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Leaving Cert Applied Maths: First-order difference equations

How often First-order difference equations comes up on the Applied Maths papers, every year it was asked, and questions to try.

HL Asked on 2 of the last 3 Higher Level papers, most recently in 2024. most years

OL Asked on 3 of the last 3 Ordinary Level papers, most recently in 2025. banker

First-order difference equations, Higher Level(8 marks)

Quick ones on First-order difference equations.

(a)Solution of the difference equation uₙ₊₁ = a·uₙ?
  1. uₙ = n·a·u₀
  2. uₙ = aⁿ·u₀
  3. uₙ = a·u₀ⁿ
(b)Steady state (equilibrium value) of uₙ₊₁ = 0.5uₙ + 30?
  1. 15
  2. 30
  3. 60
(c)A population grows 5% per year and 40 people emigrate each year. Difference equation?
  1. Pₙ₊₁ = 1.05Pₙ − 40
  2. Pₙ₊₁ = 1.05Pₙ + 40
  3. Pₙ₊₁ = 0.05Pₙ − 40
Show the answers

(a) uₙ = aⁿ·u₀

(b) 60

(c) Pₙ₊₁ = 1.05Pₙ − 40

Your turn: Higher Level questions on First-order difference equations.

Higher Level

Asked on 2 of the last 3 Higher Level papers, most recently in 2024. most years

Every paper, year by year

YearWhere it came up
2025Not asked
2024Q6
2023Q10

Links open the State Examinations Commission’s paper for that year.

More First-order difference equations questions

First-order difference equations, 3 marks

General solution of uₙ₊₁ = 1.5uₙ − 10?

  1. uₙ = A(1.5)ⁿ − 20
  2. uₙ = A(1.5)ⁿ + 10
  3. uₙ = A(1.5)ⁿ + 20
Show the answer

uₙ = A(1.5)ⁿ + 20

The constant part is the steady state: c = 1.5c − 10, so c = 20. Add the free part A(1.5)ⁿ. Check: 1.5(A(1.5)ⁿ + 20) − 10 = A(1.5)ⁿ⁺¹ + 20.

First-order difference equations, 3 marks

uₙ₊₁ = −0.5uₙ + 6 with u₀ = 0. Long-term behaviour?

  1. It oscillates with growing size
  2. It oscillates and converges to 4
  3. It increases steadily to 4
Show the answer

It oscillates and converges to 4

Terms: 0, 6, 3, 4.5, 3.75, … Steady state c = −0.5c + 6 gives c = 4. The negative multiplier makes it jump above and below 4; |−0.5| < 1 makes the jumps shrink.

First-order difference equations, 3 marks

A stock of 1000 falls by 10% a year. After how many whole years is it first below 500?

  1. 7
  2. 5
  3. 6
Show the answer

7

uₙ = 1000(0.9)ⁿ. 0.9⁶ = 0.531 gives 531, still above 500; 0.9⁷ = 0.478 gives 478. Solving 0.9ⁿ < 0.5 gives n > 6.58, so 7 years.

Ordinary Level

Asked on 3 of the last 3 Ordinary Level papers, most recently in 2025. banker

Every paper, year by year

YearWhere it came up
2025Q2, Q8
2024Q3, Q10
2023Q1, Q8

Links open the State Examinations Commission’s paper for that year.

More First-order difference equations questions

First-order difference equations, 2 marks

Which of these is a first-order difference equation?

  1. uₙ₊₂ = uₙ₊₁ + uₙ
  2. uₙ = n² + 1
  3. uₙ₊₁ = 3uₙ + 2
Show the answer

uₙ₊₁ = 3uₙ + 2

First order means each term depends only on the term just before it. The second links three terms (second order); the third is a formula in n, not a difference equation.

First-order difference equations, 3 marks

Find the steady state (fixed point) of uₙ₊₁ = 0.5uₙ + 10.

  1. 5
  2. 20
  3. 10
Show the answer

20

At a steady state the value does not change: L = 0.5L + 10, so 0.5L = 10 and L = 20. Check: 0.5 × 20 + 10 = 20.

First-order difference equations, 3 marks

500 fish grow 10% during the year; 30 are caught at the end of the year. Number after 1 year?

  1. 520
  2. 550
  3. 470
Show the answer

520

Growth first: 500 × 1.1 = 550. Then subtract the catch: 550 − 30 = 520. So Pₙ₊₁ = 1.1Pₙ − 30.

Other Applied Maths topics

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