Question 6 - A-Level Maths

CAIE FP1 2016 November — Question 6 9 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2016
Session	November
Marks	9
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 second-order linear ODE with constant coefficients requiring both complementary function (solving auxiliary equation with real roots) and particular integral (using trial solution for sinusoidal forcing). The additional asymptotic behavior question requires understanding that the CF decays exponentially, leaving only the PI for large t. This combines multiple techniques and conceptual understanding beyond routine exercises.
Spec	4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

6 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 7 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 10 x = 116 \sin 2 t$$ State an approximate solution for large positive values of $t$.

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks	Guidance
Answer	Marks	Guidance
$m^2 + 7m + 10 = 0 \Rightarrow (m+2)(m+5) = 0 \Rightarrow m = -2$ or $-5$	M1
CF: $Ae^{-2t} + Be^{-5t}$	A1
PI: $x = p\sin 2t + q\cos 2t \Rightarrow \dot{x} = 2p\cos 2t - 2q\sin 2t \Rightarrow \ddot{x} = -4p\sin 2t - 4q\cos 2t$	M1A1
Substituting: $14p + 6q = 0$ and $6p - 14q = 116$	M1
Solving: $p = 3$, $q = -7 \Rightarrow x = 3\sin 2t - 7\cos 2t$	M1A1
GS: $x = Ae^{-2t} + Be^{-5t} + 3\sin 2t - 7\cos 2t$	A1✓
For large positive values of $t$, $x \approx 3\sin 2t - 7\cos 2t$	B1✓

## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $m^2 + 7m + 10 = 0 \Rightarrow (m+2)(m+5) = 0 \Rightarrow m = -2$ or $-5$ | M1 | |
| CF: $Ae^{-2t} + Be^{-5t}$ | A1 | |
| PI: $x = p\sin 2t + q\cos 2t \Rightarrow \dot{x} = 2p\cos 2t - 2q\sin 2t \Rightarrow \ddot{x} = -4p\sin 2t - 4q\cos 2t$ | M1A1 | |
| Substituting: $14p + 6q = 0$ and $6p - 14q = 116$ | M1 | |
| Solving: $p = 3$, $q = -7 \Rightarrow x = 3\sin 2t - 7\cos 2t$ | M1A1 | |
| GS: $x = Ae^{-2t} + Be^{-5t} + 3\sin 2t - 7\cos 2t$ | A1✓ | |
| For large positive values of $t$, $x \approx 3\sin 2t - 7\cos 2t$ | B1✓ | |

---

Show LaTeX source

6 Find the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 7 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 10 x = 116 \sin 2 t$$

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

\hfill \mbox{\textit{CAIE FP1 2016 Q6 [9]}}

This paper (12 questions)

View full paper

Q1 5 Q2 6 Q3 7 Q4 6 Q5 8 Q6 9 Q7 11 Q8 11 Q9 11 Q10 12 Q11 EITHER Q11 OR