| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Homogeneous equation (y = vx substitution) |
| Difficulty | Standard +0.3 This is a standard homogeneous differential equation requiring the routine y=ux substitution (explicitly given), followed by straightforward separation of variables and integration. While it's a Further Maths topic, the technique is mechanical with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| \((y = xu \Rightarrow) \frac{dy}{dx} = x\frac{du}{dx} + u\) | B1 | For a correct statement |
| \(x\frac{du}{dx} + u = \frac{2 + u^2}{u}\) | M1 | For using the substitution to eliminate \(y\) (If B0, then \(y\) must be eliminated from LHS, but \(\frac{d(uv)}{dx}\) sufficient) |
| \(\Rightarrow x\frac{du}{dx} = \frac{2}{u}\) | A1 | For correct equation AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int u \, du = \int \frac{2}{x} dx\) | M1 | For separating variables and writing/attempting integrals |
| \(\Rightarrow \frac{1}{2}u^2 = 2\ln(kx)\) OR \(\frac{1}{2}u^2 = 2\ln x (+c)\) | A1 | For correct integration both sides (\(k\) or \(c\) not required here) |
| \(\Rightarrow \frac{1}{2}\left(\frac{y}{x}\right)^2 = 2\ln(kx)\) OR \(\frac{1}{2}\left(\frac{y}{x}\right)^2 = 2\ln x + c\) | M1 | For substituting for \(u\) into integrated terms with constant (on either side) |
| \(\Rightarrow y^2 = 4x^2\ln(kx)\) OR \(y^2 = 4x^2\ln x + Cx^2\) | A1 | For correct solution AEF \(y^2 = f(x)\) Do not penalise "\(c\)" being used for different constants e.g. \(2\ln x + c = 2\ln(cx)\) |
| [4] |
**(i)**
| $(y = xu \Rightarrow) \frac{dy}{dx} = x\frac{du}{dx} + u$ | B1 | For a correct statement |
| $x\frac{du}{dx} + u = \frac{2 + u^2}{u}$ | M1 | For using the substitution to eliminate $y$ (If B0, then $y$ must be eliminated from LHS, but $\frac{d(uv)}{dx}$ sufficient) |
| $\Rightarrow x\frac{du}{dx} = \frac{2}{u}$ | A1 | For correct equation AG |
| | [3] | |
**(ii)**
| $\int u \, du = \int \frac{2}{x} dx$ | M1 | For separating variables and writing/attempting integrals |
| $\Rightarrow \frac{1}{2}u^2 = 2\ln(kx)$ OR $\frac{1}{2}u^2 = 2\ln x (+c)$ | A1 | For correct integration both sides ($k$ or $c$ not required here) |
| $\Rightarrow \frac{1}{2}\left(\frac{y}{x}\right)^2 = 2\ln(kx)$ OR $\frac{1}{2}\left(\frac{y}{x}\right)^2 = 2\ln x + c$ | M1 | For substituting for $u$ into integrated terms with constant (on either side) |
| $\Rightarrow y^2 = 4x^2\ln(kx)$ OR $y^2 = 4x^2\ln x + Cx^2$ | A1 | For correct solution AEF $y^2 = f(x)$ Do not penalise "$c$" being used for different constants e.g. $2\ln x + c = 2\ln(cx)$ |
| | [4] | |1 The variables $x$ and $y$ are related by the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ^ { 2 } + y ^ { 2 } } { x y } .$$
(i) Use the substitution $y = u x$, where $u$ is a function of $x$, to obtain the differential equation
$$x \frac { \mathrm {~d} u } { \mathrm {~d} x } = \frac { 2 } { u } .$$
(ii) Hence find the general solution of the differential equation (A), giving your answer in the form $y ^ { 2 } = \mathrm { f } ( x )$.
\hfill \mbox{\textit{OCR FP3 2012 Q1 [7]}}