Question 7 - A-Level Maths

Edexcel FP2 2010 June — Question 7 12 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2010
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Bernoulli equation
Difficulty	Challenging +1.2 This is a standard Bernoulli equation problem from FP2 with a guided transformation already provided. Part (a) requires straightforward substitution and chain rule application, part (b) is a routine integrating factor problem, and part (c) involves back-substitution. While it requires multiple techniques and is Further Maths content, the step-by-step structure and provided transformation make it more accessible than typical FP2 questions requiring independent insight.
Spec	1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates 4.10c Integrating factor: first order equations

7. (a) Show that the transformation $z = y ^ { \frac { 1 } { 2 } }$ transforms the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - 4 y \tan x = 2 y ^ { \frac { 1 } { 2 } }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 2 z \tan x = 1$$ (b) Solve the differential equation (II) to find $z$ as a function of $x$.
(c) Hence obtain the general solution of the differential equation (I).

Show mark scheme Show mark scheme source

Question 7:

Part (a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\frac{dy}{dx} = \frac{dy}{dz}\cdot\frac{dz}{dx}$ and $\frac{dy}{dz} = 2z$ so $\frac{dy}{dx} = 2z\cdot\frac{dz}{dx}$	M1 M1 A1
Substituting to get $2z\cdot\frac{dz}{dx} - 4z^2\tan x = 2z$ and thus $\frac{dz}{dx} - 2z\tan x = 1$ ★	M1 A1	(5 marks total)

Part (b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\text{I.F.} = e^{\int -2\tan x\, dx} = e^{2\ln\cos x} = \cos^2 x$	M1 A1
$\therefore \frac{d}{dx}(z\cos^2 x) = \cos^2 x \therefore z\cos^2 x = \int \cos^2 x\, dx$	M1
$z\cos^2 x = \int\frac{1}{2}(\cos 2x + 1)\,dx = \frac{1}{4}\sin 2x + \frac{1}{2}x + c$	M1 A1
$\therefore z = \frac{1}{2}\tan x + \frac{1}{2}x\sec^2 x + c\sec^2 x$	A1	(6 marks total)

Part (c):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\therefore y = \left(\frac{1}{2}\tan x + \frac{1}{2}x\sec^2 x + c\sec^2 x\right)^2$	B1ft	(1 mark total)

## Question 7:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy}{dz}\cdot\frac{dz}{dx}$ and $\frac{dy}{dz} = 2z$ so $\frac{dy}{dx} = 2z\cdot\frac{dz}{dx}$ | M1 M1 A1 | |
| Substituting to get $2z\cdot\frac{dz}{dx} - 4z^2\tan x = 2z$ and thus $\frac{dz}{dx} - 2z\tan x = 1$ ★ | M1 A1 | (5 marks total) |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{I.F.} = e^{\int -2\tan x\, dx} = e^{2\ln\cos x} = \cos^2 x$ | M1 A1 | |
| $\therefore \frac{d}{dx}(z\cos^2 x) = \cos^2 x \therefore z\cos^2 x = \int \cos^2 x\, dx$ | M1 | |
| $z\cos^2 x = \int\frac{1}{2}(\cos 2x + 1)\,dx = \frac{1}{4}\sin 2x + \frac{1}{2}x + c$ | M1 A1 | |
| $\therefore z = \frac{1}{2}\tan x + \frac{1}{2}x\sec^2 x + c\sec^2 x$ | A1 | (6 marks total) |

### Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\therefore y = \left(\frac{1}{2}\tan x + \frac{1}{2}x\sec^2 x + c\sec^2 x\right)^2$ | B1ft | (1 mark total) |

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7. (a) Show that the transformation $z = y ^ { \frac { 1 } { 2 } }$ transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } - 4 y \tan x = 2 y ^ { \frac { 1 } { 2 } }$$

into the differential equation

$$\frac { \mathrm { d } z } { \mathrm {~d} x } - 2 z \tan x = 1$$

(b) Solve the differential equation (II) to find $z$ as a function of $x$.\\
(c) Hence obtain the general solution of the differential equation (I).\\

\hfill \mbox{\textit{Edexcel FP2 2010 Q7 [12]}}

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