| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Session | Specimen |
| Marks | 11 |
| 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 standard first-order linear ODE requiring integrating factor method, but involves non-trivial integration (∫tan x dx = ln|sec x|) and manipulation of trigonometric expressions. The 11 marks and two-part structure indicate moderate complexity. While the method is routine for Further Maths students, the algebraic manipulation and application of the boundary condition require careful execution, placing it above average difficulty but not exceptionally challenging. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Integrating factor \(= e^{\int 2\tan\!xdx}\) | M1 | AO1 |
| \(= e^{2\ln\sec x}\) | A1 | AO1 |
| \(= e^{\ln\sec^2 x}\) | A1 | AO1 |
| \(= \sec^2 x\) | A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sec^2 x \frac{dy}{dx} + 2y\tan x \sec^2 x = \sin x \sec^2 x\) | M1 | AO1 |
| \(= \frac{\sin x}{\cos^2 x}\) (or \(\sec x \tan x\)) | A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y\sec^2 x = \sec x + C\) | A1, A1 | AO1 |
| \(0 = \sqrt{2} + C\) | M1 | AO1 |
| \(C = -\sqrt{2}\) | A1 | AO1 |
| The solution is \(y = \cos x - \sqrt{2}\cos^2 x\) | A1 | AO1 |
## Part (a)
Integrating factor $= e^{\int 2\tan\!xdx}$ | M1 | AO1
$= e^{2\ln\sec x}$ | A1 | AO1
$= e^{\ln\sec^2 x}$ | A1 | AO1
$= \sec^2 x$ | A1 | AO1
## Part (b)
Applying the integrating factor:
$\sec^2 x \frac{dy}{dx} + 2y\tan x \sec^2 x = \sin x \sec^2 x$ | M1 | AO1
$= \frac{\sin x}{\cos^2 x}$ (or $\sec x \tan x$) | A1 | AO1
Integrating:
$y\sec^2 x = \sec x + C$ | A1, A1 | AO1
$0 = \sqrt{2} + C$ | M1 | AO1
$C = -\sqrt{2}$ | A1 | AO1
The solution is $y = \cos x - \sqrt{2}\cos^2 x$ | A1 | AO1
**Total: [11]**
---Consider the differential equation
$$\frac{dy}{dx} + 2y \tan x = \sin x, \quad 0 < x < \frac{\pi}{2}.$$
\begin{enumerate}[label=(\alph*)]
\item Find an integrating factor for this differential equation. [4]
\item Solve the differential equation given that $y = 0$ when $x = \frac{\pi}{4}$, giving your answer in the form $y = f(x)$. [7]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 Q10 [11]}}