Question 5 - A-Level Maths

CAIE Further Paper 2 2022 June — Question 5 9 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2022
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Standard linear first order - variable coefficients
Difficulty	Standard +0.8 This is a first-order linear differential equation requiring division by x(x+7) to get standard form, finding an integrating factor (which involves partial fractions for ∫7/[x(x+7)]dx), then integration and applying initial conditions. While the method is standard for Further Maths, the algebraic manipulation with partial fractions and the specific initial condition make it moderately challenging, placing it slightly above average difficulty.
Spec	4.10c Integrating factor: first order equations

5 Find the solution of the differential equation $$x ( x + 7 ) \frac { d y } { d x } + 7 y = x$$ for which $y = 7$ when $x = 1$. Give your answer in the form $y = f ( x )$.

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
Answer	Marks	Guidance
$\dfrac{dy}{dx} + \dfrac{7y}{x(x+7)} = \dfrac{1}{x+7}$	B1	Divides through by $x(x+7)$
$\dfrac{7}{x(x+7)} = \dfrac{1}{x} - \dfrac{1}{x+7}$ or $\dfrac{7}{x(x+7)} = \dfrac{7}{\left(x+\frac{7}{2}\right)^2 - \frac{49}{4}}$	B1	Converts to partial fractions or completes the square
$e^{\int 7x^{-1}(x+7)^{-1}dx} = e^{\ln x - \ln(x+7)} = \dfrac{x}{x+7}$	M1 A1	Finds integrating factor. Accept non-zero multiple of $\dfrac{x}{x+7}$
$\dfrac{d}{dx}\!\left(\dfrac{yx}{x+7}\right) = \dfrac{x}{(x+7)^2} = \dfrac{1}{x+7} - \dfrac{7}{(x+7)^2}$	M1	Correct form on LHS and attempt to integrate RHS. RHS must be of the form $\dfrac{a}{x+7} + \dfrac{b}{(x+7)^2}$
$\dfrac{yx}{x+7} = \ln(x+7) + 7(x+7)^{-1} \quad (+C)$	A1
$\dfrac{7}{8} = \ln 8 + \dfrac{7}{8} + C$	M1	Finds $C$, substitutes into their expression
$y = \left(1+\dfrac{7}{x}\right)\ln\!\left(\dfrac{x+7}{8}\right) + \dfrac{7}{x}$	M1 A1	Divides their expression through by coefficient of $y$
Total: 9

## Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} + \dfrac{7y}{x(x+7)} = \dfrac{1}{x+7}$ | B1 | Divides through by $x(x+7)$ |
| $\dfrac{7}{x(x+7)} = \dfrac{1}{x} - \dfrac{1}{x+7}$ or $\dfrac{7}{x(x+7)} = \dfrac{7}{\left(x+\frac{7}{2}\right)^2 - \frac{49}{4}}$ | B1 | Converts to partial fractions or completes the square |
| $e^{\int 7x^{-1}(x+7)^{-1}dx} = e^{\ln x - \ln(x+7)} = \dfrac{x}{x+7}$ | M1 A1 | Finds integrating factor. Accept non-zero multiple of $\dfrac{x}{x+7}$ |
| $\dfrac{d}{dx}\!\left(\dfrac{yx}{x+7}\right) = \dfrac{x}{(x+7)^2} = \dfrac{1}{x+7} - \dfrac{7}{(x+7)^2}$ | M1 | Correct form on LHS and attempt to integrate RHS. RHS must be of the form $\dfrac{a}{x+7} + \dfrac{b}{(x+7)^2}$ |
| $\dfrac{yx}{x+7} = \ln(x+7) + 7(x+7)^{-1} \quad (+C)$ | A1 | |
| $\dfrac{7}{8} = \ln 8 + \dfrac{7}{8} + C$ | M1 | Finds $C$, substitutes into their expression |
| $y = \left(1+\dfrac{7}{x}\right)\ln\!\left(\dfrac{x+7}{8}\right) + \dfrac{7}{x}$ | M1 A1 | Divides their expression through by coefficient of $y$ |
| **Total: 9** | | |

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5 Find the solution of the differential equation

$$x ( x + 7 ) \frac { d y } { d x } + 7 y = x$$

for which $y = 7$ when $x = 1$. Give your answer in the form $y = f ( x )$.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q5 [9]}}

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Q1 5 Q2 8 Q3 7 Q4 9 Q5 9 Q6 10 Q7 11 Q8 16