| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Implicit differential equation series solution |
| Difficulty | Standard +0.8 This is a standard Further Maths FP2 question on finding Taylor series from differential equations through successive differentiation. While it requires careful algebraic manipulation and implicit differentiation to find higher derivatives, the method is well-practiced in FP2. The multi-part structure guides students through the process systematically. It's moderately challenging for Further Maths but follows a predictable template, placing it somewhat above average difficulty overall. |
| Spec | 4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\dfrac{d^2y}{dx^2}\right)_0 = \dfrac{1}{4}\) | B1 | 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\dfrac{dy}{dx}\right)_0 \approx \dfrac{y_1 - y_{-1}}{2h} \Rightarrow \dfrac{1}{2} \approx \dfrac{y_1 - y_{-1}}{0.2} \Rightarrow y_1 - y_{-1} \approx 0.1\) | M1A1 | |
| \(\left(\dfrac{d^2y}{dx^2}\right)_0 \approx \dfrac{y_1 - 2y_0 + y_{-1}}{h^2} \Rightarrow \dfrac{1}{4} \approx \dfrac{y_1 - 2 + y_{-1}}{0.01}\) | M1 | |
| \(\Rightarrow y_1 + y_{-1} \approx 2.0025\) | A1 | |
| Adding to give \(y_1 \approx 1.05125\) | M1A1 | 6 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Diff: \(4(1+x^2)\dfrac{d^3y}{dx^3} + 8x\dfrac{d^2y}{dx^2} + 4x\dfrac{d^2y}{dx^2} + 4\dfrac{dy}{dx} = \dfrac{dy}{dx}\) | M1A1 | |
| Substituting appropriate values \(\Rightarrow 4\left(\dfrac{d^3y}{dx^3}\right)_0 = -\dfrac{3}{2} \Rightarrow \left(\dfrac{d^3y}{dx^3}\right)_0 = -\dfrac{3}{8}\) | M1A1 | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = y_0 + y_0'x + \dfrac{y_0''}{2!}x^2 + \dfrac{y_0'''}{3!}x^3 + \cdots = 1 + \dfrac{1}{2}x + \dfrac{1}{8}x^2 - \dfrac{1}{16}x^3 + \cdots\) | M1A1ft2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.05119\) | A1 | 1 mark total [14] |
# Question 11:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\dfrac{d^2y}{dx^2}\right)_0 = \dfrac{1}{4}$ | B1 | 1 mark total |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\dfrac{dy}{dx}\right)_0 \approx \dfrac{y_1 - y_{-1}}{2h} \Rightarrow \dfrac{1}{2} \approx \dfrac{y_1 - y_{-1}}{0.2} \Rightarrow y_1 - y_{-1} \approx 0.1$ | M1A1 | |
| $\left(\dfrac{d^2y}{dx^2}\right)_0 \approx \dfrac{y_1 - 2y_0 + y_{-1}}{h^2} \Rightarrow \dfrac{1}{4} \approx \dfrac{y_1 - 2 + y_{-1}}{0.01}$ | M1 | |
| $\Rightarrow y_1 + y_{-1} \approx 2.0025$ | A1 | |
| Adding to give $y_1 \approx 1.05125$ | M1A1 | 6 marks total |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Diff: $4(1+x^2)\dfrac{d^3y}{dx^3} + 8x\dfrac{d^2y}{dx^2} + 4x\dfrac{d^2y}{dx^2} + 4\dfrac{dy}{dx} = \dfrac{dy}{dx}$ | M1A1 | |
| Substituting appropriate values $\Rightarrow 4\left(\dfrac{d^3y}{dx^3}\right)_0 = -\dfrac{3}{2} \Rightarrow \left(\dfrac{d^3y}{dx^3}\right)_0 = -\dfrac{3}{8}$ | M1A1 | 4 marks total |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = y_0 + y_0'x + \dfrac{y_0''}{2!}x^2 + \dfrac{y_0'''}{3!}x^3 + \cdots = 1 + \dfrac{1}{2}x + \dfrac{1}{8}x^2 - \dfrac{1}{16}x^3 + \cdots$ | M1A1ft2 | |
## Part (e)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.05119$ | A1 | 1 mark total **[14]** |11. The variable $y$ satisfies the differential equation
$$4 \left( 1 + x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = y$$
At $x = 0 , y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 }$.\\
(a) Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at $x = 0$.\\
(1) (c) Find the value of $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$ at $x = 0$\\
(d) Express $y$ as a series, in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.\\
(e) Find the value that the series gives for $y$ at $x = 0.1$, giving your answer to 5 decimal places.\\
(1)(Total 14 marks)
\hfill \mbox{\textit{Edexcel FP2 2005 Q11 [14]}}