Question 8 - A-Level Maths

CAIE FP1 2011 June — Question 8 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2011
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Asymptotic behavior for large values
Difficulty	Standard +0.8 This is a standard second-order linear differential equation with constant coefficients and a sinusoidal forcing term, requiring the auxiliary equation method, particular integral by undetermined coefficients, application of initial conditions, and analysis of asymptotic behavior. While methodical, it involves multiple techniques and the damped oscillation analysis for large t requires understanding of complementary function decay, placing it moderately above average difficulty.
Spec	4.10e Second order non-homogeneous: complementary + particular integral 4.10g Damped oscillations: model and interpret

8 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 10 \sin t$$ Find the particular solution, given that $x = 5$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 2$ when $t = 0$. State an approximate solution for large positive values of $t$.

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Question 8:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$m^2 + 2m + 5 = 0 \Rightarrow m = -1 \pm 2i$	M1	Forms and solves AQE
CF: $e^{-t}(A\cos 2t + B\sin 2t)$ (OE)	A1	States CF
PI: $x = p\cos t + q\sin t \Rightarrow \dot{x} = -p\sin t + q\cos t$	M1	States form for PI
$\Rightarrow \ddot{x} = -p\cos t - q\sin t$
$-p\cos t - q\sin t - 2p\sin t + 2q\cos t + 5p\cos t + 5q\sin t = 10\sin t$		Substitutes in equation
$4p + 2q = 0$ and $-2p + 4q = 10$	M1	Obtains values for $p$ and $q$ by comparing coefficients
$\Rightarrow p = -1,\; q = 2$	A1
GS: $x = e^{-t}(A\cos 2t + B\sin 2t) + 2\sin t - \cos t$ (OE)	A1	States GS. Part total: 6
$t=0,\; x=5 \Rightarrow A = 6$	B1	Uses initial conditions
$\dot{x} = -e^{-t}(A\cos 2t + B\sin 2t) + e^{-t}(-2A\sin 2t + 2B\cos 2t) + 2\cos t + \sin t$	M1
$2 = -6 + 2B + 2 \Rightarrow B = 3$	A1
$x = e^{-t}(6\cos 2t + 3\sin 2t) + 2\sin t - \cos t$ (OE)	A1	States particular solution. Part total: 4
As $t\to\infty$, $x \approx 2\sin t - \cos t$	B1	Gives approximate solution. Final mark independent of $A$ and $B$. Part total: 1
Total: [11]

## Question 8:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $m^2 + 2m + 5 = 0 \Rightarrow m = -1 \pm 2i$ | M1 | Forms and solves AQE |
| CF: $e^{-t}(A\cos 2t + B\sin 2t)$ (OE) | A1 | States CF |
| PI: $x = p\cos t + q\sin t \Rightarrow \dot{x} = -p\sin t + q\cos t$ | M1 | States form for PI |
| $\Rightarrow \ddot{x} = -p\cos t - q\sin t$ | | |
| $-p\cos t - q\sin t - 2p\sin t + 2q\cos t + 5p\cos t + 5q\sin t = 10\sin t$ | | Substitutes in equation |
| $4p + 2q = 0$ and $-2p + 4q = 10$ | M1 | Obtains values for $p$ and $q$ by comparing coefficients |
| $\Rightarrow p = -1,\; q = 2$ | A1 | |
| GS: $x = e^{-t}(A\cos 2t + B\sin 2t) + 2\sin t - \cos t$ (OE) | A1 | States GS. **Part total: 6** |
| $t=0,\; x=5 \Rightarrow A = 6$ | B1 | Uses initial conditions |
| $\dot{x} = -e^{-t}(A\cos 2t + B\sin 2t) + e^{-t}(-2A\sin 2t + 2B\cos 2t) + 2\cos t + \sin t$ | M1 | |
| $2 = -6 + 2B + 2 \Rightarrow B = 3$ | A1 | |
| $x = e^{-t}(6\cos 2t + 3\sin 2t) + 2\sin t - \cos t$ (OE) | A1 | States particular solution. **Part total: 4** |
| As $t\to\infty$, $x \approx 2\sin t - \cos t$ | B1 | Gives approximate solution. Final mark independent of $A$ and $B$. **Part total: 1** |
| | | **Total: [11]** |

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8 Find the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 10 \sin t$$

Find the particular solution, given that $x = 5$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 2$ when $t = 0$.

State an approximate solution for large positive values of $t$.

\hfill \mbox{\textit{CAIE FP1 2011 Q8 [11]}}

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Q1 5 Q2 5 Q3 6 Q4 8 Q5 8 Q6 8 Q7 11 Q8 11 Q9 11 Q10 13 Q11 EITHER Q11 OR