| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| 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 Further Maths F2 question requiring the integrating factor method with non-trivial functions (tan x, sec³x, e^(4x)). While the method is standard, the integration after applying the integrating factor (which involves sec³x × e^(4x)) is significantly more challenging than typical A-level examples, and finding the particular solution adds another layer. This is harder than average but still within expected Further Maths scope. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e^{-3\int\tan x\,dx} = e^{-3\ln\sec x} = \sec^{-3}x\) or \(\cos^3 x\) | M1A1 | Attempt IF including integration of \((-3)\tan x\); correct simplified IF. |
| \(\frac{d}{dx}(y\cos^3 x) = e^{4x} \Rightarrow y\cos^3 x = \int e^{4x}\,dx\) | M1 | Multiply by IF and recognise exact derivative on LHS. |
| \(y\cos^3 x = \frac{1}{4}e^{4x}\ (+c)\) | M1 | Integrate RHS; constant not needed yet. |
| \(y=\left(\frac{1}{4}e^{4x}+c\right)\sec^3 x\) or \(\left(\frac{1}{4}e^{4x}+c\right)\cos^{-3}x\) | A1 (5) | Correct result including \(y=\ldots\) and constant. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y=4,\ x=0\): \(4=\left(\frac{1}{4}+c\right)\), so \(c=\frac{15}{4}\) | M1 | Use initial conditions to find \(c\). |
| \(y=\frac{1}{4}(e^{4x}+15)\sec^3 x\) or \(\frac{1}{4}(e^{4x}+15)\cos^{-3}x\) | A1 (2) | Fully correct final answer including \(y=\ldots\) |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e^{-3\int\tan x\,dx} = e^{-3\ln\sec x} = \sec^{-3}x$ or $\cos^3 x$ | M1A1 | Attempt IF including integration of $(-3)\tan x$; correct simplified IF. |
| $\frac{d}{dx}(y\cos^3 x) = e^{4x} \Rightarrow y\cos^3 x = \int e^{4x}\,dx$ | M1 | Multiply by IF and recognise exact derivative on LHS. |
| $y\cos^3 x = \frac{1}{4}e^{4x}\ (+c)$ | M1 | Integrate RHS; constant not needed yet. |
| $y=\left(\frac{1}{4}e^{4x}+c\right)\sec^3 x$ or $\left(\frac{1}{4}e^{4x}+c\right)\cos^{-3}x$ | A1 (5) | Correct result including $y=\ldots$ and constant. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y=4,\ x=0$: $4=\left(\frac{1}{4}+c\right)$, so $c=\frac{15}{4}$ | M1 | Use initial conditions to find $c$. |
| $y=\frac{1}{4}(e^{4x}+15)\sec^3 x$ or $\frac{1}{4}(e^{4x}+15)\cos^{-3}x$ | A1 (2) | Fully correct final answer including $y=\ldots$ |
---\begin{enumerate}
\item (a) Determine the general solution of the differential equation
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y \tan x = \mathrm { e } ^ { 4 x } \sec ^ { 3 } x$$
giving your answer in the form $y = \mathrm { f } ( x )$\\
(b) Determine the particular solution for which $y = 4$ at $x = 0$
\hfill \mbox{\textit{Edexcel F2 2022 Q4 [7]}}