Question 3 - A-Level Maths

OCR FP3 2012 June — Question 3 9 marks

Exam Board	OCR
Module	FP3 (Further Pure Mathematics 3)
Year	2012
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 +0.3 This is a standard integrating factor question from Further Maths requiring identification of P(x) = cot x, finding the integrating factor sin x, integrating 2x sin x by parts, and applying the initial condition. While it involves multiple steps and integration by parts, it follows a completely routine procedure with no novel insight required. Being Further Maths places it slightly above average difficulty, but it remains a textbook example of the method.
Spec	4.10c Integrating factor: first order equations

3 Find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + y \cot x = 2 x$$ for which $y = 2$ when $x = \frac { 1 } { 6 } \pi$. Give your answer in the form $y = \mathrm { f } ( x )$.

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Question 3:

Answer	Marks	Guidance
Answer	Marks	Guidance
Integrating factor $= e^{\int\cot x\,dx} = e^{\ln\sin x} = \sin x$	M1	For IF $= e^{\pm\ln\sin x}$ OR $e^{\pm\ln\cos x}$
A1	For simplified IF
$\Rightarrow \frac{d}{dx}(y\sin x) = 2x\sin x$	M1	For $\frac{d}{dx}(y\cdot\text{their IF}) = 2x\cdot\text{their IF}$
$\Rightarrow y\sin x = -2x\cos x + \int 2\cos x\,dx$	M1*	For attempt to integrate RHS using parts for $\int x\begin{cases}\sin x\\\cos x\end{cases}dx$. Must use $u=(2)x$
A1	For correct RHS 1st stage
$\Rightarrow y\sin x = -2x\cos x + 2\sin x\,(+c)$	A1	oe
$\left(\frac{1}{6}\pi, 2\right)\Rightarrow c = \frac{1}{6}\pi\sqrt{3}$	M1dep*	For substituting $\left(\frac{1}{6}\pi,2\right)$ into their GS (with $c$). $c=0.907$
A1FT	For correctly finding $c$ (FT from GS)
$\Rightarrow y = -2x\cot x + 2 + \frac{1}{6}\pi\sqrt{3}\operatorname{cosec} x$	A1	For correct solution AEF of standard notation $y=f(x)$

# Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrating factor $= e^{\int\cot x\,dx} = e^{\ln\sin x} = \sin x$ | M1 | For IF $= e^{\pm\ln\sin x}$ OR $e^{\pm\ln\cos x}$ |
| | A1 | For simplified IF |
| $\Rightarrow \frac{d}{dx}(y\sin x) = 2x\sin x$ | M1 | For $\frac{d}{dx}(y\cdot\text{their IF}) = 2x\cdot\text{their IF}$ |
| $\Rightarrow y\sin x = -2x\cos x + \int 2\cos x\,dx$ | M1* | For attempt to integrate RHS using parts for $\int x\begin{cases}\sin x\\\cos x\end{cases}dx$. Must use $u=(2)x$ |
| | A1 | For correct RHS 1st stage |
| $\Rightarrow y\sin x = -2x\cos x + 2\sin x\,(+c)$ | A1 | oe |
| $\left(\frac{1}{6}\pi, 2\right)\Rightarrow c = \frac{1}{6}\pi\sqrt{3}$ | M1dep* | For substituting $\left(\frac{1}{6}\pi,2\right)$ into their GS (with $c$). $c=0.907$ |
| | A1FT | For correctly finding $c$ (FT from GS) |
| $\Rightarrow y = -2x\cot x + 2 + \frac{1}{6}\pi\sqrt{3}\operatorname{cosec} x$ | A1 | For correct solution AEF of standard notation $y=f(x)$ |

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3 Find the solution of the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + y \cot x = 2 x$$

for which $y = 2$ when $x = \frac { 1 } { 6 } \pi$. Give your answer in the form $y = \mathrm { f } ( x )$.

\hfill \mbox{\textit{OCR FP3 2012 Q3 [9]}}

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