| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| 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 integrating factor problem from Further Maths FP2, requiring students to rearrange to standard form, identify and calculate the integrating factor involving ln|sec x|, then integrate 3cos(2x)sin(x) which requires a trigonometric identity and substitution. While the method is routine for FP2 students, the trigonometric manipulation and integration steps elevate it slightly above average difficulty. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \cot x\, dx = \ln | \sin x | \), IF \(= \sin x\) |
| \(\sin x\frac{dy}{dx} + y\cos x = 3\cos 2x \sin x\) | ||
| \(y\sin x = \int 3\cos 2x \sin x\, dx\) | M1, A1 | M1: Multiply through by IF and integrate LHS. A1: correct so far |
| \(y\sin x = \int 3(2\cos^2 x - 1)\sin x\, dx\) | dM1 | dep on previous M mark: integrate RHS using double angle or factor formula. \(k\cos^2 x\sin x \to \pm\cos^3 x\), \(k\sin^2 x\cos x \to k\sin^3 x\), \(\cos 3x \to \pm\frac{1}{3}\sin 3x\), \(\sin 3x \to \pm\frac{1}{3}\cos 3x\) |
| \(y\sin x = 3\!\left[-\frac{2}{3}\cos^3 x + \cos x\right](+c)\) | A1 | All correct so far, constant not needed |
| \(y = \frac{3\cos x - 2\cos^3 x + c'}{\sin x}\) oe | B1ft | Obtain answer in form \(y = \ldots\) any equivalent form. Constant must be included and dealt with correctly. Award if correctly obtained from previous line |
# Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \cot x\, dx = \ln|\sin x|$, IF $= \sin x$ | M1 | Divide by $\tan x$ and attempt IF $e^{\int \cot x\, dx}$ or equivalent needed |
| $\sin x\frac{dy}{dx} + y\cos x = 3\cos 2x \sin x$ | | |
| $y\sin x = \int 3\cos 2x \sin x\, dx$ | M1, A1 | M1: Multiply through by IF and integrate LHS. A1: correct so far |
| $y\sin x = \int 3(2\cos^2 x - 1)\sin x\, dx$ | dM1 | dep on previous M mark: integrate RHS using double angle or factor formula. $k\cos^2 x\sin x \to \pm\cos^3 x$, $k\sin^2 x\cos x \to k\sin^3 x$, $\cos 3x \to \pm\frac{1}{3}\sin 3x$, $\sin 3x \to \pm\frac{1}{3}\cos 3x$ |
| $y\sin x = 3\!\left[-\frac{2}{3}\cos^3 x + \cos x\right](+c)$ | A1 | All correct so far, constant not needed |
| $y = \frac{3\cos x - 2\cos^3 x + c'}{\sin x}$ oe | B1ft | Obtain answer in form $y = \ldots$ any equivalent form. Constant must be included and dealt with correctly. Award if correctly obtained from previous line |
---\begin{enumerate}
\item Find, in the form $y = \mathrm { f } ( x )$, the general solution of the differential equation
\end{enumerate}
$$\tan x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 3 \cos 2 x \tan x , \quad 0 < x < \frac { \pi } { 2 }$$
\hfill \mbox{\textit{Edexcel FP2 2015 Q3 [6]}}