| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - variable coefficients |
| Difficulty | Standard +0.3 This is a structured, guided question where students verify given solutions rather than derive them independently. Part (a) requires substitution and simplification, part (b) involves straightforward implicit differentiation, and part (c) applies the standard principle of superposition (general solution = particular integral + complementary function). While it covers Further Maths content, the heavy scaffolding and verification-based approach make it easier than a typical unguided first-order ODE problem. |
| Spec | 4.10a General/particular solutions: of differential equations4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = x^3 - x \Rightarrow y'(x) = 3x^2 - 1\) | B1 | Accept general cubic |
| \(\frac{dy}{dx} + \frac{2xy}{x^2-1} = 3x^2 - 1 + \frac{2x(x^3-x)}{x^2-1}\) | M1 | Substitution into LHS of DE |
| \(= 3x^2 - 1 + \frac{2x^2(x^2-1)}{x^2-1} = 5x^2-1\) | A1 | 3 marks; Completion. If using general cubic all unknown constants must be found |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d}{dx}\left[(x^2-1)y\right] = 2xy + (x^2-1)\frac{dy}{dx}\) | M1A1 | |
| Differentiating \((x^2-1)y = c\) wrt \(x\) leads to \(2xy + (x^2-1)\frac{dy}{dx} = 0\) | SC: Differentiated but not implicitly, give max of 1/3 for complete solution | |
| \(\Rightarrow y = \frac{c}{x^2-1}\) is a solution of \(\frac{dy}{dx} + \frac{2xy}{x^2-1} = 0\) | A1 | 3 marks; Be generous |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\Rightarrow y = \frac{c}{x^2-1}\) is a solution with one arbitrary constant of \(\frac{dy}{dx} + \frac{2xy}{x^2-1} = 0\) | ||
| \(\Rightarrow y = \frac{c}{x^2-1}\) is a CF of the DE | ||
| GS is CF + PI | M1 | Must be using 'hence'; CF and PI functions of \(x\) only; CSO |
| \(y = \frac{c}{x^2-1} + x^3 - x\) | A1 | 2 marks; Must have explicitly considered the link between one arbitrary constant and the GS of a first order DE |
# Question 3:
## Part 3(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = x^3 - x \Rightarrow y'(x) = 3x^2 - 1$ | B1 | Accept general cubic |
| $\frac{dy}{dx} + \frac{2xy}{x^2-1} = 3x^2 - 1 + \frac{2x(x^3-x)}{x^2-1}$ | M1 | Substitution into LHS of DE |
| $= 3x^2 - 1 + \frac{2x^2(x^2-1)}{x^2-1} = 5x^2-1$ | A1 | 3 marks; Completion. If using general cubic all unknown constants must be found |
## Part 3(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}\left[(x^2-1)y\right] = 2xy + (x^2-1)\frac{dy}{dx}$ | M1A1 | |
| Differentiating $(x^2-1)y = c$ wrt $x$ leads to $2xy + (x^2-1)\frac{dy}{dx} = 0$ | | SC: Differentiated but not implicitly, give max of 1/3 for complete solution |
| $\Rightarrow y = \frac{c}{x^2-1}$ is a solution of $\frac{dy}{dx} + \frac{2xy}{x^2-1} = 0$ | A1 | 3 marks; Be generous |
## Part 3(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\Rightarrow y = \frac{c}{x^2-1}$ is a solution with one arbitrary constant of $\frac{dy}{dx} + \frac{2xy}{x^2-1} = 0$ | | |
| $\Rightarrow y = \frac{c}{x^2-1}$ is a CF of the DE | | |
| GS is CF + PI | M1 | Must be using 'hence'; CF and PI functions of $x$ only; CSO |
| $y = \frac{c}{x^2-1} + x^3 - x$ | A1 | 2 marks; Must have explicitly considered the link between one arbitrary constant and the GS of a first order DE |
---3
\begin{enumerate}[label=(\alph*)]
\item Show that $y = x ^ { 3 } - x$ is a particular integral of the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x y } { x ^ { 2 } - 1 } = 5 x ^ { 2 } - 1$$
\item By differentiating $\left( x ^ { 2 } - 1 \right) y = c$ implicitly, where $y$ is a function of $x$ and $c$ is a constant, show that $y = \frac { c } { x ^ { 2 } - 1 }$ is a solution of the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x y } { x ^ { 2 } - 1 } = 0$$
\item Hence find the general solution of
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x y } { x ^ { 2 } - 1 } = 5 x ^ { 2 } - 1$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2006 Q3 [8]}}