Question 2 - A-Level Maths

AQA FP3 2015 June — Question 2 9 marks

Exam Board	AQA
Module	FP3 (Further Pure Mathematics 3)
Year	2015
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Standard linear first order - variable coefficients
Difficulty	Standard +0.8 This is a Further Maths FP3 question requiring the integrating factor method with non-trivial trigonometric functions. While the method is standard, finding the integrating factor involves integrating tan x (giving -ln\|cos x\| = ln\|sec x\|), and the subsequent integration of tan³x sec²x requires trigonometric manipulation or substitution. The initial condition adds another step. This is moderately challenging for Further Maths students, above average difficulty overall.
Spec	4.10c Integrating factor: first order equations

2 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \tan x ) y = \tan ^ { 3 } x \sec x$$ given that $y = 2$ when $x = \frac { \pi } { 3 }$.
[0pt] [9 marks]

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Question 2:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Integrating factor: $e^{\int \tan x \, dx} = e^{\ln \sec x} = \sec x$	M1 A1	Correct IF obtained
$\frac{d}{dx}(y \sec x) = \tan^3 x \sec x \cdot \sec x = \tan^3 x \sec^2 x$	M1	Multiply through by IF
$\int \tan^3 x \sec^2 x \, dx = \frac{\tan^4 x}{4}$	M1 A1	Correct integration
$y \sec x = \frac{\tan^4 x}{4} + C$	A1	Correct equation before applying IC
$x = \frac{\pi}{3}$: $2 \sec\frac{\pi}{3} = \frac{\tan^4(\pi/3)}{4} + C \Rightarrow 4 = \frac{81}{4} \cdot \frac{1}{4}... \Rightarrow 4 = \frac{9}{4} + C$	M1	Substituting initial condition
$C = 4 - \frac{9}{4} = \frac{7}{4}$	A1	Correct value of $C$
$y = \frac{\tan^4 x}{4}\cos x + \frac{7}{4}\cos x$	A1	Correct final answer

# Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrating factor: $e^{\int \tan x \, dx} = e^{\ln \sec x} = \sec x$ | M1 A1 | Correct IF obtained |
| $\frac{d}{dx}(y \sec x) = \tan^3 x \sec x \cdot \sec x = \tan^3 x \sec^2 x$ | M1 | Multiply through by IF |
| $\int \tan^3 x \sec^2 x \, dx = \frac{\tan^4 x}{4}$ | M1 A1 | Correct integration |
| $y \sec x = \frac{\tan^4 x}{4} + C$ | A1 | Correct equation before applying IC |
| $x = \frac{\pi}{3}$: $2 \sec\frac{\pi}{3} = \frac{\tan^4(\pi/3)}{4} + C \Rightarrow 4 = \frac{81}{4} \cdot \frac{1}{4}... \Rightarrow 4 = \frac{9}{4} + C$ | M1 | Substituting initial condition |
| $C = 4 - \frac{9}{4} = \frac{7}{4}$ | A1 | Correct value of $C$ |
| $y = \frac{\tan^4 x}{4}\cos x + \frac{7}{4}\cos x$ | A1 | Correct final answer |

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2 By using an integrating factor, find the solution of the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \tan x ) y = \tan ^ { 3 } x \sec x$$

given that $y = 2$ when $x = \frac { \pi } { 3 }$.\\[0pt]
[9 marks]

\hfill \mbox{\textit{AQA FP3 2015 Q2 [9]}}

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Q1 5 Q2 9 Q3 8 Q4 7 Q5 11 Q6 17 Q7 18