Question 8 - A-Level Maths

CAIE FP1 2012 June — Question 8 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2012
Session	June
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 differential equation with constant coefficients requiring both complementary function (involving complex roots from characteristic equation) and particular integral (using trial solution with exponential), followed by applying two initial conditions to find constants. While systematic, it involves multiple non-trivial steps including complex number manipulation, making it moderately harder than average A-level questions.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

8 Find the particular solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 10 \mathrm { e } ^ { - 2 x }$$ given that $y = 5$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ when $x = 0$.

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Question 8:

Part: Forms AQE and solves

Answer	Marks
$m^2+2m+5=0 \Rightarrow m=-1\pm 2i$	M1A1

Part: Writes CF

Answer	Marks
CF: $y=e^{-x}(A\cos 2x + B\sin 2x)$	A1

Part: Correct form for PI and differentiates twice

Answer	Marks
$y=ke^{-2x} \Rightarrow y'=-2ke^{-2x} \Rightarrow y''=4e^{-2x}$	M1

Part: Substitutes

Answer	Marks
$\Rightarrow 4k-4k+5k=10 \Rightarrow k=2$	M1

Part: Writes PI

Answer	Marks
PI: $y=2e^{-2x}$	A1

Part: Writes GS

Answer	Marks
GS: $y=e^{-x}(A\cos 2x+B\sin 2x)+2e^{-2x}$	A1

Part: Uses y(0)=5 to find A

Answer	Marks
$y=5, x=0 \Rightarrow 5=A+2 \Rightarrow A=3$	B1

Part: Uses y'(0)=1 to find B

$y'=-e^{-x}(A\cos 2x+B\sin 2x)+e^{-x}(-2A\sin 2x+2B\cos 2x)-4e^{-2x}$

Answer	Marks
$y'=1, x=0 \Rightarrow 1=-3+2B-4 \Rightarrow B=4$	M1A1

Part: Writes particular solution

Answer	Marks	Guidance
$\therefore y=e^{-x}(3\cos 2x+4\sin 2x)+2e^{-2x}$	A1 CAO	11 marks

# Question 8:

## Part: Forms AQE and solves
$m^2+2m+5=0 \Rightarrow m=-1\pm 2i$ | M1A1 |

## Part: Writes CF
CF: $y=e^{-x}(A\cos 2x + B\sin 2x)$ | A1 |

## Part: Correct form for PI and differentiates twice
$y=ke^{-2x} \Rightarrow y'=-2ke^{-2x} \Rightarrow y''=4e^{-2x}$ | M1 |

## Part: Substitutes
$\Rightarrow 4k-4k+5k=10 \Rightarrow k=2$ | M1 |

## Part: Writes PI
PI: $y=2e^{-2x}$ | A1 |

## Part: Writes GS
GS: $y=e^{-x}(A\cos 2x+B\sin 2x)+2e^{-2x}$ | A1 |

## Part: Uses y(0)=5 to find A
$y=5, x=0 \Rightarrow 5=A+2 \Rightarrow A=3$ | B1 |

## Part: Uses y'(0)=1 to find B
$y'=-e^{-x}(A\cos 2x+B\sin 2x)+e^{-x}(-2A\sin 2x+2B\cos 2x)-4e^{-2x}$

$y'=1, x=0 \Rightarrow 1=-3+2B-4 \Rightarrow B=4$ | M1A1 |

## Part: Writes particular solution
$\therefore y=e^{-x}(3\cos 2x+4\sin 2x)+2e^{-2x}$ | A1 CAO | 11 marks

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8 Find the particular solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 10 \mathrm { e } ^ { - 2 x }$$

given that $y = 5$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ when $x = 0$.

\hfill \mbox{\textit{CAIE FP1 2012 Q8 [11]}}

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