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A Level Maths 9709 · Paper 3 Pure Mathematics 3 · 3.8 · ~5 min read

Differential equations

Forming a differential equation from a rate-of-change statement, solving first-order equations by separating the variables, using a condition to find the particular solution, and interpreting the result in context.

Forming equationsProportionalitySeparating variablesGeneral solutionParticular solutionExponential growth & decayPartial fractions in DEsInterpreting solutions
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • A differential equation links a rate to a quantity. "The rate of decrease of N is proportional to N" means dN/dt = −kN with k > 0.
  • Separate the variables: get all y-terms with dy on one side and all x-terms with dx on the other, then integrate both sides.
  • One constant is enough. Add + c on one side only, then use the given condition straight away.
  • Interpret: check the long-term behaviour (what happens as t → ∞) and answer in the context's units.

By the end you can

  • Form a simple differential equation from a statement about rates of change.
  • Solve first-order equations with separable variables, using any integration technique from 3.5.
  • Find a particular solution from given conditions.
  • Interpret the solution and its behaviour in context.

1Forming the equation

Words into symbols

StatementEquation
The rate of increase of P is proportional to PdP/dt = kP
The temperature θ falls at a rate proportional to its excess over 20 °Cdθ/dt = −k(θ − 20)
The volume decreases at a rate proportional to √VdV/dt = −k√V
The gradient of the curve at (x, y) is xy²dy/dx = xy²
  • Constants: state that k > 0 when the quantity decreases, and put the minus sign in the equation.
Common mistakeMissing the minus sign for a decreasing quantity, or writing dV/dx when the variable is time.

2Separating the variables

Method

dy/dx = f(x)g(y) ⇒ ∫ 1/g(y) dy = ∫ f(x) dx
then add one constant
Worked example

Solve dy/dx = 2xy, given y = 3 when x = 0.

  1. ∫(1/y) dy = ∫2x dx M1
  2. ln y = x² + c A1
  3. x = 0, y = 3 ⇒ c = ln 3 M1
  4. y = ex² + ln 3 = 3ex² A1
Worked example

Solve x dy/dx = y² + 1 for x > 0, given y = 1 when x = 1. Give y in terms of x.

  1. ∫1/(y² + 1) dy = ∫1/x dx M1
  2. tan⁻¹y = ln x + c A1 A1
  3. x = 1, y = 1: π/4 = 0 + c M1
  4. y = tan(ln x + π/4) A1
Worked example

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

  1. ∫(1/y) dy = ∫1/[(x + 1)(x + 2)] dx = ∫[1/(x + 1) − 1/(x + 2)] dx M1 A1
  2. ln y = ln(x + 1) − ln(x + 2) + c A1
  3. ln 2 = ln ½ + c ⇒ c = ln 4 M1
  4. y = 4(x + 1)/(x + 2) A1
  • Removing logs: write the constant as ln A so you can combine: ln y = ln[A(x + 1)/(x + 2)]. It's neater than ec.
Common mistakeWriting ex² + c = ex² + ec. It is ex² × ec. The 2024 Paper 3 report highlighted this exponential error.

3In context

Growth, decay and cooling

xydy/dx = xyeach dash hasgradient xyRed and blue: the solutions y = Ae^(x²/2) for two values of A.
Every solution curve follows the slope field; the initial condition picks one curve.
Worked example

A liquid cools so that dθ/dt = −k(θ − 20), where θ °C is the temperature after t minutes. θ = 90 when t = 0 and θ = 60 when t = 10. Find how long it takes to reach 30 °C.

  1. ∫1/(θ − 20) dθ = −∫k dt ⇒ ln(θ − 20) = −kt + c M1 A1
  2. t = 0: c = ln 70, so θ = 20 + 70e−kt A1
  3. t = 10: 40 = 70e−10k ⇒ k = ln(7/4)/10 = 0.0560 M1 A1
  4. 10 = 70e−kt ⇒ t = ln 7/k = 34.8 minutes A1
  • Long term: as t → ∞, e−kt → 0 and θ → 20 °C, the room temperature. Questions often ask for this "what happens eventually".
Common mistakeRounding k too early. Keep k exact (ln(7/4)/10) or in the calculator memory; rounding to 0.06 changes the final answer to 32.4.
Linking questionThe integrations here are from Further integration, the partial fractions from Algebra, and the exponential and log rules from Logarithms and exponentials.

Test yourself — cover the answers first

Quick check

  1. Write "the rate of increase of x is proportional to x³" as an equation.
  2. Separate: dy/dx = y cos x.
  3. How many arbitrary constants does a first-order DE solution need?
  4. Simplify eln 5 + 2x.
  5. In dθ/dt = −k(θ − 15), what does θ approach in the long term?

Exam-style

  1. Solve dy/dx = ex − y, given y = 0 when x = 0.
  2. Solve dy/dx = (y + 1)/x, given y = 2 when x = 1.
  3. A population P satisfies dP/dt = 0.02P with P = 500 at t = 0. Find t when P = 1000.
Show answers
  1. dx/dt = kx³
  2. ∫(1/y) dy = ∫cos x dx
  3. one
  4. 5e2x
  5. 15
  6. ∫ey dy = ∫ex dx ⇒ ey = ex + c, c = 0 ⇒ y = x
  7. ln(y + 1) = ln x + ln 3 ⇒ y = 3x − 1
  8. P = 500e0.02t; e0.02t = 2 ⇒ t = 50 ln 2 = 34.7
Practise this chapter
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.