| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Substitution changing independent variable |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear scaffolding through parts (a) and (b) that guide students to the solution. Part (a) involves routine chain rule applications to verify given results. Part (b) is straightforward substitution using the results from (a). Part (c) requires solving a standard constant-coefficient second-order ODE and back-substituting. While it's a multi-step problem requiring careful algebra and is from FP3, the heavy scaffolding and standard techniques make it more accessible than typical Further Maths questions requiring genuine insight. |
| Spec | 4.10a General/particular solutions: of differential equations4.10d Second order homogeneous: auxiliary equation method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dt} = 2t\) | B1 | PI or for \(\frac{dt}{dx} = \frac{1}{2}x^{-\frac{1}{2}}\) |
| \(\frac{dx}{dt}\cdot\frac{dy}{dx} = \frac{dy}{dt}\) | M1 | OE Chain rule \(\frac{dy}{dx} = \ldots\) or \(\frac{dy}{dt} = \ldots\) |
| \(2t\frac{dy}{dx} = \frac{dy}{dt}\) so \(2\sqrt{x}\,\frac{dy}{dx} = \frac{dy}{dt}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d}{dx}\!\left(2\sqrt{x}\,\frac{dy}{dx}\right) = \frac{d}{dx}\!\left(\frac{dy}{dt}\right) = \frac{dt}{dx}\cdot\frac{d}{dt}\!\left(\frac{dy}{dt}\right)\) | M1 | \(\frac{d}{dx}(f(t)) = \frac{dt}{dx}\cdot\frac{d}{dt}(f(t))\) OE |
| \(2\sqrt{x}\,\frac{d^2y}{dx^2} + x^{-\frac{1}{2}}\frac{dy}{dx} = \frac{1}{2t}\frac{d^2y}{dt^2}\) | M1 | Product rule OE |
| \(4t\sqrt{x}\,\frac{d^2y}{dx^2} + 2tx^{-\frac{1}{2}}\frac{dy}{dx} = \frac{d^2y}{dt^2}\) | ||
| \(\Rightarrow 4x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = \frac{d^2y}{dt^2}\) | A1 | AG, Completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4x\frac{d^2y}{dx^2} + 2(1+2\sqrt{x})\frac{dy}{dx} - 3y = 0\) becomes: | ||
| \(\left(4x\frac{d^2y}{dx^2} + 2\frac{dy}{dx}\right) + 2\left(2\sqrt{x}\,\frac{dy}{dx}\right) - 3y = 0\) | M1 | Use of either (a)(i) or (a)(ii) |
| \(\frac{d^2y}{dt^2} + 2\frac{dy}{dt} - 3y = 0\) | A1 | AG, Completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Auxiliary equation: \(m^2 + 2m - 3 = 0\) | M1 | PI |
| \((m+3)(m-1) = 0 \Rightarrow m = -3\) and \(1\) | A1 | PI |
| General solution: \(y = Ae^{-3t} + Be^{t}\) | M1 | \(Ae^{-3x} + Be^x\) scores M0 |
| \(\Rightarrow y = Ae^{-3\sqrt{x}} + Be^{\sqrt{x}}\) | A1 |
# Question 8:
## Part 8(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 2t$ | B1 | PI or for $\frac{dt}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ |
| $\frac{dx}{dt}\cdot\frac{dy}{dx} = \frac{dy}{dt}$ | M1 | OE Chain rule $\frac{dy}{dx} = \ldots$ or $\frac{dy}{dt} = \ldots$ |
| $2t\frac{dy}{dx} = \frac{dy}{dt}$ so $2\sqrt{x}\,\frac{dy}{dx} = \frac{dy}{dt}$ | A1 | AG | Total: 3 marks |
## Part 8(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}\!\left(2\sqrt{x}\,\frac{dy}{dx}\right) = \frac{d}{dx}\!\left(\frac{dy}{dt}\right) = \frac{dt}{dx}\cdot\frac{d}{dt}\!\left(\frac{dy}{dt}\right)$ | M1 | $\frac{d}{dx}(f(t)) = \frac{dt}{dx}\cdot\frac{d}{dt}(f(t))$ OE |
| $2\sqrt{x}\,\frac{d^2y}{dx^2} + x^{-\frac{1}{2}}\frac{dy}{dx} = \frac{1}{2t}\frac{d^2y}{dt^2}$ | M1 | Product rule OE |
| $4t\sqrt{x}\,\frac{d^2y}{dx^2} + 2tx^{-\frac{1}{2}}\frac{dy}{dx} = \frac{d^2y}{dt^2}$ | | |
| $\Rightarrow 4x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = \frac{d^2y}{dt^2}$ | A1 | AG, Completion | Total: 3 marks |
## Part 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4x\frac{d^2y}{dx^2} + 2(1+2\sqrt{x})\frac{dy}{dx} - 3y = 0$ becomes: | | |
| $\left(4x\frac{d^2y}{dx^2} + 2\frac{dy}{dx}\right) + 2\left(2\sqrt{x}\,\frac{dy}{dx}\right) - 3y = 0$ | M1 | Use of either (a)(i) or (a)(ii) |
| $\frac{d^2y}{dt^2} + 2\frac{dy}{dt} - 3y = 0$ | A1 | AG, Completion | Total: 2 marks |
## Part 8(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Auxiliary equation: $m^2 + 2m - 3 = 0$ | M1 | PI |
| $(m+3)(m-1) = 0 \Rightarrow m = -3$ and $1$ | A1 | PI |
| General solution: $y = Ae^{-3t} + Be^{t}$ | M1 | $Ae^{-3x} + Be^x$ scores M0 |
| $\Rightarrow y = Ae^{-3\sqrt{x}} + Be^{\sqrt{x}}$ | A1 | | Total: 4 marks |8
\begin{enumerate}[label=(\alph*)]
\item Given that $x = t ^ { 2 }$, where $t \geqslant 0$, and that $y$ is a function of $x$, show that:
\begin{enumerate}[label=(\roman*)]
\item $2 \sqrt { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }$;
\item $\quad 4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }$.
\end{enumerate}\item Hence show that the substitution $x = t ^ { 2 }$, where $t \geqslant 0$, transforms the differential equation
$$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$
into
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 3 y = 0$$
(2 marks)
\item Hence find the general solution of the differential equation
$$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$
giving your answer in the form $y = \mathrm { g } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2009 Q8 [12]}}