A sentence about a rate AB & BC

Three words in a problem statement mean "take a derivative": rate of change, grows, changes. If a sentence describes how fast something is moving, the left side of your equation is $\frac{dy}{dt}$, not $y$.

That sounds too obvious to state, and it is the single most common error in this topic. "A population grows at a rate proportional to its size" becomes

$$\frac{dP}{dt} = kP,$$

and the wrong version, $P = kt$, is a sentence about the population rather than about its rate. The two are not variations of one idea. $P = kt$ says the population equals a constant times the time elapsed, which is a claim about how big the colony is; $\frac{dP}{dt} = kP$ says nothing at all about how big it is and everything about how fast it is changing.

Identify the variables before writing anything. Name the quantity, name what it depends on, and only then decide which of them the rate is taken with respect to. A tank problem with a depth $h$ and a time $t$ has a rate $\frac{dh}{dt}$; a cost problem with a production level $x$ has a rate $\frac{dC}{dx}$.

One more thing worth saying at the start: a differential equation is not solved in this topic and does not need to be. The model is the deliverable.

"Proportional to" means "equal to a constant times", and the only question is times what. Whatever phrase follows those words is what $k$ multiplies, in full.

1. Proportional to $y$ gives $\frac{dy}{dt} = ky$.
2. Proportional to the difference between $T$ and $70$ gives $\frac{dT}{dt} = k(T - 70)$. The parentheses are not decoration. Written as $kT - 70$ it says something else entirely, and it is no longer proportional to anything.
3. Inversely proportional to $y$ gives $\frac{dy}{dt} = \frac{k}{y}$.
4. Proportional to the square root of $h$ gives $\frac{dh}{dt} = k\sqrt{h}$.
5. Proportional to the product of $y$ and $M - y$ gives $\frac{dy}{dt} = ky(M - y)$. Both factors sit inside the product.

A cheap test that catches the parenthesis error every time: a proportionality has to survive scaling. If $\frac{dT}{dt} = k(T - 70)$ and the difference $T - 70$ doubles, the rate doubles. If instead $\frac{dT}{dt} = kT - 70$, doubling that difference does nothing so tidy, and the rate can even be negative when the difference is positive. Only the first is a statement of proportionality.

Watch for the phrase that names the object indirectly. "The rate is proportional to the amount remaining" in a $500$-unit process means $\frac{dA}{dt} = k(500 - A)$, since the amount remaining is what is left, not what is present.

In $\frac{dy}{dt} = ky$, the constant $k$ decides direction. If $k > 0$ the rate has the same sign as $y$, so a positive quantity grows. If $k < 0$ it decays. A model of a decaying substance with a positive $k$ is not a small slip; it describes the opposite phenomenon.

Two conventions appear and both are correct as long as you are consistent. Some write $\frac{dA}{dt} = -kA$ with $k > 0$, putting the sign in the equation; others write $\frac{dA}{dt} = kA$ and let $k$ come out negative. Read which one a problem has set up before reporting a value of $k$.

Three things $k$ is not:

1. Not the starting amount. That is a separate constant, and it does not appear in the differential equation at all. It arrives with the initial condition.
2. Not an amount per unit time. $k = 0.05$ per year in $\frac{dP}{dt} = 0.05P$ means the population grows at $5\%$ of its current size each year, so the actual rate changes as $P$ does. It does not mean five individuals per year, and it does not mean five of anything.
3. Not the doubling time or the half-life. Those are consequences of $k$ and are computed from it with a logarithm, never by dividing into it.

The units settle it. In $\frac{dP}{dt} = kP$ the left side is people per year and $P$ is people, so $k$ carries units of $1 / \text{year}$. A constant whose units are a pure reciprocal time cannot be a count of anything.

Given an equation and no story, you can still say a great deal about the behaviour, and AP asks for exactly this.

Take $\frac{dy}{dt} = 0.3(200 - y)$. Read the right side as a sign machine. When $y < 200$ the factor $200 - y$ is positive, so $y$ increases. When $y > 200$ it is negative, so $y$ decreases. When $y = 200$ the rate is zero and $y$ holds still forever: that constant function is an equilibrium solution. Every solution is therefore pulled toward $200$ from whichever side it starts on.

Notice what the rate does near the equilibrium. As $y$ approaches $200$ the factor $200 - y$ approaches zero, so the approach slows down. Fastest change happens far from equilibrium, not close to it, and that is why cooling curves flatten.

The same reading applies to $\frac{dP}{dt} = kP$ with $k > 0$: the rate is proportional to what is already there, so growth accelerates, and there is no equilibrium other than $P = 0$. Two models, two shapes, and both were read off the right side without solving anything.

Last check before moving on. Substitute one plausible value and ask whether the sign that comes out matches the story. A cooling coffee at $T = 130$ in a $70$ degree room must give a negative $\frac{dT}{dt}$. If your model gives a positive one, the model is wrong, and you found it in ten seconds.