One constant, and everything it decides AB & BC

$\frac{dy}{dt} = ky$ separates to $\frac{dy}{y} = k\,dt$, integrates to $\ln|y| = kt + C$, and solves to

$$y = y_{0}e^{kt},$$

where $y_{0}$ is the value at $t = 0$. Every proportional-rate problem lands here, and it is worth recognising the form rather than rederiving it each time.

The constant carries the whole behaviour. If $k > 0$ the quantity grows and the growth accelerates, since a bigger $y$ makes a bigger rate. If $k < 0$ it decays toward zero without reaching it. If $k = 0$ nothing changes.

Two things $k$ is not, restated because they cost marks every year. It is not a fixed amount per unit time: $k = 0.05$ per year means the quantity grows by five percent of its current size each year, so the amount added rises as the quantity does. And it is not the doubling time or the half-life, which come out of it through a logarithm.

The units settle both claims. In $\frac{dy}{dt} = ky$ the left side has units of $y$ per unit time and the right has units of $k$ times $y$, so $k$ has units of one over time. A pure reciprocal time cannot be a count and cannot be a duration.

Both come from the same one-line calculation. Set the final value to a multiple of the starting value and solve for $t$.

Doubling. $2y_{0} = y_{0}e^{kt}$ gives $2 = e^{kt}$, so $t = \frac{\ln 2}{k}$.

Halving. $\tfrac{1}{2}y_{0} = y_{0}e^{kt}$ gives $t = \frac{\ln 2}{-k} = -\frac{\ln 2}{k}$, positive because $k$ is negative in a decay model.

The starting value cancels in both. That is why a half-life does not depend on how much you started with, which is the fact these problems are usually testing.

Run it the other way when the half-life is given. Carbon-14 has a half-life of $5730$ years, so $k = -\frac{\ln 2}{5730} \approx -1.21 \times 10^{-4}$ per year. Dividing $1$ by $5730$ instead is the error to avoid: that number is a rate of a different kind and it is about forty-four percent too large in size.

After $n$ half-lives the fraction remaining is $\left(\frac{1}{2}\right)^{n}$, so three half-lives leave $12.5\%$ rather than the $100\% - 3(50\%)$ that a linear instinct suggests.

Newton's law of cooling says the rate is proportional to the difference from the surroundings: $\frac{dT}{dt} = k(T - T_{a})$ with $k < 0$. Separating means dividing by $T - T_{a}$, which is legal only when $T \neq T_{a}$. Then

$$\ln|T - T_{a}| = kt + C, \qquad T = T_{a} + Ce^{kt}.$$

Read what that says. The difference from the ambient value decays exponentially; the temperature itself does not. Applying the decay to $T$ rather than to $T - T_{a}$ predicts coffee colder than the room it is sitting in, which is the most common error here.

The constant is set by the start: at $t = 0$, $C = T_{0} - T_{a}$. Coffee at $190$ degrees in a $70$ degree room gives $T = 70 + 120e^{kt}$, and the same formula handles a cold drink warming up, where $C$ comes out negative.

And the equilibrium. Dividing by $T - T_{a}$ discarded the constant solution $T = T_{a}$: a cup already at room temperature stays there. Here the family recovers it at $C = 0$, so nothing is permanently lost, but the check is the same three seconds it always is, and other models are not as forgiving.

The same shape appears whenever a quantity approaches a limiting value: $\frac{dy}{dt} = k(y - L)$ has solution $y = L + Ce^{kt}$, and $y = L$ is its equilibrium.

Four habits that turn a correct model into full marks.

1. Say which variable is which. "$P$ is the population in thousands and $t$ is years since 2020" costs one line and prevents every unit error downstream.
2. Keep $k$ exact until the end. $k = \frac{\ln 1.6}{3}$ rounded to $0.16$ and then multiplied by $10$ drifts noticeably. Store the exact expression and round once.
3. Answer the question asked. "How long until it doubles" wants a time; "how much after ten years" wants an amount. The same model answers both and the arithmetic differs.
4. Sanity-check the direction. A decay answer larger than the starting amount, or a cooling answer below room temperature, is a sign error you can find in five seconds.

One structural check worth having: given two data points and no $k$, use them to find $k$ first. From $y(0) = 500$ and $y(3) = 800$, $800 = 500e^{3k}$ gives $k = \frac{\ln 1.6}{3}$, and that value then answers everything else about the model.