변수분리형 미분방정식

(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.