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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 y when x = 0.5.
(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 \({x^2}\) .
(i) Solve the differential equation.
(ii) Find the value of a for which \(y \to \infty \) as \(x \to a\).
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]
(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]
(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]
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).
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).
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).
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 \(y = f\left( x \right)\).
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 \({x^{p – 1}} = \frac{1}{p}\).
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.
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]
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]
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]
Examiners report
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