CAIE FP1 2010 June — Question 8 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeAsymptotic behavior for large values
DifficultyChallenging +1.2 This is a standard second-order linear differential equation with constant coefficients and a particular integral involving trigonometric functions. While it requires multiple techniques (auxiliary equation, particular integral by undetermined coefficients, and understanding that complementary function terms decay for large x), these are all routine Further Maths procedures. The asymptotic behavior follows directly from recognizing exponential decay, making this a straightforward multi-step question slightly above average difficulty due to the Further Maths context and multiple parts.
Spec1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc4.10e Second order non-homogeneous: complementary + particular integral

8 Obtain the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 10 \sin 3 x - 20 \cos 3 x$$ Show that, for large positive \(x\) and independently of the initial conditions, $$y \approx R \sin ( 3 x + \phi )$$ where the constants \(R\) and \(\phi\), such that \(R > 0\) and \(0 < \phi < 2 \pi\), are to be determined correct to 2 decimal places.

AnswerMarks Guidance
Complementary function is \(Ae^{-x} + Be^{-4x}\)M1A1
Particular integral of form \(P \sin 3x + Q \cos 3x\), so that
\(-5P + 15Q = 10\) and \(15P - 5Q = -20\)M1
\(\Rightarrow P = -7/5, Q = -1/5\)A1
General solution is \(y = Ae^{-x} + Be^{-4x} - 1.4 \sin 3x - 0.2 \cos 3x\)A1 [5]
\(Ae^{-x} + Be^{-4x} \to 0\) as \(x \to +\infty\)M1
\(\tan \phi = \frac{1}{7}\)M1
\(R = \sqrt{2} = 1.41, \phi = \pi + \arctan(1/7) = 3.28\)B1A1
Accept \(R = \sqrt{2}\), but must be positive.[4]
Complementary function is $Ae^{-x} + Be^{-4x}$ | M1A1 |

Particular integral of form $P \sin 3x + Q \cos 3x$, so that | |

$-5P + 15Q = 10$ and $15P - 5Q = -20$ | M1 |

$\Rightarrow P = -7/5, Q = -1/5$ | A1 |

General solution is $y = Ae^{-x} + Be^{-4x} - 1.4 \sin 3x - 0.2 \cos 3x$ | A1 | [5] |

$Ae^{-x} + Be^{-4x} \to 0$ as $x \to +\infty$ | M1 |

$\tan \phi = \frac{1}{7}$ | M1 |

$R = \sqrt{2} = 1.41, \phi = \pi + \arctan(1/7) = 3.28$ | B1A1 |

Accept $R = \sqrt{2}$, but must be positive. | [4] |

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8 Obtain the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 10 \sin 3 x - 20 \cos 3 x$$

Show that, for large positive $x$ and independently of the initial conditions,

$$y \approx R \sin ( 3 x + \phi )$$

where the constants $R$ and $\phi$, such that $R > 0$ and $0 < \phi < 2 \pi$, are to be determined correct to 2 decimal places.

\hfill \mbox{\textit{CAIE FP1 2010 Q8 [9]}}