Question 6 - A-Level Maths

CAIE Further Paper 2 2023 June — Question 6 11 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2023
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Particular solution with initial conditions
Difficulty	Challenging +1.2 This is a standard second-order linear differential equation with constant coefficients requiring complementary function (repeated root λ=6), particular integral (trying x=Asin t + Bcos t), and applying initial conditions. While it involves multiple steps and algebraic manipulation, it follows a completely routine procedure taught in Further Maths with no novel problem-solving required. The repeated root and the arithmetic make it slightly above average difficulty.
Spec	4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

6 Find the particular solution of the differential equation $$\frac { d ^ { 2 } x } { d t ^ { 2 } } - 12 \frac { d x } { d t } + 36 x = 37 \sin t$$ given that, when $t = 0 , x = \frac { d x } { d t } = 0$.

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Question 6:

Answer	Marks	Guidance
Answer	Marks	Guidance
$m^2 - 12m + 36 = 0$	M1	Auxiliary equation
$x = e^{6t}(At + B)$	A1	Complementary function. Allow "$x=$" missing
$x = p\sin t + q\cos t \Rightarrow x' = p\cos t - q\sin t \Rightarrow x'' = -p\sin t - q\cos t$	M1 A1	Particular integral and its derivatives
$-p + 12q + 36p = 37 \quad -q - 12p + 36q = 0$	M1	Substitutes and equates coefficients. Must have two unknowns
$p = \frac{35}{37} \quad q = \frac{12}{37}$	A1
$x = e^{6t}(At + B) + \frac{35}{37}\sin t + \frac{12}{37}\cos t$	A1 FT	Must have "$x=$". FT on CF
$x' = Ae^{6t} + 6e^{6t}(At + B) + \frac{35}{37}\cos t - \frac{12}{37}\sin t$	M1*	Differentiates using product rule. Must have their PI differentiated
$B = -\frac{12}{37} \quad A + 6B + \frac{35}{37} = 0 \Rightarrow A = 1$	DM1 A1	Forms simultaneous equations using initial conditions
$x = e^{6t}\!\left(t - \frac{12}{37}\right) + \frac{35}{37}\sin t + \frac{12}{37}\cos t$	A1	Must have "$x=$"

# Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $m^2 - 12m + 36 = 0$ | M1 | Auxiliary equation |
| $x = e^{6t}(At + B)$ | A1 | Complementary function. Allow "$x=$" missing |
| $x = p\sin t + q\cos t \Rightarrow x' = p\cos t - q\sin t \Rightarrow x'' = -p\sin t - q\cos t$ | M1 A1 | Particular integral and its derivatives |
| $-p + 12q + 36p = 37 \quad -q - 12p + 36q = 0$ | M1 | Substitutes and equates coefficients. Must have two unknowns |
| $p = \frac{35}{37} \quad q = \frac{12}{37}$ | A1 | |
| $x = e^{6t}(At + B) + \frac{35}{37}\sin t + \frac{12}{37}\cos t$ | A1 FT | Must have "$x=$". FT on CF |
| $x' = Ae^{6t} + 6e^{6t}(At + B) + \frac{35}{37}\cos t - \frac{12}{37}\sin t$ | M1* | Differentiates using product rule. Must have their PI differentiated |
| $B = -\frac{12}{37} \quad A + 6B + \frac{35}{37} = 0 \Rightarrow A = 1$ | DM1 A1 | Forms simultaneous equations using initial conditions |
| $x = e^{6t}\!\left(t - \frac{12}{37}\right) + \frac{35}{37}\sin t + \frac{12}{37}\cos t$ | A1 | Must have "$x=$" |

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6 Find the particular solution of the differential equation

$$\frac { d ^ { 2 } x } { d t ^ { 2 } } - 12 \frac { d x } { d t } + 36 x = 37 \sin t$$

given that, when $t = 0 , x = \frac { d x } { d t } = 0$.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q6 [11]}}

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