| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Substitution reducing to first order linear ODE |
| Difficulty | Standard +0.8 This is a Further Maths FP3 question requiring recognition that a non-standard substitution simplifies the equation, followed by solving a separable differential equation and back-substitution. Part (i) involves careful algebraic manipulation with the given substitution (not student-chosen), while part (ii) requires standard integration and rearrangement. The substitution being provided reduces difficulty, but the multi-step nature and Further Maths context place it moderately above average. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = u - \frac{1}{x} \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{1}{x^2}\) | M1, A1 | For differentiating substitution; For correct expression |
| \(x^3\left(\frac{du}{dx} + \frac{1}{x^2}\right) = x\left(u - \frac{1}{x}\right) + x + 1\) | M1 | For substituting \(y\) and \(\frac{dy}{dx}\) into DE |
| \(\Rightarrow x^2\frac{du}{dx} = u\) | A1 4 | For obtaining correct equation AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\frac{1}{u}du = \int\frac{1}{x^2}dx \Rightarrow \ln ku = -\frac{1}{x}\) | M1, A1 | For separating variables and attempt at integration; For correct integration (\(k\) not required here) |
| \(ku = e^{-1/x} \Rightarrow k\left(y + \frac{1}{x}\right) = e^{-1/x}\) | M1, M1 | For any 2 of: \(k\) seen, exponentiating; For all 3 of: substituting for \(u\) |
| \(\Rightarrow y = Ae^{-1/x} - \frac{1}{x}\) | A1 5 | For correct solution AEF in form \(y = f(x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{du}{dx} - \frac{1}{x^2}u = 0 \Rightarrow\) I.F. \(e^{\int -1/x^2\,dx} = e^{1/x}\) | M1 | For attempt to find I.F. |
| \(\Rightarrow \frac{d}{dx}\left(ue^{1/x}\right) = 0\) | A1 | For correct result |
| \(ue^{1/x} = k \Rightarrow y + \frac{1}{x} = ke^{-1/x}\) | M1, M1 | From \(\left\lvert u \times \text{I.F.} =\right\rvert\), for \(k\) seen; for substituting for \(u\) (in either order) |
| \(\Rightarrow y = ke^{-1/x} - \frac{1}{x}\) | A1 5 | For correct solution AEF in form \(y = f(x)\) |
# Question 5:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = u - \frac{1}{x} \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{1}{x^2}$ | M1, A1 | For differentiating substitution; For correct expression |
| $x^3\left(\frac{du}{dx} + \frac{1}{x^2}\right) = x\left(u - \frac{1}{x}\right) + x + 1$ | M1 | For substituting $y$ and $\frac{dy}{dx}$ into DE |
| $\Rightarrow x^2\frac{du}{dx} = u$ | A1 **4** | For obtaining correct equation **AG** |
## Part (ii) Method 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{1}{u}du = \int\frac{1}{x^2}dx \Rightarrow \ln ku = -\frac{1}{x}$ | M1, A1 | For separating variables and attempt at integration; For correct integration ($k$ not required here) |
| $ku = e^{-1/x} \Rightarrow k\left(y + \frac{1}{x}\right) = e^{-1/x}$ | M1, M1 | For any 2 of: $k$ seen, exponentiating; For all 3 of: substituting for $u$ |
| $\Rightarrow y = Ae^{-1/x} - \frac{1}{x}$ | A1 **5** | For correct solution **AEF** in form $y = f(x)$ |
## Part (ii) Method 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{du}{dx} - \frac{1}{x^2}u = 0 \Rightarrow$ I.F. $e^{\int -1/x^2\,dx} = e^{1/x}$ | M1 | For attempt to find I.F. |
| $\Rightarrow \frac{d}{dx}\left(ue^{1/x}\right) = 0$ | A1 | For correct result |
| $ue^{1/x} = k \Rightarrow y + \frac{1}{x} = ke^{-1/x}$ | M1, M1 | From $\left\lvert u \times \text{I.F.} =\right\rvert$, for $k$ seen; for substituting for $u$ (in either order) |
| $\Rightarrow y = ke^{-1/x} - \frac{1}{x}$ | A1 **5** | For correct solution **AEF** in form $y = f(x)$ |
**Total: 9 marks**
---5 The variables $x$ and $y$ are related by the differential equation
$$x ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = x y + x + 1 .$$
(i) Use the substitution $y = u - \frac { 1 } { x }$, where $u$ is a function of $x$, to show that the differential equation may be written as
$$x ^ { 2 } \frac { \mathrm {~d} u } { \mathrm {~d} x } = u .$$
(ii) Hence find the general solution of the differential equation (A), giving your answer in the form $y = \mathrm { f } ( x )$.
\hfill \mbox{\textit{OCR FP3 2009 Q5 [9]}}