Question 6 - A-Level Maths

CAIE Further Paper 2 2020 November — Question 6 11 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2020
Session	November
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Particular solution with initial conditions
Difficulty	Standard +0.8 This is a second-order linear ODE with constant coefficients requiring both complementary function (solving auxiliary equation with real roots) and particular integral (using trial solution for cosine forcing term, involving algebraic manipulation to find coefficients), followed by applying two initial conditions. While methodical, it requires multiple techniques and careful algebra, making it moderately challenging for Further Maths students.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

6 Find the particular solution of the differential equation $$\frac { d ^ { 2 } x } { d t ^ { 2 } } + 8 \frac { d x } { d t } + 15 x = 102 \cos 3 t$$ given that, when $t = 0 , x = 1$ and $\frac { \mathrm { dx } } { \mathrm { dt } } = 0$.

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Question 6:

Answer	Marks	Guidance
$m^2 + 8m + 15 = 0 \Rightarrow m = -3, -5$	M1	Auxiliary equation
$x = Ae^{-5t} + Be^{-3t}$	A1	Complementary function
$x = p\sin 3t + q\cos 3t \Rightarrow \dot{x} = 3p\cos 3t - 3q\sin 3t \Rightarrow \ddot{x} = -9p\sin 3t - 9q\cos 3t$	M1 A1	Particular integral and its derivatives
$-9p - 24q + 15p = 0 \Rightarrow 6p - 24q = 0$ and $-9q + 24p + 15q = 102 \Rightarrow 4p + q = 17$	M1	Substitutes and equates coefficients
$p = 4,\quad q = 1$	A1
$x = Ae^{-5t} + Be^{-3t} + 4\sin 3t + \cos 3t$	A1
$\dot{x} = -5Ae^{-5t} - 3Be^{-3t} + 12\cos 3t - 3\sin 3t$	M1	Differentiating the correct form
$1 = A + B + 1$ and $0 = -5A - 3B + 12 \Rightarrow A = 6,\ B = -6$	M1 A1	Forms simultaneous equations using initial conditions
$x = 6e^{-5t} - 6e^{-3t} + 4\sin 3t + \cos 3t$	A1

## Question 6:

| $m^2 + 8m + 15 = 0 \Rightarrow m = -3, -5$ | M1 | Auxiliary equation |
|---|---|---|
| $x = Ae^{-5t} + Be^{-3t}$ | A1 | Complementary function |
| $x = p\sin 3t + q\cos 3t \Rightarrow \dot{x} = 3p\cos 3t - 3q\sin 3t \Rightarrow \ddot{x} = -9p\sin 3t - 9q\cos 3t$ | M1 A1 | Particular integral and its derivatives |
| $-9p - 24q + 15p = 0 \Rightarrow 6p - 24q = 0$ and $-9q + 24p + 15q = 102 \Rightarrow 4p + q = 17$ | M1 | Substitutes and equates coefficients |
| $p = 4,\quad q = 1$ | A1 | |
| $x = Ae^{-5t} + Be^{-3t} + 4\sin 3t + \cos 3t$ | A1 | |
| $\dot{x} = -5Ae^{-5t} - 3Be^{-3t} + 12\cos 3t - 3\sin 3t$ | M1 | Differentiating the correct form |
| $1 = A + B + 1$ and $0 = -5A - 3B + 12 \Rightarrow A = 6,\ B = -6$ | M1 A1 | Forms simultaneous equations using initial conditions |
| $x = 6e^{-5t} - 6e^{-3t} + 4\sin 3t + \cos 3t$ | A1 | |

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6 Find the particular solution of the differential equation

$$\frac { d ^ { 2 } x } { d t ^ { 2 } } + 8 \frac { d x } { d t } + 15 x = 102 \cos 3 t$$

given that, when $t = 0 , x = 1$ and $\frac { \mathrm { dx } } { \mathrm { dt } } = 0$.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q6 [11]}}

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Q1 5 Q2 6 Q3 4 Q4 8 Q5 8 Q6 11 Q7 7 Q8 10 Q9 16

Answer	Marks	Guidance
\(m^2 + 8m + 15 = 0 \Rightarrow m = -3, -5\)	M1	Auxiliary equation
\(x = Ae^{-5t} + Be^{-3t}\)	A1	Complementary function
\(x = p\sin 3t + q\cos 3t \Rightarrow \dot{x} = 3p\cos 3t - 3q\sin 3t \Rightarrow \ddot{x} = -9p\sin 3t - 9q\cos 3t\)	M1 A1	Particular integral and its derivatives
\(-9p - 24q + 15p = 0 \Rightarrow 6p - 24q = 0\) and \(-9q + 24p + 15q = 102 \Rightarrow 4p + q = 17\)	M1	Substitutes and equates coefficients
\(p = 4,\quad q = 1\)	A1
\(x = Ae^{-5t} + Be^{-3t} + 4\sin 3t + \cos 3t\)	A1
\(\dot{x} = -5Ae^{-5t} - 3Be^{-3t} + 12\cos 3t - 3\sin 3t\)	M1	Differentiating the correct form
\(1 = A + B + 1\) and \(0 = -5A - 3B + 12 \Rightarrow A = 6,\ B = -6\)	M1 A1	Forms simultaneous equations using initial conditions
\(x = 6e^{-5t} - 6e^{-3t} + 4\sin 3t + \cos 3t\)	A1