Question 4 - A-Level Maths

CAIE Further Paper 2 2021 June — Question 4 9 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2021
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	Challenging +1.2 This is a standard integrating factor problem for Further Maths, requiring division by sin θ to get standard form, finding integrating factor cosec θ (with the integral helpfully provided), then integration and applying initial conditions. The tan(θ/2) term and half-angle substitution add modest algebraic complexity beyond routine examples, but the method is entirely standard with no novel insight required.
Spec	1.08h Integration by substitution 4.10c Integrating factor: first order equations

4 Find the solution of the differential equation $$\sin \theta \frac { d y } { d \theta } + y = \tan \frac { 1 } { 2 } \theta$$ where $0 < \theta < \pi$, given that $y = 1$ when $\theta = \frac { 1 } { 2 } \pi$. Give your answer in the form $y = \mathrm { f } ( \theta )$. [You may use without proof the result that $\int \operatorname { cosec } \theta d \theta = \ln \tan \frac { 1 } { 2 } \theta$.]

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Question 4:

Answer	Marks	Guidance
$\frac{dy}{d\theta} + \cosec\theta\, y = \frac{\tan\frac{1}{2}\theta}{\sin\theta}$	B1	Divides through by $\sin\theta$.
$e^{\int \cosec\theta\, d\theta} = e^{\ln\tan\frac{1}{2}\theta} = \tan\tfrac{1}{2}\theta$	M1 A1	Finds integrating factor.
$\frac{d}{d\theta}\left(y\tan\tfrac{1}{2}\theta\right) = \frac{\tan^2\frac{1}{2}\theta}{\sin\theta} = \tfrac{1}{2}\tan\tfrac{1}{2}\theta\sec^2\tfrac{1}{2}\theta \left[= \frac{\frac{1}{2}\sin\frac{1}{2}\theta}{\cos^3\frac{1}{2}\theta}\right]$	M1	Correct form on LHS and uses an appropriate identity.
$y\tan\tfrac{1}{2}\theta = \tfrac{1}{2}\tan^2\tfrac{1}{2}\theta + C \quad \left(\text{or } \tfrac{1}{2}\sec^2\tfrac{1}{2}\theta + C\right)$	M1 A1	Integrates RHS.
$1 = \frac{1}{2} + C$	M1	Substitutes initial conditions.
$y = \tfrac{1}{2}\left(\tan\tfrac{1}{2}\theta + \cot\tfrac{1}{2}\theta\right) [= \cosec\theta]$	M1 A1	Divides through by their integrating factor.

## Question 4:

$\frac{dy}{d\theta} + \cosec\theta\, y = \frac{\tan\frac{1}{2}\theta}{\sin\theta}$ | **B1** | Divides through by $\sin\theta$.

$e^{\int \cosec\theta\, d\theta} = e^{\ln\tan\frac{1}{2}\theta} = \tan\tfrac{1}{2}\theta$ | **M1 A1** | Finds integrating factor.

$\frac{d}{d\theta}\left(y\tan\tfrac{1}{2}\theta\right) = \frac{\tan^2\frac{1}{2}\theta}{\sin\theta} = \tfrac{1}{2}\tan\tfrac{1}{2}\theta\sec^2\tfrac{1}{2}\theta \left[= \frac{\frac{1}{2}\sin\frac{1}{2}\theta}{\cos^3\frac{1}{2}\theta}\right]$ | **M1** | Correct form on LHS and uses an appropriate identity.

$y\tan\tfrac{1}{2}\theta = \tfrac{1}{2}\tan^2\tfrac{1}{2}\theta + C \quad \left(\text{or } \tfrac{1}{2}\sec^2\tfrac{1}{2}\theta + C\right)$ | **M1 A1** | Integrates RHS.

$1 = \frac{1}{2} + C$ | **M1** | Substitutes initial conditions.

$y = \tfrac{1}{2}\left(\tan\tfrac{1}{2}\theta + \cot\tfrac{1}{2}\theta\right) [= \cosec\theta]$ | **M1 A1** | Divides through by their integrating factor.

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4 Find the solution of the differential equation

$$\sin \theta \frac { d y } { d \theta } + y = \tan \frac { 1 } { 2 } \theta$$

where $0 < \theta < \pi$, given that $y = 1$ when $\theta = \frac { 1 } { 2 } \pi$. Give your answer in the form $y = \mathrm { f } ( \theta )$. [You may use without proof the result that $\int \operatorname { cosec } \theta d \theta = \ln \tan \frac { 1 } { 2 } \theta$.]\\

\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q4 [9]}}

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Q1 5 Q2 7 Q3 10 Q4 9 Q5 10 Q6 11 Q7 10 Q8 13

Answer	Marks	Guidance
\(\frac{dy}{d\theta} + \cosec\theta\, y = \frac{\tan\frac{1}{2}\theta}{\sin\theta}\)	B1	Divides through by \(\sin\theta\).
\(e^{\int \cosec\theta\, d\theta} = e^{\ln\tan\frac{1}{2}\theta} = \tan\tfrac{1}{2}\theta\)	M1 A1	Finds integrating factor.
\(\frac{d}{d\theta}\left(y\tan\tfrac{1}{2}\theta\right) = \frac{\tan^2\frac{1}{2}\theta}{\sin\theta} = \tfrac{1}{2}\tan\tfrac{1}{2}\theta\sec^2\tfrac{1}{2}\theta \left[= \frac{\frac{1}{2}\sin\frac{1}{2}\theta}{\cos^3\frac{1}{2}\theta}\right]\)	M1	Correct form on LHS and uses an appropriate identity.
\(y\tan\tfrac{1}{2}\theta = \tfrac{1}{2}\tan^2\tfrac{1}{2}\theta + C \quad \left(\text{or } \tfrac{1}{2}\sec^2\tfrac{1}{2}\theta + C\right)\)	M1 A1	Integrates RHS.
\(1 = \frac{1}{2} + C\)	M1	Substitutes initial conditions.
\(y = \tfrac{1}{2}\left(\tan\tfrac{1}{2}\theta + \cot\tfrac{1}{2}\theta\right) [= \cosec\theta]\)	M1 A1	Divides through by their integrating factor.