Question 4 - A-Level Maths

Edexcel F2 2021 October — Question 4 9 marks

Exam Board	Edexcel
Module	F2 (Further Pure Mathematics 2)
Year	2021
Session	October
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 +0.3 This is a standard integrating factor question from Further Maths with straightforward algebra. While it requires knowing the integrating factor method and involves exponential integration, the structure is routine: divide to standard form, find integrating factor μ = 1/(x+1), integrate e^(3x)/(x+1), and apply initial conditions. The algebra is clean with no particularly tricky integration or manipulation, making it slightly easier than an average A-level question overall but typical for this Further Maths topic.
Spec	4.10a General/particular solutions: of differential equations 4.10c Integrating factor: first order equations

4. (a) Determine the general solution of the differential equation $$( x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } - x y = \mathrm { e } ^ { 3 x } \quad x > - 1$$ giving your answer in the form $y = \mathrm { f } ( x )$.
(b) Determine the particular solution of the differential equation for which $y = 5$ when $x = 0$

Show mark scheme Show mark scheme source

Question 4(a):

Answer	Marks	Guidance
Working/Answer	Mark	Notes
$\frac{dy}{dx} - \frac{xy}{(x+1)} = \frac{e^{3x}}{(x+1)}$	B1	Correctly rearranged equation
$I = e^{\int \frac{-x}{x+1}dx} = e^{\int\left(-1+\frac{1}{x+1}\right)dx}$	M1	Correct strategy for integrating factor including attempt at integration
$= e^{-x+\ln(x+1)}$	A1	For $-x + \ln(x+1)$
$= (x+1)e^{-x}$	A1	Correct integrating factor
$y(x+1)e^{-x} = \int \frac{e^{3x}}{x+1} \times (x+1)e^{-x}\,dx$	M1	Uses integrating factor to reach form $yI = \int QI\,dx$
$y(x+1)e^{-x} = \frac{1}{2}e^{2x} + c$	A1	Correct equation (with or without $+c$)
$y = \frac{e^{3x}}{2(x+1)} + \frac{ce^x}{(x+1)}$	A1	Correct answer (allow equivalent forms). Must have $y=\ldots$

Question 4(b):

Answer	Marks	Guidance
Working/Answer	Mark	Notes
$x=0,\ y=5 \Rightarrow 5 = \frac{1}{2}+c \Rightarrow c = \frac{9}{2}$	M1	Substitutes $x=0$ and $y=5$ and attempts to find $c$
$y = \frac{e^{3x}}{2(x+1)} + \frac{9e^x}{2(x+1)}$	A1	cao (oe). Must have $y=\ldots$

## Question 4(a):

| Working/Answer | Mark | Notes |
|---|---|---|
| $\frac{dy}{dx} - \frac{xy}{(x+1)} = \frac{e^{3x}}{(x+1)}$ | B1 | Correctly rearranged equation |
| $I = e^{\int \frac{-x}{x+1}dx} = e^{\int\left(-1+\frac{1}{x+1}\right)dx}$ | M1 | Correct strategy for integrating factor including attempt at integration |
| $= e^{-x+\ln(x+1)}$ | A1 | For $-x + \ln(x+1)$ |
| $= (x+1)e^{-x}$ | A1 | Correct integrating factor |
| $y(x+1)e^{-x} = \int \frac{e^{3x}}{x+1} \times (x+1)e^{-x}\,dx$ | M1 | Uses integrating factor to reach form $yI = \int QI\,dx$ |
| $y(x+1)e^{-x} = \frac{1}{2}e^{2x} + c$ | A1 | Correct equation (with or without $+c$) |
| $y = \frac{e^{3x}}{2(x+1)} + \frac{ce^x}{(x+1)}$ | A1 | Correct answer (allow equivalent forms). Must have $y=\ldots$ |

## Question 4(b):

| Working/Answer | Mark | Notes |
|---|---|---|
| $x=0,\ y=5 \Rightarrow 5 = \frac{1}{2}+c \Rightarrow c = \frac{9}{2}$ | M1 | Substitutes $x=0$ and $y=5$ and attempts to find $c$ |
| $y = \frac{e^{3x}}{2(x+1)} + \frac{9e^x}{2(x+1)}$ | A1 | cao (oe). Must have $y=\ldots$ |

Show LaTeX source

4. (a) Determine the general solution of the differential equation

$$( x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } - x y = \mathrm { e } ^ { 3 x } \quad x > - 1$$

giving your answer in the form $y = \mathrm { f } ( x )$.\\
(b) Determine the particular solution of the differential equation for which $y = 5$ when $x = 0$

\hfill \mbox{\textit{Edexcel F2 2021 Q4 [9]}}

This paper (9 questions)

View full paper

Q1 4 Q2 8 Q3 6 Q4 9 Q5 8 Q6 9 Q7 11 Q8 11 Q9 9