Question 2 - A-Level Maths

CAIE Further Paper 2 2023 June — Question 2 7 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2023
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Substitution z = x + y or similar linear combination
Difficulty	Challenging +1.2 This is a Further Maths differential equations question requiring a guided substitution (z = x + y) to transform into a separable equation, then integration and applying initial conditions. While it involves multiple steps and algebraic manipulation, the substitution is explicitly given and the method is standard for this topic. It's harder than typical A-level questions due to being Further Maths content, but straightforward once the substitution is applied.
Spec	1.08k Separable differential equations: dy/dx = f(x)g(y)4.10a General/particular solutions: of differential equations

2 Use the substitution $z = x + y$ to find the solution of the differential equation $$\frac { d y } { d x } = \frac { 1 + 3 x + 3 y } { 3 x + 3 y - 1 }$$ for which $y = 0$ when $x = 1$. Give your answer in the form $\operatorname { aln } ( \mathrm { x } + \mathrm { y } ) + \mathrm { b } ( \mathrm { x } - \mathrm { y } ) + \mathrm { c } = 0$, where $a , b$ and $c$ are constants to be determined.

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Question 2:

Answer	Marks	Guidance
$\frac{dy}{dx} = \frac{dz}{dx} - 1$	B1
$\frac{dz}{dx} - 1 = \frac{1+3z}{3z-1}$	M1	Substitutes and derives first order separable equation
$\frac{dz}{dx} = \frac{1+3z}{3z-1} + 1 = \frac{6z}{3z-1}$	A1
$\int \frac{1}{2} - \frac{1}{6}z^{-1}\,dz = \int 1\,dx$	M1	Separates variables and integrates both sides
$\frac{1}{2}z - \frac{1}{6}\ln z = x + C \Rightarrow -\frac{1}{2}x + \frac{1}{2}y - \frac{1}{6}\ln(x+y) = C$	A1
$C = -\frac{1}{2}$	M1	Substitutes initial conditions into their expression
$\frac{1}{6}\ln(x+y) + \frac{1}{2}(x-y) - \frac{1}{2} = 0$	A1	OE

## Question 2:

| $\frac{dy}{dx} = \frac{dz}{dx} - 1$ | B1 | |
|---|---|---|
| $\frac{dz}{dx} - 1 = \frac{1+3z}{3z-1}$ | M1 | Substitutes and derives first order separable equation |
| $\frac{dz}{dx} = \frac{1+3z}{3z-1} + 1 = \frac{6z}{3z-1}$ | A1 | |
| $\int \frac{1}{2} - \frac{1}{6}z^{-1}\,dz = \int 1\,dx$ | M1 | Separates variables and integrates both sides |
| $\frac{1}{2}z - \frac{1}{6}\ln z = x + C \Rightarrow -\frac{1}{2}x + \frac{1}{2}y - \frac{1}{6}\ln(x+y) = C$ | A1 | |
| $C = -\frac{1}{2}$ | M1 | Substitutes initial conditions into their expression |
| $\frac{1}{6}\ln(x+y) + \frac{1}{2}(x-y) - \frac{1}{2} = 0$ | A1 | OE |

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2 Use the substitution $z = x + y$ to find the solution of the differential equation

$$\frac { d y } { d x } = \frac { 1 + 3 x + 3 y } { 3 x + 3 y - 1 }$$

for which $y = 0$ when $x = 1$. Give your answer in the form $\operatorname { aln } ( \mathrm { x } + \mathrm { y } ) + \mathrm { b } ( \mathrm { x } - \mathrm { y } ) + \mathrm { c } = 0$, where $a , b$ and $c$ are constants to be determined.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q2 [7]}}

This paper (8 questions)

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Q1 5 Q2 7 Q3 8 Q4 9 Q5 10 Q6 11 Q7 11 Q8 14

Answer	Marks	Guidance
\(\frac{dy}{dx} = \frac{dz}{dx} - 1\)	B1
\(\frac{dz}{dx} - 1 = \frac{1+3z}{3z-1}\)	M1	Substitutes and derives first order separable equation
\(\frac{dz}{dx} = \frac{1+3z}{3z-1} + 1 = \frac{6z}{3z-1}\)	A1
\(\int \frac{1}{2} - \frac{1}{6}z^{-1}\,dz = \int 1\,dx\)	M1	Separates variables and integrates both sides
\(\frac{1}{2}z - \frac{1}{6}\ln z = x + C \Rightarrow -\frac{1}{2}x + \frac{1}{2}y - \frac{1}{6}\ln(x+y) = C\)	A1
\(C = -\frac{1}{2}\)	M1	Substitutes initial conditions into their expression
\(\frac{1}{6}\ln(x+y) + \frac{1}{2}(x-y) - \frac{1}{2} = 0\)	A1	OE