Question 7 - A-Level Maths

CAIE FP1 2019 June — Question 7 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 second-order linear differential equation with constant coefficients requiring the complementary function (solving the auxiliary equation), a particular integral (using trial solution for polynomial RHS), and 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

7 Find the particular solution of the differential equation $$10 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 3 \frac { \mathrm {~d} x } { \mathrm {~d} t } - x = t + 2$$ given that when $t = 0 , x = 0$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0$.

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Question 7:

Answer	Marks	Guidance
$10u^2 + 3u - 1 = 0 \Rightarrow (2u+1)(5u-1) = 0$	M1	Auxiliary equation
CF: $x = Ae^{-\frac{1}{2}t} + Be^{\frac{1}{5}t}$	A1
PI: $x = p + qt \Rightarrow \dot{x} = q \Rightarrow \ddot{x} = 0$	M1	Forms PI and differentiates
$3q - p - qt = t + 2 \Rightarrow q = -1,\ p = -5$	M1 A1	Substitutes
GS: $x = Ae^{-\frac{1}{2}t} + Be^{\frac{1}{5}t} - t - 5$	A1	States general solution
$\dot{x} = -\frac{1}{2}Ae^{-\frac{1}{2}t} + \frac{1}{5}Be^{\frac{1}{5}t} - 1$	M1	Differentiates
$A + B = 5$ and $-\frac{1}{2}A + \frac{1}{5}B = 1$	M1	Forms simultaneous equations
$A = 0,\ B = 5$	A1
$x = 5e^{\frac{1}{5}t} - t - 5$	A1	States PS

## Question 7:

| $10u^2 + 3u - 1 = 0 \Rightarrow (2u+1)(5u-1) = 0$ | M1 | Auxiliary equation |
|---|---|---|
| CF: $x = Ae^{-\frac{1}{2}t} + Be^{\frac{1}{5}t}$ | A1 | |
| PI: $x = p + qt \Rightarrow \dot{x} = q \Rightarrow \ddot{x} = 0$ | M1 | Forms PI and differentiates |
| $3q - p - qt = t + 2 \Rightarrow q = -1,\ p = -5$ | M1 A1 | Substitutes |
| GS: $x = Ae^{-\frac{1}{2}t} + Be^{\frac{1}{5}t} - t - 5$ | A1 | States general solution |
| $\dot{x} = -\frac{1}{2}Ae^{-\frac{1}{2}t} + \frac{1}{5}Be^{\frac{1}{5}t} - 1$ | M1 | Differentiates |
| $A + B = 5$ and $-\frac{1}{2}A + \frac{1}{5}B = 1$ | M1 | Forms simultaneous equations |
| $A = 0,\ B = 5$ | A1 | |
| $x = 5e^{\frac{1}{5}t} - t - 5$ | A1 | States PS |

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7 Find the particular solution of the differential equation

$$10 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 3 \frac { \mathrm {~d} x } { \mathrm {~d} t } - x = t + 2$$

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

\hfill \mbox{\textit{CAIE FP1 2019 Q7 [10]}}

This paper (12 questions)

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Q1 6 Q2 5 Q3 8 Q4 8 Q5 8 Q6 9 Q7 10 Q8 10 Q9 10 Q10 12 Q11 EITHER Q11 OR

Answer	Marks	Guidance
\(10u^2 + 3u - 1 = 0 \Rightarrow (2u+1)(5u-1) = 0\)	M1	Auxiliary equation
CF: \(x = Ae^{-\frac{1}{2}t} + Be^{\frac{1}{5}t}\)	A1
PI: \(x = p + qt \Rightarrow \dot{x} = q \Rightarrow \ddot{x} = 0\)	M1	Forms PI and differentiates
\(3q - p - qt = t + 2 \Rightarrow q = -1,\ p = -5\)	M1 A1	Substitutes
GS: \(x = Ae^{-\frac{1}{2}t} + Be^{\frac{1}{5}t} - t - 5\)	A1	States general solution
\(\dot{x} = -\frac{1}{2}Ae^{-\frac{1}{2}t} + \frac{1}{5}Be^{\frac{1}{5}t} - 1\)	M1	Differentiates
\(A + B = 5\) and \(-\frac{1}{2}A + \frac{1}{5}B = 1\)	M1	Forms simultaneous equations
\(A = 0,\ B = 5\)	A1
\(x = 5e^{\frac{1}{5}t} - t - 5\)	A1	States PS