Lesson 11: Differential Equations (Intro)

Differential equations describe how things change — the language of physics, biology, and economics. Where Lesson 6 integrated known functions, differential equations ask: which function has this behaviour? This introductory lesson covers forming differential equations from descriptions, solving by separating variables, particular solutions from initial conditions, and modelling with exponential and logistic growth. The theme is translation: words → equation → solution → interpretation — the full modelling cycle in miniature.

What is a differential equation?

A differential equation (DE) relates a function to its derivatives: dy/dx = 3x² − 2, d²y/dx² + y = 0, dP/dt = kP. The order is the highest derivative; the general solution contains as many arbitrary constants as the order (each integration introduces one). Solving dy/dx = 3x² − 2 gives y = x³ − 2x + c — a family of curves; an initial condition like y(0) = 5 selects one: c = 5. First-order DE, one constant — the pattern holds generally.

Forming DEs from descriptions is the key modelling skill. 'The rate of increase of a population is proportional to the population': dP/dt = kP. 'A cup of tea cools at a rate proportional to the temperature difference with the room': dT/dt = −k(T − 20). 'A rumour spreads at a rate proportional to the product of those who know and those who do not': dN/dt = kN(P − N) — the logistic equation. Translate carefully: 'rate of change of y' → dy/dt; 'proportional to' → = k×; 'difference' → subtraction.

Terminology: general solution (with constants), particular solution (constants fixed by conditions), initial/boundary conditions. A first-order DE needs one condition; second-order needs two. Examiners test formation ('form a differential equation to model...') as often as solution — defining variables with units is part of the answer.

Separation of variables

If dy/dx = f(x) g(y), separate: (1/g(y)) dy = f(x) dx, then integrate both sides: ∫ dy/g(y) = ∫ f(x) dx. Worked example: dy/dx = xy, y(0) = 3. Separate: dy/y = x dx → ln|y| = x²/2 + c → y = A e^{x²/2} (A = ±e^c, absorbing the sign). Initial condition: 3 = A e⁰ → A = 3, so y = 3e^{x²/2}. The modulus handling: ln|y| = ... → |y| = e^{...} → y = ±e^{...}; the initial condition's sign selects the branch — write y = A e^{...} and let A carry the sign.

Worked example: dy/dx = (x + 1)/(y²), y(0) = 2. Separate: y² dy = (x + 1) dx → y³/3 = x²/2 + x + c → y³ = (3/2)x² + 3x + C. y(0) = 2 → 8 = C → y = ∛(1.5x² + 3x + 8). Check by differentiating implicitly — good practice. Partial fractions appear: dy/dx = y(1 − y) → dy/(y(1 − y)) = dx → (1/y + 1/(1 − y)) dy = dx → ln|y| − ln|1 − y| = x + c → y/(1 − y) = A eˣ → y = A eˣ/(1 + A eˣ) — the logistic curve, derived!

Worked example (separation with a square). Solve dy/dx = 2x/(3y²) with y(0) = 1. Separate: 3y² dy = 2x dx. Integrate: y³ = x² + c. Condition: 1 = 0 + c → c = 1. Solution: y = ∛(x² + 1). Verify by implicit differentiation: 3y² y' = 2x → y' = 2x/(3y²) ✓. The verify step costs twenty seconds and guarantees the answer — build it into every DE solution.

Common errors: forgetting +c (loses the family — and the mark); dividing by g(y) without noting g(y) = 0 gives equilibrium solutions (for dy/dx = y(1 − y), y = 0 and y = 1 are valid constant solutions the separation misses — mention them); and mishandling the modulus-to-±A step. Examiners award marks for stating equilibrium solutions explicitly.

Exponential and logistic models

The DE dN/dt = kN gives N = N₀ e^{kt} — exponential growth/decay, the solution of Lesson 8's modelling. But unlimited growth is unrealistic: the logistic DE dN/dt = kN(1 − N/K) adds a carrying capacity K. Solution (from above): N = K/(1 + A e^{−kt}) — S-shaped: exponential at first, flattening toward K. A: from N(0) = N₀, A = (K − N₀)/N₀.

Worked example: bacteria in a dish, K = 10,000, k = 0.8 h⁻¹, N₀ = 500. Then A = (10000 − 500)/500 = 19, N(t) = 10000/(1 + 19e^{−0.8t}). At t = 5: 19e^{−4} ≈ 0.348 → N ≈ 10000/1.348 ≈ 7418. When does N reach 9000? 9000 = 10000/(1 + 19e^{−0.8t}) → 1 + 19e^{−0.8t} = 1.111 → e^{−0.8t} = 0.00585 → t ≈ 6.43 h. The inflection point (fastest growth) is at N = K/2 — prove by setting d²N/dt² = 0.

In context: UK COVID modelling used logistic-like curves for cumulative cases; fisheries set quotas from logistic fish-stock models; a viral video's views follow dV/dt = kV(1 − V/M) with M the potential audience. The logistic DE is the simplest model with saturation — recognise 'growth proportional to current size times remaining capacity' as its verbal signature.

Setting up and solving — full problems

Full exam problem: 'A tank contains 1000 litres of brine. Fresh water flows in at 10 L/min, mixed brine flows out at 10 L/min. Initially 50 kg of salt. Form and solve a DE for the salt mass S(t).' Rate in: 0. Rate out: (S/1000 kg/L) × 10 L/min = S/100 kg/min. So dS/dt = −S/100 → S = 50e^{−t/100} kg. When is 10 kg left? 10 = 50e^{−t/100} → t = 100 ln 5 ≈ 161 min. Every word translated: 'flows in at' → rate, 'mixed' → uniform concentration assumption (state it!).

Another: 'A parachutist (80 kg) falls with air resistance proportional to velocity: dv/dt = g − kv.' With v(0) = 0: separate: dv/(g − kv) = dt → −(1/k) ln|g − kv| = t + c → v = (g/k)(1 − e^{−kt}). Terminal velocity g/k: as t → ∞, v → g/k. If terminal velocity is 55 m/s, k = 9.8/55 ≈ 0.178 s⁻¹. Time to reach 50 m/s: 50 = 55(1 − e^{−0.178t}) → e^{−0.178t} = 1/11 → t ≈ 13.5 s. The DE predicts the whole fall from one proportionality assumption.

Exam technique: (1) define variables with units first; (2) write the DE in words-to-symbols, justifying each term; (3) solve by separation showing the +c; (4) apply conditions to find constants (exact forms, e.g. A = 19, not decimals, until asked); (5) interpret: state what the solution predicts in context, with units; (6) comment on model limitations (the tank assumes perfect mixing; the parachutist ignores the chute opening!). Modelling marks reward the interpretation, not just the algebra.

Second-order glimpse and connections

Second-order DEs need two integrations: d²y/dx² = 6x → dy/dx = 3x² + A → y = x³ + Ax + B. Two constants, two conditions needed (e.g. y(0) = 1, y'(0) = 2 give B = 1, A = 2). The equation d²y/dx² = −ω²y has solutions y = A sin(ωt) + B cos(ωt) — simple harmonic motion, the heartbeat of the physics course (oscillations). Verify: y'' = −ω² A sin(ωt) − ω² B cos(ωt) = −ω²y ✓. Recognising this DE means recognising oscillation.

Connections across the course: DEs use integration (Lesson 6), partial fractions (Lesson 1), exponentials (Lesson 8), and numerical methods when exact solution fails (Euler's method: yₙ₊₁ = yₙ + h·f(xₙ, yₙ) — follows the gradient in small steps; a numerical DE solver in one line). Mechanics (Lessons 18–20) is applied differential equations: F = ma is m · d²x/dt² = F. This lesson is the hinge between pure techniques and the physical world.

Common mistakes

• Forgetting the +c, then 'finding' the condition anyway. Without the constant there is no family for the condition to select. +c appears at the integration line, before any substitution.
• Losing equilibrium solutions. Separating dy/dx = y(1 − y) divides by y(1 − y); the constant solutions y = 0 and y = 1 are valid and examiners expect them stated.
• Sign chaos with decay. dT/dt = −k(T − 20) with k > 0: the minus sign is what makes T fall toward 20. Dropping it predicts a cooling cup heating up — a quick sense check catches it.
• Not sorting the modulus. ln|y| = x²/2 + c gives y = A e^{x²/2} with A = ±e^c; presenting |y| = ... as the final answer, or fixing A from the condition before removing the modulus, loses the thread.
• Modelling answers with no interpretation. 'y = 3e^{x²/2}' earns the algebra marks only. Add the sentence the context demands: what happens, when, and whether the model stays believable.

Exam Experience

Questions in the style of the A-level pure paper. Attempt them before reading the mark scheme.

1. Solve the differential equation dy/dx = 2xy, given that y = 4 when x = 0. Give your answer in the form y = f(x). [4]

2. A cup of coffee at 80°C is placed in a room at 20°C. Its temperature T°C after t minutes satisfies dT/dt = −k(T − 20), and after 10 minutes T = 60. (a) Show that k = (ln(3/2))/10. [3] (b) Find the temperature after 25 minutes, to one decimal place. [2]

Mark scheme.

• 1. Separate: (1/y) dy = 2x dx (M1). Integrate: ln|y| = x² + c (M1) (the +c is required here). So y = A e^{x²} (A1) for exponentiating correctly with the constant absorbed into A. y(0) = 4 → A = 4: y = 4e^{x²} (A1).
• 2(a) Separate: dT/(T − 20) = −k dt → ln|T − 20| = −kt + c → T − 20 = A e^{−kt} (M1). T(0) = 80 → A = 60, so T − 20 = 60 e^{−kt} (M1). At t = 10, T = 60: 40 = 60 e^{−10k} → e^{−10k} = 2/3 → −10k = ln(2/3) → k = (ln(3/2))/10 (A1) — target reached exactly.
• 2(b) T − 20 = 60 e^{−25k} = 60 × (2/3)^{2.5} (M1) (using e^{−10k} = 2/3 rather than a rounded k keeps precision). = 60 × 0.3629 ≈ 21.8, so T ≈ 41.8°C (A1).