausible, since a small algebra error can easily turn into a big error in your solution.

Let 100 years ago be the time $t=0$. Then if $Q(t)$ is the amount of Radium-226 in the sample, $Q(0)=1$, and

$$\begin{aligned} Q(t)&=e^{-kt} \\ \text{for some positive constant $k$. When $t=100$, the amount of Radium-226 left is 0.9576 grams, so} \\ 0.9576=Q(100)&=e^{-k\cdot 100}=\left(e^{-k}\right)^{100}\\ e^{-k}&=0.9576^{\tfrac{1}{100}}\\ \text{This tells us} \\ Q(t)&=0.9576^{\tfrac{t}{100}} \\ \text{So, if half the original amount of Radium-226 is left,} \\ \frac{1}{2}&=0.9576^{\tfrac{t}{100}}\\ \log\left(\frac{1}{2}\right)&=\log\left(0.9576^{\tfrac{t}{100}}\right)\\ -\log 2&=\frac{t}{100}\log(0.9576)\\ t&=-100\frac{\log 2}{\log 0.9576}\approx 1600 \end{aligned}$$

So, the half life of Radium-226 is about 1600 years.

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.

According to the Malthusian Model, if the ship wrecked at year $t=0$ and 2 rats washed up on the island, then $t$ years after the wreck, the population of rats will be

$$\begin{aligned} P(t)&=2e^{bt} \\ \text{for some constant $b$. We want to get rid of this extraneous variable $b$, so we use the given information. If 1928 is $a$ years after the wreck:} \\ 1000=P(a)&=2e^{ba}\\ 1500=P(a+1)&=2e^{b(a+1)}=2e^{ba}e^b \\ \text{So,} \\ (1000)\left(e^b\right)&=\left(2e^{ba}\right)\left(e^b\right)=1500 \\ \text{Which tells us} \\ e^b&=\frac{1500}{1000}=1.5 \\ \text{Now, our model is complete:} \\ P(t)&=2\left(e^b\right)^t=2\cdot1.5^t \\ \text{Since $P(a)=1000$, we can find $a$:} \\ 1000=P(a)&=2\cdot1.5^a\\ 500&=1.5^a\\ \log(500)&=\log\left(1.5^a\right)=a\log(1.5)\\ a&=\frac{\log(500)}{\log(1.5)}\approx 15.3 \end{aligned}$$

So, the year 1928 was about 15.3 years after the shipwreck. Since we aren't given a month when the rats reached exactly 1000 in number, that puts the shipwreck at the year 1912 or 1913.

Let $Q(t)$ denote the amount of radioactive material after $t$ days. Then $Q(t)=Q(0)e^{kt}$. We are told

$$\begin{aligned} Q(8)&=0.8\,Q(0) \\ \text{So,} \\ Q(0)e^{8k}&=0.8\,Q(0)\\ e^{8k}&=0.8\\ e^k&=0.8^{\tfrac{1}{8}} \\ \text{If $Q(0)=100$, the time $t$ at which $Q(t)=40$ is determined by} \\ 40=Q(t)&=Q(0)e^{kt}=100e^{kt}=100\left(0.8^{\tfrac{1}{8}}\right)^t=100\cdot 0.8^{\tfrac{t}{8}} \\ \text{Solving for $t$:} \\ \frac{40}{100}&=0.8^{\tfrac{t}{8}}\\ \log\left(0.4\right)&=\log\left(0.8^{\tfrac{t}{8}}\right)=\frac{t}{8}\log(0.8)\\ t&=\frac{8\log(0.4)}{\log(0.8)}\approx 32.85\text{ days} \end{aligned}$$

100 grams will decay to 40 grams in about 32.85 days.

(a) If $y=5(e^x-3x^2-6x-6)$, then $\frac{\mathrm{d}y}{\mathrm{d}x} = 5(e^x-6x-6)$. Let's see whether this is equal to $y+15x^2$:

$$\begin{aligned} y+15x^2&=5(e^x-3x^2-6x-6)+15x^2\\ &=5(e^x-3x^2-6x-6+3x^2)\\ &=5(e^x-6x-6)\\ &=\frac{\mathrm{d}y}{\mathrm{d}x} \end{aligned}$$

So, $y=5(e^x-3x^2-6x-6)$ is indeed a solution to the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y+15x^2$. (b) If $y=\dfrac{-2}{x^2+1}$, then $\frac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4x}{(x+1)^2}$. Let's see whether this is equal to $xy^2$:

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

So, $y=\dfrac{-2}{x^2+1}$ is indeed a solution to the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=xy^2$. (c) If $y=x^{3/2}+x$, then $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{2}\sqrt{x}+1$.

$$\begin{aligned} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+\frac{\mathrm{d}y}{\mathrm{d}x}&=\left(\frac{3}{2}\sqrt{x}+1\right)^2+\frac{3}{2}\sqrt{x}+1\\ &=\frac{9}{4}x+\frac{9}{2}\sqrt{x}+2\\ &\neq \frac{\mathrm{d}y}{\mathrm{d}x} \end{aligned}$$

So, $y=x^{3/2}+x$ is not a solution to the differential equation $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+\frac{\mathrm{d}y}{\mathrm{d}x}=y$.

Rearranging, we have:

$$\begin{aligned} e^y\,\mathrm{d}y &= 2x\,\mathrm{d}x. \\ \text{ Integrating both sides: } \\ \int e^y\,\mathrm{d}y &= \int 2x\,\mathrm{d}x\\ e^y &= x^2+C \\ \text{Since $y=\log2$ when $x=0$, we have} \\ e^{\log 2} &= 0^2+C \\ 2 &= C, \\ \text{and therefore} \\ e^{y} &= x^2+2 \\ y &= \log(x^2+2) \end{aligned}$$

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

$$\begin{aligned} y'&=e^{\frac{y}{3}}\cos t\\ e^{-y/3}\mathrm{d}y&=\cos t\,\mathrm{d}t\\ \int e^{-y/3}\mathrm{d}y&=\int\cos t\,\mathrm{d}t\\ -3e^{-y/3}&=\sin t +C\\ \frac{1}{e^{y/3}}&=\frac{\sin t +C}{-3}\\ e^{y/3} &=\frac{-3}{C+\sin t} \\ \frac{y}{3}&=\log\left(\frac{-3}{C+\sin t}\right) \\y(t)&=3\log\left(\frac{-3}{C+\sin t}\right) \end{aligned}$$

for any constant $C$.

Since the domain of logarithm is $(0,\infty)$, the solution only exists when $C+\sin t<0$.

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

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= x e^{x^2-\log(y^2)} =\frac{x e^{x^2}}{y^2}\\ y^2\mathrm{d}y&=x e^{x^2}\,\mathrm{d}x\\ \int y^2\mathrm{d}y&=\int x e^{x^2}\,\mathrm{d}x \\ \text{We can guess the antiderivative of $xe^{x^2}$, or use the substitution $u=x^2$, $\mathrm{d}u=2x\mathrm{d}x$.} \\ \frac{y^3}{3}&=\frac{1}{2} e^{x^2}+C'\\ y^3&=\frac{3}{2} e^{x^2}+3C' \\ \text{Since $C'$ can be any constant in $(-\infty,\infty)$, then also $3C'$ can be any constant in $(-\infty,\infty)$, so we replace $3C'$ with the arbitrary constant $C$.} \\ y^3&=\frac{3}{2} e^{x^2}+C \\y&=\sqrt[3]{\frac{3}{2} e^{x^2}+C} \end{aligned}$$