OCR FP3 2009 January — Question 4 9 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2009
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeStandard non-homogeneous with trigonometric RHS
DifficultyStandard +0.8 This is a standard Further Maths second-order linear ODE with constant coefficients and trigonometric forcing term. Students must find the complementary function (solving auxiliary equation with complex roots) and particular integral (using undetermined coefficients with both sin and cos terms). While methodical, it requires multiple techniques and careful algebra, placing it moderately above average difficulty.
Spec4.10e Second order non-homogeneous: complementary + particular integral

4 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 65 \sin 2 x$$

Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(m^2 + 4m + 5 (= 0) \Rightarrow m = \frac{-4 \pm \sqrt{16-20}}{2}\)M1 For attempt to solve correct auxiliary equation
\(= -2 \pm i\)A1 For correct roots
\(CF = e^{-2x}(C\cos x + D\sin x)\)A1\(\sqrt{}\) For correct CF (here or later). f.t. from \(m\). AEtrig but not forms including \(e^{ix}\)
\(PI = p\sin 2x + q\cos 2x\)B1 For stating a trial PI of the correct form
\(y' = 2p\cos 2x - 2q\sin 2x\), \(y'' = -4p\sin 2x - 4q\cos 2x\)M1 For differentiating PI twice and substituting into the DE
\(\cos 2x(-4q + 8p + 5q) + \sin 2x(-4p - 8q + 5p) = 65\sin 2x\)A1 For correct equation
\(8p + q = 0\), \(p - 8q = 65\), \(p = 1\), \(q = -8\)M1 For equating coefficients of \(\cos 2x\) and \(\sin 2x\) and attempting to solve for \(p\) and/or \(q\)
\(PI = \sin 2x - 8\cos 2x\)A1 For correct \(p\) and \(q\)
\(\Rightarrow y = e^{-2x}(C\cos x + D\sin x) + \sin 2x - 8\cos 2x\)B1\(\sqrt{}\) For using \(GS = CF + PI\), with 2 arbitrary constants in CF and none in PI
Total: 9 marks
# Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2 + 4m + 5 (= 0) \Rightarrow m = \frac{-4 \pm \sqrt{16-20}}{2}$ | M1 | For attempt to solve correct auxiliary equation |
| $= -2 \pm i$ | A1 | For correct roots |
| $CF = e^{-2x}(C\cos x + D\sin x)$ | A1$\sqrt{}$ | For correct CF (here or later). f.t. from $m$. **AEtrig** but not forms including $e^{ix}$ |
| $PI = p\sin 2x + q\cos 2x$ | B1 | For stating a trial PI of the correct form |
| $y' = 2p\cos 2x - 2q\sin 2x$, $y'' = -4p\sin 2x - 4q\cos 2x$ | M1 | For differentiating PI twice and substituting into the DE |
| $\cos 2x(-4q + 8p + 5q) + \sin 2x(-4p - 8q + 5p) = 65\sin 2x$ | A1 | For correct equation |
| $8p + q = 0$, $p - 8q = 65$, $p = 1$, $q = -8$ | M1 | For equating coefficients of $\cos 2x$ and $\sin 2x$ and attempting to solve for $p$ and/or $q$ |
| $PI = \sin 2x - 8\cos 2x$ | A1 | For correct $p$ and $q$ |
| $\Rightarrow y = e^{-2x}(C\cos x + D\sin x) + \sin 2x - 8\cos 2x$ | B1$\sqrt{}$ | For using $GS = CF + PI$, with 2 arbitrary constants in CF and none in PI |

**Total: 9 marks**

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4 Find the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 65 \sin 2 x$$

\hfill \mbox{\textit{OCR FP3 2009 Q4 [9]}}