Edexcel F2 2021 January — Question 5 9 marks

Exam BoardEdexcel
ModuleF2 (Further Pure Mathematics 2)
Year2021
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeFind higher derivatives from equation
DifficultyChallenging +1.2 This is a Further Maths F2 question requiring differentiation of an implicit differential equation and series solution construction. While it involves careful algebraic manipulation and multiple derivatives, it follows a standard algorithmic procedure: differentiate the given equation using product/chain rules, substitute given values sequentially to find coefficients, then write the Maclaurin series. The techniques are routine for Further Maths students, though the algebra requires care. Slightly above average difficulty due to the multi-step nature and Further Maths context, but not requiring novel insight.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08a Maclaurin series: find series for function
5. Given that $$\left( 2 - x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } = 3 y$$
  1. show that $$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 1 } { \left( 2 - x ^ { 2 } \right) } \left( 2 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \left( 1 - 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) - 5 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right)$$ Given also that \(y = 3\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 4 }\) at \(x = 0\)
  2. obtain a series solution for \(y\) in ascending powers of \(x\) with simplified coefficients, up to and including the term in \(x ^ { 3 }\)
Question 5:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(-2x\frac{d^2y}{dx^2} + (2-x^2)\frac{d^3y}{dx^3}\)M1 Differentiate \((2-x^2)\frac{d^2y}{dx^2}\) using product rule
\(+5\left(\frac{dy}{dx}\right)^2 + 5x \times 2\frac{dy}{dx}\frac{d^2y}{dx^2} = 3\frac{dy}{dx}\)M1A1, B1 Differentiate \(5x\left(\frac{dy}{dx}\right)^2\) using product and chain rule; A1 correct derivative of \(5x\left(\frac{dy}{dx}\right)^2\); B1 correct derivative of \(3y\)
\(\frac{d^3y}{dx^3} = \frac{1}{(2-x^2)}\left(2x\frac{d^2y}{dx^2}\left(1-5\frac{dy}{dx}\right)-5\left(\frac{dy}{dx}\right)^2+3\frac{dy}{dx}\right)\) ✱A1✱ Correct result obtained from fully correct working
ALT 1 (Rearrange and use quotient rule):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\frac{d^2y}{dx^2} = \frac{3y-5x\left(\frac{dy}{dx}\right)^2}{(2-x^2)}\)
\(\frac{d^3y}{dx^3} = \frac{\left[3\frac{dy}{dx}-5\left(\frac{dy}{dx}\right)^2-5x\times 2\frac{dy}{dx}\frac{d^2y}{dx^2}\right](2-x^2)-\left[3y-5x\left(\frac{dy}{dx}\right)^2\right](-2x)}{(2-x^2)^2}\)M1M1A1 Use quotient rule (denom must be \((2-x^2)^2\), numerator difference of 2 terms); differentiate \(\left[3y-5x\left(\frac{dy}{dx}\right)^2\right]\) using product and chain rule; fully correct differentiation
Replace \(3y\) with \((2-x^2)\frac{d^2y}{dx^2}+5x\frac{dy}{dx}\)M1 NB: B1 on ePEN
\(\frac{d^3y}{dx^3} = \frac{1}{(2-x^2)}\left(2x\frac{d^2y}{dx^2}\left(1-5\frac{dy}{dx}\right)-5\left(\frac{dy}{dx}\right)^2+3\frac{dy}{dx}\right)\) ✱A1✱ Correct result from fully correct working
ALT 2 (Rearrange, separate into 2 fractions, then use quotient rule):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\frac{d^2y}{dx^2} = \frac{3y}{(2-x^2)} - \frac{5x\left(\frac{dy}{dx}\right)^2}{(2-x^2)}\)
Differentiate both fractions using quotient ruleM1M1A1 Denominators must be \((2-x^2)^2\), numerator of each to be difference of 2 terms; differentiate \(3y\) using chain rule AND differentiate \(5x\left(\frac{dy}{dx}\right)^2\) using product and chain rule; fully correct differentiation
Replace \(3y\) with \((2-x^2)\frac{d^2y}{dx^2}+5x\frac{dy}{dx}\)M1 NB: B1 on ePEN
\(\frac{d^3y}{dx^3} = \frac{1}{(2-x^2)}\left(2x\frac{d^2y}{dx^2}\left(1-5\frac{dy}{dx}\right)-5\left(\frac{dy}{dx}\right)^2+3\frac{dy}{dx}\right)\) ✱A1✱ Correct result from fully correct working
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(x=0 \Rightarrow 2\frac{d^2y}{dx^2} = 9 \Rightarrow \frac{d^2y}{dx^2} = \frac{9}{2}\)B1 Correct value of \(\frac{d^2y}{dx^2}\)
\(\frac{d^3y}{dx^3} = \frac{1}{2}\left(-5\left(\frac{dy}{dx}\right)^2+3\frac{dy}{dx}\right) = \frac{1}{2}\left(-5\times\frac{1}{16}+\frac{3}{4}\right) = \frac{7}{32}\)M1 Use given result from (a) to obtain value for \(\frac{d^3y}{dx^3}\)
\(y = 3 + \frac{1}{4}x + \frac{9}{2}\cdot\frac{x^2}{2!} + \frac{7}{32}\cdot\frac{x^3}{3!}\)M1 Taylor's series formed using their values for the derivatives (accept \(2!\) or \(2\) and \(3!\) or \(6\))
\(y = 3 + \frac{1}{4}x + \frac{9}{4}x^2 + \frac{7}{192}x^3\)A1 (4) Correct series, must start (or end) \(y=\ldots\) but accept \(f(x)\) provided \(y=f(x)\) defined somewhere 
# Question 5:

## Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $-2x\frac{d^2y}{dx^2} + (2-x^2)\frac{d^3y}{dx^3}$ | M1 | Differentiate $(2-x^2)\frac{d^2y}{dx^2}$ using product rule |
| $+5\left(\frac{dy}{dx}\right)^2 + 5x \times 2\frac{dy}{dx}\frac{d^2y}{dx^2} = 3\frac{dy}{dx}$ | M1A1, B1 | Differentiate $5x\left(\frac{dy}{dx}\right)^2$ using product and chain rule; A1 correct derivative of $5x\left(\frac{dy}{dx}\right)^2$; B1 correct derivative of $3y$ |
| $\frac{d^3y}{dx^3} = \frac{1}{(2-x^2)}\left(2x\frac{d^2y}{dx^2}\left(1-5\frac{dy}{dx}\right)-5\left(\frac{dy}{dx}\right)^2+3\frac{dy}{dx}\right)$ ✱ | A1✱ | Correct result obtained from fully correct working |

**ALT 1** (Rearrange and use quotient rule):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = \frac{3y-5x\left(\frac{dy}{dx}\right)^2}{(2-x^2)}$ | | |
| $\frac{d^3y}{dx^3} = \frac{\left[3\frac{dy}{dx}-5\left(\frac{dy}{dx}\right)^2-5x\times 2\frac{dy}{dx}\frac{d^2y}{dx^2}\right](2-x^2)-\left[3y-5x\left(\frac{dy}{dx}\right)^2\right](-2x)}{(2-x^2)^2}$ | M1M1A1 | Use quotient rule (denom must be $(2-x^2)^2$, numerator difference of 2 terms); differentiate $\left[3y-5x\left(\frac{dy}{dx}\right)^2\right]$ using product and chain rule; fully correct differentiation |
| Replace $3y$ with $(2-x^2)\frac{d^2y}{dx^2}+5x\frac{dy}{dx}$ | M1 | NB: B1 on ePEN |
| $\frac{d^3y}{dx^3} = \frac{1}{(2-x^2)}\left(2x\frac{d^2y}{dx^2}\left(1-5\frac{dy}{dx}\right)-5\left(\frac{dy}{dx}\right)^2+3\frac{dy}{dx}\right)$ ✱ | A1✱ | Correct result from fully correct working |

**ALT 2** (Rearrange, separate into 2 fractions, then use quotient rule):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = \frac{3y}{(2-x^2)} - \frac{5x\left(\frac{dy}{dx}\right)^2}{(2-x^2)}$ | | |
| Differentiate both fractions using quotient rule | M1M1A1 | Denominators must be $(2-x^2)^2$, numerator of each to be difference of 2 terms; differentiate $3y$ using chain rule AND differentiate $5x\left(\frac{dy}{dx}\right)^2$ using product and chain rule; fully correct differentiation |
| Replace $3y$ with $(2-x^2)\frac{d^2y}{dx^2}+5x\frac{dy}{dx}$ | M1 | NB: B1 on ePEN |
| $\frac{d^3y}{dx^3} = \frac{1}{(2-x^2)}\left(2x\frac{d^2y}{dx^2}\left(1-5\frac{dy}{dx}\right)-5\left(\frac{dy}{dx}\right)^2+3\frac{dy}{dx}\right)$ ✱ | A1✱ | Correct result from fully correct working |

## Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $x=0 \Rightarrow 2\frac{d^2y}{dx^2} = 9 \Rightarrow \frac{d^2y}{dx^2} = \frac{9}{2}$ | B1 | Correct value of $\frac{d^2y}{dx^2}$ |
| $\frac{d^3y}{dx^3} = \frac{1}{2}\left(-5\left(\frac{dy}{dx}\right)^2+3\frac{dy}{dx}\right) = \frac{1}{2}\left(-5\times\frac{1}{16}+\frac{3}{4}\right) = \frac{7}{32}$ | M1 | Use given result from (a) to obtain value for $\frac{d^3y}{dx^3}$ |
| $y = 3 + \frac{1}{4}x + \frac{9}{2}\cdot\frac{x^2}{2!} + \frac{7}{32}\cdot\frac{x^3}{3!}$ | M1 | Taylor's series formed using their values for the derivatives (accept $2!$ or $2$ and $3!$ or $6$) |
| $y = 3 + \frac{1}{4}x + \frac{9}{4}x^2 + \frac{7}{192}x^3$ | A1 (4) | Correct series, must start (or end) $y=\ldots$ but accept $f(x)$ provided $y=f(x)$ defined somewhere |

---
5. Given that

$$\left( 2 - x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } = 3 y$$
\begin{enumerate}[label=(\alph*)]
\item show that

$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 1 } { \left( 2 - x ^ { 2 } \right) } \left( 2 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \left( 1 - 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) - 5 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right)$$

Given also that $y = 3$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 4 }$ at $x = 0$
\item obtain a series solution for $y$ in ascending powers of $x$ with simplified coefficients, up to and including the term in $x ^ { 3 }$\\

\begin{center}

\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel F2 2021 Q5 [9]}}
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