Logistic Models with Differential Equations BC only
The logistic equation appears as $\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right)$ or as $\frac{dP}{dt} = kP(M - P)$, and the reliable way to read the carrying capacity off either is to set the right side to zero: the equilibria are $P = 0$, which repels, and $P = M$, which attracts. Two facts then follow with no solving. Every solution with $P(0) > 0$ approaches $M$, from below or from above; and the rate, being a downward parabola in $P$ with zeros at $0$ and $M$, is largest at $P = \frac{M}{2}$, where the solution curve inflects.
Growth is SLOWEST near the carrying capacity, not fastest, and the population never reaches $M$ in finite time. Solving needs the partial fractions of Topic 6.12: $\frac{1}{P(M - P)} = \frac{1}{M}\left(\frac{1}{P} + \frac{1}{M - P}\right)$, which integrates to logarithms and rearranges to $P = \frac{M}{1 + Ae^{-kt}}$ with $A = \frac{M - P_{0}}{P_{0}}$. That constant is the ratio of the room left to the amount present, not $\frac{M}{P_{0}}$, and substituting $t = 0$ checks it in one line.
The work
3 ways in · any order
Lesson
Logistic Models with Differential Equations
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Reads the carrying capacity off both standard forms by finding the equilibria, establishes that every positive solution approaches it and that growth peaks at half of it where the curve inflects, then solves the equation with the partial fractions of Topic 6.12 and assembles the constant from the room left over the amount present.
Diagnostic
10-item topic check
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Ten items on logistic models: identifying the carrying capacity from either form, locating the fastest growth at half capacity rather than at the ceiling, reading long-run behaviour from above and below, and assembling the solved form and its constant correctly.
Targeted Practice
Drill a single misconception
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Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.