| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Substitution reducing to first order linear ODE |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with a guided substitution provided. Part (i) requires careful algebraic manipulation of derivatives (dy/dx = -1/u² · du/dx) but is straightforward verification. Part (ii) is a standard integrating factor problem once transformed. While it's Further Maths content requiring multiple techniques, the scaffolding and routine nature of each step place it moderately above average difficulty. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dx} = -u^{-2}\dfrac{du}{dx}\) | M1 | Differentiate |
| \(2u - xu^2\left(-u^{-2}\dfrac{du}{dx}\right) = \dfrac{1}{x^2}\) | M1 | Substitute |
| \(x\dfrac{du}{dx} + 2u = \dfrac{1}{x^2}\) | A1 | ag Convincingly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{du}{dx} + \dfrac{2u}{x} = \dfrac{1}{x^3}\) | ||
| \(I = \exp\left(\int\dfrac{2}{x}dx\right) = e^{2\ln x}\) | M1 | \(e^{k\ln x}\) |
| \(= x^2\) | A1 | Incorrect IF means no further marks can be gained; if RHS not multiplied by IF no further marks; or \(\ldots = \ln kx\) |
| \(\dfrac{d}{dx}(x^2 u) = x^{-1}\) | M1 | For LHS, multiply and recognise derivative |
| \(x^2 u = \ln x + A\) | A1 | |
| \(u = (\ln x + A)/x^2\) | ||
| \(y = x^2/(\ln x + A)\) | M1 | For \(y =\) reciprocal of 'their \(u\)' |
| \(x=1, y=1 \Rightarrow 1 = \dfrac{1}{0+A} \Rightarrow A=1\) | M1 | or \(k=e\) |
| \(y = \dfrac{x^2}{\ln x + 1}\) | A1 | oe without fractions within fractions; or \(y=\dfrac{x^2}{\ln ex}\) |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = -u^{-2}\dfrac{du}{dx}$ | M1 | Differentiate |
| $2u - xu^2\left(-u^{-2}\dfrac{du}{dx}\right) = \dfrac{1}{x^2}$ | M1 | Substitute |
| $x\dfrac{du}{dx} + 2u = \dfrac{1}{x^2}$ | A1 | ag Convincingly shown |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{du}{dx} + \dfrac{2u}{x} = \dfrac{1}{x^3}$ | | |
| $I = \exp\left(\int\dfrac{2}{x}dx\right) = e^{2\ln x}$ | M1 | $e^{k\ln x}$ |
| $= x^2$ | A1 | Incorrect IF means no further marks can be gained; if RHS not multiplied by IF no further marks; or $\ldots = \ln kx$ |
| $\dfrac{d}{dx}(x^2 u) = x^{-1}$ | M1 | For LHS, multiply and recognise derivative |
| $x^2 u = \ln x + A$ | A1 | |
| $u = (\ln x + A)/x^2$ | | |
| $y = x^2/(\ln x + A)$ | M1 | For $y =$ reciprocal of 'their $u$' |
| $x=1, y=1 \Rightarrow 1 = \dfrac{1}{0+A} \Rightarrow A=1$ | M1 | or $k=e$ |
| $y = \dfrac{x^2}{\ln x + 1}$ | A1 | oe without fractions within fractions; or $y=\dfrac{x^2}{\ln ex}$ |
---3 The differential equation
$$\frac { 2 } { y } - \frac { x } { y ^ { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { 1 } { x ^ { 2 } }$$
is to be solved subject to the condition $y = 1$ when $x = 1$.\\
(i) Show that $y = \frac { 1 } { u }$ transforms the differential equation into
$$x \frac { \mathrm {~d} u } { \mathrm {~d} x } + 2 u = \frac { 1 } { x ^ { 2 } } .$$
(ii) Find $y$ in terms of $x$.
\hfill \mbox{\textit{OCR FP3 2016 Q3 [10]}}