| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Explicit differential equation series solution |
| Difficulty | Challenging +1.2 This is a standard FP2 series solution question requiring systematic differentiation and substitution. While it involves multiple derivatives and careful algebra, the method is algorithmic: differentiate the given equation, substitute x=0 and y=1/2 to find coefficients, then construct the series. The technique is well-practiced in FP2 and requires no novel insight, just methodical execution across 11 marks. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07f Convexity/concavity: points of inflection4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Correct method for producing 2nd order differential equation | M1 | |
| e.g. \(\frac{d}{dx}\left[(1 + 2x)\frac{dy}{dx}\right] = \frac{d}{dx}\left[x + 4y^2\right]\) attempted | ||
| \((1 + 2x)\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 1 + 8y\frac{dy}{dx}\) seen + conclusion | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \((1 + 2y)\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} - 8y\frac{d^2y}{dx^2} - 8\left(\frac{dy}{dx}\right)^2 - 2\frac{d^2y}{dx^2}\) or equiv. | M1, A2, 1, 0 | 3 |
| [e.g. \((1 + 2y)\frac{d^3y}{dx^3} = 8\left(\frac{dy}{dx}\right)^2 + 4(2y - 1)\frac{d^2y}{dx^2}\)] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx}\) (at \(x = 0\)) \(= 1\) | B1 | |
| Finding \(\frac{d^2y}{dx^2}\) (at \(x = 0\)) \(= 3\) | M1 | |
| Finding \(\frac{d^3y}{dx^3}\) at \(x = 0\); \(= 8\) [A1 f.t. is on part (c) values only] | M1, A1ft | |
| \(y = \frac{1}{2} + x + \frac{3}{2}x^2 + \frac{4}{3}x^3 + ...\) | M1, A1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Polynomial for \(y\): \(y = \frac{1}{2} + ax + bx^2 + cx^3 + ...\) | M1 | |
| In given d.e.: \((1 + 2y)(a + 2bx + 3cx^2 + ...) \equiv x + 4(y + ax + bx^2 + cx^3 + ...)^2\) | M1A1 | |
| \(a = 1\) | B1 | Complete method for other coefficients M1, answer A1 |
**Part (a)**
| Correct method for producing 2nd order differential equation | M1 | |
| e.g. $\frac{d}{dx}\left[(1 + 2x)\frac{dy}{dx}\right] = \frac{d}{dx}\left[x + 4y^2\right]$ attempted | | |
| $(1 + 2x)\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 1 + 8y\frac{dy}{dx}$ seen + conclusion | A1 | 2 |
**Part (b)**
| $(1 + 2y)\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} - 8y\frac{d^2y}{dx^2} - 8\left(\frac{dy}{dx}\right)^2 - 2\frac{d^2y}{dx^2}$ or equiv. | M1, A2, 1, 0 | 3 |
| [e.g. $(1 + 2y)\frac{d^3y}{dx^3} = 8\left(\frac{dy}{dx}\right)^2 + 4(2y - 1)\frac{d^2y}{dx^2}$] | | |
**Part (c)**
| $\frac{dy}{dx}$ (at $x = 0$) $= 1$ | B1 | |
| Finding $\frac{d^2y}{dx^2}$ (at $x = 0$) $= 3$ | M1 | |
| Finding $\frac{d^3y}{dx^3}$ at $x = 0$; $= 8$ [A1 f.t. is on part (c) values only] | M1, A1ft | |
| $y = \frac{1}{2} + x + \frac{3}{2}x^2 + \frac{4}{3}x^3 + ...$ | M1, A1 | 6 |
[11]
**Alternative (c):**
| Polynomial for $y$: $y = \frac{1}{2} + ax + bx^2 + cx^3 + ...$ | M1 | |
| In given d.e.: $(1 + 2y)(a + 2bx + 3cx^2 + ...) \equiv x + 4(y + ax + bx^2 + cx^3 + ...)^2$ | M1A1 | |
| $a = 1$ | B1 | Complete method for other coefficients M1, answer A1 |7.
$$( 1 + 2 x ) \frac { \mathrm { d } y } { \mathrm {~d} x } = x + 4 y ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$( 1 + 2 x ) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 1 + 2 ( 4 y - 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x }$$
\item Differentiate equation 1 with respect to $x$ to obtain an equation involving
$$\frac { \mathrm { d } ^ { 3 } } { \mathrm {~d} x ^ { 3 } } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x } , \quad x \text { and } y .$$
Given that $y = \frac { 1 } { 2 }$ at $x = 0$,
\item find a series solution for $y$, in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.\\
(6)(Total 11 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2006 Q7 [11]}}