Standard difficulty · 20 questions · 40 marks
Show your working for Sections B and C. Answers are on the last page — try every question first!
Differential Equations · AW-12M-DEQ-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.
| Q | Answer | Working |
|---|---|---|
| 1 | (a) order 1, degree 3 | highest derivative y′, raised to the power 3 |
| 2 | (c) y = x2 + C | integrate both sides |
| 3 | (c) 1/x | e^(∫P dx) with P = −1/x |
| 4 | (b) 3 | one constant for each order |
| 5 | (c) the highest derivative in it | — |
| 6 | (c) y = 5 sin 4x | (sin 4x)″ = −16 sin 4x |
| 7 | (c) homogeneous | every term has degree 2, so dy/dx is a function of y/x |
| 8 | (d) A is false, but R is true. | A is false: the solution is y = Ce9x, not Ce−9x. R is true. |
| 9 | (9/4)y4 = (7/3)x3 + C | 9y3 dy = 7x2 dx |
| 10 | y = 8e5x | y = Ce5x; x = 0 gives C = 8 |
| 11 | y = 4 + Ce−x | integrating factor ex; yex = 4ex + C |
| 12 | y = (1/6)x5 + C/x | integrating factor x; d/dx(xy) = x5 |
| 13 | y″ + 9y = 0 | differentiate twice and eliminate the constants |
| 14 | y = −7x log|x| + Cx | put y = vx: x dv/dx = −7, so v = −7 log|x| + C |
| 15 | y = 8x3 + 4x − 6 | y = 8x3 + 4x + C; C = −6 |
| 16 | 64 years | P = P0e^(kt) with e^(16k) = 2, so 16 = 24 needs t = 4 × 16 |
| 17 | y = −(3/2)e2x + (5/2)e−4x | integrating factor e4x: ye4x = −(3/2)e6x + C; C = 1 − (−3/2) = 5/2 |
| 18 | y = 5x2 + 2x | divide by x2: d/dx(y/x) = 5, so y/x = 5x + C; C = 2 |
| 19 | y2 − 14y = x2 + 16x | (y − 7) dy = (x + 8) dx ⇒ y2/2 − 7y = x2/2 + 8x + C; C = 0 |
| 20 | 1/8 remains after 165 years; 1/8 remains after 165 years | A = A0e^(−λt) with e^(−55λ) = 1/2, so A = A0(1/2)^(t/55) |