| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| 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 question from Further Maths, requiring students to rearrange to standard form, identify the integrating factor as sec x, and integrate (cos x)(ln x) by parts. While the method is routine for FM students, the integration by parts with ln x and careful algebraic manipulation elevate it slightly above average difficulty. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} + y\frac{\sin x}{\cos x} = \cos x \ln x\) | M1 | Attempt to divide through by \(\cos x\); must see at least 2 terms divided |
| \(I = e^{\int \frac{\sin x}{\cos x}dx} = e^{-\ln\cos x}\) | dM1 | \(e^{\pm\text{their }P(x)(dx)}\); dependent on first method mark |
| \(= \frac{1}{\cos x}\) | A1 | \(\frac{1}{\cos x}\) or \((\cos x)^{-1}\) or \(\sec x\) |
| \(\frac{y}{\cos x} = \int \ln x \, dx\) or \(\frac{d}{dx}\left(\frac{y}{\cos x}\right) = \ln x\) | M1A1 | M1: \(y \times\) their \(I = \int Q(x) \times\) their \(I \, dx\) |
| \(\frac{y}{\cos x} = x\ln x - x + C\) | M1 | Attempts \(\int \ln x \, dx\) by parts correctly; correct sign needed |
| \(y = (x\ln x - x + C)\cos x\) | A1 | Any equivalent with constant correctly placed; "\(y=\)" must appear |
## Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} + y\frac{\sin x}{\cos x} = \cos x \ln x$ | M1 | Attempt to divide through by $\cos x$; must see at least 2 terms divided |
| $I = e^{\int \frac{\sin x}{\cos x}dx} = e^{-\ln\cos x}$ | dM1 | $e^{\pm\text{their }P(x)(dx)}$; dependent on first method mark |
| $= \frac{1}{\cos x}$ | A1 | $\frac{1}{\cos x}$ or $(\cos x)^{-1}$ or $\sec x$ |
| $\frac{y}{\cos x} = \int \ln x \, dx$ or $\frac{d}{dx}\left(\frac{y}{\cos x}\right) = \ln x$ | M1A1 | M1: $y \times$ their $I = \int Q(x) \times$ their $I \, dx$ |
| $\frac{y}{\cos x} = x\ln x - x + C$ | M1 | Attempts $\int \ln x \, dx$ by parts correctly; correct sign needed |
| $y = (x\ln x - x + C)\cos x$ | A1 | Any equivalent with constant correctly placed; "$y=$" must appear |
**Note: Failure to divide by $\cos x$ at start means only the 3rd method mark is available.**
---\begin{enumerate}
\item Find the general solution of the differential equation
\end{enumerate}
$$\cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y \sin x = \left( \cos ^ { 2 } x \right) \ln x , \quad 0 < x < \frac { \pi } { 2 }$$
Give your answer in the form $y = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{Edexcel F2 2017 Q6 [8]}}