Question 3 - A-Level Maths

Edexcel FP2 2013 June — Question 3 5 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	June
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Series solution from differential equation
Difficulty	Standard +0.8 This is a Further Maths FP2 question requiring series solution method for a second-order differential equation with initial conditions. Students must differentiate repeatedly, substitute into the DE to find successive derivatives, then construct the Maclaurin series. While systematic, it requires careful algebraic manipulation and understanding of the series solution technique, placing it moderately above average difficulty.
Spec	4.08a Maclaurin series: find series for function 4.10d Second order homogeneous: auxiliary equation method

3. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y - \sin x = 0$$ Given that $y = \frac { 1 } { 2 }$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 8 }$ at $x = 0$, find a series expansion for $y$ in terms of $x$, up to and including the term in $x ^ { 3 }$.

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Question 3:

Answer	Marks	Guidance
Answer	Mark	Guidance
$(x=0)$: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \sin 0 - 4 \times \dfrac{1}{2} = -2$	B1	For $\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 = -2$ wherever seen
$\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} - \cos x = 0$	M1	For attempting differentiation of the given equation to obtain $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + k\dfrac{\mathrm{d}y}{\mathrm{d}x} \pm \cos x = 0$
$(x=0)$: $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \cos 0 - 4 \times \dfrac{1}{8} = \dfrac{1}{2}$	A1	For substituting $x=0$ to obtain $\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 = \dfrac{1}{2}$
$y = y_0 + x\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0 + \dfrac{x^2}{2!}\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 + \dfrac{x^3}{3!}\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 + \ldots$	M1	For using the expansion with their values for $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}$ and $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$. Factorial can be omitted in $x^2$ term but must be shown explicitly in $x^3$ term or implied by further working
$y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}$	A1 cao (5)	Ignore any higher powers included. Exact decimals allowed. Must include $y=\ldots$

Alternative:

Answer	Marks	Guidance
Answer	Mark	Guidance
$y = \dfrac{1}{2} + \dfrac{x}{8} + ax^2 + bx^3 + \ldots$	B1
$y'' = 2a + 6bx + \ldots$	M1	Differentiate twice (may not be completely correct)
$2a + 6bx + \ldots = \sin x - \left(\dfrac{1}{2} + \dfrac{x}{8} + ax^2 + bx^3\ldots\right)$	A1	Correct differentiation and using the given equation and expansion of $\sin x$
$2a + 2 = 0 \Rightarrow a = -1$	M1	For equating coefficients to obtain a value for $a$ or $b$
$6b + \dfrac{1}{2} = 1 \Rightarrow b = \dfrac{1}{12}$
$y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}$	A1 cao	Ignore any higher powers included

## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| $(x=0)$: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \sin 0 - 4 \times \dfrac{1}{2} = -2$ | B1 | For $\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 = -2$ wherever seen |
| $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} - \cos x = 0$ | M1 | For attempting differentiation of the given equation to obtain $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + k\dfrac{\mathrm{d}y}{\mathrm{d}x} \pm \cos x = 0$ |
| $(x=0)$: $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \cos 0 - 4 \times \dfrac{1}{8} = \dfrac{1}{2}$ | A1 | For substituting $x=0$ to obtain $\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 = \dfrac{1}{2}$ |
| $y = y_0 + x\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0 + \dfrac{x^2}{2!}\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 + \dfrac{x^3}{3!}\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 + \ldots$ | M1 | For using the expansion with their values for $\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}$ and $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$. Factorial can be omitted in $x^2$ term but must be shown explicitly in $x^3$ term or implied by further working |
| $y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}$ | A1 cao (5) | Ignore any higher powers included. Exact decimals allowed. Must include $y=\ldots$ |

*Alternative:*
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \dfrac{1}{2} + \dfrac{x}{8} + ax^2 + bx^3 + \ldots$ | B1 | |
| $y'' = 2a + 6bx + \ldots$ | M1 | Differentiate twice (may not be completely correct) |
| $2a + 6bx + \ldots = \sin x - \left(\dfrac{1}{2} + \dfrac{x}{8} + ax^2 + bx^3\ldots\right)$ | A1 | Correct differentiation and using the given equation and expansion of $\sin x$ |
| $2a + 2 = 0 \Rightarrow a = -1$ | M1 | For equating coefficients to obtain a value for $a$ or $b$ |
| $6b + \dfrac{1}{2} = 1 \Rightarrow b = \dfrac{1}{12}$ | | |
| $y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}$ | A1 cao | Ignore any higher powers included |

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3.

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y - \sin x = 0$$

Given that $y = \frac { 1 } { 2 }$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 8 }$ at $x = 0$, find a series expansion for $y$ in terms of $x$, up to and including the term in $x ^ { 3 }$.\\

\hfill \mbox{\textit{Edexcel FP2 2013 Q3 [5]}}

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