| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Homogeneous equation (y = vx substitution) |
| Difficulty | Challenging +1.2 This is a standard FP3 homogeneous differential equation requiring the substitution y=xz, which is explicitly given. The solution involves routine separation of variables and integration using a standard formula from MF1. While it requires multiple steps and is from Further Maths (inherently harder), the technique is textbook-standard with no novel insight needed. The particular solution in part (ii) is straightforward substitution. |
| Spec | 4.10a General/particular solutions: of differential equations4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = xz = \frac{dy}{dx} = z + x\frac{dz}{dx}\) | B1 | For correct differentiation of substitution |
| \(\Rightarrow xz + x^2\frac{dz}{dx} = xz \cos z \Rightarrow x\frac{dz}{dx} = x\cos z\) | M1 | For substituting into DE |
| \(\Rightarrow \frac{dz}{dx} = \cos z\) | A1 | For DE in variables separable form |
| \(\Rightarrow \int\sec z \, dz = \int\frac{1}{x}dx\) | M1 | For attempt at integration to ln form on LHS |
| \(= \ln(\sec z + \tan z) = \ln kx\) | A1 | For correct integration (\(k\) not required here) |
| OR \(\ln\tan\left(\frac{z}{2} + \frac{\pi}{4}\right) = \ln kx\) | ||
| \(\Rightarrow \sec\left(\frac{y}{x}\right) + \tan\left(\frac{y}{x}\right) = kx\) | A1 6 | For correct solution |
| AEF including RHS \(= e^{\ln x + c}\) | ||
| OR \(\tan\left(\frac{y}{2x} + \frac{\pi}{4}\right) = kx\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((4, \pi): \sec\frac{\pi}{4} + \tan\frac{\pi}{4} + \pi + 1 = 4k\) | M1 | For substituting \((4, \pi)\) into their solution (with \(k\)) |
| OR \(\tan\left(\frac{\pi}{8} + \frac{\pi}{4}\right) = 4k\) | ||
| \(\Rightarrow \sec\left(\frac{y}{x}\right) + \tan\left(\frac{y}{x}\right) = \frac{1}{4}(1+\sqrt{2})x\) | A1 2 | For correct solution AEF |
| Allow decimal equivalent \(0.60355 x\) | ||
| Allow \(e^{\ln x}\) for \(x\) | ||
| OR \(\tan\left(\frac{y}{2x} + \frac{\pi}{4}\right) = \left(\frac{1}{4}\tan\frac{\pi}{8}\right)x\) or \(\frac{1}{4}(1+\sqrt{2})x\) |
## (i)
$y = xz = \frac{dy}{dx} = z + x\frac{dz}{dx}$ | B1 | For correct differentiation of substitution
$\Rightarrow xz + x^2\frac{dz}{dx} = xz \cos z \Rightarrow x\frac{dz}{dx} = x\cos z$ | M1 | For substituting into DE
$\Rightarrow \frac{dz}{dx} = \cos z$ | A1 | For DE in variables separable form
$\Rightarrow \int\sec z \, dz = \int\frac{1}{x}dx$ | M1 | For attempt at integration to ln form on LHS
$= \ln(\sec z + \tan z) = \ln kx$ | A1 | For correct integration ($k$ not required here)
OR $\ln\tan\left(\frac{z}{2} + \frac{\pi}{4}\right) = \ln kx$ | |
$\Rightarrow \sec\left(\frac{y}{x}\right) + \tan\left(\frac{y}{x}\right) = kx$ | A1 6 | For correct solution
| | AEF including RHS $= e^{\ln x + c}$
OR $\tan\left(\frac{y}{2x} + \frac{\pi}{4}\right) = kx$ | |
## (ii)
$(4, \pi): \sec\frac{\pi}{4} + \tan\frac{\pi}{4} + \pi + 1 = 4k$ | M1 | For substituting $(4, \pi)$ into their solution (with $k$)
OR $\tan\left(\frac{\pi}{8} + \frac{\pi}{4}\right) = 4k$ | |
$\Rightarrow \sec\left(\frac{y}{x}\right) + \tan\left(\frac{y}{x}\right) = \frac{1}{4}(1+\sqrt{2})x$ | A1 2 | For correct solution AEF
| | Allow decimal equivalent $0.60355 x$
| | Allow $e^{\ln x}$ for $x$
OR $\tan\left(\frac{y}{2x} + \frac{\pi}{4}\right) = \left(\frac{1}{4}\tan\frac{\pi}{8}\right)x$ or $\frac{1}{4}(1+\sqrt{2})x$ | |
---\begin{enumerate}[label=(\roman*)]
\item Use the substitution $y = xz$ to find the general solution of the differential equation
$$x \frac{dy}{dx} - y = x \cos \left(\frac{y}{x}\right),$$
giving your answer in a form without logarithms. (You may quote an appropriate result given in the List of Formulae (MF1).) [6]
\item Find the solution of the differential equation for which $y = \pi$ when $x = 4$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2010 Q4 [8]}}