Differential Equations

This lesson covers first-order differential equations solved by separating variables — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to be able to separate variables, integrate both sides, find general and particular solutions, and form differential equations from contextual problems.

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What Is a Differential Equation?

A differential equation is an equation that involves derivatives (rates of change). For example:

• dy/dx = 3x²
• dy/dx = 2y
• dy/dx = xy + 1

At A-Level, you focus on first-order differential equations (involving dy/dx only, not d²y/dx²).

General Solution vs Particular Solution

• The general solution contains the constant of integration c and represents a family of curves.
• A particular solution is found by using a boundary condition (an initial condition or known point) to determine c. It represents one specific curve from the family.

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Separating Variables

The method of separation of variables works when the differential equation can be written in the form:

dy/dx = f(x) × g(y)

That is, the right-hand side can be expressed as a product of a function of x only and a function of y only.

Step-by-Step Method

1. Rearrange so that all y-terms are on one side and all x-terms are on the other: (1/g(y)) dy = f(x) dx

2. Integrate both sides: ∫ (1/g(y)) dy = ∫ f(x) dx

3. Evaluate both integrals (include one constant of integration).

4. If possible, rearrange to express y in terms of x.

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Worked Examples

Example 1: dy/dx = 3x²y

Step 1: Separate variables. (1/y) dy = 3x² dx

Step 2: Integrate both sides. ∫ (1/y) dy = ∫ 3x² dx ln|y| = x³ + c

Step 3: Express y in terms of x. |y| = e^(x³ + c) = e^c × e^(x³)

Since e^c is a positive constant, write it as A: y = Ae^(x³) (where A is a constant, A ≠ 0)

Example 2: dy/dx = (x + 1)/y²

Step 1: Separate variables. y² dy = (x + 1) dx

Step 2: Integrate both sides. ∫ y² dy = ∫ (x + 1) dx y³/3 = x²/2 + x + c

Step 3: Rearrange. y³ = 3x²/2 + 3x + 3c

We can write 3c as a single constant C: y³ = (3x² + 6x)/2 + C

Or simply: y³ = (3/2)x² + 3x + C

Example 3: Finding a Particular Solution

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

Step 1: Separate. 1/(y + 1) dy = 2x dx

Step 2: Integrate. ∫ 1/(y + 1) dy = ∫ 2x dx ln|y + 1| = x² + c

Step 3: Use the boundary condition y = 2 when x = 0. ln|3| = 0 + c → c = ln 3

Step 4: Write the particular solution. ln|y + 1| = x² + ln 3 ln|y + 1| - ln 3 = x² ln|(y + 1)/3| = x² (y + 1)/3 = e^(x²) (taking the positive case since y + 1 > 0) y + 1 = 3e^(x²) y = 3e^(x²) - 1

Example 4: dy/dx = y cos x

Step 1: Separate. (1/y) dy = cos x dx

Step 2: Integrate. ln|y| = sin x + c

Step 3: General solution. y = Ae^(sin x) (where A = ±e^c)

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Forming Differential Equations

Exam questions may give you a real-world scenario and ask you to form (set up) the differential equation before solving it.

Example: Population Growth

A population P grows at a rate proportional to its current size. Write down a differential equation for P in terms of time t.

"Rate proportional to P" means dP/dt = kP for some positive constant k.

Solving: Separate variables. (1/P) dP = k dt ln P = kt + c P = P₀e^(kt) (where P₀ is the initial population)

Example: Newton's Law of Cooling

The temperature T of a hot object cools at a rate proportional to the difference between its temperature and the surrounding temperature T_s.

dT/dt = -k(T - T_s) (where k > 0)

Solving: Let u = T - T_s, so du/dt = dT/dt = -ku. (1/u) du = -k dt ln|u| = -kt + c u = Ae^(-kt) T - T_s = Ae^(-kt) T = T_s + Ae^(-kt)

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Exponential Growth and Decay Models

Many differential equations at A-Level model exponential growth or decay:

Model	Differential Equation	Solution
Exponential growth	dy/dx = ky (k > 0)	y = Ae^(kx)
Exponential decay	dy/dx = -ky (k > 0)	y = Ae^(-kx)

Key Features

• Growth (k > 0): The quantity increases without bound as x → ∞.
• Decay (k > 0 in dy/dx = -ky): The quantity decreases towards 0 as x → ∞.
• The half-life is the time taken for the quantity to halve.
• The doubling time is the time taken for the quantity to double.

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Connected Rates of Change

Sometimes differential equations arise from chain rule connections:

If you know dV/dt and you need dr/dt, and V is a function of r, then:

dr/dt = (dr/dV) × (dV/dt)

Example: A spherical balloon is inflated at a rate of 10 cm³/s. Find the rate of increase of the radius when r = 5 cm.

V = (4/3)πr³, so dV/dr = 4πr²

dr/dt = (dr/dV)(dV/dt) = (1/(4πr²)) × 10 = 10/(4π × 25) = 10/(100π) = 1/(10π) cm/s

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Differential Equations with Given Substitutions

Sometimes the question provides a substitution to help solve the equation.

Example: Use the substitution y = vx to solve x(dy/dx) = x + y.

Step 1: y = vx → dy/dx = v + x(dv/dx) (by the product rule)

Step 2: Substitute into the DE: x(v + x(dv/dx)) = x + vx xv + x²(dv/dx) = x + vx x²(dv/dx) = x dv/dx = 1/x

Step 3: Integrate: v = ln|x| + c

Step 4: Replace v = y/x: y/x = ln|x| + c y = x ln|x| + cx

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Common Mistakes

1. Not separating variables correctly — Make sure ALL y-terms are on one side and ALL x-terms on the other before integrating.
2. Forgetting the constant of integration — The general solution must include a constant.
3. Incorrect handling of ln — When you have ln|y| = f(x) + c, remember that y = ±e^c × e^(f(x)). Combine ±e^c into a single constant A.
4. Not using the boundary condition — To find a particular solution, substitute the given values to find c.
5. Sign errors — Be especially careful when the DE has negative signs.

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Summary

• A differential equation involves derivatives (rates of change).
• Separation of variables: rearrange to (1/g(y)) dy = f(x) dx, then integrate both sides.
• The general solution contains an arbitrary constant; the particular solution uses a boundary condition to find it.
• Common models: exponential growth (dy/dx = ky) and decay (dy/dx = -ky).
• Context problems (populations, cooling, rates) often require you to form the DE first.
• Always include a constant of integration in the general solution.

A-Level Deep Dive: Differential Equations

Spec mapping

Edexcel 9MA0-02 specification section 11 — Differential equations: "Construct simple differential equations in pure mathematics and in context (contexts may include kinematics, population growth and modelling the relationship between price and demand). Evaluate the analytical solution of simple first-order differential equations with separable variables, including finding particular solutions; interpret the solution of a differential equation in the context of solving a problem, including identifying limitations of the solution." This is Year 2 content sitting at the apex of pure: it draws on integration (section 8), exponentials and logarithms (section 6), partial fractions (section 4) and modelling reasoning (AO3). Differential equations appear on Paper 2 (Pure 2) and increasingly within Paper 3 (Mechanics) when the question is set up as $\frac{dv}{dt} = f(v)$dtdv​=f(v). The Edexcel formula booklet provides no integral results specific to ODEs — students must recognise the standard $\int \frac{1}{x}\,dx = \ln|x| + C$∫x1​dx=ln∣x∣+C and exponential families unaided.

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Worked example with full mark scheme

Question (8 marks): Solve the differential equation $\dfrac{dy}{dx} = xy$dxdy​=xy given that $y = 2$y=2 when $x = 0$x=0. Find the general solution and the particular solution.

Solution with mark scheme:

Step 1 — separate the variables.

The equation $\dfrac{dy}{dx} = xy$dxdy​=xy is first-order and separable: the right-hand side factorises as a function of $x$x multiplied by a function of $y$y. Divide both sides by $y$y (valid provided $y \neq 0$y=0) and multiply both sides by $dx$dx:

$\dfrac{1}{y}\,dy = x\,dx$y1​dy=xdx

M1 — correctly separating into the form $g(y)\,dy = f(x)\,dx$g(y)dy=f(x)dx. A common error is to write $\dfrac{dy}{y} = x\,dx + \dfrac{dy}{dx}$ydy​=xdx+dxdy​ or to leave $y$y on the right-hand side; this earns nothing.

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 — applying $\int \frac{1}{y}\,dy = \ln|y|$∫y1​dy=ln∣y∣ on the left. M1 — applying the power rule $\int x\,dx = \tfrac{1}{2}x^2$∫xdx=21​x2 on the right. A1 — both integrals correct with a single constant of integration $C$C on one side. Writing two constants $C_1$C1​ and $C_2$C2​ is not penalised provided they are combined immediately; leaving them separate throughout loses the A1.

Step 3 — exponentiate to obtain the general solution.

$|y| = e^{x^2/2 + C} = e^C \cdot e^{x^2/2}$∣y∣=ex2/2+C=eC⋅ex2/2

Writing $A = \pm e^C$A=±eC (an arbitrary non-zero constant absorbing the sign):

$y = A e^{x^2/2}$y=Aex2/2

M1 — exponentiating both sides correctly, with the sign collapsed into a single arbitrary constant. A1 — general solution $y = A e^{x^2/2}$y=Aex2/2.

Step 4 — apply the boundary condition $y = 2$y=2 when $x = 0$x=0.

$2 = A e^{0} = A \cdot 1 = A$2=Ae0=A⋅1=A

So $A = 2$A=2.

M1 — substituting the boundary condition into the general solution. A1 — particular solution $y = 2 e^{x^2/2}$y=2ex2/2.

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

A worth-noting subtlety: the case $y = 0$y=0 (the constant zero function) is also a solution of $\frac{dy}{dx} = xy$dxdy​=xy that is lost when dividing by $y$y. For an A* answer, note explicitly that the boundary condition $y(0) = 2 \neq 0$y(0)=2=0 rules this branch out, so $y = A e^{x^2/2}$y=Aex2/2 is the complete family relevant here.

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): A population of bacteria $P$P at time $t$t hours satisfies the differential equation

$\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{2000}\right)$dtdP​=kP(1−2000P​)

where $k > 0$k>0 is a constant. At $t = 0$t=0 the population is $P = 200$P=200.

(a) Without solving the equation, explain what happens to $\frac{dP}{dt}$dtdP​ as $P$P approaches $2000$2000, and interpret this in context. (2)

(b) State, with a reason, the long-term behaviour of $P$P as $t \to \infty$t→∞. (2)

(c) The biologist suggests that the model breaks down for very large $t$t. Give one reason why this might be the case in real bacterial populations. (2)

Mark scheme decomposition by AO:

(a)

• B1 (AO2.4) — observing that as $P \to 2000$P→2000, the bracket $(1 - P/2000) \to 0$(1−P/2000)→0, so $\frac{dP}{dt} \to 0$dtdP​→0.
• B1 (AO3.2a) — interpreting in context: the population stops growing as it approaches the carrying capacity of $2000$2000.

(b)

• B1 (AO2.2a) — stating that $P \to 2000$P→2000 as $t \to \infty$t→∞.
• B1 (AO2.4) — reasoning: starting from $P = 200 < 2000$P=200<2000 with $k > 0$k>0 gives positive growth that decelerates as $P$P rises, asymptotically reaching the carrying capacity.

(c)

• B1 (AO3.5b) — identifying a limitation, e.g. nutrient depletion, accumulation of waste products, environmental change, or death/decay not modelled by the equation.
• B1 (AO3.5b) — extending: the model assumes a fixed carrying capacity, but real environments fluctuate, so long-term predictions are unreliable.

Total: 6 marks split AO2 = 3, AO3 = 3. This is an AO3-heavy modelling question — Edexcel uses logistic-style ODEs precisely to test interpretation, not just solution. Notice that part (a) asks candidates to reason about behaviour without integrating: this is deliberate, and writing out a separation-of-variables solution earns no marks for parts (a) or (b).

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Synoptic links

Connects to:

1. Section 8 — Integration: every separable ODE reduces to two integrals, one in $y$y and one in $x$x. Confidence with $\int \frac{1}{y}\,dy = \ln|y|$∫y1​dy=ln∣y∣, $\int e^{kx}\,dx = \tfrac{1}{k}e^{kx}$∫ekxdx=k1​ekx and the chain rule in reverse is non-negotiable. A weak Section 8 student cannot solve Section 11 questions, however cleanly they separate.

2. Section 6 — Exponential modelling, growth and decay: the equation $\frac{dy}{dt} = ky$dtdy​=ky has solution $y = A e^{kt}$y=Aekt, the simplest exponential growth/decay model. Radioactive decay ($k < 0$k<0), continuously compounded interest ($k > 0$k>0) and unrestricted bacterial growth all sit here. A Section 11 question asking "show that $y = A e^{kt}$y=Aekt satisfies $\frac{dy}{dt} = ky$dtdy​=ky" tests differentiation of exponentials; the converse (derive $y = A e^{kt}$y=Aekt by separation) tests Section 11.

3. Newton's law of cooling: $\frac{dT}{dt} = -k(T - T_{\text{room}})$dtdT​=−k(T−Troom​) is the canonical ODE of Year 2 modelling. The substitution $u = T - T_{\text{room}}$u=T−Troom​ converts it to $\frac{du}{dt} = -ku$dtdu​=−ku, which has solution $u = A e^{-kt}$u=Ae−kt, hence $T = T_{\text{room}} + A e^{-kt}$T=Troom​+Ae−kt. Every coffee-cooling question on 9MA0-02 reduces to this template.

4. Logistic equation: $\frac{dP}{dt} = kP(1 - P/M)$dtdP​=kP(1−P/M) has carrying capacity $M$M and is solved by separation combined with partial fractions (section 4 of the spec). The integral $\int \frac{dP}{P(1 - P/M)}$∫P(1−P/M)dP​ requires decomposition into $\frac{1}{P} + \frac{1/M}{1 - P/M}$P1​+1−P/M1/M​. Logistic models appear regularly on Paper 2 in the highest-mark questions.

5. Section 4 — Partial fractions: any ODE whose right-hand side factorises with multiple linear roots in $y$y, e.g. $\frac{dy}{dx} = (y - 1)(y - 3)$dxdy​=(y−1)(y−3), demands partial-fraction decomposition before integrating. This is the synoptic link Edexcel exploits most often when stretching marks above 8.

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Mark-scheme literacy

Differential-equations questions on 9MA0-02 split AO marks more evenly than algebra-and-functions questions, because modelling and interpretation matter: