| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Session | Specimen |
| 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 structured Bernoulli equation problem with clear guidance through each step. Part (a) is routine verification of a given substitution, part (b) is standard integrating factor technique, and part (c) is back-substitution. While it requires multiple techniques and careful algebra with trigonometric functions, the scaffolding makes it more accessible than an unguided problem. Slightly above average difficulty due to the multi-step nature and Further Maths context. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\frac{dy}{dx} = \frac{dy}{dz}\cdot\frac{dz}{dx}\) 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) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\text{I.F.} = e^{\int -2\tan x\, dx} = e^{2\ln\cos x} = \cos^2 x\) | 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) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\therefore y = \left(\frac{1}{2}\tan x + \frac{1}{2}x\sec^2 x + c\sec^2 x\right)^2\) | B1ft | (1) |
# Question 7:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy}{dz}\cdot\frac{dz}{dx}$ 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) |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\text{I.F.} = e^{\int -2\tan x\, dx} = e^{2\ln\cos x} = \cos^2 x$ | 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) |
## Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\therefore y = \left(\frac{1}{2}\tan x + \frac{1}{2}x\sec^2 x + c\sec^2 x\right)^2$ | B1ft | (1) |
---\begin{enumerate}
\item (a) Show that the transformation $z = y ^ { \frac { 1 } { 2 } }$ transforms the differential equation
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} x } - 4 y \tan x = 2 y ^ { \frac { 1 } { 2 } }$$
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).\\
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\end{array} \right.$$
\hfill \mbox{\textit{Edexcel F2 Q7 [12]}}