CAIE Further Paper 2 2024 November — Question 5 10 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeParticular solution with initial conditions
DifficultyStandard +0.8 This is a standard second-order linear ODE with constant coefficients requiring both complementary function (solving auxiliary equation with complex roots) and particular integral (trying y = ax² + bx + c), followed by applying two initial conditions. While methodical, it involves multiple techniques and careful algebra across 10 marks, placing it moderately above average difficulty for Further Maths students.
Spec4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral
Find the particular solution of the differential equation $$3\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = x^2,$$ given that, when \(x = 0\), \(y = \frac{dy}{dx} = 0\). [10]
Question 5:
AnswerMarks Guidance
53m2 +2m+1=0m=−1 2i
3 3M1 Auxiliary equation. M1 for correct type of
roots (complex).
y=e −1 3 x( Acos 2 x+Bsin 2 x )
AnswerMarks Guidance
3 3A1 Complementary function.
y= px2 +qx+r y'=2px+q y''=2pB1 Particular integral and its derivatives.
p=1 4p+q=0 6p+2q+r=0M1 Substitutes and equates coefficients.
q=−4 r=2A1
y=e −1 3 x( Acos 2 x+Bsin 2 x ) +x2 −4x+2
AnswerMarks Guidance
3 3A1FT General solution. Must have ‘y=’. FT on
CF.
y'=e −1 3 x( − 2 Asin 2 x+ 2Bcos 2 x ) −1e −1 3 x( Acos 2 x+Bsin 2 x ) +2x−4
AnswerMarks Guidance
3 3 3 3 3 3 3*M1 Differentiates. GS must be in correct form.
A+2=0 2B−1A−4=0  A=−2,B=5 2
AnswerMarks Guidance
3 3DM1A1 Uses initial conditions.
y=e −1 3 x( −2cos 2 x+5 2sin 2 x ) +x2 −4x+2
AnswerMarks Guidance
3 3A1 Must have ‘ y=’.
10
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
5 | 3m2 +2m+1=0m=−1 2i
3 3 | M1 | Auxiliary equation. M1 for correct type of
roots (complex).
y=e −1 3 x( Acos 2 x+Bsin 2 x )
3 3 | A1 | Complementary function.
y= px2 +qx+r y'=2px+q y''=2p | B1 | Particular integral and its derivatives.
p=1 4p+q=0 6p+2q+r=0 | M1 | Substitutes and equates coefficients.
q=−4 r=2 | A1
y=e −1 3 x( Acos 2 x+Bsin 2 x ) +x2 −4x+2
3 3 | A1FT | General solution. Must have ‘y=’. FT on
CF.
y'=e −1 3 x( − 2 Asin 2 x+ 2Bcos 2 x ) −1e −1 3 x( Acos 2 x+Bsin 2 x ) +2x−4
3 3 3 3 3 3 3 | *M1 | Differentiates. GS must be in correct form.
A+2=0 2B−1A−4=0  A=−2,B=5 2
3 3 | DM1A1 | Uses initial conditions.
y=e −1 3 x( −2cos 2 x+5 2sin 2 x ) +x2 −4x+2
3 3 | A1 | Must have ‘ y=’.
10
Question | Answer | Marks | Guidance
Find the particular solution of the differential equation
$$3\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = x^2,$$
given that, when $x = 0$, $y = \frac{dy}{dx} = 0$. [10]

\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q5 [10]}}
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Q1 4 Q2 7 Q3 7 Q4 10 Q5 10 Q6 13 Q7 10 Q8 14