IB DP Maths Topic 9.5 Variables separable. HL Paper 3

 

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Question

(a)     Show that the solution of the differential equation

\[\frac{{{\text{d}}y}}{{{\text{d}}x}} = \cos x{\cos ^2}y{\text{,}}\]

given that \(y = \frac{\pi }{4}{\text{ when }}x = \pi {\text{, is }}y = \arctan (1 + \sin x){\text{.}}\)

(b)     Determine the value of the constant a for which the following limit exists

\[\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\arctan (1 + \sin x) – a}}{{{{\left( {x – \frac{\pi }{2}} \right)}^2}}}\]

and evaluate that limit.

Answer/Explanation

Markscheme

(a)     this separable equation has general solution

\(\int {{{\sec }^2}y{\text{d}}y = \int {\cos x{\text{d}}x} } \)     (M1)(A1)

\(\tan y = \sin x + c\)     A1

the condition gives

\(\tan \frac{\pi }{4} = \sin \pi  + c \Rightarrow c = 1\)     M1

the solution is \(\tan y = 1 + \sin x\)     A1

\(y = \arctan (1 + \sin x)\)     AG

[5 marks]

 

(b)     the limit cannot exist unless \(a = \arctan \left( {1 + \sin \frac{\pi }{2}} \right) = \arctan 2\)     R1A1

in that case the limit can be evaluated using l’Hopital’s rule (twice) limit is

\(\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\left( {\arctan (1 + \sin x)} \right)}^\prime }}}{{2\left( {x – \frac{\pi }{2}} \right)}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y’}}{{2\left( {x – \frac{\pi }{2}} \right)}}\)     M1A1

where y is the solution of the differential equation

the numerator has zero limit (from the factor \(\cos x\) in the differential equation)     R1

so required limit is

\(\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y”}}{2}\)     M1A1

finally,

\(y” = – \sin x{\cos ^2}y – 2\cos x\cos y\sin y \times y'(x)\)     M1A1

since \(\cos y\left( {\frac{\pi }{2}} \right) = \frac{1}{{\sqrt 5 }}\)     A1

\(y” = – \frac{1}{5}{\text{ at }}x = \frac{\pi }{2}\)     A1

the required limit is \( – \frac{1}{{10}}\)     A1

[12 marks]

Total [17 marks]

Examiners report

Many candidates successfully obtained the displayed solution of the differential equation in part(a). Few complete solutions to part(b) were seen which used the result in part(a). The problem can, however, be solved by direct differentiation although this is algebraically more complicated. Some successful solutions using this method were seen.

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.
Answer/Explanation

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.

Question

Find the general solution of the differential equation \(t\frac{{{\text{d}}y}}{{{\text{d}}t}} = \cos t – 2y\) , for t > 0 .

Answer/Explanation

Markscheme

recognise equation as first order linear and attempt to find the IF     M1

\({\text{IF}} = {{\text{e}}^{\int {\frac{2}{t}{\text{d}}t} }} = {t^2}\)     A1

solution \(y{t^2} = \int {t\cos t{\text{d}}t} \)     M1A1

using integration by parts with the correct choice of u and v     (M1)

\(\int {t\cos t{\text{d}}t = t\sin t + \cos t( + C)} \)     A1

obtain \(y = \frac{{\sin t}}{t} + \frac{{\cos t + C}}{{{t^2}}}\)     A1

[7 marks]

Examiners report

Perhaps a small number of candidates were put off by the unusual choice of variables but in most instances it seemed that candidates who recognised the need for an integration factor could make a good attempt at this problem. Candidates who were not able to simplify the integrating factor from \({e^{2\ln t}}\) to \({t^2}\) rarely gained full marks. A significant number of candidates did not gain the final mark due to a lack of an arbitrary constant or not dividing the constant by the integration factor.

Question

Consider the differential equation \(x\frac{{{\text{d}}y}}{{{\text{d}}x}} – y = {x^p} + 1\) where \(x \in \mathbb{R},\,x \ne 0\) and \(p\) is a positive integer, \(p > 1\).

Solve the differential equation given that \(y =  – 1\) when \(x = 1\). Give your answer in the form \(y = f\left( x \right)\).

[8]
a.

Show that the \(x\)-coordinate(s) of the points on the curve \(y = f\left( x \right)\) where \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) satisfy the equation \({x^{p – 1}} = \frac{1}{p}\).

[2]
b.i.

Deduce the set of values for \(p\) such that there are two points on the curve \(y = f\left( x \right)\) where \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\). Give a reason for your answer.

[2]
b.ii.
Answer/Explanation

Markscheme

METHOD 1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} = {x^{p – 1}} + \frac{1}{x}\)    (M1)

integrating factor \( = {{\text{e}}^{\int { – \frac{1}{x}{\text{d}}x} }}\)     M1

\({\text{ = }}{{\text{e}}^{ – {\text{ln}}\,x}}\)     (A1)

= \(\frac{1}{x}\)     A1

\(\frac{1}{x}\frac{{{\text{d}}y}}{{{\text{d}}x}} – \frac{y}{{{x^2}}} = {x^{p – 2}} + \frac{1}{{{x^2}}}\)     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{y}{x}} \right) = {x^{p – 2}} + \frac{1}{{{x^2}}}\)

\(\frac{y}{x} = \frac{1}{{p – 1}}{x^{p – 1}} – \frac{1}{x} + C\)    A1

Note: Condone the absence of C.

\(y = \frac{1}{{p – 1}}{x^p} + Cx – 1\)

substituting \(x = 1\), \(y =  – 1 \Rightarrow C =  – \frac{1}{{p – 1}}\)    M1 

Note: Award M1 for attempting to find their value of C.

\(y = \frac{1}{{p – 1}}\left( {{x^p} – x} \right) – 1\)      A1

[8 marks]

METHOD 2

put \(y = vx\) so that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)    M1(A1)

substituting,       M1 

\(x\left( {v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}} \right) – vx = {x^p} + 1\)     (A1)

\(x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p – 1}} + \frac{1}{x}\)      M1

\(\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p – 2}} + \frac{1}{{{x^2}}}\)

\(v = \frac{1}{{p – 1}}{x^{p – 1}} – \frac{1}{x} + C\)     A1

Note: Condone the absence of C.

\(y = \frac{1}{{p – 1}}{x^p} + Cx – 1\)

substituting \(x = 1\), \(y =  – 1 \Rightarrow C =  – \frac{1}{{p – 1}}\)    M1 

Note: Award M1 for attempting to find their value of C.

\(y = \frac{1}{{p – 1}}\left( {{x^p} – x} \right) – 1\)      A1

[8 marks]

a.

METHOD 1

find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) and solve \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) for \(x\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{p – 1}}\left( {p{x^{p – 1}} – 1} \right)\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow p{x^{p – 1}} – 1 = 0\)     A1

\(p{x^{p – 1}} = 1\)

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

\({x^{p – 1}} = \frac{1}{p}\)     AG

METHOD 2

substitute \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) and their \(y\) into the differential equation and solve for \(x\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow  – \left( {\frac{{{x^p} – x}}{{p – 1}}} \right) + 1 = {x^p} + 1\)     M1

\({x^p} – x = {x^p} – p{x^p}\)     A1

\(p{x^{p – 1}} = 1\)

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

\({x^{p – 1}} = \frac{1}{p}\)     AG

[2 marks]

b.i.

there are two solutions for \(x\) when \(p\) is odd (and \(p > 1\)     A1

if \(p – 1\) is even there are two solutions (to \({x^{p – 1}} = \frac{1}{p}\))

and if \(p – 1\) is odd there is only one solution (to \({x^{p – 1}} = \frac{1}{p}\))   R1

Note: Only award the R1 if both cases are considered.

[4 marks]

b.ii.

Examiners report

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a.

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b.i.

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b.ii.

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