| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Euler-Cauchy equations via exponential substitution |
| Difficulty | Standard +0.8 This is a standard Further Maths FP3 Euler-Cauchy equation question requiring the exponential substitution x=e^t. Part (a) involves routine chain rule applications to derive standard results, while part (b) requires converting to constant coefficients and solving. It's methodical rather than insightful, but the multi-step nature and Further Maths content place it moderately above average difficulty. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(\frac{dt}{d\ell} = e^t\) \([\ell = x]\) | B1 | |
| \(x\frac{dy}{dx} = x\frac{dy}{dt} \cdot \frac{dt}{dx} = x\frac{dy}{dt} \cdot \frac{1}{e^t} = \frac{dy}{dt}\) | M1 | |
| \(= \frac{dy}{d\ell}\) | A1 | Total: 3 |
| (ii) \(\frac{d^2y}{dt^2} = \frac{d}{dt}\left(x\frac{dy}{dx}\right)\) | M1 | |
| \(= \frac{dx}{dt}\frac{dy}{dx} + x\frac{d}{dt}\left(\frac{dy}{dx}\right)\) | M1 | |
| \(= \frac{dy}{dt} + x\frac{d}{dt}\left(\frac{dy}{dx}\right)\) | M1 | |
| \(\ldots = \frac{dy}{dt} + x^2\left(\frac{d^2y}{dx^2}\right)\) | A1 | Total: 3 |
| \(\Rightarrow x^2\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}\) | ||
| (b) \(x^2\frac{d^2y}{dx^2} - 6x\frac{dy}{dx} + 6y = 0\) | ||
| \(\Rightarrow \frac{d^2y}{dt^2} - 7\frac{dy}{dt} + 6y = 0\) | M1 | |
| Auxl eqn \(m^2 - 7m + 6 = 0\) | m1 | |
| \((m-6)(m-1) = 0\) | A1 | |
| \(m = 1\) and \(6\) | ||
| \(y = Ae^t + Be^t\) | M1 | |
| \(y = Ax^6 + Bx\) | A1/ | Total: 5 |
**(a)(i)** $\frac{dt}{d\ell} = e^t$ $[\ell = x]$ | B1 | |
$x\frac{dy}{dx} = x\frac{dy}{dt} \cdot \frac{dt}{dx} = x\frac{dy}{dt} \cdot \frac{1}{e^t} = \frac{dy}{dt}$ | M1 | | Chain rule
$= \frac{dy}{d\ell}$ | A1 | Total: 3 | Completion. AG
**(ii)** $\frac{d^2y}{dt^2} = \frac{d}{dt}\left(x\frac{dy}{dx}\right)$ | M1 | |
$= \frac{dx}{dt}\frac{dy}{dx} + x\frac{d}{dt}\left(\frac{dy}{dx}\right)$ | M1 | | Product rule
$= \frac{dy}{dt} + x\frac{d}{dt}\left(\frac{dy}{dx}\right)$ | M1 | |
$\ldots = \frac{dy}{dt} + x^2\left(\frac{d^2y}{dx^2}\right)$ | A1 | Total: 3 | Condone leaving in this form
$\Rightarrow x^2\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}$ | | | AG
**(b)** $x^2\frac{d^2y}{dx^2} - 6x\frac{dy}{dx} + 6y = 0$ | | |
$\Rightarrow \frac{d^2y}{dt^2} - 7\frac{dy}{dt} + 6y = 0$ | M1 | | Using results in (a) to reach DE of this form
Auxl eqn $m^2 - 7m + 6 = 0$ | m1 | | PI
$(m-6)(m-1) = 0$ | A1 | | PI
$m = 1$ and $6$ | | |
$y = Ae^t + Be^t$ | M1 | | Must be solving the 'correct' DE. (Give M1A0 for $y = Ae^{6x} + Be^x$)
$y = Ax^6 + Bx$ | A1/ | Total: 5 | Ft a minor slip only if previous A0 and all three method marks gained
## **TOTAL: 75 marks**8
\begin{enumerate}[label=(\alph*)]
\item Given that $x = \mathrm { e } ^ { t }$ and that $y$ is a function of $x$, show that:
\begin{enumerate}[label=(\roman*)]
\item $x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }$;
\item $\quad x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t }$.
\end{enumerate}\item Hence find the general solution of the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 0$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2008 Q8 [11]}}