미분방정식 전체 문제

In the beginning of this section, the text says “A differential equation is an equation for an unknown function that involves the derivative of the unknown function." Our unknown function is $y$, so a differential equation is an equation that relates $y$ and $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$. This applies to (a) and (b), but not (c), (d), or (e).

Note that $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}x}=1$: this is the derivative of $x$ with respect to $x$.

What we're asked to find is when

$$\begin{aligned} Q(t)&=0 \\ \text{That is,} \\ Ce^{-kt}&=0 \end{aligned}$$

If $C=0$, then this is the case for all $t$. There was no isotope to begin with, and there will continue not being any undecayed isotope forever.

If $C>0$, then since $e^{-kt}>0$, also $Q(t)>0$: so $Q(t)$ is never 0 for any value of $t$. (But as $t$ gets bigger and bigger, $Q(t)$ gets closer and closer to 0.)

Remark: The last result is somewhat disturbing: surely at some point the last atom has decayed. The differential equation we use is a model that assumes $Q$ runs continuously. This is a good approximation only when there is a very large number of atoms. In practice, that is almost always the case.

We want to know when

$$\begin{aligned} [T(0)-A]e^{kt}+A&=A \\ \text{That is, when} \\ [T(0)-A]e^{kt}&=0 \\ \text{Since $e^{kt}>0$ for all values of $k$ and $t$, this happens exactly when} \\ T(0)-A=0 \end{aligned}$$

So: if the initial temperature of the object is not the same as the ambient temperature, then according to this model, it never will be! (However, as $t$ gets larger and larger, $T(t)$ gets closer and closer to $A$–it just never exactly reaches there.)

If the initial temperature of the object starts out the same as the ambient temperature, then $T(t)=A$ for all values of $t$.

Since $b$ is a positive constant, $\displaystyle \lim_{t \to \infty}e^{bt}=\infty$. Therefore:

$$\begin{aligned} \lim_{t \to \infty}P(t)&=\lim_{t \to \infty} P(0)e^{bt} =\left\{\begin{array}{cc} 0&\text{ if }P(0)=0\\ \infty&\text{ if }P(0)>0 \end{array}\right. \end{aligned}$$

If $P(0)=0$, then the model simply says “a population that starts with no individuals continues to have no individuals indefinitely," which certainly makes sense. If $P(0) \neq 0$, then (since we can't have a negative population) $P(0)>0$, and the model says “a population that starts out with some individuals will end up with any gigantically huge number you can think of, given enough time." This one doesn't make so much sense. Populations only grow to a certain finite amount, due to scarcity of resources and such. In the derivation of the Malthusian model, we assume a constant net birth rate–that the birth and death rates (per individual) don't depend on the population, which is not a reasonable assumption long-term.

(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 $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+\frac{\mathrm{d}y}{\mathrm{d}x}=y$.

To say $y=f(x)+C$ is a solution to the differential equation means:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{f(x)+C\}&=x(f(x)+C) \\ \text{Since $y=f(x)$ is a solution, we know $\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}=xf(x)$. Also, $\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)+C\}=\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}$. So, $\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)+C\}=xf(x)$.} \\ xf(x)&=x(f(x)+C)\\ 0&=xC \end{aligned}$$

Our equation should hold for all $x$ in our domain, and for the derivative to $y$ with respect to $x$ to make sense, our domain should not be a single point. So, there is some $x$ in our domain such that $x \neq 0$. Therefore, the $C$ must be zero. So, $f(x)+C$ is not a solution to the differential equation for any constant $C$.

When we're finding a general antiderivative, we add “$+C$" at the end. When we're finding a general solution to a differential equation, the “$+C$" gets added when we antidifferentiate–we don't add another one at the end of our work.

(a) Since $|y| \geq 0$ no matter what $y$ is, we see $Cx \ge 0$ for all $x$ in the domain of $f(x)$. Since $C$ is positive, that means the domain of $f(x)$ only includes nonnegative numbers. So, the largest possible domain of $f(x)$ is $[0,\infty)$. (b) None exists.

The graph of $Cx$ is given below for some positive constant $C$, also with the graph of $-Cx$. If $y=f(x)$ were sometimes the top function, and other times the bottom function, then there would be a jump discontinuity where it switched. Then the derivative of $f(x)$ would not exist, violating the second property.

A tiny technical note is that it's possible that $f(x)=Cx$ when $x=0$ and $f(x)=-Cx$ when $x>0$ (or vice-versa). This would not introduce a jump discontinuity, but it also does not satisfy that $f(x)>0$ for some values of $x$.

Remark: in several instances below, solving a differential equation will lead us to conclude something like $|y|=g(x)$. In these cases, we choose either $y=g(x)$, or $y=-g(x)$, but not $y=\pm g(x)$ (which is not a function) or that $y$ is sometimes $g(x)$, and other times $-g(x)$. The reasoning above somewhat explains this choice: if $y$ were sometimes positive and sometimes negative, then $\frac{\mathrm{d}y}{\mathrm{d}x}$ would not exist at the values of $x$ where the sign of $y$ switches, unless that switch occurrs at a root of $g(x)$. Since that's a pretty specific occurrence, we usually feel safe ignoring it to avoid getting bogged down in technical details.

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.