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Date May 2012 Marks available 8 Reference code 12M.3ca.hl.TZ0.2
Level HL only Paper Paper 3 Calculus Time zone TZ0
Command term Find and Show that Question number 2 Adapted from N/A

Question

Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2}}}{{1 + x}}\), where x > −1 and y = 1 when x = 0 .

Use Euler’s method, with a step length of 0.1, to find an approximate value of when x = 0.5.

[7]
a.

(i)     Show that \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{2{y^3} - {y^2}}}{{{{(1 + x)}^2}}}\).

(ii)     Hence find the Maclaurin series for y, up to and including the term in \({x^2}\) .

[8]
b.

(i)     Solve the differential equation.

(ii)     Find the value of a for which \(y \to \infty \) as \(x \to a\).

[6]
c.

Markscheme

attempt the first step of

\({y_{n + 1}} = {y_n} + (0.1)f({x_n},\,{y_n})\) with \({y_0} = 1,{\text{ }}{x_0} = 0\)     (M1)

\({y_1} = 1.1\)     A1

\({y_2} = 1.1 + (0.1)\frac{{{{1.1}^2}}}{{1.1}} = 1.21\)     (M1)A1

\({y_3} = 1.332(0)\)     (A1)

\({y_4} = 1.4685\)     (A1)

\({y_5} = 1.62\)     A1

[7 marks]

a.

(i)     recognition of both quotient rule and implicit differentiation     M1

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{(1 + x)2y\frac{{{\text{d}}y}}{{{\text{d}}x}} - {y^2} \times 1}}{{{{(1 + x)}^2}}}\)     A1A1 

Note: Award A1 for first term in numerator, A1 for everything else correct.

 

\( = \frac{{(1 + x)2y\frac{{{y^2}}}{{1 + x}} - {y^2} \times 1}}{{{{(1 + x)}^2}}}\)     M1A1

\( = \frac{{2{y^3} - {y^2}}}{{{{(1 + x)}^2}}}\)     AG

 

(ii)     attempt to use \(y = y(0) + x\frac{{{\text{d}}y}}{{{\text{d}}x}}(0) + \frac{{{x^2}}}{{2!}}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}(0) + ...\)     (M1)

\( = 1 + x + \frac{{{x^2}}}{2}\)     A1A1 

Note: Award A1 for correct evaluation of \(y(0),{\text{ }}\frac{{dy}}{{dx}}(0),{\text{ }}\frac{{{d^2}y}}{{d{x^2}}}(0)\), A1 for correct series.

 

[8 marks]

b.

(i)     separating the variables \(\int {\frac{1}{{{y^2}}}{\text{d}}y = \int {\frac{1}{{1 + x}}{\text{d}}x} } \)     M1

obtain \( - \frac{1}{y} = \ln (1 + x) + (c)\)     A1

impose initial condition \( - 1 = \ln 1 + c\)     M1

obtain \(y = \frac{1}{{1 - \ln (1 + x)}}\)     A1

 

(ii)     \(y \to \infty \) if \(\ln (1 + x) \to 1\) , so a = e – 1     (M1)A1 

Note: To award A1 must see either \(x \to e - 1\) or a = e – 1 . Do not accept x = e – 1.

 

[6 marks]

c.

Examiners report

Most candidates had a good knowledge of Euler’s method and were confident in applying it to the differential equation in part (a). A few candidates who knew the Euler’s method completed one iteration too many to arrive at an incorrect answer but this was rare. Nearly all candidates who applied the correct technique in part (a) correctly calculated the answer. Most candidates were able to attempt part (b) but some lost marks due to a lack of rigour by not clearly showing the implicit differentiation in the first line of working. Part (c) was reasonably well attempted by many candidates and many could solve the integrals although some did not find the arbitrary constant meaning that it was not possible to solve (ii) of the part (c).

a.

Most candidates had a good knowledge of Euler’s method and were confident in applying it to the differential equation in part (a). A few candidates who knew the Euler’s method completed one iteration too many to arrive at an incorrect answer but this was rare. Nearly all candidates who applied the correct technique in part (a) correctly calculated the answer. Most candidates were able to attempt part (b) but some lost marks due to a lack of rigour by not clearly showing the implicit differentiation in the first line of working. Part (c) was reasonably well attempted by many candidates and many could solve the integrals although some did not find the arbitrary constant meaning that it was not possible to solve (ii) of the part (c).

b.

Most candidates had a good knowledge of Euler’s method and were confident in applying it to the differential equation in part (a). A few candidates who knew the Euler’s method completed one iteration too many to arrive at an incorrect answer but this was rare. Nearly all candidates who applied the correct technique in part (a) correctly calculated the answer. Most candidates were able to attempt part (b) but some lost marks due to a lack of rigour by not clearly showing the implicit differentiation in the first line of working. Part (c) was reasonably well attempted by many candidates and many could solve the integrals although some did not find the arbitrary constant meaning that it was not possible to solve (ii) of the part (c).

c.

Syllabus sections

Topic 9 - Option: Calculus » 9.6 » Taylor series developed from differential equations.

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