Edexcel FP2 2013 June — Question 4 9 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeImplicit differential equation series solution
DifficultyChallenging +1.2 This is a standard FP2 implicit differentiation question requiring systematic differentiation of the given equation and substitution of initial conditions. While it involves multiple derivatives and careful algebraic manipulation, it follows a well-established procedure taught explicitly in Further Maths courses with no novel problem-solving required.
Spec1.07q Product and quotient rules: differentiation4.08a Maclaurin series: find series for function
  1. Given that
$$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } + 5 y = 0$$
  1. find \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) in terms of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x }\) and \(y\). Given that \(y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2\) at \(x = 0\)
  2. find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).
Question 4
4(a)
M1 for using the product rule to differentiate \(y\frac{d^2y}{dx^2}\)
M1 for differentiating \(5y\) and using the product rule or chain rule to differentiate \(\left(\frac{dy}{dx}\right)^2\)
\[\frac{d^3y}{dx^3} = \frac{-5\frac{dy}{dx} - 3\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}}{y}\]
A2,1,0 for above expression. Give A1A1 if fully correct, A1A0 if one error and A0A0 if more than one error. If there are two sign errors and no other error then give A1A0.
Do NOT deduct if the two terms are shown separately.
Alternative to Q4(a):
Can be re-arranged first and then differentiated.
M1M1 for differentiating, product and chain rule both needed (or quotient rule as an alternative to product rule)
\[\frac{d^3y}{dx^3} = \frac{1}{y^2}\left(\frac{dy}{dx}\right)^3 - \frac{2}{y}\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx^2}\right)\]
A2,1,0 for above expression. Give A1A1 if fully correct, A1A0 if one error and A0A0 if more than one error
4(b)
M1 for substituting \(\frac{dy}{dx}=2\) and \(y=2\) in the equation to obtain a numerical value for \(\frac{d^2y}{dx^2}\)
A1 for \(\frac{d^2y}{dx^2}=-7\)
A1 for obtaining the correct value, \(16\), for \(\frac{d^3y}{dx^3}\)
M1 for using the series \(y=f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\ldots\) (The general series may be shown explicitly or implied by their substitution)
A1 for \(y=2+2x-\frac{7}{2}x^2+\frac{8}{3}x^3\) or equivalent. Must have \(y=\ldots\) and be in ascending powers of \(x\).
Alternative to Q4(b):
M1 for setting \(y=2+2x+ax^2+bx^3\)
M1 for \((2+2x+ax^2+bx^3)(2a+6bx)+(2+2ax+3bx^2\ldots)^2+5(2+2x+ax^2+bx^3)=0\)
A1 for equating constant terms to get \(a=-\frac{7}{2}\)
A1 for equating coefficients of \(x\) to get \(b=\frac{8}{3}\)
A1 for \(y=2+2x-\frac{7}{2}x^2+\frac{8}{3}x^3\)
NOTES
Accept the dash notation in this question
# Question 4

## 4(a)

M1 for using the product rule to differentiate $y\frac{d^2y}{dx^2}$

M1 for differentiating $5y$ and using the product rule or chain rule to differentiate $\left(\frac{dy}{dx}\right)^2$

$$\frac{d^3y}{dx^3} = \frac{-5\frac{dy}{dx} - 3\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}}{y}$$

A2,1,0 for above expression. Give A1A1 if fully correct, A1A0 if one error and A0A0 if more than one error. If there are two sign errors and no other error then give A1A0.

Do NOT deduct if the two terms are shown separately.

**Alternative to Q4(a):**

Can be re-arranged first and then differentiated.

M1M1 for differentiating, product and chain rule both needed (or quotient rule as an alternative to product rule)

$$\frac{d^3y}{dx^3} = \frac{1}{y^2}\left(\frac{dy}{dx}\right)^3 - \frac{2}{y}\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx^2}\right)$$

A2,1,0 for above expression. Give A1A1 if fully correct, A1A0 if one error and A0A0 if more than one error

## 4(b)

M1 for substituting $\frac{dy}{dx}=2$ and $y=2$ in the equation to obtain a numerical value for $\frac{d^2y}{dx^2}$

A1 for $\frac{d^2y}{dx^2}=-7$

A1 for obtaining the correct value, $16$, for $\frac{d^3y}{dx^3}$

M1 for using the series $y=f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\ldots$ (The general series may be shown explicitly or implied by their substitution)

A1 for $y=2+2x-\frac{7}{2}x^2+\frac{8}{3}x^3$ or equivalent. Must have $y=\ldots$ and be in ascending powers of $x$.

**Alternative to Q4(b):**

M1 for setting $y=2+2x+ax^2+bx^3$

M1 for $(2+2x+ax^2+bx^3)(2a+6bx)+(2+2ax+3bx^2\ldots)^2+5(2+2x+ax^2+bx^3)=0$

A1 for equating constant terms to get $a=-\frac{7}{2}$

A1 for equating coefficients of $x$ to get $b=\frac{8}{3}$

A1 for $y=2+2x-\frac{7}{2}x^2+\frac{8}{3}x^3$

**NOTES**

Accept the dash notation in this question
\begin{enumerate}
  \item Given that
\end{enumerate}

$$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } + 5 y = 0$$

(a) find $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$ in terms of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x }$ and $y$.

Given that $y = 2$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2$ at $x = 0$\\
(b) find a series solution for $y$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.

\hfill \mbox{\textit{Edexcel FP2 2013 Q4 [9]}}
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