Question 4 - A-Level Maths

CAIE FP1 2014 June — Question 4 6 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2014
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Standard non-homogeneous with trigonometric RHS
Difficulty	Standard +0.8 This is a standard second-order linear non-homogeneous differential equation requiring both complementary function (solving auxiliary equation with complex roots) and particular integral (using undetermined coefficients with trigonometric RHS). While methodical, it involves multiple techniques and careful algebraic manipulation, placing it moderately above average difficulty for A-level, though routine for Further Maths students who have practiced this topic.
Spec	4.10a General/particular solutions: of differential equations 4.10e Second order non-homogeneous: complementary + particular integral

4 Obtain the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 25 x = 195 \sin 2 t$$

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Question 4:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$m^2 - 6m + 25 = 0 \Rightarrow m = 3 \pm 4\text{i}$	M1
CF: $x = e^{3t}(A\cos 4t + B\sin 4t)$	A1
PI: $x - p\sin 2t + q\cos 2t \Rightarrow \dot{x} = 2p\cos 2t - 2q\sin 2t \Rightarrow \ddot{x} = -4p\sin 2t - 4q\cos 2t$	M1
$\Rightarrow 21p + 12q = 195$	M1
$-12p + 21q = 0$
$\Rightarrow p = 7, q = 4$	A1
GS: $x = e^{3t}(A\cos 4t + B\sin 4t) + 7\sin 2t + 4\cos 2t$	A1

Total: [6]

## Question 4:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $m^2 - 6m + 25 = 0 \Rightarrow m = 3 \pm 4\text{i}$ | M1 | |
| CF: $x = e^{3t}(A\cos 4t + B\sin 4t)$ | A1 | |
| PI: $x - p\sin 2t + q\cos 2t \Rightarrow \dot{x} = 2p\cos 2t - 2q\sin 2t \Rightarrow \ddot{x} = -4p\sin 2t - 4q\cos 2t$ | M1 | |
| $\Rightarrow 21p + 12q = 195$ | M1 | |
| $-12p + 21q = 0$ | | |
| $\Rightarrow p = 7, q = 4$ | A1 | |
| GS: $x = e^{3t}(A\cos 4t + B\sin 4t) + 7\sin 2t + 4\cos 2t$ | A1 | |

**Total: [6]**

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4 Obtain the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 25 x = 195 \sin 2 t$$

\hfill \mbox{\textit{CAIE FP1 2014 Q4 [6]}}

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