Question 9 - A-Level Maths

WJEC Further Unit 4 2023 June — Question 9 8 marks

Exam Board	WJEC
Module	Further Unit 4 (Further Unit 4)
Year	2023
Session	June
Marks	8
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 ODE requiring identification of integrating factor μ = (x+1)^5, integration of a polynomial term, and application of initial conditions. While the method is standard for Further Maths, the algebraic manipulation and integration steps are non-trivial, placing it moderately above average difficulty.
Spec	4.10c Integrating factor: first order equations

Consider the differential equation $$\left(x+1\right)\frac{\mathrm{d}y}{\mathrm{d}x} + 5y = (x+1)^2, \quad x > -1.$$ Given that $y = \frac{1}{4}$ when $x = 1$, find the value of $y$ when $x = 0$. [8]

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Answer	Marks	Guidance
Dividing both sides by $(x+1)$: $\frac{dy}{dx} + \frac{5y}{x+1} = x + 1$	M1
Integrating factor: $e^{\int\frac{5}{x+1}dx} = e^{5\ln(x+1)} = (x+1)^5$	m1, A1	cao
Multiplying both sides: $(x+1)^5\frac{dy}{dx} + 5y(x+1)^4 = (x+1)^6$	M1
Integrating: $(x+1)^5y = \frac{(x+1)^7}{7} + c$	m1, A1	Or equivalent, cao
Substituting $32 \times \frac{1}{4} = \frac{128}{7} + c \to c = -\frac{72}{7}$	B1
Solution: $(x+1)^5y = \frac{(x+1)^7}{7} - \frac{72}{7}$
When $x = 0$: $y = \frac{1}{7} - \frac{72}{7} = -\frac{71}{7}$	B1
$\to y = -\frac{71}{7}$ $(-10.142857)$
[8]

| Dividing both sides by $(x+1)$: $\frac{dy}{dx} + \frac{5y}{x+1} = x + 1$ | M1 | |
| | | |
| Integrating factor: $e^{\int\frac{5}{x+1}dx} = e^{5\ln(x+1)} = (x+1)^5$ | m1, A1 | cao |
| | | |
| Multiplying both sides: $(x+1)^5\frac{dy}{dx} + 5y(x+1)^4 = (x+1)^6$ | M1 | |
| Integrating: $(x+1)^5y = \frac{(x+1)^7}{7} + c$ | m1, A1 | Or equivalent, cao |
| | | |
| Substituting $32 \times \frac{1}{4} = \frac{128}{7} + c \to c = -\frac{72}{7}$ | B1 | |
| | | |
| Solution: $(x+1)^5y = \frac{(x+1)^7}{7} - \frac{72}{7}$ | | |
| | | |
| When $x = 0$: $y = \frac{1}{7} - \frac{72}{7} = -\frac{71}{7}$ | B1 | |
| $\to y = -\frac{71}{7}$ $(-10.142857)$ | | |
| | [8] | |

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Consider the differential equation
$$\left(x+1\right)\frac{\mathrm{d}y}{\mathrm{d}x} + 5y = (x+1)^2, \quad x > -1.$$

Given that $y = \frac{1}{4}$ when $x = 1$, find the value of $y$ when $x = 0$. [8]

\hfill \mbox{\textit{WJEC Further Unit 4 2023 Q9 [8]}}

This paper (13 questions)

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Q1 5 Q2 7 Q3 9 Q4 5 Q5 7 Q6 16 Q7 7 Q8 11 Q9 8 Q10 8 Q11 14 Q12 6 Q13 17

Answer	Marks	Guidance
Dividing both sides by \((x+1)\): \(\frac{dy}{dx} + \frac{5y}{x+1} = x + 1\)	M1
Integrating factor: \(e^{\int\frac{5}{x+1}dx} = e^{5\ln(x+1)} = (x+1)^5\)	m1, A1	cao
Multiplying both sides: \((x+1)^5\frac{dy}{dx} + 5y(x+1)^4 = (x+1)^6\)	M1
Integrating: \((x+1)^5y = \frac{(x+1)^7}{7} + c\)	m1, A1	Or equivalent, cao
Substituting \(32 \times \frac{1}{4} = \frac{128}{7} + c \to c = -\frac{72}{7}\)	B1
Solution: \((x+1)^5y = \frac{(x+1)^7}{7} - \frac{72}{7}\)
When \(x = 0\): \(y = \frac{1}{7} - \frac{72}{7} = -\frac{71}{7}\)	B1
\(\to y = -\frac{71}{7}\) \((-10.142857)\)
[8]