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Leaving Cert Applied Maths: Separable differential equations

How often Separable differential equations comes up on the Applied Maths papers, every year it was asked, and questions to try.

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

Separable differential equations, Higher Level(6 marks)

Quick ones on Separable differential equations.

(a)When is a first-order differential equation separable?
  1. When it has no constant term
  2. When it can be written as g(y) dy = f(x) dx
  3. When its solution is a straight line
(b)What does the differential equation dN/dt = kN model?
  1. Growth to a fixed limit
  2. Linear growth
  3. Exponential growth or decay
(c)∫ (1/y) dy = ?
  1. ln|y| + c
  2. −1/y² + c
  3. y²/2 + c
Show the answers

(a) When it can be written as g(y) dy = f(x) dx

(b) Exponential growth or decay

(c) ln|y| + c

Your turn: Higher Level questions on Separable differential equations.

Higher Level

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

Every paper, year by year

YearWhere it came up
2025Q8
2024Q4, Q10
2023Q7

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

More Separable differential equations questions

Separable differential equations, 2 marks

Solve dy/dx = 3x², given y = 2 when x = 0.

  1. y = 6x + 2
  2. y = x³
  3. y = x³ + 2
Show the answer

y = x³ + 2

Integrate: y = x³ + c. Put x = 0, y = 2 to get c = 2. Differentiating 3x² gives 6x, which is the wrong direction. Dropping c loses the starting value.

Separable differential equations, 3 marks

Solve dy/dx = y², given y = 1 when x = 0.

  1. y = eˣ
  2. y = 1/(1 − x)
  3. y = 1/(1 + x)
Show the answer

y = 1/(1 − x)

Separate: ∫ y⁻² dy = ∫ dx gives −1/y = x + c. At x = 0, y = 1, so c = −1: −1/y = x − 1, y = 1/(1 − x). eˣ solves dy/dx = y, not y².

Separable differential equations, 3 marks

dN/dt = −0.1N (t in days). Half-life, 2 d.p.?

  1. 6.93 days
  2. 5.00 days
  3. 0.07 days
Show the answer

6.93 days

N = N₀e^(−0.1t). Half when e^(−0.1t) = ½, so t = ln 2 ÷ 0.1 = 6.93 days. The half-life does not depend on N₀.

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