| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| 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 integrating factor problem from Further Maths FP2, requiring division by cos x to get standard form, identifying integrating factor sec x, integrating 2cos²x sin x + sec x (requiring substitution u = cos x), then applying initial conditions with exact trigonometric values. While methodical, it involves multiple technical steps with potential algebraic pitfalls and exact value manipulation, placing it moderately above average difficulty. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx}+y\tan x=2\cos^2 x\sin x+\frac{1}{\cos x}\) | M1 | Divides by \(\cos x\); LHS both terms divided; RHS minimum 1 term divided |
| \(I=e^{\int\tan x\,dx}=e^{\ln\sec x}=\sec x\) | dM1A1 | M1: Attempt integrating factor \(e^{\int\tan x\,dx}\) needed; A1: Correct integrating factor \(\sec x\) or \(\frac{1}{\cos x}\) |
| \(y\sec x=\int(2\sin x\cos x+\sec^2 x)\,dx\) | M1 | Multiply through by IF and integrate LHS; \(yI=\int(\text{their RHS})I\,dx\) |
| \(y\sec x=-\frac{1}{2}\cos 2x+\tan x(+c)\) | M1A1A1 | M1: Attempt integration of at least one term on RHS (both sides multiplied by IF); OR \(\sec^2 x\to K\tan x\); A1: \(-\frac{1}{2}\cos 2x\) or equivalent integration of \(2\sin x\cos x\) (\(\sin^2 x\) or \(-\cos^2 x\)); A1: \(\tan x\), constant not needed |
| \(y=\left(-\frac{1}{2}\cos 2x+\tan x+c\right)\cos x\) | A1ft | Include constant and deal with it correctly; must start \(y=...\); or equivalent e.g. \(y=-\frac{1}{2}\cos 2x\cos x+\sin x+c\cos x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=\frac{\pi}{4}\Rightarrow 5\sqrt{2}=...\Rightarrow c=...\) | M1 | Substitutes for \(x\) and \(y\) and solves for \(c\); if substitution not shown, award for at least one term evaluated correctly |
| \(x=\frac{\pi}{6}\Rightarrow y=..........\) | M1 | Substitutes \(x=\frac{\pi}{6}\) to find value for \(y\) |
| \(y=\frac{1}{2}+\frac{35}{8}\sqrt{3}\) or \(y=0.5+4.375\sqrt{3}\) | A1cao | Must be in given form; equivalent fractions allowed |
# Question 7:
$$\cos x\frac{dy}{dx}+y\sin x=2\cos^3 x\sin x+1$$
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx}+y\tan x=2\cos^2 x\sin x+\frac{1}{\cos x}$ | M1 | Divides by $\cos x$; LHS both terms divided; RHS minimum 1 term divided |
| $I=e^{\int\tan x\,dx}=e^{\ln\sec x}=\sec x$ | dM1A1 | M1: Attempt integrating factor $e^{\int\tan x\,dx}$ needed; A1: Correct integrating factor $\sec x$ or $\frac{1}{\cos x}$ |
| $y\sec x=\int(2\sin x\cos x+\sec^2 x)\,dx$ | M1 | Multiply through by IF and integrate LHS; $yI=\int(\text{their RHS})I\,dx$ |
| $y\sec x=-\frac{1}{2}\cos 2x+\tan x(+c)$ | M1A1A1 | M1: Attempt integration of at least one term on RHS (both sides multiplied by IF); OR $\sec^2 x\to K\tan x$; A1: $-\frac{1}{2}\cos 2x$ or equivalent integration of $2\sin x\cos x$ ($\sin^2 x$ or $-\cos^2 x$); A1: $\tan x$, constant not needed |
| $y=\left(-\frac{1}{2}\cos 2x+\tan x+c\right)\cos x$ | A1ft | Include constant and deal with it correctly; must start $y=...$; or equivalent e.g. $y=-\frac{1}{2}\cos 2x\cos x+\sin x+c\cos x$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\frac{\pi}{4}\Rightarrow 5\sqrt{2}=...\Rightarrow c=...$ | M1 | Substitutes for $x$ and $y$ and solves for $c$; if substitution not shown, award for at least one term evaluated correctly |
| $x=\frac{\pi}{6}\Rightarrow y=..........$ | M1 | Substitutes $x=\frac{\pi}{6}$ to find value for $y$ |
| $y=\frac{1}{2}+\frac{35}{8}\sqrt{3}$ or $y=0.5+4.375\sqrt{3}$ | A1cao | Must be in given form; equivalent fractions allowed |
NB: If no working shown due to calculator use: final answer correct in required form (no decimals instead of $\sqrt{3}$) scores 3/3; final answer incorrect or decimals instead of $\sqrt{3}$ scores 0/3.
---7. (a) Find, in the form $y = \mathrm { f } ( x )$, the general solution of the equation
$$\cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y \sin x = 2 \cos ^ { 3 } x \sin x + 1 , \quad 0 < x < \frac { \pi } { 2 }$$
Given that $y = 5 \sqrt { 2 }$ when $x = \frac { \pi } { 4 }$\\
(b) find the value of $y$ when $x = \frac { \pi } { 6 }$, giving your answer in the form $a + b \sqrt { 3 }$, where $a$ and $b$ are rational numbers to be found.
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\hfill \mbox{\textit{Edexcel FP2 2017 Q7 [11]}}