| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| 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 Taylor series question requiring systematic differentiation of an implicit differential equation. While it involves multiple derivatives and careful algebraic manipulation, the method is algorithmic: differentiate the given equation, substitute x=0 and known values at each stage. The technique is well-practiced in FP2 and doesn't require novel insight, making it moderately above average difficulty but routine for Further Maths students. |
| Spec | 4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y\dfrac{d^3y}{dx^3} + \dfrac{dy}{dx}\dfrac{d^2y}{dx^2}\) | M1, A1 | M1: Use of Product Rule on \(y\dfrac{d^2y}{dx^2}\), 2 terms added with at least one term correct. A1: Fully correct derivative of \(y\dfrac{d^2y}{dx^2}\) |
| \(+3x\dfrac{d^2y}{dx^2} + 3\dfrac{dy}{dx}\) | B1 | Correct derivative of \(3x\dfrac{dy}{dx}\) |
| \(-6y\dfrac{dy}{dx}\) | B1 | oe |
| At \(x=0\): \(2\dfrac{d^2y}{dx^2} + 3(0)(1) - 3(4) = 0 \Rightarrow \dfrac{d^2y}{dx^2} = \ldots\) and \(2\dfrac{d^3y}{dx^3} + (1)(6) + 3(1) - 6(2)(1) = 0 \Rightarrow \dfrac{d^3y}{dx^3} = \ldots\) | M1 | Sub \(x=0\), \(y=2\) and \(\dfrac{dy}{dx}=1\) (must use these values) leading to numerical values for \(\dfrac{d^2y}{dx^2}\) and \(\dfrac{d^3y}{dx^3}\) |
| \(\dfrac{d^3y}{dx^3} = \dfrac{3}{2}\) | A1cso(6) | Given answer** cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((y =)\ 2 + x\) | B1 | Use the given values to form the first 2 terms of the series |
| \((y =)\ 2 + x + \dfrac{6}{2!}x^2 + \dfrac{\frac{3}{2}}{3!}x^3\ (+\ldots)\) | M1 | Find a numerical value for \(\dfrac{d^2y}{dx^2}\) (may be seen in (a)) and use with the given value of \(\dfrac{d^3y}{dx^3}\) to form the \(x^2\) and \(x^3\) terms of the series expansion |
| \(y = 2 + x + 3x^2 + \dfrac{1}{4}x^3\ (+\ldots)\) | A1ft | Follow through their value of \(\dfrac{d^2y}{dx^2}\) used correctly. Must start \(y = \ldots\) Allow \(f(x)\) only if this has been defined anywhere in the question to be equal to \(y\) |
## Question 5:
$y\dfrac{d^2y}{dx^2} + 3x\dfrac{dy}{dx} - 3y^2 = 0$
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y\dfrac{d^3y}{dx^3} + \dfrac{dy}{dx}\dfrac{d^2y}{dx^2}$ | M1, A1 | M1: Use of Product Rule on $y\dfrac{d^2y}{dx^2}$, 2 terms added with at least one term correct. A1: Fully correct derivative of $y\dfrac{d^2y}{dx^2}$ |
| $+3x\dfrac{d^2y}{dx^2} + 3\dfrac{dy}{dx}$ | B1 | Correct derivative of $3x\dfrac{dy}{dx}$ |
| $-6y\dfrac{dy}{dx}$ | B1 | oe |
| At $x=0$: $2\dfrac{d^2y}{dx^2} + 3(0)(1) - 3(4) = 0 \Rightarrow \dfrac{d^2y}{dx^2} = \ldots$ and $2\dfrac{d^3y}{dx^3} + (1)(6) + 3(1) - 6(2)(1) = 0 \Rightarrow \dfrac{d^3y}{dx^3} = \ldots$ | M1 | Sub $x=0$, $y=2$ and $\dfrac{dy}{dx}=1$ (must use these values) leading to numerical values for $\dfrac{d^2y}{dx^2}$ **and** $\dfrac{d^3y}{dx^3}$ |
| $\dfrac{d^3y}{dx^3} = \dfrac{3}{2}$** | A1cso(6) | **Given answer** cso |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y =)\ 2 + x$ | B1 | Use the given values to form the first 2 terms of the series |
| $(y =)\ 2 + x + \dfrac{6}{2!}x^2 + \dfrac{\frac{3}{2}}{3!}x^3\ (+\ldots)$ | M1 | Find a numerical value for $\dfrac{d^2y}{dx^2}$ (may be seen in (a)) and use with the **given** value of $\dfrac{d^3y}{dx^3}$ to form the $x^2$ and $x^3$ terms of the series expansion |
| $y = 2 + x + 3x^2 + \dfrac{1}{4}x^3\ (+\ldots)$ | A1ft | Follow through their value of $\dfrac{d^2y}{dx^2}$ used correctly. Must start $y = \ldots$ Allow $f(x)$ **only** if this has been defined anywhere in the question to be equal to $y$ |
**Total: 9**5.
$$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y ^ { 2 } = 0$$
Given that at $x = 0 , y = 2$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$
\begin{enumerate}[label=(\alph*)]
\item show that, at $x = 0 , \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 3 } { 2 }$
\item Find a series solution for $y$ up to and including the term in $x ^ { 3 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2018 Q5 [9]}}