CAIE Further Paper 2 2023 November — Question 4 10 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeParticular solution with initial conditions
DifficultyStandard +0.3 This is a standard second-order linear differential equation with constant coefficients and polynomial forcing term. Students follow a routine procedure: find complementary function (complex roots requiring exponential-trigonometric form), find particular integral (polynomial trial solution with undetermined coefficients), then apply initial conditions. While it requires multiple steps and careful algebra, it's a textbook exercise with no novel insight required, making it slightly easier than average for Further Maths.
Spec4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral

Find the particular solution of the differential equation $$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 3y = 27x^2,$$ given that, when \(x = 0\), \(y = 2\) and \(\frac{dy}{dx} = -8\). [10]

Question 4:
AnswerMarks Guidance
4m2 +2m+3=0 M1
[y=]e−x( )
AnswerMarks Guidance
Acos 2x+Bsin 2xA1 Complementary function. Allow with “y=” missing.
y= px2 +qx+r y'=2px+q y''=2pB1 Particular integral and its derivatives.
3p=27 3q+4p=0 2p+2q+3r=0M1 Substitutes and equates coefficients.
QuestionAnswer Marks
4p=9 q=−12 r=2 A1
y=e−x( )
AnswerMarks Guidance
Acos 2x+Bsin 2x +9x2 −12x+2A1 General solution. Must have “y=”.
y'=e−x( ) −e−x( )
AnswerMarks Guidance
− 2Asin 2x+ 2Bcos 2x Acos 2x+Bsin 2x +18x−12M1* Differentiates. Must use product rule.
A+2=2 2B−A−12=−8
AnswerMarks Guidance
 A=0,B=2 2DM1
A1Uses initial conditions
y=2 2e−xsin 2x+9x2 −12x+2A1 Must have “y=”.
10
AnswerMarks Guidance
QuestionAnswer Marks
Question 4:
4 | m2 +2m+3=0 | M1 | Auxiliary equation.
[y=]e−x( )
Acos 2x+Bsin 2x | A1 | Complementary function. Allow with “y=” missing.
y= px2 +qx+r y'=2px+q y''=2p | B1 | Particular integral and its derivatives.
3p=27 3q+4p=0 2p+2q+3r=0 | M1 | Substitutes and equates coefficients.
Question | Answer | Marks | Guidance
4 | p=9 q=−12 r=2 | A1
y=e−x( )
Acos 2x+Bsin 2x +9x2 −12x+2 | A1 | General solution. Must have “y=”.
y'=e−x( ) −e−x( )
− 2Asin 2x+ 2Bcos 2x Acos 2x+Bsin 2x +18x−12 | M1* | Differentiates. Must use product rule.
A+2=2 2B−A−12=−8
 A=0,B=2 2 | DM1
A1 | Uses initial conditions
y=2 2e−xsin 2x+9x2 −12x+2 | A1 | Must have “y=”.
10
Question | Answer | Marks | Guidance
Find the particular solution of the differential equation
$$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 3y = 27x^2,$$
given that, when $x = 0$, $y = 2$ and $\frac{dy}{dx} = -8$. [10]

\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q4 [10]}}