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Date	May 2022	Marks available	1	Reference code	22M.2.AHL.TZ1.12
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	12	Adapted from	N/A

Question

Consider the differential equation $x^{2} \frac{d y}{d x} = y^{2} - 2 x^{2}$ for $x > 0$ and $y > 2 x$ . It is given that $y = 3$ when $x = 1$ .

Use Euler’s method, with a step length of $0.1$ , to find an approximate value of $y$ when $x = 1.5$ .

[4]

Use the substitution $y = v x$ to show that $x \frac{d v}{d x} = v^{2} - v - 2$ .

[3]

By solving the differential equation, show that $y = \frac{8 x + x^{4}}{4 - x^{3}}$ .

[10]

c.i.

Find the actual value of $y$ when $x = 1.5$ .

[1]

c.ii.

Using the graph of $y = \frac{8 x + x^{4}}{4 - x^{3}}$ , suggest a reason why the approximation given by Euler’s method in part (a) is not a good estimate to the actual value of $y$ at $x = 1.5$ .

[1]

c.iii.

Markscheme

attempt to use Euler’s method (M1)

$x_{n + 1} = x_{n} + 0.1; y_{n + 1} = y_{n} + 0.1 \times \frac{d y}{d x}$ , where $\frac{d y}{d x} = \frac{y^{2} - 2 x^{2}}{x^{2}}$

correct intermediate $y$ -values (A1)(A1)

$3.7, 4.63140 \dots, 5.92098, 7.79542 \dots$

Note: A1 for any two correct $y$ -values seen

$y = 10.6958 \dots$

$y = 10.7$ A1

Note: For the final A1, the value $10.7$ must be the last value in a table or a list, or be given as a final answer, not just embedded in a table which has further lines.

[4 marks]

$y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ (A1)

replacing $y$ with $v x$ and $\frac{d y}{d x}$ with $v + x \frac{d v}{d x}$ M1

$x^{2} \frac{d y}{d x} = y^{2} - 2 x^{2} \Rightarrow x^{2} (v + x \frac{d v}{d x}) = v^{2} x^{2} - 2 x^{2}$ A1

$v + x \frac{d v}{d x} = v^{2} - 2$ (since $x > 0$ )

$x \frac{d v}{d x} = v^{2} - v - 2$ AG

[3 marks]

attempt to separate variables $v$ and $x$ (M1)

$\int \frac{d v}{v^{2} - v - 2} = \int \frac{d x}{x}$

$\int \frac{d v}{(v - 2) (v + 1)} = \int \frac{d x}{x}$ (A1)

attempt to express in partial fraction form M1

$\frac{1}{(v - 2) (v + 1)} \equiv \frac{A}{v - 2} + \frac{B}{v + 1}$

$\frac{1}{(v - 2) (v + 1)} = \frac{1}{3} (\frac{1}{v - 2} - \frac{1}{v + 1})$ A1

$\frac{1}{3} \int (\frac{1}{v - 2} - \frac{1}{v + 1}) d v = \int \frac{d x}{x}$

$\frac{1}{3} (\ln |v - 2| - \ln |v + 1|) = \ln |x| (+ c)$ A1

Note: Condone absence of modulus signs throughout.

EITHER

attempt to find $c$ using $x = 1, y = 3, v = 3$ M1

$c = \frac{1}{3} \ln \frac{1}{4}$

$\frac{1}{3} (\ln |v - 2| - \ln |v + 1|) = \ln |x| + \frac{1}{3} \ln \frac{1}{4}$

expressing both sides as a single logarithm (M1)

$\ln |\frac{v - 2}{v + 1}| = \ln (\frac{{|x|}^{3}}{4})$

expressing both sides as a single logarithm (M1)

$\ln |\frac{v - 2}{v + 1}| = \ln (A {|x|}^{3})$

attempt to find $A$ using $x = 1, y = 3, v = 3$ M1

$A = \frac{1}{4}$

THEN

$|\frac{v - 2}{v + 1}| = \frac{1}{4} x^{3}$ (since $x > 0$ )

substitute $v = \frac{y}{x}$ (seen anywhere) M1

$\frac{\frac{y}{x} - 2}{\frac{y}{x} + 1} = \frac{1}{4} x^{3}$ (since $y > 2 x$ )

$(\Rightarrow \frac{y - 2 x}{y + x} = \frac{1}{4} x^{3})$

attempt to make $y$ the subject M1

$y - \frac{x^{3} y}{4} = 2 x + \frac{x^{4}}{4}$ A1

$y = \frac{8 x + x^{4}}{4 - x^{3}}$ AG

[10 marks]

c.i.

actual value at $y (1.5) = 27.3$ A1

[1 mark]

c.ii.

gradient changes rapidly (during the interval considered) OR

the curve has a vertical asymptote at $x = \sqrt[3]{4} (= 1.5874 \dots)$ R1

[1 mark]

c.iii.

Examiners report

Most candidates showed evidence of an attempt to use Euler's method in part a), although very few explicitly wrote down the formulae, they used in order to calculate successive y-value. In addition, many seemed to take a step-by-step approach rather than using the recursive capabilities of the graphical display calculator.

There were many good attempts at part b), but not all candidates recognised that this would help them to solve part c).

Part c) was done very well by many candidates, although there were a significant number who failed to recognise the need for partial fractions and could not make further progress. A common error was to integrate without a constant of integration, which meant that the initial condition could not be used. The reasoning given for the estimate being poor was often too vague and did not address the specific nature of the function given clearly enough.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

Syllabus sections

Topic 1—Number and algebra » AHL 1.11—Partial fractions

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