Growth with a ceiling BC only

The logistic equation appears in two dresses, and reading the carrying capacity off each is the first skill.

1. $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{M}\right)$. Here $M$ is written into the equation directly.
2. $\dfrac{dP}{dt} = kP(M - P)$. Here $M$ is again the number subtracted from $P$, but the constant out front is not the same $k$ as in the first form. The two are related by a factor of $M$.

The reliable move is to ignore the packaging and find the equilibria: set the right side to zero and solve. Both forms give $P = 0$ and $P = M$, so for $\frac{dP}{dt} = 0.002P(500 - P)$ the carrying capacity is $500$, not $0.002$ and not $1$.

That also settles what the leading constant means. In the second form its units involve one over a population and one over a time, so it is not a growth rate you can quote as a percentage. Reading $0.002$ as "point two percent per year" is a misreading of a different equation.

The two equilibria behave in opposite ways. $P = 0$ is repelling, so any positive starting population moves away from it. $P = M$ is attracting, so every positive solution approaches it. Nothing crosses either, which is Topic 7.4 applied here.

One: the long-run value. For any $P(0) > 0$, $\displaystyle\lim_{t \to \infty} P(t) = M$. Starting below the ceiling the population rises to it; starting above, it falls to it. Either way the answer is $M$ and no solving is required to say so.

Two: where growth is fastest. The rate $kP\left(1 - \frac{P}{M}\right)$ is a downward parabola in $P$ with zeros at $0$ and $M$, so it peaks halfway between them, at

$$P = \frac{M}{2}.$$

This is the fact students invert. Growth is slowest near the carrying capacity, not fastest, because the second factor is nearly zero there. It is also slow near $P = 0$, because the first factor is. The maximum rate is $k \cdot \frac{M}{2} \cdot \frac{1}{2} = \frac{kM}{4}$ in the first form.

Half capacity is also where the solution curve has its inflection point: the population is still increasing, but it stops increasing faster and starts increasing more slowly. Below $\frac{M}{2}$ the curve is concave up; above it, concave down. That is the whole S shape in one sentence.

A third fact worth stating because it is quietly examined: the population never reaches $M$ in finite time. It approaches it, and a claim that a colony hits its carrying capacity in the eleventh year describes a different model.

Separation puts $\frac{dP}{P(M - P)}$ on one side, and that integrand is exactly the Topic 6.12 form: one constant over each distinct linear factor. Decompose

$$\frac{1}{P(M - P)} = \frac{1}{M}\left(\frac{1}{P} + \frac{1}{M - P}\right),$$

integrate both pieces to logarithms, combine them, exponentiate, and solve for $P$. The result is

$$P = \frac{M}{1 + Ae^{-kt}}, \qquad A = \frac{M - P_{0}}{P_{0}}.$$

Three checks on that formula, all quick. At $t = 0$ it gives $\frac{M}{1 + A} = P_{0}$. As $t \to \infty$ the exponential vanishes and it gives $M$. And $A$ is positive when the population starts below the ceiling and negative when it starts above, which is what puts the curve on the right side of $M$.

The constant is $\frac{M - P_{0}}{P_{0}}$, not $\frac{M}{P_{0}}$. That off-by-one is the most common slip in assembling the solution, and substituting $t = 0$ catches it immediately.

A logistic solution starting below the ceiling has a shape worth being able to describe in words:

1. Early. $P$ is small, the second factor is close to $1$, and the equation behaves like $\frac{dP}{dt} = kP$. Growth is nearly exponential and accelerating.
2. At $\frac{M}{2}$. The rate peaks. The curve inflects.
3. Late. The second factor shrinks toward zero, the rate follows it, and the curve flattens against $P = M$ without touching it.

Starting above the ceiling the picture is simpler: the second factor is negative, the population falls, and it flattens onto $M$ from above. There is no inflection point on that side, since the rate is largest at the start and shrinks all the way.

On a slope field the two equilibria show as horizontal lines at $P = 0$ and $P = M$, with the steepest segments along $P = \frac{M}{2}$ and shallower ones as you move toward either line. Matching a field to a logistic equation means checking those three heights, not the general S impression.