Question 8 - A-Level Maths

CAIE FP1 2008 November — Question 8 9 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2008
Session	November
Marks	9
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 requiring finding both complementary function (complex roots from auxiliary equation) and particular integral (polynomial form), then applying two initial conditions to find constants. While systematic, it involves multiple techniques and careful algebra, making it moderately challenging for Further Maths students.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

8 Find $y$ in terms of $t$, given that $$5 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 5 y = 15 + 12 t + 5 t ^ { 2 }$$ and that $y = \frac { \mathrm { d } y } { \mathrm {~d} t } = 0$ when $t = 0$.

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Answer	Marks
AQE has roots $-3/5 \pm (4/5)i$	M1
CF: $e^{-3t/5}[A\cos(4t/5) + B\sin(4t/5)]$	A1
PI $= af^2 + bt + c \Rightarrow 10a + 6(2at + b) + 5(at^2 + bt + c) \equiv 5t^2 + 12t + 15$	M1
$5a = 5, \quad 12a + 5b = 12, \quad 10a + 6b + 5c = 15$	A1
$\Rightarrow a = 1, \quad b = 0, \quad c = 1$	A1
GS: $y = e^{-3t/5}[A\cos(4t/5) + B\sin(4t/5)] + t^2 + 1$	A1
$y(0) = 0 \Rightarrow 0 = A + 1 \Rightarrow A = -1$	B1
$\dot{y} = -(3/5 e^{-3t/5})[A\cos(4t/5) + B\sin(4t/5)] + e^{-3t/5}[(-4A/5)\sin(4t/5) + (4B/5)\cos(4t/5)] + 2t$
$\Rightarrow \dot{y}(0) = 0 \Rightarrow -3A/5 + 4B/5 = 0$	M1
$\Rightarrow B = -3/4 \Rightarrow y = -(1/4)e^{-3t/5}[(4\cos(4t/5) + 3\sin(4t/5)] + t^2 + 1$	A1
or $-1.25\cos(0.8t - 0.64)$
or $1.25\cos(0.8t + 2.50)$ etc.

AQE has roots $-3/5 \pm (4/5)i$ | M1 |

CF: $e^{-3t/5}[A\cos(4t/5) + B\sin(4t/5)]$ | A1 |

PI $= af^2 + bt + c \Rightarrow 10a + 6(2at + b) + 5(at^2 + bt + c) \equiv 5t^2 + 12t + 15$ | M1 |

$5a = 5, \quad 12a + 5b = 12, \quad 10a + 6b + 5c = 15$ | A1 |

$\Rightarrow a = 1, \quad b = 0, \quad c = 1$ | A1 |

GS: $y = e^{-3t/5}[A\cos(4t/5) + B\sin(4t/5)] + t^2 + 1$ | A1 |

$y(0) = 0 \Rightarrow 0 = A + 1 \Rightarrow A = -1$ | B1 |

$\dot{y} = -(3/5 e^{-3t/5})[A\cos(4t/5) + B\sin(4t/5)] + e^{-3t/5}[(-4A/5)\sin(4t/5) + (4B/5)\cos(4t/5)] + 2t$ | |

$\Rightarrow \dot{y}(0) = 0 \Rightarrow -3A/5 + 4B/5 = 0$ | M1 |

$\Rightarrow B = -3/4 \Rightarrow y = -(1/4)e^{-3t/5}[(4\cos(4t/5) + 3\sin(4t/5)] + t^2 + 1$ | A1 |

or $-1.25\cos(0.8t - 0.64)$ | |

or $1.25\cos(0.8t + 2.50)$ etc. | |

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8 Find $y$ in terms of $t$, given that

$$5 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 5 y = 15 + 12 t + 5 t ^ { 2 }$$

and that $y = \frac { \mathrm { d } y } { \mathrm {~d} t } = 0$ when $t = 0$.

\hfill \mbox{\textit{CAIE FP1 2008 Q8 [9]}}

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Q1 5 Q2 6 Q3 6 Q4 6 Q5 7 Q6 7 Q7 8 Q8 9 Q9 10 Q10 10 Q11 12 Q12 EITHER Q12 OR

Answer	Marks
AQE has roots \(-3/5 \pm (4/5)i\)	M1
CF: \(e^{-3t/5}[A\cos(4t/5) + B\sin(4t/5)]\)	A1
PI \(= af^2 + bt + c \Rightarrow 10a + 6(2at + b) + 5(at^2 + bt + c) \equiv 5t^2 + 12t + 15\)	M1
\(5a = 5, \quad 12a + 5b = 12, \quad 10a + 6b + 5c = 15\)	A1
\(\Rightarrow a = 1, \quad b = 0, \quad c = 1\)	A1
GS: \(y = e^{-3t/5}[A\cos(4t/5) + B\sin(4t/5)] + t^2 + 1\)	A1
\(y(0) = 0 \Rightarrow 0 = A + 1 \Rightarrow A = -1\)	B1
\(\dot{y} = -(3/5 e^{-3t/5})[A\cos(4t/5) + B\sin(4t/5)] + e^{-3t/5}[(-4A/5)\sin(4t/5) + (4B/5)\cos(4t/5)] + 2t\)
\(\Rightarrow \dot{y}(0) = 0 \Rightarrow -3A/5 + 4B/5 = 0\)	M1
\(\Rightarrow B = -3/4 \Rightarrow y = -(1/4)e^{-3t/5}[(4\cos(4t/5) + 3\sin(4t/5)] + t^2 + 1\)	A1
or \(-1.25\cos(0.8t - 0.64)\)
or \(1.25\cos(0.8t + 2.50)\) etc.