Quick Recall
Quick check — pick the correct option in each question before moving on.
Separating variables: $\dfrac{\text{d}N}{N} = 0.4\,\text{d}t$, so $\ln N = 0.4t + C$ and $N = Ae^{0.4t}$ where $A = e^C$.
Substitute $t = 0$, $N = 50$ into $N = Ae^{0.4t}$: $50 = A \cdot e^0 = A$, so the particular solution is $N = 50e^{0.4t}$.
$k$ is the constant of proportionality: the growth rate per unit population per unit time. Larger $k$ means the population grows faster.
The Wall
A biologist uses the model $N = 50e^{0.4t}$ for a bacteria colony (in millions), with $t$ in hours. She computes $N$ at $t = 24$ hours and gets $N \approx 7.4 \times 10^5$ million (about 740 billion bacteria). She then computes $N$ at $t = 100$ hours and gets $N \approx 1.2 \times 10^{19}$ million — roughly $10^{25}$ bacteria, which would weigh about 10 million metric tons, more than 60,000 blue whales.
The algebra is correct. The particular solution is correct. And yet the answer is physically absurd. Why? Because the equation $N = 50e^{0.4t}$ grows without bound, but the real world has limits (food, space, waste). The math lies when the model is pushed too far. Today's lesson is about reading the solution — not just computing it.
The Need
We need to learn to read a solution, not just write it. Today we ask four questions of every DE solution:
▷ What is $N_0$? (the starting value — set by the initial condition)
▷ What is $k$? (the rate constant — how fast the change happens)
▷ What happens as $t \to \infty$? (the long-term behaviour)
▷ Is the answer sensible? (does the model break down?)
💡 The Bridge (click to reveal)
A solution like $N = 50e^{0.4t}$ contains three pieces of information: the initial population ($N_0 = 50$), the growth rate ($k = 0.4$), and the long-term behaviour ($N \to \infty$). The first two match the real world; the third does not — which tells us the model is only valid for short times.
Today we will formulate, solve, and — crucially — interpret DEs from real contexts: cooling coffee, growing bacteria, accelerating objects, spreading oil. In every case the algebra gives a curve; interpretation turns the curve into a story.
A solution of a DE is more than a formula — it is a prediction about a real quantity. Three families dominate the syllabus: exponential growth, exponential decay, and approach to equilibrium. Drag the sliders below to feel how each parameter changes the story.
For a quantity whose rate of change is proportional to itself, $\dfrac{\text{d}N}{\text{d}t} = kN$ with $k > 0$, the solution is:
$$N = N_0 e^{kt}$$$\bullet$ $N_0$ is the initial value (set by the initial condition $N(0) = N_0$).
$\bullet$ $k$ is the rate constant — larger $k$ means faster growth.
$\bullet$ As $t \to \infty$, $N \to \infty$ — the model predicts unbounded growth, which is unrealistic for real populations.
For a quantity whose rate of decrease is proportional to itself, $\dfrac{\text{d}N}{\text{d}t} = -kN$ with $k > 0$, the solution is:
$$N = N_0 e^{-kt}$$$\bullet$ $N_0$ is the initial amount (e.g. initial mass of radioactive substance).
$\bullet$ $k$ is the decay constant — larger $k$ means faster decay.
$\bullet$ The half-life is $t_{1/2} = \dfrac{\ln 2}{k}$.
$\bullet$ As $t \to \infty$, $N \to 0$ — the quantity vanishes, but never in finite time.
Three common patterns appear in CIE 9709:
$\bullet$ Pure growth $N = N_0 e^{kt}$: $\;\lim_{t \to \infty} N = \infty$ (model breaks down).
$\bullet$ Pure decay $N = N_0 e^{-kt}$: $\;\lim_{t \to \infty} N = 0$.
$\bullet$ Approach to equilibrium $\theta = \theta_a + (\theta_0 - \theta_a)e^{-kt}$: $\;\lim_{t \to \infty} \theta = \theta_a$ (ambient value).
The limit is found by inspecting which terms vanish as $t \to \infty$ — any term containing $e^{-kt}$ with $k > 0$ disappears.
Play in pairs. One of you is the Describer, the other the Guesser. The Describer may look at the secret curve and must describe it using mathematical terms only — the Guesser never sees it. Fewer clues used means more points: 2 clues = 2 points, 3 clues = 1 point. Three rounds, then swap roles. The curves stay on screen, so a precise description is the only way to win.
Scoring: 2 clues and a correct guess = 2 points; 3 clues and correct = 1 point; 4 clues = 0 points (too easy). A wrong guess scores nothing — the Guesser has to reason from the words alone.
On your own. Play with the sliders first and watch what the curve does. Then probe with the two questions. Finally predict the two challenge values — you may set the sliders to check yourself, but commit to an answer first.
Dotted amber line = the starting value $N_{0}$. In decay mode the curve always flattens out towards $0$.
A cup of coffee cools at a rate proportional to the temperature of the coffee above that of the surrounding air. Initially, the coffee is at a temperature of $95^{\circ} \text{C}$ and is cooling at a rate of $0.5^{\circ} \text{C s}^{-1}$. The surrounding air is at $15^{\circ} \text{C}$. Find a differential equation to model this situation.
👁 Reveal Solution
Let $\theta(t)$ be the coffee temperature in $^{\circ}\text{C}$ at time $t$ seconds, and let $\theta_a = 15^{\circ}\text{C}$ be the surrounding air temperature. “Cools at a rate proportional to the temperature above ambient” means the rate of decrease of $\theta$ is $k(\theta - 15)$, so:
$$\boxed{\dfrac{\text{d}\theta}{\text{d}t} = -k(\theta - 15)}$$
Initially $\theta = 95$ and $\dfrac{\text{d}\theta}{\text{d}t} = -0.5$ (negative because the temperature is falling). Substituting:
$-0.5 = -k(95 - 15) = -80k \;\Rightarrow\; k = \dfrac{0.5}{80} = \dfrac{1}{160}$.
So the model is $\dfrac{\text{d}\theta}{\text{d}t} = -\dfrac{1}{160}(\theta - 15)$.
Separate: $\dfrac{\text{d}\theta}{\theta - 15} = -\dfrac{1}{160}\,\text{d}t$.
Integrate: $\ln|\theta - 15| = -\dfrac{t}{160} + C$.
Exponentiate: $\theta - 15 = Ae^{-t/160}$ where $A = e^C$.
Apply $\theta(0) = 95$: $95 - 15 = Ae^{0} = A$, so $A = 80$.
$$\boxed{\theta = 15 + 80e^{-t/160}}$$
▷ Initial temperature: $\theta(0) = 15 + 80 = 95^{\circ}\text{C}$ (matches the data — good sanity check).
▷ Initial cooling rate: $\dot\theta(0) = -\dfrac{80}{160} = -0.5^{\circ}\text{C s}^{-1}$ (matches the data).
▷ Ambient temperature: the “$15$” in the formula is the room temperature — the value the coffee tends towards.
▷ Initial temperature gap: $A = 80 = 95 - 15$ is the initial gap between coffee and room, not the initial coffee temperature. This is the meaning of the constant $A$ in Newton's law of cooling.
▷ Long-term behaviour: as $t \to \infty$, $e^{-t/160} \to 0$, so $\theta \to 15^{\circ}\text{C}$. The coffee approaches room temperature asymptotically — it never quite reaches it in finite time.
▷ Time scale: the time constant is $\tau = \dfrac{1}{k} = 160\,\text{s} \approx 2\text{ min }40\text{ s}$; after this time the gap has shrunk to $e^{-1} \approx 37\%$ of its initial value.
The rate of increase of bacteria is modelled as being proportional to the number of bacteria at any time during their initial growth phase. When the bacteria number $2 \times 10^6$ they are increasing at a rate of $10^5$ per day. Find a differential equation to model this situation.
👁 Reveal Solution
Let $N(t)$ be the number of bacteria at time $t$ days. “Rate of increase proportional to the number” gives:
$$\boxed{\dfrac{\text{d}N}{\text{d}t} = kN}$$
Substitute $\dfrac{\text{d}N}{\text{d}t} = 10^5$ and $N = 2 \times 10^6$:
$10^5 = k \cdot (2 \times 10^6) \;\Rightarrow\; k = \dfrac{10^5}{2 \times 10^6} = \dfrac{1}{20} = 0.05\,\text{day}^{-1}$.
So the model is $\dfrac{\text{d}N}{\text{d}t} = 0.05N$.
Separate: $\dfrac{\text{d}N}{N} = 0.05\,\text{d}t$.
Integrate: $\ln|N| = 0.05t + C$.
Exponentiate: $N = Ae^{0.05t}$ where $A = e^C$.
The data point $N = 2 \times 10^6$ at $t = 0$ gives $A = 2 \times 10^6$:
$$\boxed{N = 2\times 10^6 \cdot e^{0.05t}}$$
▷ Initial population: $N(0) = 2\times 10^6$ (matches data).
▷ Initial growth rate: $\dot N(0) = 0.05 \times 2\times 10^6 = 10^5$ per day (matches data).
▷ Doubling time: $t_{\text{double}} = \dfrac{\ln 2}{k} = \dfrac{0.693}{0.05} \approx 13.9$ days. The colony roughly doubles every two weeks.
▷ Long-term behaviour: as $t \to \infty$, $N \to \infty$. This is the breakdown.
▷ When does the model become unrealistic? At $t = 100$ days, $N \approx 2\times 10^6 \cdot e^{5} \approx 3\times 10^8$. The question says “initial growth phase” — this is the clue. Beyond that phase, food and space limit growth, the rate slows, and a logistic (S-curve) model would be needed instead.
▷ Take-away: an unbounded solution is not a flaw in the algebra; it is a signal that the DE only models a limited time window. Always check whether the long-term behaviour matches the physical situation.
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest. Each one asks you to interpret the solution, not just compute it.
The acceleration (i.e. the rate of change of velocity) of a moving object under a particular force is inversely proportional to the square root of its velocity. When the speed is $4\,\text{m s}^{-1}$ the acceleration is $2\,\text{m s}^{-2}$. Find a differential equation to model this situation.
💡 Hint
Acceleration is the rate of change of velocity: $a = \dfrac{\text{d}v}{\text{d}t}$.
“Inversely proportional to $\sqrt{v}$” means $a = \dfrac{k}{\sqrt{v}}$ for some constant $k$.
Substitute the data point $v = 4$, $a = 2$ to find $k$.
👁 Reveal Solution
Let $v(t)$ be the velocity in $\text{m s}^{-1}$ at time $t$ seconds. Acceleration $a = \dfrac{\text{d}v}{\text{d}t}$, and “inversely proportional to $\sqrt{v}$” means:
$$\dfrac{\text{d}v}{\text{d}t} = \dfrac{k}{\sqrt{v}} = \dfrac{k}{v^{1/2}}$$
Substitute $v = 4$ and $\dfrac{\text{d}v}{\text{d}t} = 2$:
$2 = \dfrac{k}{\sqrt{4}} = \dfrac{k}{2} \;\Rightarrow\; k = 4$.
$$\boxed{\dfrac{\text{d}v}{\text{d}t} = \dfrac{4}{\sqrt{v}}}$$
Solve by separation: $\sqrt{v}\,\text{d}v = 4\,\text{d}t$, so $\tfrac{2}{3}v^{3/2} = 4t + C$, i.e. $v^{3/2} = 6t + C'$.
▷ Initial growth: at $v = 4$, $\dfrac{\text{d}v}{\text{d}t} = 2\,\text{m s}^{-2}$ — the object accelerates.
▷ As $v$ increases, $a$ decreases: at $v = 16$, $a = \dfrac{4}{4} = 1\,\text{m s}^{-2}$ — the force becomes less effective at higher speed.
▷ Long-term behaviour: $v \to \infty$ as $t \to \infty$, but $v$ grows like $t^{2/3}$ (slower than linear) — the object keeps speeding up, but at a diminishing rate.
▷ What $k = 4$ means: the constant $k$ captures how strongly the force couples to $1/\sqrt{v}$; a larger $k$ means the force is more effective at every speed.
The radius of a circular patch of oil is increasing at a rate inversely proportional to its area $A$. Find an expression for $\dfrac{dA}{dt}$.
💡 Hint
The radius $r$ increases at a rate inversely proportional to the area $A$:
$$\dfrac{\text{d}r}{\text{d}t} = \dfrac{k}{A}$$
The area is $A = \pi r^2$, so use the chain rule $\dfrac{\text{d}A}{\text{d}t} = \dfrac{\text{d}A}{\text{d}r} \cdot \dfrac{\text{d}r}{\text{d}t}$.
Convert the final answer to be in terms of $A$ (not $r$) using $r = \sqrt{A/\pi}$.
👁 Reveal Solution
Let $r(t)$ be the radius and $A(t) = \pi r^2$ the area. “Radius increasing at a rate inversely proportional to the area” means:
$$\dfrac{\text{d}r}{\text{d}t} = \dfrac{k}{A}$$
where $k > 0$ is the constant of proportionality.
Since $A = \pi r^2$, we have $\dfrac{\text{d}A}{\text{d}r} = 2\pi r$. By the chain rule:
$\dfrac{\text{d}A}{\text{d}t} = \dfrac{\text{d}A}{\text{d}r} \cdot \dfrac{\text{d}r}{\text{d}t} = 2\pi r \cdot \dfrac{k}{A} = \dfrac{2\pi r k}{\pi r^2} = \dfrac{2k}{r}$.
Since $A = \pi r^2$, we have $r = \sqrt{\dfrac{A}{\pi}}$, so $\dfrac{1}{r} = \sqrt{\dfrac{\pi}{A}}$. Substituting:
$$\boxed{\dfrac{\text{d}A}{\text{d}t} = 2k \sqrt{\dfrac{\pi}{A}} = \dfrac{2k\sqrt{\pi}}{\sqrt{A}} = 2k\sqrt{\pi}\, A^{-1/2}}$$
▷ Sign: $\dfrac{\text{d}A}{\text{d}t} > 0$ — the patch grows, as expected.
▷ Power law: $\dfrac{\text{d}A}{\text{d}t} \propto \dfrac{1}{\sqrt{A}}$ — the larger the patch, the slower its area grows.
▷ Long-term behaviour: solving $\dfrac{\text{d}A}{\text{d}t} = \dfrac{C}{\sqrt{A}}$ gives $\sqrt{A}\,\text{d}A = C\,\text{d}t$, so $\tfrac{2}{3}A^{3/2} = Ct + C'$, i.e. $A \propto t^{2/3}$ — the area grows without bound, but more and more slowly. The patch keeps spreading forever (an idealised model with no surface-tension limit).
▷ What $k$ means: $k$ captures how quickly the oil spreads on the surface; a larger $k$ means faster spreading at any given area.
The variables $x$ and $\theta$ satisfy the differential equation $2\cos^2 \theta \dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \sqrt{(2x + 1)}$, and $x = 0$ when $\theta = \frac{1}{4}\pi$. Solve the differential equation and obtain an expression for $x$ in terms of $\theta$.
💡 Hint
Separate the variables: divide both sides by $\sqrt{2x+1}$ and multiply by $\text{d}\theta$:
$$\dfrac{\text{d}x}{\sqrt{2x+1}} = \dfrac{\text{d}\theta}{2\cos^2\theta} = \tfrac{1}{2}\sec^2\theta\,\text{d}\theta$$
Integrate both sides. Recall $\displaystyle\int \sec^2\theta\,\text{d}\theta = \tan\theta$.
Apply $x = 0$, $\theta = \pi/4$ to find the constant.
👁 Reveal Solution
Divide by $\sqrt{2x+1}$ and rearrange:
$\dfrac{\text{d}x}{\sqrt{2x+1}} = \dfrac{\text{d}\theta}{2\cos^2\theta} = \dfrac{1}{2}\sec^2\theta\,\text{d}\theta$.
LHS: let $u = 2x + 1$, $\text{d}u = 2\,\text{d}x$:
$\displaystyle\int \dfrac{\text{d}x}{\sqrt{2x+1}} = \int \dfrac{\text{d}u/2}{\sqrt{u}} = \sqrt{u} = \sqrt{2x+1}$.
RHS:
$\displaystyle\int \tfrac{1}{2}\sec^2\theta\,\text{d}\theta = \tfrac{1}{2}\tan\theta$.
Therefore: $\sqrt{2x+1} = \tfrac{1}{2}\tan\theta + C$.
Substitute: $\sqrt{2(0)+1} = \tfrac{1}{2}\tan(\pi/4) + C \;\Rightarrow\; 1 = \tfrac{1}{2}(1) + C \;\Rightarrow\; C = \tfrac{1}{2}$.
So $\sqrt{2x+1} = \tfrac{1}{2}\tan\theta + \tfrac{1}{2} = \tfrac{1}{2}(\tan\theta + 1)$.
Square both sides: $2x + 1 = \dfrac{1}{4}(\tan\theta + 1)^2$.
$2x = \dfrac{1}{4}(\tan\theta + 1)^2 - 1$.
$$\boxed{x = \dfrac{1}{8}(\tan\theta + 1)^2 - \dfrac{1}{2}}$$
Verification: at $\theta = \pi/4$, $x = \tfrac{1}{8}(1+1)^2 - \tfrac{1}{2} = \tfrac{4}{8} - \tfrac{1}{2} = 0$. ✓
▷ Initial value: $x = 0$ when $\theta = \pi/4$ — matches the data.
▷ $x$ as a function of $\theta$: $x$ grows as $\tan\theta$ grows, i.e. as $\theta \to \tfrac{\pi}{2}^-$.
▷ Long-term (boundary) behaviour: as $\theta \to \tfrac{\pi}{2}^-$, $\tan\theta \to \infty$, so $x \to \infty$ — the variable $x$ blows up at the vertical asymptote of $\tan\theta$. The model is meaningful only for $\theta \in [\pi/4, \pi/2)$.
▷ What the constant $\tfrac{1}{2}$ means: it is the value of $\sqrt{2x+1} - \tfrac{1}{2}\tan\theta$ at the initial point — the “offset” that aligns the curve with the initial condition. Without it (i.e. if $C = 0$), the curve would pass through $\theta = \arctan(-1) = -\pi/4$ at $x = 0$, not $\theta = \pi/4$.
The variables $x$ and $y$ are related by the differential equation $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{5}xy^{\frac{1}{2}}\sin\left(\dfrac{1}{3}x\right)$.
(i) Find the general solution, giving $y$ in terms of $x$.
(ii) Given that $y = 100$ when $x = 0$, find the value of $y$ when $x = 25$.
💡 Hint 1
Separate: divide both sides by $y^{1/2}$ and multiply by $\text{d}x$:
$$y^{-1/2}\,\text{d}y = \tfrac{1}{5}x\sin\!\left(\tfrac{x}{3}\right)\text{d}x$$
LHS integrates to $2y^{1/2}$. RHS needs integration by parts: let $u = x$, $\text{d}v = \sin(x/3)\,\text{d}x$.
💡 Hint 2
For the RHS, $\displaystyle\int x\sin\!\left(\tfrac{x}{3}\right)\text{d}x$: let $u = x$, $\text{d}v = \sin(x/3)\,\text{d}x$, so $\text{d}u = \text{d}x$ and $v = -3\cos(x/3)$.
$$\int u\,\text{d}v = uv - \int v\,\text{d}u = -3x\cos(x/3) + 3\!\int\!\cos(x/3)\,\text{d}x = -3x\cos(x/3) + 9\sin(x/3)$$
For part (ii): apply $y = 100$ at $x = 0$ to find the constant, then substitute $x = 25$. Recall $25/3 \approx 8.33$ rad; reduce modulo $2\pi$ before evaluating sin/cos.
👁 Reveal Solution
Separate: $y^{-1/2}\,\text{d}y = \tfrac{1}{5}x\sin(x/3)\,\text{d}x$.
Integrate LHS: $\displaystyle\int y^{-1/2}\,\text{d}y = 2y^{1/2}$.
Integrate RHS by parts: let $u = x$, $\text{d}v = \sin(x/3)\,\text{d}x$. Then $\text{d}u = \text{d}x$, $v = -3\cos(x/3)$:
$\displaystyle\int x\sin(x/3)\,\text{d}x = -3x\cos(x/3) + 3\int\cos(x/3)\,\text{d}x = -3x\cos(x/3) + 9\sin(x/3)$.
Therefore: $2y^{1/2} = \dfrac{1}{5}\left[-3x\cos(x/3) + 9\sin(x/3)\right] + C$.
Divide by 2 (and re-label the constant as $C'$):
$$y^{1/2} = \dfrac{-3x\cos(x/3) + 9\sin(x/3)}{10} + C'$$
$$\boxed{y = \left[\dfrac{-3x\cos(x/3) + 9\sin(x/3)}{10} + C'\right]^2}$$
At $x = 0$: $y^{1/2} = 10$, $\cos 0 = 1$, $\sin 0 = 0$. So:
$10 = \dfrac{0 + 0}{10} + C' \;\Rightarrow\; C' = 10$.
Therefore $y^{1/2} = \dfrac{-3x\cos(x/3) + 9\sin(x/3)}{10} + 10$.
At $x = 25$: $x/3 = 25/3 \approx 8.3333$ rad. Reduce: $25/3 - 2\pi \approx 2.0501$ rad, so:
$\sin(25/3) \approx 0.8873$, $\;\cos(25/3) \approx -0.4612$.
Compute the bracket:
$-3(25)\cos(25/3) + 9\sin(25/3) \approx -75(-0.4612) + 9(0.8873) \approx 34.59 + 7.99 \approx 42.58$.
So $y^{1/2} \approx \dfrac{42.58}{10} + 10 \approx 4.258 + 10 \approx 14.258$.
$$\boxed{y(25) \approx 14.258^2 \approx 203 \text{ (3 s.f.)}}$$
▷ Initial value: $y(0) = (0 + 10)^2 = 100$ — matches the data.
▷ Growth: $y$ increases from 100 to about 203 as $x$ goes from 0 to 25 — the rate of growth is driven by the $x\sin(x/3)$ factor.
▷ Oscillation: because the RHS contains $\sin(x/3)$, the rate $\dfrac{\text{d}y}{\text{d}x}$ changes sign periodically (every $3\pi \approx 9.42$ in $x$). The variable $y$ does not simply grow — it speeds up, slows down, and may even decrease when $\sin(x/3) < 0$. Over the interval $[0, 25]$, $\sin(x/3)$ is positive for $x \in (0, 3\pi)$ and negative for $x \in (3\pi, 6\pi)$; since $25 \approx 7.96 < 3\pi \approx 9.42$, we are still in the “growth” phase at $x = 25$.
▷ Long-term behaviour: because $\sin(x/3)$ oscillates, $y^{1/2}$ oscillates about the value $10$ with growing amplitude ($\propto x$). The model is not monotonic and does not have a simple limit as $x \to \infty$ — it is a driven oscillation rather than a growth or decay.
▷ What $C' = 10$ means: the constant sets the baseline around which $y^{1/2}$ oscillates. With $C' = 0$, the solution would have $y(0) = 0$; the initial condition lifts this baseline to $10$, giving $y(0) = 100$.
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.