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Leaving Cert Applied Maths: Reducing second-order DEs

How often Reducing second-order DEs 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

Reducing second-order DEs, Higher Level(8 marks)

Quick ones on Reducing second-order DEs.

(a)Why write acceleration as v dv/ds in some motion problems?
  1. To make the equation linear in v
  2. To link v and s directly, with no t
  3. Because dv/dt is zero in that case
(b)For d²y/dx² = f(x), putting p = dy/dx turns it into…?
  1. dp/dy = f(x)
  2. p dp/dx = f(x)
  3. dp/dx = f(x)
(c)For d²y/dx² = f(y), with p = dy/dx, d²y/dx² is written as…?
  1. p dp/dy
  2. y dp/dx
  3. dp/dy
Show the answers

(a) To link v and s directly, with no t

(b) dp/dx = f(x)

(c) p dp/dy

Your turn: Higher Level questions on Reducing second-order DEs.

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
2025Q2
2024Q6
2023Q4

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

More Reducing second-order DEs questions

Reducing second-order DEs, 3 marks

d²y/dx² = 1/x². With p = dy/dx and p = 0 at x = 1, find p.

  1. p = 1/x − 1
  2. p = −2/x³
  3. p = 1 − 1/x
Show the answer

p = 1 − 1/x

dp/dx = x⁻², so p = −1/x + c. p = 0 at x = 1 gives c = 1, so p = 1 − 1/x. Differentiating x⁻² instead of integrating gives −2/x³.

Reducing second-order DEs, 3 marks

A particle moves with a = 2v (v > 0) and v = 1 at x = 0. Using a = v dv/dx, find v in terms of x.

  1. v = x² + 1
  2. v = 2x + 1
  3. v = e^(2x)
Show the answer

v = 2x + 1

v dv/dx = 2v, so dv/dx = 2 and v = 2x + c, with c = 1. Using dv/dt = 2v instead gives v = e^(2t), which is in terms of time, not distance.

Reducing second-order DEs, 3 marks

d²y/dx² = 2y, and p = dy/dx = 2 when y = 1. Using p dp/dy, find p².

  1. p² = 2y² + 2
  2. p² = y² + 3
  3. p² = 4y²
Show the answer

p² = 2y² + 2

p dp/dy = 2y, so ½p² = y² + c, i.e. p² = 2y² + 2c. At y = 1, p² = 4, so 2c = 2 and p² = 2y² + 2. The other options fit the starting value but not the equation.

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