변수분리형 미분방정식

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.