Question 8 - A-Level Maths

CAIE FP1 2019 June — Question 8 10 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2019
Session	June
Marks	10
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 standard second-order linear ODE with constant coefficients requiring the complementary function (repeated root from auxiliary equation 9m²+6m+1=0), particular integral (trial solution for sin t forcing term), and application of two initial conditions. While methodical, it involves multiple techniques and careful algebra, placing it moderately above average difficulty for Further Maths students.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

8 Find the particular solution of the differential equation $$9 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 50 \sin t$$ given that when $t = 0 , x = 0$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0$.

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Question 8:

Answer	Marks	Guidance
Answer	Marks	Guidance
$9u^2 + 6u + 1 = 0 \Rightarrow (3u+1)^2 = 0$	M1	Auxiliary equation
CF: $x = (At+B)e^{-\frac{1}{3}t}$	A1	States CF
$x = p\sin t + q\cos t \Rightarrow \dot{x} = p\cos t - q\sin t \Rightarrow \ddot{x} = -p\sin t - q\cos t$	M1	Forms PI and differentiates
$-9p - 6q + p = 50$; $-9q + 6p + q = 0$	M1	Substitutes
$q = -3,\ p = -4$	A1
$x = (At+B)e^{-\frac{1}{3}t} - 4\sin t - 3\cos t$	A1	States general solution. FT if correct form of CF/PI
$\dot{x} = -\frac{1}{3}(At+B)e^{-\frac{1}{3}t} + Ae^{-\frac{1}{3}t} - 4\cos t + 3\sin t$	M1	Differentiates and forms equations using initial conditions
$B = 3$	B1
$-\frac{1}{3}B + A - 4 = 0 \Rightarrow A = 5$	A1
$x = (5t+3)e^{-\frac{1}{3}t} - 4\sin t - 3\cos t$	A1	States PS
Total: 10

## Question 8:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $9u^2 + 6u + 1 = 0 \Rightarrow (3u+1)^2 = 0$ | M1 | Auxiliary equation |
| CF: $x = (At+B)e^{-\frac{1}{3}t}$ | A1 | States CF |
| $x = p\sin t + q\cos t \Rightarrow \dot{x} = p\cos t - q\sin t \Rightarrow \ddot{x} = -p\sin t - q\cos t$ | M1 | Forms PI and differentiates |
| $-9p - 6q + p = 50$; $-9q + 6p + q = 0$ | M1 | Substitutes |
| $q = -3,\ p = -4$ | A1 | |
| $x = (At+B)e^{-\frac{1}{3}t} - 4\sin t - 3\cos t$ | A1 | States general solution. FT if correct form of CF/PI |
| $\dot{x} = -\frac{1}{3}(At+B)e^{-\frac{1}{3}t} + Ae^{-\frac{1}{3}t} - 4\cos t + 3\sin t$ | M1 | Differentiates and forms equations using initial conditions |
| $B = 3$ | B1 | |
| $-\frac{1}{3}B + A - 4 = 0 \Rightarrow A = 5$ | A1 | |
| $x = (5t+3)e^{-\frac{1}{3}t} - 4\sin t - 3\cos t$ | A1 | States PS |
| **Total: 10** | | |

Show LaTeX source

8 Find the particular solution of the differential equation

$$9 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 50 \sin t$$

given that when $t = 0 , x = 0$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0$.\\

\hfill \mbox{\textit{CAIE FP1 2019 Q8 [10]}}

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