Differential Equations

A differential equation relates a function to its derivatives. At A-Level you focus on first-order differential equations that can be solved by separating variables.

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What Is a Differential Equation?

A differential equation contains derivatives such as dy/dx, d²y/dx², etc.

Order = the highest derivative present. A first-order ODE involves only dy/dx.

Examples

Equation	Order	Type
dy/dx = 3x²	1	Directly integrable
dy/dx = 2y	1	Separable
dy/dx = x/y	1	Separable
d²y/dx² + 4y = 0	2	Beyond A-Level Pure

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Separation of Variables

If a first-order ODE can be written as:

dy/dx = f(x) g(y)

then we can separate and integrate:

∫(1/g(y)) dy = ∫f(x) dx

Worked Example 1

Solve dy/dx = 3x²y.

Separate: (1/y) dy = 3x² dx

Integrate: ln|y| = x³ + c

So: y = Ae^(x³) where A = ±e^c.

Worked Example 2

Solve dy/dx = (x + 1)/y², given y = 2 when x = 0.

Separate: y² dy = (x + 1) dx

Integrate: y³/3 = x²/2 + x + c

Apply initial condition: 8/3 = 0 + 0 + c → c = 8/3

y³/3 = x²/2 + x + 8/3

= y³ = 3x²/2 + 3x + 8

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Exponential Growth and Decay

The equation dy/dt = ky has the general solution:

y = Aeᵏᵗ

• k > 0: exponential growth
• k < 0: exponential decay

Worked Example 3

A radioactive substance decays at a rate proportional to the amount present. Initially there are 500 g and after 10 years there are 400 g.

(a) Set up and solve the differential equation.

dN/dt = −λN → N = N₀ e^(−λt) = 500 e^(−λt)

At t = 10: 400 = 500 e^(−10λ) → e^(−10λ) = 0.8 → −10λ = ln 0.8 → λ = −ln 0.8/10 ≈ 0.02231

N = 500 e^(−0.02231t)

(b) Find the half-life.

250 = 500 e^(−0.02231t) → e^(−0.02231t) = 0.5 → t = ln 2/0.02231

≈ 31.1 years

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Newton’s Law of Cooling

The rate of cooling is proportional to the temperature difference between the object and its surroundings:

dT/dt = −k(T − Tₛ)

Worked Example 4

A cup of tea at 90°C cools in a room at 20°C. After 5 minutes the tea is 70°C.

dT/dt = −k(T − 20)

Separate: ∫1/(T − 20) dT = ∫−k dt

ln|T − 20| = −kt + c → T − 20 = Ae^(−kt)

At t = 0: 90 − 20 = A → A = 70
At t = 5: 70 − 20 = 70e^(−5k) → 50 = 70e^(−5k) → e^(−5k) = 5/7 → k = −ln(5/7)/5 ≈ 0.0673

T = 20 + 70e^(−0.0673t)

When will the tea reach 40°C?

40 = 20 + 70e^(−0.0673t) → 20/70 = e^(−0.0673t) → t = −ln(2/7)/0.0673

≈ 18.6 minutes

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Population Models

Logistic Growth

dP/dt = kP(1 − P/M) models a population P with carrying capacity M. This is separable but requires partial fractions to solve.

Worked Example 5

dP/dt = 0.5P(1 − P/100), P(0) = 10.

Using partial fractions and separation, the solution is:

P = 100/(1 + 9e^(−0.5t))

As t → ∞, P → 100 (the carrying capacity).

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Forming Differential Equations

Many problems require you to set up the differential equation from a word problem.

Worked Example 6

A tank initially holds 200 litres of water. Water leaks at a rate proportional to the square root of the volume. Set up a differential equation.

Let V = volume at time t. The rate of decrease is proportional to √V:

dV/dt = −k√V where k > 0.

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General Solutions vs Particular Solutions

• A general solution contains an arbitrary constant (e.g. y = Ae^(2x)).
• A particular solution uses an initial or boundary condition to determine the constant (e.g. y = 3e^(2x)).

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Exam Tips

• Always separate variables completely before integrating — all y-terms on one side, all x-terms on the other.
• Do not forget the constant of integration — apply the initial condition to find it.
• For exponential models, identify whether the problem describes growth (positive rate) or decay (negative rate).
• Show the separation step clearly — marks are often awarded here.
• Check your solution by differentiating and substituting back into the original equation.
• Clearly state initial conditions when forming a particular solution.

A-Level Deep Dive: First-Order Differential Equations

Spec mapping

AQA 7357 specification, Pure Mathematics, Section H — Integration (Year 2 content) covers simple differential equations in pure mathematics and in context, (contexts may include kinematics, population growth and modelling the relationship between price and demand). Find general and particular solutions of first-order differential equations with separable variables (refer to the official specification document for exact wording). This is examined on Paper 2 (Pure) and frequently appears as a Section C 8–12 mark question, often with an AO3 modelling context. The AQA formula booklet provides standard integrals ($\int 1/x \, dx$∫1/xdx, $\int e^{kx} \, dx$∫ekxdx etc.) but does not state the separation-of-variables technique itself — that procedural step must be deployed without prompt.

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Worked example with full mark scheme

Question (8 marks): A curve passes through the point $(0, 2)$(0,2) and satisfies the differential equation

$\dfrac{dy}{dx} = xy, \qquad y > 0.$dxdy​=xy,y>0.

(a) Solve the differential equation, expressing $y$y explicitly in terms of $x$x. (6)

(b) Hence find the value of $y$y when $x = 2$x=2, giving your answer to 3 significant figures. (2)

Solution with mark scheme:

(a) Step 1 — separate the variables.

$\dfrac{1}{y}\,dy = x\,dx$y1​dy=xdx

M1 — separation written with the $y$y-terms on one side and the $x$x-terms on the other, each multiplied by their own differential. A common error is to write $dy/y = x\,dx$dy/y=xdx and then "cancel" $y$y — separation is a formal manipulation, not algebraic cancellation, but the layout earns the M1.

Step 2 — integrate both sides.

$\int \dfrac{1}{y}\,dy = \int x\,dx$∫y1​dy=∫xdx $\ln|y| = \dfrac{x^2}{2} + C$ln∣y∣=2x2​+C

M1 — integrating both sides correctly. A1 — both integrals correct, with a single arbitrary constant $C$C on one side only. Writing $C_1$C1​ and $C_2$C2​ on each side and combining later is acceptable, but examiners prefer one constant from the outset.

Step 3 — apply the initial condition.

When $x = 0$x=0, $y = 2$y=2, so $\ln 2 = 0 + C$ln2=0+C, giving $C = \ln 2$C=ln2.

M1 — substituting the initial condition into the general solution to find $C$C. Critical: the constant must be evaluated before exponentiating, otherwise the algebra becomes painful.

Step 4 — solve for $y$y explicitly.

$\ln y = \dfrac{x^2}{2} + \ln 2 \implies y = e^{x^2/2 + \ln 2} = 2 e^{x^2/2}$lny=2x2​+ln2⟹y=ex2/2+ln2=2ex2/2

(Since $y > 0$y>0, the modulus drops cleanly.)

A1 — particular solution $y = 2 e^{x^2/2}$y=2ex2/2 in explicit form.

(b) Step 5 — evaluate.

$y(2) = 2 e^{4/2} = 2 e^2 \approx 2 \cdot 7.389 = 14.778\ldots$y(2)=2e4/2=2e2≈2⋅7.389=14.778…

M1 A1 — $y \approx 14.8$y≈14.8 to 3 s.f.

Total: 8 marks (M5 A3, split as shown).

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks, AO3-heavy): A cup of coffee is left to cool in a room. The temperature $T$T $\degree$°C of the coffee at time $t$t minutes satisfies

$\dfrac{dT}{dt} = -k(T - 20),$dtdT​=−k(T−20),

where $k > 0$k>0 is a constant and $20\degree$20°C is the room temperature. Initially the coffee is at $80\degree$80°C, and after 5 minutes it has cooled to $60\degree$60°C.

(a) Find $T$T in terms of $t$t. (4)

(b) State, with justification, the long-term temperature predicted by this model, and comment briefly on the realism of this prediction. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO3.1a) — separating: $\dfrac{1}{T - 20}\,dT = -k\,dt$T−201​dT=−kdt.
• M1 (AO1.1b) — integrating: $\ln|T - 20| = -kt + C$ln∣T−20∣=−kt+C.
• M1 (AO3.4) — applying both conditions: $T(0) = 80$T(0)=80 gives $C = \ln 60$C=ln60; $T(5) = 60$T(5)=60 gives $\ln 40 = -5k + \ln 60$ln40=−5k+ln60, so $k = \tfrac{1}{5}\ln(3/2)$k=51​ln(3/2).
• A1 (AO2.5) — explicit form: $T = 20 + 60 \cdot (2/3)^{t/5}$T=20+60⋅(2/3)t/5 or equivalently $T = 20 + 60 e^{-kt}$T=20+60e−kt with $k$k stated.

(b)

• B1 (AO3.2a) — as $t \to \infty$t→∞, $(2/3)^{t/5} \to 0$(2/3)t/5→0, so $T \to 20\degree$T→20°C, the room temperature.
• B1 (AO3.5b) — realistic limitation noted: e.g. the room temperature is treated as constant (it isn't precisely), or the cooling rate $k$k depends weakly on temperature in real fluids (Newton's law is a linearisation).

Total: 6 marks split AO1 = 1, AO2 = 1, AO3 = 4. This is an AO3-dominated question — separable ODEs in modelling contexts are AQA's preferred vehicle for Year-2 AO3 marks because they bundle reasoning, problem-solving, and critique of assumptions into a single chain.

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Synoptic links

Connects to:

1. Section G — Integration: every separable ODE collapses to two integrations, one in $y$y and one in $x$x. Confidence with $\int 1/x\,dx = \ln|x| + C$∫1/xdx=ln∣x∣+C, $\int e^{kx}\,dx = \tfrac{1}{k}e^{kx} + C$∫ekxdx=k1​ekx+C, and integration by partial fractions is the entire toolkit. A weak integrator cannot solve a separable ODE.

2. Section F — Exponentials and logarithms: the classic ODE $\dfrac{dy}{dt} = ky$dtdy​=ky produces $y = A e^{kt}$y=Aekt, which is the defining differential equation of the exponential function. Population growth, radioactive decay, continuously compounded interest, and unrestricted bacterial reproduction all reduce to this single ODE.

3. Newton's law of cooling: $\dfrac{dT}{dt} = -k(T - T_{\text{room}})$dtdT​=−k(T−Troom​) is the canonical "linear in difference" separable ODE. It generalises to heat transfer, charging capacitors ($RC$RC circuits: $\dfrac{dV}{dt} = -\tfrac{1}{RC}V$dtdV​=−RC1​V), and even some financial bond-pricing models.

4. Logistic equation: $\dfrac{dP}{dt} = rP(1 - P/K)$dtdP​=rP(1−P/K) models bounded population growth. It's separable, but the integration step requires partial fractions (Section B) — this is the classic cross-topic synoptic question. Splitting $\dfrac{1}{P(1 - P/K)} = \dfrac{1}{P} + \dfrac{1/K}{1 - P/K}$P(1−P/K)1​=P1​+1−P/K1/K​ is the pinch-point.

5. Partial fractions (Section B): any separable ODE whose left side is $\dfrac{1}{(y - a)(y - b)}$(y−a)(y−b)1​ requires partial-fraction decomposition before integrating. AQA loves this combination — a partial-fractions question disguised as a differential equation.

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Mark-scheme literacy

Differential-equation questions on 7357 split AO marks across all three objectives more evenly than most pure topics:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	35–45%	Separating variables, performing the integrations, manipulating logs and exponentials
AO2 (reasoning / interpretation)	20–30%	Justifying the dropped modulus ($y > 0$y>0), interpreting the constant $C$C, expressing the answer in the requested form
AO3 (problem-solving / modelling)	30–40%	Setting up the ODE from a description, applying initial conditions, interpreting long-term behaviour, critiquing model assumptions

Examiner-rewarded phrasing: "separating variables gives …"; "since $y > 0$y>0 we may drop the modulus"; "as $t \to \infty$t→∞ the exponential tends to zero, so $T \to T_{\text{room}}$T→Troom​"; "the model assumes the surroundings remain at constant temperature, which is a reasonable approximation over the timescale considered". Phrases that lose marks: writing the general solution as $y = e^{x^2/2} + C$y=ex2/2+C (the constant should appear inside the exponent of the integrated form, additive, then reappear multiplicatively after exponentiating — placing it outside is muddled algebra); forgetting the constant of integration entirely; substituting initial conditions before integrating; treating the modulus as decoration.

A specific AQA pattern to watch: questions phrased "form a differential equation to describe …" demand that you set up the ODE — usually by recognising "rate of change of $X$X is proportional to $Y$Y" as $dX/dt = kY$dX/dt=kY. The proportionality constant is unknown until a second condition is applied. Skipping straight to a solved form without showing the ODE first loses the AO3 setup marks even if the final answer is correct.

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Grade-band model answers

3-mark question

Question: Find the general solution of $\dfrac{dy}{dx} = \dfrac{x}{y}$dxdy​=yx​ for $y \neq 0$y=0.

Grade C response (~150 words):

Separate: $y\,dy = x\,dx$ydy=xdx. Integrate both sides: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$2y2​=2x2​+C. So $y^2 = x^2 + 2C$y2=x2+2C, or $y^2 - x^2 = k$y2−x2=k where $k = 2C$k=2C is a constant.

Examiner commentary: Full marks (3/3). The candidate correctly separates, integrates, and combines constants into a single arbitrary constant $k$k. Re-labelling $2C$2C as $k$k is good practice — examiners accept either form provided the relabelling is explicit. The implicit form $y^2 - x^2 = k$y2−x2=k is acceptable here because the question says "general solution" and does not demand explicit $y = f(x)$y=f(x).

Grade A response (~210 words):*