Question 8 - A-Level Maths

CAIE FP1 2017 June — Question 8 10 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2017
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 both complementary function (repeated root case) and particular integral (polynomial trial solution), followed by applying 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 solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 x = 18 t ^ { 2 } + 6 t + 1$$ given that, when $t = 0 , x = 3$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0$.

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Question 8:

Answer	Marks
$m^2 + 6m + 9 = 0 \Rightarrow m = -3$	M1
CF: $x = Ae^{-3t} + Bte^{-3t}$	A1
$x = pt^2 + qt + r \Rightarrow \dot{x} = 2pt + q$ and $\ddot{x} = 2p$	M1
$\Rightarrow 2p + 6(2pt+q) + 9(pt^2+qt+r) = 18t^2 + 6t + 1$	M1
$\Rightarrow 9pt^2 + (9q+12p)t + (2p+6q+9r) = 18t^2+6t+1$ $\Rightarrow p=2, q=-2, r=1$ whence PI: $2t^2-2t+1$	A1
GS: $x = Ae^{-3t} + Bte^{-3t} + 2t^2 - 2t + 1$	A1FT
$x = 3$ when $t=0 \Rightarrow A = 2$	B1
$\dot{x} = -3Ae^{-3t} - 3Bte^{-3t} + Be^{-3t} + 4t - 2$	M1
$\dot{x} = 0$ when $t=0 \Rightarrow B = 8$	A1
Hence $x = 2e^{-3t} + 8te^{-3t} + 2t^2 - 2t + 1$	A1

Total: 10

## Question 8:

| $m^2 + 6m + 9 = 0 \Rightarrow m = -3$ | M1 | |
|---|---|---|
| CF: $x = Ae^{-3t} + Bte^{-3t}$ | A1 | |
| $x = pt^2 + qt + r \Rightarrow \dot{x} = 2pt + q$ and $\ddot{x} = 2p$ | M1 | |
| $\Rightarrow 2p + 6(2pt+q) + 9(pt^2+qt+r) = 18t^2 + 6t + 1$ | M1 | |
| $\Rightarrow 9pt^2 + (9q+12p)t + (2p+6q+9r) = 18t^2+6t+1$ $\Rightarrow p=2, q=-2, r=1$ whence PI: $2t^2-2t+1$ | A1 | |
| GS: $x = Ae^{-3t} + Bte^{-3t} + 2t^2 - 2t + 1$ | A1FT | |
| $x = 3$ when $t=0 \Rightarrow A = 2$ | B1 | |
| $\dot{x} = -3Ae^{-3t} - 3Bte^{-3t} + Be^{-3t} + 4t - 2$ | M1 | |
| $\dot{x} = 0$ when $t=0 \Rightarrow B = 8$ | A1 | |
| Hence $x = 2e^{-3t} + 8te^{-3t} + 2t^2 - 2t + 1$ | A1 | |

**Total: 10**

Show LaTeX source

8 Find the solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 x = 18 t ^ { 2 } + 6 t + 1$$

given that, when $t = 0 , x = 3$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0$.\\

\hfill \mbox{\textit{CAIE FP1 2017 Q8 [10]}}

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Q1 5 Q2 6 Q3 6 Q4 7 Q5 8 Q6 10 Q7 10 Q8 10 Q9 11 Q10 13 Q11 EITHER Q11 OR

Answer	Marks
\(m^2 + 6m + 9 = 0 \Rightarrow m = -3\)	M1
CF: \(x = Ae^{-3t} + Bte^{-3t}\)	A1
\(x = pt^2 + qt + r \Rightarrow \dot{x} = 2pt + q\) and \(\ddot{x} = 2p\)	M1
\(\Rightarrow 2p + 6(2pt+q) + 9(pt^2+qt+r) = 18t^2 + 6t + 1\)	M1
\(\Rightarrow 9pt^2 + (9q+12p)t + (2p+6q+9r) = 18t^2+6t+1\) \(\Rightarrow p=2, q=-2, r=1\) whence PI: \(2t^2-2t+1\)	A1
GS: \(x = Ae^{-3t} + Bte^{-3t} + 2t^2 - 2t + 1\)	A1FT
\(x = 3\) when \(t=0 \Rightarrow A = 2\)	B1
\(\dot{x} = -3Ae^{-3t} - 3Bte^{-3t} + Be^{-3t} + 4t - 2\)	M1
\(\dot{x} = 0\) when \(t=0 \Rightarrow B = 8\)	A1
Hence \(x = 2e^{-3t} + 8te^{-3t} + 2t^2 - 2t + 1\)	A1