Model unrestricted growth and decay with $\frac{dN}{dt} = k N$ and solve the resulting separable differential equation

What this dot point is asking

NESA wants you to recognise scenarios that follow the unrestricted exponential model $\frac{dN}{dt} = k N$, write down and (when asked) verify the general solution $N = N_0 e^{kt}$, use given data to find the constants $N_0$ and $k$, and solve doubling, halving and prediction problems. The Extension 1 course then extends this to the modified law $\frac{dN}{dt} = k(N - P)$, where the rate depends on the gap between $N$ and a fixed value $P$, which models population control, cooling and any approach to an equilibrium.

The phrase to listen for is "rate of change proportional to". If the rate is proportional to the amount $N$, you have pure exponential growth or decay. If it is proportional to the difference from a fixed level $(N - P)$, you have the modified law. Nothing else in the syllabus has the defining property that its derivative is a constant multiple of itself (or of itself-minus-a-constant), which is exactly why an exponential is the answer in both cases.

The answer

The unrestricted model

If a quantity $N(t)$ grows or decays at a rate proportional to itself,

$k$ is the proportionality (or rate) constant. Its sign decides growth versus decay and its size sets the speed.

General solution and how to establish it

Treat it as a separable differential equation. Separate variables and integrate:

where $N_0 = N(0)$ is the initial value (it absorbs $e^{C}$). If a question says "show that" or "verify that $N = N_0 e^{kt}$ satisfies $\frac{dN}{dt} = kN$", do not separate variables; instead differentiate the given solution: $\frac{dN}{dt} = k N_0 e^{kt} = kN$, which matches, and note $N(0) = N_0$. Reading the verb correctly is worth a mark.

Finding $k$ from data

Given the initial value $N_0$ and one later reading $N(t_1) = N_1$:

For decay ($N_1 < N_0$) the logarithm is negative, so $k < 0$, as it must be. Keep $k$ exact (for example $k = \frac{\ln 0.8}{10}$) until the final substitution to avoid rounding error.

Doubling time and half-life

Doubling time for growth: solve $2 N_0 = N_0 e^{k T_d}$, cancel $N_0$, and take logs to get $T_d = \frac{\ln 2}{k}$.

Half-life for decay: solve $\frac{1}{2} N_0 = N_0 e^{k T_h}$ to get $T_h = \frac{\ln \frac{1}{2}}{k} = -\frac{\ln 2}{k}$, which is positive because $k < 0$.

Both cancel $N_0$, which is the algebraic reason they are independent of the starting amount.

Continuous compound interest

A bank account earning interest at a continuous rate $r$ satisfies $\frac{dA}{dt} = r A$, so $A(t) = A_0 e^{r t}$. This is the same model with $N \to A$ and $k \to r$.

The modified law dN/dt = k(N - P), stage by stage

The Extension 1 step up from pure exponentials is the law in which the rate is proportional to the difference between $N$ and a fixed value $P$:

It models a population held towards a sustainable level $P$ by culling or immigration, a body cooling towards room temperature $P$, or any quantity drawn towards an equilibrium. Its solution (by separating variables, or by substituting $M = N - P$ so $\frac{dM}{dt} = kM$) is

The four panels show what this forces the graph to do, and why $P$ acts as a horizontal asymptote whichever side you start.

Stage 1, the pure-exponential baseline. With $P = 0$ the law is just $\frac{dN}{dt} = kN$. For $k > 0$ the solution $N = N_0 e^{kt}$ climbs forever, getting steeper as it rises, and there is no level it settles at. Comparing this with the modified law shows exactly what the $-P$ term changes.

Stage 2, introduce the equilibrium value $P$. In the modified law, $\frac{dN}{dt} = 0$ exactly when $N = P$, because $k(N - P) = 0$ there. So the horizontal line $N = P$ is an equilibrium: a quantity sitting at $P$ has zero rate of change and stays there forever. This single line organises the whole picture.

Stage 3, starting above $P$. If $N_0 > P$ then $N_0 - P > 0$, and (with $k < 0$, the usual case for a controlled population or a cooling body) the term $(N_0 - P)e^{kt}$ decays from a positive value to $0$. So $N$ falls from $N_0$ and levels off at $P$ from above, never crossing it. The gap $N - P$ is what decays exponentially, exactly like a pure decay but measured from the line $N = P$ instead of from zero.

Stage 4, starting below $P$. If $N_0 < P$ then $N_0 - P < 0$, so $(N_0 - P)e^{kt}$ is a negative quantity decaying to $0$: $N$ rises from $N_0$ up to $P$. Either way the curve approaches the same horizontal asymptote $N = P$, which is the hallmark of the modified law. Whether it climbs or falls is decided purely by whether $N_0$ starts below or above $P$.

Newton's law of cooling is the same law with $P = T_a$, the ambient temperature: $\frac{dT}{dt} = k(T - T_a)$ solves to $T = T_a + (T_0 - T_a)e^{kt}$, and the temperature levels off at $T_a$ rather than at zero. See the separable differential equations dot point for the full derivation.

Out of scope: logistic growth

You may see the logistic model $\frac{dN}{dt} = kN(M - N)$, which caps growth at a carrying capacity $M$ using a product rather than a difference. That belongs to Extension 2, not Extension 1; the Extension 1 "capped" model is the linear modified law above.

How exam questions ask about exponential models

The wording pinpoints which step is being tested:

• "The rate of change is proportional to ... Set up a differential equation." Translate the words: proportional to the amount gives $\frac{dN}{dt} = kN$; proportional to the difference from a fixed level gives $\frac{dN}{dt} = k(N - P)$. Recognising the phrasing is the whole task.
• "Show that / verify that $N = N_0 e^{kt}$ satisfies $\frac{dN}{dt} = kN$." Differentiate the given solution and substitute; do not separate variables. State the initial condition too.
• "After $t_1$, the amount is ... Find $k$ / the growth rate / the decay constant." Substitute the second data point and solve with a logarithm. Decay and cooling give $k < 0$.
• "Find the amount after ... / how much remains." Substitute $t$ into the fully fitted model.
• "How long until it doubles / halves / reaches a given value?" Set the model equal to the target and solve for $t$ with a logarithm; or use $T_d = \frac{\ln 2}{k}$, $T_h = \frac{\ln 2}{|k|}$.
• "A body cools / a population is controlled towards a level." The modified law: identify $P$ (ambient or sustainable value), quote $N = P + (N_0 - P)e^{kt}$, and remember the limit is $P$, not $0$.
• "Sketch the solution." Show the initial value, the correct growth or decay shape, and, for the modified law, the horizontal asymptote $N = P$.