| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Bernoulli equation |
| Difficulty | Standard +0.8 This is a standard FP3 Bernoulli equation problem requiring substitution to linearize, then integrating factor method. While it involves multiple techniques (substitution, differentiation using chain rule, integrating factor, and back-substitution), these are well-practiced procedures in FP3. The question guides students through each step explicitly, making it a structured exercise rather than requiring novel insight. Slightly above average difficulty due to the algebraic manipulation and multi-step nature, but routine for Further Maths students. |
| Spec | 4.10a General/particular solutions: of differential equations4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = ku^{k-1}\frac{du}{dx}\) | M1 A1 | For using chain rule; For correct \(\frac{dy}{dx}\) |
| \(\Rightarrow xku^{k-1}\frac{du}{dx} + 3u^k = x^2u^{2k}\) | M1 | For substituting for \(y\) and \(\frac{dy}{dx}\) |
| \(\Rightarrow \frac{du}{dx} - \frac{3}{kx}u = \frac{1}{k}xu^{k+1}\) | A1 4 | For correct equation AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = -1\) | B1 1 | For correct \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{du}{dx} - \frac{3}{x}u = -\int\frac{-3\ln x}{x^3}\) ... IF \(e^{\int -\frac{3}{x}dx} = \frac{\sin\sqrt{x}}{x^3}\) | f.t. for IF \(x^{-1}\) using \(k\) or their numerical value for \(k\) | |
| \(\Rightarrow \frac{d}{dx}\left(u \cdot \frac{1}{x^3}\right) = -\frac{1}{x^2}\) | M1 | For \(\frac{d}{dx}(u.\text{their IF}) = -x.\text{their IF}\) |
| \(\Rightarrow u \cdot \frac{1}{x^3} = -\int(+c) \Rightarrow y = \frac{1}{cx^3+x^2}\) | A1 A1 4 | For correct integration both sides; For correct solution for \(y\) |
## (i)
$\frac{dy}{dx} = ku^{k-1}\frac{du}{dx}$ | M1 A1 | For using chain rule; For correct $\frac{dy}{dx}$
$\Rightarrow xku^{k-1}\frac{du}{dx} + 3u^k = x^2u^{2k}$ | M1 | For substituting for $y$ and $\frac{dy}{dx}$
$\Rightarrow \frac{du}{dx} - \frac{3}{kx}u = \frac{1}{k}xu^{k+1}$ | A1 4 | For correct equation AG
## (ii)
$k = -1$ | B1 1 | For correct $k$
## (iii)
$\frac{du}{dx} - \frac{3}{x}u = -\int\frac{-3\ln x}{x^3}$ ... IF $e^{\int -\frac{3}{x}dx} = \frac{\sin\sqrt{x}}{x^3}$ | | f.t. for IF $x^{-1}$ using $k$ or their numerical value for $k$
$\Rightarrow \frac{d}{dx}\left(u \cdot \frac{1}{x^3}\right) = -\frac{1}{x^2}$ | M1 | For $\frac{d}{dx}(u.\text{their IF}) = -x.\text{their IF}$
$\Rightarrow u \cdot \frac{1}{x^3} = -\int(+c) \Rightarrow y = \frac{1}{cx^3+x^2}$ | A1 A1 4 | For correct integration both sides; For correct solution for $y$
---The substitution $y = u^k$, where $k$ is an integer, is to be used to solve the differential equation
$$x \frac{dy}{dx} + 3y = x^2 y^2 \qquad (A)$$
by changing it into an equation (B) in the variables $u$ and $x$.
\begin{enumerate}[label=(\roman*)]
\item Show that equation (B) may be written in the form
$$\frac{du}{dx} + \frac{3}{kx} u = \frac{1}{k} x u^{k+1}.$$ [4]
\item Write down the value of $k$ for which the integrating factor method may be used to solve equation (B). [1]
\item Using this value of $k$, solve equation (B) and hence find the general solution of equation (A), giving your answer in the form $y = f(x)$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q5 [9]}}