Question 5 - A-Level Maths

CAIE Further Paper 2 2022 November — Question 5 10 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2022
Session	November
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Particular solution with initial conditions
Difficulty	Standard +0.8 This is a second-order linear ODE requiring both complementary function (solving auxiliary equation with complex roots) and particular integral (trying polynomial form with undetermined coefficients), followed by applying two initial conditions. While methodical, it involves multiple techniques and careful algebra, placing it moderately above average difficulty for Further Maths students.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

5 Find the particular solution of the differential equation $$2 \frac { d ^ { 2 } y } { d x ^ { 2 } } + 2 \frac { d y } { d x } + y = 4 x ^ { 2 } + 3 x + 3$$ given that, when $x = 0 , y = \frac { d y } { d x } = 0$.

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Question 5:

Answer	Marks	Guidance
Answer	Marks	Guidance
$2m^2 + 2m + 1 = 0 \Rightarrow m = -\frac{1}{2} \pm \frac{1}{2}i$	M1	Correct auxiliary equation
$y = e^{-\frac{1}{2}x}(A\cos\frac{1}{2}x + B\sin\frac{1}{2}x)$	A1	Complementary function. Accept "$y=$" missing
$y = px^2 + qx + r$ leading to $y' = 2px + q$ leading to $y'' = 2p$	B1	Particular integral and its derivatives
$p = 4$, $4p + q = 3$, $4p + 2q + r = 3$	M1	Substitutes and equates coefficients
$q = -13$, $r = 13$	A1
$y = e^{-\frac{1}{2}x}(A\cos\frac{1}{2}x + B\sin\frac{1}{2}x) + 4x^2 - 13x + 13$	A1	General solution
$y' = e^{-\frac{1}{2}x}(-\frac{1}{2}A\sin\frac{1}{2}x + \frac{1}{2}B\cos\frac{1}{2}x) - \frac{1}{2}e^{-\frac{1}{2}x}(A\cos\frac{1}{2}x + B\sin\frac{1}{2}x) + 8x - 13$	M1	Full use of product rule. $y$ must be of form shown in general solution above with $p \neq 0$
$A + 13 = 0$, $\frac{1}{2}B - \frac{1}{2}A - 13 = 0$, $A = -13$, $B = 13$	M1 A1	Uses initial conditions to derive two linear equations in two unknowns
$y = e^{-\frac{1}{2}x}(-13\cos\frac{1}{2}x + 13\sin\frac{1}{2}x) + 4x^2 - 13x + 13$	A1

## Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2m^2 + 2m + 1 = 0 \Rightarrow m = -\frac{1}{2} \pm \frac{1}{2}i$ | M1 | Correct auxiliary equation |
| $y = e^{-\frac{1}{2}x}(A\cos\frac{1}{2}x + B\sin\frac{1}{2}x)$ | A1 | Complementary function. Accept "$y=$" missing |
| $y = px^2 + qx + r$ leading to $y' = 2px + q$ leading to $y'' = 2p$ | B1 | Particular integral and its derivatives |
| $p = 4$, $4p + q = 3$, $4p + 2q + r = 3$ | M1 | Substitutes and equates coefficients |
| $q = -13$, $r = 13$ | A1 | |
| $y = e^{-\frac{1}{2}x}(A\cos\frac{1}{2}x + B\sin\frac{1}{2}x) + 4x^2 - 13x + 13$ | A1 | General solution |
| $y' = e^{-\frac{1}{2}x}(-\frac{1}{2}A\sin\frac{1}{2}x + \frac{1}{2}B\cos\frac{1}{2}x) - \frac{1}{2}e^{-\frac{1}{2}x}(A\cos\frac{1}{2}x + B\sin\frac{1}{2}x) + 8x - 13$ | M1 | Full use of product rule. $y$ must be of form shown in general solution above with $p \neq 0$ |
| $A + 13 = 0$, $\frac{1}{2}B - \frac{1}{2}A - 13 = 0$, $A = -13$, $B = 13$ | M1 A1 | Uses initial conditions to derive two linear equations in two unknowns |
| $y = e^{-\frac{1}{2}x}(-13\cos\frac{1}{2}x + 13\sin\frac{1}{2}x) + 4x^2 - 13x + 13$ | A1 | |

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5 Find the particular solution of the differential equation

$$2 \frac { d ^ { 2 } y } { d x ^ { 2 } } + 2 \frac { d y } { d x } + y = 4 x ^ { 2 } + 3 x + 3$$

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

\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q5 [10]}}

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Q1 5 Q2 7 Q3 6 Q4 12 Q5 10 Q6 11 Q7 10 Q8 14