Question 4 - A-Level Maths

Edexcel FP2 — Question 4 8 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Taylor series
Type	Explicit differential equation series solution
Difficulty	Challenging +1.2 This is a systematic Taylor series application to a nonlinear ODE with clear initial conditions. While it requires careful differentiation and substitution through multiple derivatives, the method is algorithmic and well-practiced in FP2. The nonlinearity adds some algebraic complexity beyond routine questions, but the structure is standard for this topic.
Spec	4.08a Maclaurin series: find series for function

4. Use the Taylor Series method to find the series solution, ascending up to and including the term in $x ^ { 3 }$, of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } = 3 x + 4$$ given that $\frac { \mathrm { dy } } { \mathrm { dx } } = y = 1$ at $x = 0$.
(Total 8 marks)

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Question 4:

Answer	Marks	Guidance
Working	Marks	Notes
$\frac{d^2y}{dx^2} + 1 + 1 = 4$ at $x=0$, $\therefore \frac{d^2y}{dx^2} = 2$	B1
Differentiate to give $\frac{d^3y}{dx^3} + \left[\left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2}\right] + 2y\frac{dy}{dx} = 3$	M1 [M1 A1]
At $x=0$: $\frac{d^3y}{dx^3} + [1^2 + 1\times 2] + 2 = 3$ and $\frac{d^3y}{dx^3} = -2$	B1
$y = 1 + x + \frac{2x^2}{2} - \frac{2x^3}{6} + \ldots$	M1 A1

# Question 4:

| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{d^2y}{dx^2} + 1 + 1 = 4$ at $x=0$, $\therefore \frac{d^2y}{dx^2} = 2$ | B1 | |
| Differentiate to give $\frac{d^3y}{dx^3} + \left[\left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2}\right] + 2y\frac{dy}{dx} = 3$ | M1 [M1 A1] | |
| At $x=0$: $\frac{d^3y}{dx^3} + [1^2 + 1\times 2] + 2 = 3$ and $\frac{d^3y}{dx^3} = -2$ | B1 | |
| $y = 1 + x + \frac{2x^2}{2} - \frac{2x^3}{6} + \ldots$ | M1 A1 | |

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4. Use the Taylor Series method to find the series solution, ascending up to and including the term in $x ^ { 3 }$, of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } = 3 x + 4$$

given that $\frac { \mathrm { dy } } { \mathrm { dx } } = y = 1$ at $x = 0$.\\
(Total 8 marks)\\

\hfill \mbox{\textit{Edexcel FP2  Q4 [8]}}

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