Question 4 - A-Level Maths

CAIE Further Paper 2 2023 November — Question 4 9 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2023
Session	November
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Standard linear first order - constant coefficients
Difficulty	Standard +0.3 This is a standard linear first-order differential equation requiring the integrating factor method with a straightforward initial condition. While it's a Further Maths topic (slightly elevating difficulty), the question follows a completely routine procedure: identify P(x)=3, find integrating factor e^(3x), multiply through, integrate sin(x)e^(3x) by parts, and apply the initial condition. The integration by parts is mechanical rather than insightful, making this easier than average overall.
Spec	4.10c Integrating factor: first order equations

4 Find the solution of the differential equation $$\frac { d y } { d x } + 3 y = \sin x$$ for which $y = 1$ when $x = 0$. Give your answer in the form $y = f ( x )$.

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Question 4:

Answer	Marks	Guidance
Answer	Marks	Guidance
$e^{\int 3dx} = e^{3x}$	M1 A1	Finds integrating factor
$\frac{d}{dx}(ye^{3x}) = e^{3x}\sin x$	M1	Correct form on LHS and attempt to integrate RHS
Either: $\int e^{3x}\sin x\,dx = -e^{3x}\cos x + 3\int e^{3x}\cos x\,dx$ Or: $\int e^{3x}\sin x\,dx = \frac{1}{3}e^{3x}\sin x - \frac{1}{3}\int e^{3x}\cos x\,dx$	M1 A1	Integrates by parts once or uses $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$
Either: $\int e^{3x}\sin x\,dx = -e^{3x}\cos x + 3\!\left(e^{3x}\sin x - 3\int e^{3x}\sin x\,dx\right)$ Or: $\int e^{3x}\sin x\,dx = \frac{1}{3}e^{3x}\sin x - \frac{1}{3}\!\left(\frac{1}{3}e^{3x}\cos x + \frac{1}{3}\int e^{3x}\sin x\,dx\right)$	M1	Integrates by parts again or substitutes $\frac{e^{ix}-e^{-ix}}{2i}=\sin x$ and $\frac{e^{ix}+e^{-ix}}{2}=\cos x$
$ye^{3x} = \frac{1}{10}e^{3x}(3\sin x - \cos x) + C$	A1	Must not see i
$1 = -\frac{1}{10} + C$	M1	Finds $C$. Substitutes into their expression (must be integrated)
$y = \frac{3}{10}\sin x - \frac{1}{10}\cos x + \frac{11}{10}e^{-3x}$	A1	Divides through by coefficient of $y$
Total: 9

## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{\int 3dx} = e^{3x}$ | M1 A1 | Finds integrating factor |
| $\frac{d}{dx}(ye^{3x}) = e^{3x}\sin x$ | M1 | Correct form on LHS and attempt to integrate RHS |
| **Either:** $\int e^{3x}\sin x\,dx = -e^{3x}\cos x + 3\int e^{3x}\cos x\,dx$ **Or:** $\int e^{3x}\sin x\,dx = \frac{1}{3}e^{3x}\sin x - \frac{1}{3}\int e^{3x}\cos x\,dx$ | M1 A1 | Integrates by parts once or uses $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ |
| **Either:** $\int e^{3x}\sin x\,dx = -e^{3x}\cos x + 3\!\left(e^{3x}\sin x - 3\int e^{3x}\sin x\,dx\right)$ **Or:** $\int e^{3x}\sin x\,dx = \frac{1}{3}e^{3x}\sin x - \frac{1}{3}\!\left(\frac{1}{3}e^{3x}\cos x + \frac{1}{3}\int e^{3x}\sin x\,dx\right)$ | M1 | Integrates by parts again or substitutes $\frac{e^{ix}-e^{-ix}}{2i}=\sin x$ and $\frac{e^{ix}+e^{-ix}}{2}=\cos x$ |
| $ye^{3x} = \frac{1}{10}e^{3x}(3\sin x - \cos x) + C$ | A1 | Must not see i |
| $1 = -\frac{1}{10} + C$ | M1 | Finds $C$. Substitutes into their expression (must be integrated) |
| $y = \frac{3}{10}\sin x - \frac{1}{10}\cos x + \frac{11}{10}e^{-3x}$ | A1 | Divides through by coefficient of $y$ |
| **Total: 9** | | |

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4 Find the solution of the differential equation

$$\frac { d y } { d x } + 3 y = \sin x$$

for which $y = 1$ when $x = 0$. Give your answer in the form $y = f ( x )$.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q4 [9]}}

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