| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - variable coefficients |
| Difficulty | Standard +0.3 This is a standard integrating factor question from Further Maths FP2. While it requires knowing the integrating factor method and involves trigonometric functions, it follows a completely routine procedure: identify P(x) = 2tan(x), find IF = sec²(x), integrate the RHS which simplifies nicely to 2sec(x), then apply the boundary condition. The algebraic manipulation is straightforward and the question structure is typical textbook fare. It's slightly above average difficulty (0.0) only because it's Further Maths content with trig functions rather than Core content with simpler functions. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y\sec^2\frac{\pi}{6} - 2\sec^2\frac{\pi}{3} = 2\ln\left(\frac{\sec\frac{\pi}{6}}{\sec\frac{\pi}{3}}\right)\) | M1A1 | For finding difference between \(y\sec^2\frac{\pi}{6}\) and \(2\sec^2\frac{\pi}{3}\) |
| \(\frac{4}{3}y - 8 = 2\ln\frac{1}{\sqrt{3}}\) | ||
| \(y = \frac{3}{4}\left(8 + 2\ln\frac{1}{\sqrt{3}}\right) = 6 + \frac{3}{2}\ln\frac{1}{\sqrt{3}} = 6 - \frac{3}{4}\ln 3\) | M1A1 | For re-arranging to \(y = \ldots\) and simplification |
| Answer | Marks | Guidance |
|---|---|---|
| Criteria | Mark | Guidance |
| For \(e^{\int 2\tan x\,dx}\) or \(e^{\int \tan x\,dx}\) and attempting integration; \(e^{(2)\ln\sec x}\) should be seen if final result is not \(\sec^2 x\) | M1 | |
| IF \(= \sec^2 x\) | A1 | |
| For multiplying equation by their IF and attempting to integrate lhs | M1 | |
| Attempting integration of rhs; \(\sin 2x = 2\sin x\cos x\) and \(\sec x = \frac{1}{\cos x}\) needed | M1dep | Dependent on second M mark |
| \(y\sec^2 x = 2\ln\sec x\ (+c)\) all integration correct | A1cao | Constant not needed |
| Re-writing in form \(y = \ldots\) | A1ft | Accept any equivalent form but constant must be present. e.g. \(y = \frac{\ln(A\sec^2 x)}{\sec^2 x}\), \(y = \cos^2 x\left[\ln(\sec^2 x) + c\right]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Criteria | Mark | Guidance |
| Using \(y=2\), \(x=\frac{\pi}{3}\) in general solution to obtain constant of integration | M1 | |
| \(c = 8 - 2\ln 2\) or \(A = \frac{1}{4}e^8\) | A1 | Check constant is correct for their answer to (a). Answers to 3 s.f. acceptable; can include \(\cos\frac{\pi}{3}\) or \(\sec\frac{\pi}{3}\) |
| Using their constant and \(x = \frac{\pi}{6}\) in their general solution and attempting simplification | M1 | |
| \(y = 6 - \frac{3}{4}\ln 3\) \(\left(\frac{3}{4} \text{ or } 0.75\right)\) | A1cao |
## Question 5 (Alternative for part b):
| Working | Mark | Guidance |
|---------|------|----------|
| $y\sec^2\frac{\pi}{6} - 2\sec^2\frac{\pi}{3} = 2\ln\left(\frac{\sec\frac{\pi}{6}}{\sec\frac{\pi}{3}}\right)$ | M1A1 | For finding difference between $y\sec^2\frac{\pi}{6}$ and $2\sec^2\frac{\pi}{3}$ |
| $\frac{4}{3}y - 8 = 2\ln\frac{1}{\sqrt{3}}$ | | |
| $y = \frac{3}{4}\left(8 + 2\ln\frac{1}{\sqrt{3}}\right) = 6 + \frac{3}{2}\ln\frac{1}{\sqrt{3}} = 6 - \frac{3}{4}\ln 3$ | M1A1 | For re-arranging to $y = \ldots$ and simplification |
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### Question 5a Notes:
| Criteria | Mark | Guidance |
|----------|------|----------|
| For $e^{\int 2\tan x\,dx}$ or $e^{\int \tan x\,dx}$ and attempting integration; $e^{(2)\ln\sec x}$ should be seen if final result is not $\sec^2 x$ | M1 | |
| IF $= \sec^2 x$ | A1 | |
| For multiplying equation by their IF and attempting to integrate lhs | M1 | |
| Attempting integration of rhs; $\sin 2x = 2\sin x\cos x$ and $\sec x = \frac{1}{\cos x}$ needed | M1dep | Dependent on second M mark |
| $y\sec^2 x = 2\ln\sec x\ (+c)$ all integration correct | A1cao | Constant not needed |
| Re-writing in form $y = \ldots$ | A1ft | Accept any equivalent form but constant must be present. e.g. $y = \frac{\ln(A\sec^2 x)}{\sec^2 x}$, $y = \cos^2 x\left[\ln(\sec^2 x) + c\right]$ |
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### Question 5b Notes:
| Criteria | Mark | Guidance |
|----------|------|----------|
| Using $y=2$, $x=\frac{\pi}{3}$ in general solution to obtain constant of integration | M1 | |
| $c = 8 - 2\ln 2$ or $A = \frac{1}{4}e^8$ | A1 | Check constant is correct for their answer to (a). Answers to 3 s.f. acceptable; can include $\cos\frac{\pi}{3}$ or $\sec\frac{\pi}{3}$ |
| Using their constant and $x = \frac{\pi}{6}$ in their general solution and attempting simplification | M1 | |
| $y = 6 - \frac{3}{4}\ln 3$ $\left(\frac{3}{4} \text{ or } 0.75\right)$ | A1cao | |
---\begin{enumerate}
\item (a) Find, in the form $y = \mathrm { f } ( x )$, the general solution of the equation
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y \tan x = \sin 2 x , \quad 0 < x < \frac { \pi } { 2 }$$
Given that $y = 2$ at $x = \frac { \pi } { 3 }$\\
(b) find the value of $y$ at $x = \frac { \pi } { 6 }$, giving your answer in the form $a + k \ln b$, where $a$ and $b$ are integers and $k$ is rational.\\
\hfill \mbox{\textit{Edexcel FP2 2013 Q5 [10]}}