OCR FP3 2008 January — Question 5 9 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2008
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeStandard linear first order - variable coefficients
DifficultyStandard +0.8 This is a first-order linear ODE requiring integrating factor method (standard FP3 technique), but the integration of x·sin(2x) requires integration by parts twice, making it more involved than typical examples. The boundary condition application and asymptotic behavior analysis add further steps, placing this above average difficulty for A-level but within expected FP3 range.
Spec1.08h Integration by substitution4.10c Integrating factor: first order equations

  1. Find the general solution of the differential equation $$\frac{dy}{dx} - \frac{y}{x} = \sin 2x,$$ expressing \(y\) in terms of \(x\) in your answer. [6]
In a particular case, it is given that \(y = \frac{2}{\pi}\) when \(x = \frac{1}{4}\pi\).
  1. Find the solution of the differential equation in this case. [2]
  2. Write down a function to which \(y\) approximates when \(x\) is large and positive. [1]

(i)
If \(e^{\int 1 dx} = e^{\ln x} = x\)
OR \(x\frac{dy}{dx} + y = x\sin 2x\)
\(\Rightarrow \frac{d}{dx}(xy) = x\sin 2x\)
\(\Rightarrow xy = \int x\sin 2x(dx)\)
\(xy = -\frac{1}{2}x\cos 2x + \frac{1}{2}\int \cos 2x(dx)\)
\(xy = -\frac{1}{2}x\cos 2x + \frac{1}{4}\sin 2x (+c)\)
\(\Rightarrow y = -\frac{1}{2}\cos 2x + \frac{1}{4x}\sin 2x + \frac{c}{x}\)
AnswerMarks
M1For correct process for finding integrating factor OR for multiplying equation through by \(x\)
A1For writing DE in this form (may be implied)
M1For integration by parts the correct way round
A1For 1st term correct
M1For their 1st term and attempt at integration of \(\frac{\cos x}{\sin}kx\)
A1For correct expression for \(y\) (6 marks total)
(ii)
\(\left(\frac{1}{4}\pi, \frac{\pi}{2}\right) \Rightarrow \frac{2}{\pi} = \frac{1}{\pi} + \frac{4c}{\pi} \Rightarrow c = \frac{1}{4}\)
\(\Rightarrow y = -\frac{1}{2}\cos 2x + \frac{1}{4x}\sin 2x + \frac{1}{4x}\)
AnswerMarks
M1For substituting \(\left(\frac{1}{4}\pi, \frac{\pi}{2}\right)\) in solution
A1For correct solution. Requires \(y = \boxed{\phantom{x}}\) (2 marks)
(iii)
\((y') = -\frac{1}{2}\cos 2x\)
AnswerMarks
B1For correct function AEF f.t. from (ii) (1 mark)
## (i)

If $e^{\int 1 dx} = e^{\ln x} = x$

OR $x\frac{dy}{dx} + y = x\sin 2x$

$\Rightarrow \frac{d}{dx}(xy) = x\sin 2x$

$\Rightarrow xy = \int x\sin 2x(dx)$

$xy = -\frac{1}{2}x\cos 2x + \frac{1}{2}\int \cos 2x(dx)$

$xy = -\frac{1}{2}x\cos 2x + \frac{1}{4}\sin 2x (+c)$

$\Rightarrow y = -\frac{1}{2}\cos 2x + \frac{1}{4x}\sin 2x + \frac{c}{x}$

| M1 | For correct process for finding integrating factor OR for multiplying equation through by $x$ |
| A1 | For writing DE in this form (may be implied) |
| M1 | For integration by parts the correct way round |
| A1 | For 1st term correct |
| M1 | For their 1st term and attempt at integration of $\frac{\cos x}{\sin}kx$ |
| A1 | For correct expression for $y$ (6 marks total) |

## (ii)

$\left(\frac{1}{4}\pi, \frac{\pi}{2}\right) \Rightarrow \frac{2}{\pi} = \frac{1}{\pi} + \frac{4c}{\pi} \Rightarrow c = \frac{1}{4}$

$\Rightarrow y = -\frac{1}{2}\cos 2x + \frac{1}{4x}\sin 2x + \frac{1}{4x}$

| M1 | For substituting $\left(\frac{1}{4}\pi, \frac{\pi}{2}\right)$ in solution |
| A1 | For correct solution. Requires $y = \boxed{\phantom{x}}$ (2 marks) |

## (iii)

$(y') = -\frac{1}{2}\cos 2x$

| B1 | For correct function AEF f.t. from (ii) (1 mark) |

---
\begin{enumerate}[label=(\roman*)]
\item Find the general solution of the differential equation
$$\frac{dy}{dx} - \frac{y}{x} = \sin 2x,$$
expressing $y$ in terms of $x$ in your answer. [6]
\end{enumerate}

In a particular case, it is given that $y = \frac{2}{\pi}$ when $x = \frac{1}{4}\pi$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{1}
\item Find the solution of the differential equation in this case. [2]
\item Write down a function to which $y$ approximates when $x$ is large and positive. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR FP3 2008 Q5 [9]}}