| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| 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 standard Further Maths series solution question requiring differentiation of the given DE to find higher derivatives, then building a Maclaurin series using initial conditions. While it involves multiple steps and careful algebraic manipulation, the technique is routine for F2 students with no novel insight required. The method is algorithmic: differentiate the DE, substitute to find coefficients, construct the series. Slightly above average difficulty due to the algebraic care needed and being Further Maths content, but well within the standard F2 syllabus. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y''' - 2y' - 2xy'' + 2y'(=0)\ (y'''=2xy'')\) | M1A1 | M1: Attempt to differentiate including use of product rule on \(2x\frac{dy}{dx}\); equation may have been rewritten as \(y''=\ldots\) before differentiating; A1: Correct differentiation |
| \(y'''' - 2y'' - 2xy''' - 2y'' + 2y''(=0)\) | dM1A1 | M1: Second use of product rule, dependent on first M1; A1: Correct differentiation |
| \(y'''' = 2xy''' + 2y'' = 2x(2xy'') + 2y'' = (4x^2+2)y''\) | A1 | Cao and cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y_0'' = -2\) | B1 | — |
| \(y_0''' = 0,\ y_0^{iv} = -4\) | M1A1 | M1: Attempts \(y_0'''\) and \(y_0^{iv}\); A1: All correct and obtained from correct expressions |
| \((y=)1+3x-x^2-\frac{x^4}{6}\) | M1A1 | M1: Correct use of Maclaurin series; A1: Fully correct expansion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((y=)1+3(-0.2)-(-0.2)^2-\frac{(-0.2)^4}{6}\) | M1 | Use of the correct Maclaurin series and substitution of \(x=-0.2\) |
| \((y=)\ 0.3597\) | A1 | Allow awrt |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y''' - 2y' - 2xy'' + 2y'(=0)\ (y'''=2xy'')$ | M1A1 | M1: Attempt to differentiate including use of product rule on $2x\frac{dy}{dx}$; equation may have been rewritten as $y''=\ldots$ before differentiating; A1: Correct differentiation |
| $y'''' - 2y'' - 2xy''' - 2y'' + 2y''(=0)$ | dM1A1 | M1: Second use of product rule, dependent on first M1; A1: Correct differentiation |
| $y'''' = 2xy''' + 2y'' = 2x(2xy'') + 2y'' = (4x^2+2)y''$ | A1 | Cao and cso |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y_0'' = -2$ | B1 | — |
| $y_0''' = 0,\ y_0^{iv} = -4$ | M1A1 | M1: Attempts $y_0'''$ and $y_0^{iv}$; A1: All correct and obtained from correct expressions |
| $(y=)1+3x-x^2-\frac{x^4}{6}$ | M1A1 | M1: Correct use of Maclaurin series; A1: Fully correct expansion |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y=)1+3(-0.2)-(-0.2)^2-\frac{(-0.2)^4}{6}$ | M1 | Use of the correct Maclaurin series and substitution of $x=-0.2$ |
| $(y=)\ 0.3597$ | A1 | Allow awrt |5.
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = \left( a x ^ { 2 } + b \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$
where $a$ and $b$ are constants to be found.
Given that $y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$ at $x = 0$
\item find a series solution for $y$ in ascending powers of $x$ up to and including the term in $x ^ { 4 }$
\item use your series to estimate the value of $y$ at $x = - 0.2$, giving your answer to four decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2014 Q5 [12]}}