| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2002 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Implicit differential equation series solution |
| Difficulty | Challenging +1.2 This is a standard FP2 implicit differentiation question requiring systematic differentiation of the given equation, substitution of initial conditions, and series construction. While it involves multiple derivatives and careful algebraic manipulation, the method is algorithmic and well-practiced. Part (c) adds minimal difficulty as it's a straightforward comment on radius of convergence. Harder than average A-level but routine for Further Maths students who have practiced this technique. |
| Spec | 4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y\frac{d^3y}{dx^3}+\frac{dy}{dx}\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0\) | M1 A1; B1; B1 | marks can be awarded in (b) |
| \(\frac{d^3y}{dx^3}=\frac{-3\frac{dy}{dx}\frac{d^2y}{dx^2}-\frac{dy}{dx}}{y}\) or sensible correct alternative | B1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x=0\): \(\frac{d^2y}{dx^2}=-2\), and \(\frac{d^3y}{dx^3}=5\) | M1A1, A1 ft | |
| \(\therefore y=1+x-x^2+\frac{5}{6}x^3\ldots\) | M1, A1 ft | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Could use for \(x=0.2\) but not for \(x=50\) | B1 | |
| approximation is best at values close to \(x=0\) | B1 | (2) |
| (12 marks) |
# Question 10:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y\frac{d^3y}{dx^3}+\frac{dy}{dx}\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$ | M1 A1; B1; B1 | marks can be awarded in (b) |
| $\frac{d^3y}{dx^3}=\frac{-3\frac{dy}{dx}\frac{d^2y}{dx^2}-\frac{dy}{dx}}{y}$ or sensible correct alternative | B1 | **(5)** |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x=0$: $\frac{d^2y}{dx^2}=-2$, and $\frac{d^3y}{dx^3}=5$ | M1A1, A1 ft | |
| $\therefore y=1+x-x^2+\frac{5}{6}x^3\ldots$ | M1, A1 ft | **(5)** |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Could use for $x=0.2$ but not for $x=50$ | B1 | |
| approximation is best at values close to $x=0$ | B1 | **(2)** |
| | | **(12 marks)** |10.
$$y \frac { d ^ { 2 } y } { d x ^ { 2 } } + \left( \frac { d y } { d x } \right) ^ { 2 } + y = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$.
Given that $y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ at $x = 0$,
\item find the series solution for $y$, in ascending powers of $x$, up to an including the term in $x ^ { 3 }$.
\item Comment on whether it would be sensible to use your series solution to give estimates for $y$ at $x = 0.2$ and at $x = 50$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2002 Q10 [12]}}