Question 6 - A-Level Maths

CAIE FP1 2011 November — Question 6 8 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2011
Session	November
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Sketch or describe solution behavior
Difficulty	Standard +0.8 This is a standard second-order linear ODE with constant coefficients requiring both complementary function (repeated root case) and particular integral (using undetermined coefficients for sin 2t), followed by analysis of long-term behavior. While methodical, it requires multiple techniques, careful algebra with the particular integral, and conceptual understanding of how exponential decay dominates oscillatory terms as t→∞. Slightly above average for Further Maths due to the multi-step nature and behavior analysis.
Spec	4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

6 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 4 x = \sin 2 t$$ Describe the behaviour of $x$ as $t \rightarrow \infty$, justifying your answer.

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Question 6:

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
$m=-2$: $m^2+4m+4=0 \Rightarrow (m+2)^2=0$	M1	Forms auxiliary equation and factorises
CF: $Ae^{-2t}+Bte^{-2t}$	A1	States CF
PI: $x = p\sin 2t + q\cos 2t$	M1	States form of PI
$\dot{x}=2p\cos 2t - 2q\sin 2t$; $\ddot{x}=-4p\sin 2t - 4q\cos 2t$		Differentiates twice
$\Rightarrow -8q=1$; $8p=0$ $\Rightarrow q=-\frac{1}{8}$; $p=0$	M1A1	Compares coefficients and solves
GS: $x = Ae^{-2t}+Bte^{-2t}-\frac{1}{8}\cos 2t$	A1	States GS; Part marks: 6
As $t\to\infty$, $e^{-2t}$ and $te^{-2t}\to 0$	B1	Reason
Hence $x$ oscillates. (Accept $x\approx -\frac{1}{8}\cos 2t$)	B1	Behaviour; Part marks: 2; Total: [8]

## Question 6:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $m=-2$: $m^2+4m+4=0 \Rightarrow (m+2)^2=0$ | M1 | Forms auxiliary equation and factorises |
| CF: $Ae^{-2t}+Bte^{-2t}$ | A1 | States CF |
| PI: $x = p\sin 2t + q\cos 2t$ | M1 | States form of PI |
| $\dot{x}=2p\cos 2t - 2q\sin 2t$; $\ddot{x}=-4p\sin 2t - 4q\cos 2t$ | | Differentiates twice |
| $\Rightarrow -8q=1$; $8p=0$ $\Rightarrow q=-\frac{1}{8}$; $p=0$ | M1A1 | Compares coefficients and solves |
| GS: $x = Ae^{-2t}+Bte^{-2t}-\frac{1}{8}\cos 2t$ | A1 | States GS; Part marks: 6 |
| As $t\to\infty$, $e^{-2t}$ and $te^{-2t}\to 0$ | B1 | Reason |
| Hence $x$ oscillates. (Accept $x\approx -\frac{1}{8}\cos 2t$) | B1 | Behaviour; Part marks: 2; **Total: [8]** |

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6 Find the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 4 x = \sin 2 t$$

Describe the behaviour of $x$ as $t \rightarrow \infty$, justifying your answer.

\hfill \mbox{\textit{CAIE FP1 2011 Q6 [8]}}

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