| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Series solution from differential equation |
| Difficulty | Challenging +1.2 This is a structured FP2 series solution question with clear steps: differentiate the given equation to find the third derivative, then use initial conditions to build a Taylor series. While it requires careful algebraic manipulation and understanding of implicit differentiation, the method is standard and well-practiced in FP2. It's harder than typical A-level questions due to the Further Maths content, but routine within FP2. |
| Spec | 1.07s Parametric and implicit differentiation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = -\frac{2}{y}\left(\frac{dy}{dx}\right)^2 - 2\) | ||
| \(\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}\) seen | B1 | |
| \(\frac{d^3y}{dx^3} = -\frac{4}{y}\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} + \frac{2}{y^2}\left(\frac{dy}{dx}\right)^3\) | M1, A1A1 | M1: divide and differentiate; A1 each term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \(x=0\): \(\frac{d^2y}{dx^2} = \frac{1}{2}\left(-2\times\left(\frac{1}{2}\right)^2 - 4\right) = -\frac{9}{4}\) | M1A1 | |
| \(\frac{d^3y}{dx^3} = \frac{1}{2}\left(-5\times\frac{1}{2}\times\left(-\frac{9}{4}\right) - 2\times\frac{1}{2}\right) = \frac{37}{16}\) | A1 | |
| \(y = 2 + \frac{1}{2}x + \left(-\frac{9}{4}\right)\frac{x^2}{2!} + \left(\frac{37}{16}\right)\frac{x^3}{3!} + \ldots\) | M1 | Using Taylor series with their derivatives |
| \(y = 2 + \frac{1}{2}x - \frac{9}{8}x^2 + \frac{37}{96}x^3 + \ldots\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}\) seen in differentiation | B1 | Must be visible in working |
| Divide equation by \(y\) and differentiate wrt \(x\) using chain and product/quotient rules | M1 | |
| Correct expressions with \(-1\) for each error | A1A1 | Ignore simplification following differentiation; obtaining \(\frac{d^3y}{dx^3} = \ldots\) |
| ALT: B1 as above, then differentiating before dividing | M1 | |
| Rearrange to correct expression for \(\frac{d^3y}{dx^3}\) | A1A1 | \(-1\) each error |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using values for \(x\) and \(\frac{dy}{dx}\) to obtain a value for \(\frac{d^2y}{dx^2}\) | M1 | |
| Correct value for \(\frac{d^2y}{dx^2}\) | A1 | |
| Correct value for \(\frac{d^3y}{dx^3}\) | A1 | |
| Taylor's series formed using their values for the differentials | M1 | Accept \(2!\) or \(2\) and \(3!\) or \(6\) |
| Correct series, must start \(y =\) (or end \(= y\)) | A1 |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = -\frac{2}{y}\left(\frac{dy}{dx}\right)^2 - 2$ | | |
| $\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}$ seen | B1 | |
| $\frac{d^3y}{dx^3} = -\frac{4}{y}\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} + \frac{2}{y^2}\left(\frac{dy}{dx}\right)^3$ | M1, A1A1 | M1: divide and differentiate; A1 each term |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $x=0$: $\frac{d^2y}{dx^2} = \frac{1}{2}\left(-2\times\left(\frac{1}{2}\right)^2 - 4\right) = -\frac{9}{4}$ | M1A1 | |
| $\frac{d^3y}{dx^3} = \frac{1}{2}\left(-5\times\frac{1}{2}\times\left(-\frac{9}{4}\right) - 2\times\frac{1}{2}\right) = \frac{37}{16}$ | A1 | |
| $y = 2 + \frac{1}{2}x + \left(-\frac{9}{4}\right)\frac{x^2}{2!} + \left(\frac{37}{16}\right)\frac{x^3}{3!} + \ldots$ | M1 | Using Taylor series with their derivatives |
| $y = 2 + \frac{1}{2}x - \frac{9}{8}x^2 + \frac{37}{96}x^3 + \ldots$ | A1 | Correct final answer |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}$ seen in differentiation | B1 | Must be visible in working |
| Divide equation by $y$ and differentiate wrt $x$ using chain and product/quotient rules | M1 | |
| Correct expressions with $-1$ for each error | A1A1 | Ignore simplification following differentiation; obtaining $\frac{d^3y}{dx^3} = \ldots$ |
| **ALT:** B1 as above, then differentiating before dividing | M1 | |
| Rearrange to correct expression for $\frac{d^3y}{dx^3}$ | A1A1 | $-1$ each error |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using values for $x$ and $\frac{dy}{dx}$ to obtain a value for $\frac{d^2y}{dx^2}$ | M1 | |
| Correct value for $\frac{d^2y}{dx^2}$ | A1 | |
| Correct value for $\frac{d^3y}{dx^3}$ | A1 | |
| Taylor's series formed using their values for the differentials | M1 | Accept $2!$ or $2$ and $3!$ or $6$ |
| Correct series, must start $y =$ (or end $= y$) | A1 | |
---5.
$$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 2 y = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$ in terms of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x }$ and $y$.
Given that $y = 2$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0.5$ at $x = 0$,
\item find a series solution for $y$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.\\
5. $y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 2 y = 0$\\
(a) Find an expression for $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$ in terms of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x }$ and $y$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2014 Q5 [9]}}