Question
(a) Show that the solution of the homogeneous differential equation
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} + 1,{\text{ }}x > 0,\)
given that \(y = 0{\text{ when }}x = {\text{e, is }}y = x(\ln x – 1)\).
(b) (i) Determine the first three derivatives of the function \(f(x) = x(\ln x – 1)\).
(ii) Hence find the first three non-zero terms of the Taylor series for f(x) about x = 1.
▶️Answer/Explanation
Markscheme
(a) EITHER
use the substitution y = vx
\(\frac{{{\text{d}}y}}{{{\text{d}}x}}x + v = v + 1\) M1A1
\(\int {{\text{d}}v = \int {\frac{{{\text{d}}x}}{x}} } \)
by integration
\(v = \frac{y}{x} = \ln x + c\) A1
OR
the equation can be rearranged as first order linear
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} – \frac{1}{x}y = 1\) M1
the integrating factor I is
\({{\text{e}}^{\int { – \frac{1}{x}{\text{d}}x} }} = {{\text{e}}^{ – \ln x}} = \frac{1}{x}\) A1
multiplying by I gives
\(\frac{{{\text{d}}}}{{{\text{d}}x}}\left( {\frac{1}{x}y} \right) = \frac{1}{x}\)
\(\frac{1}{x}y = \ln x + c\) A1
THEN
the condition gives c = –1
so the solution is \(y = x(\ln x – 1)\) AG
[5 marks]
(b) (i) \(f'(x) = \ln x – 1 + 1 = \ln x\) A1
\(f”(x) = \frac{1}{x}\) A1
\(f”'(x) = – \frac{1}{{{x^2}}}\) A1
(ii) the Taylor series about x = 1 starts
\(f(x) \approx f(1) + f'(1)(x – 1) + f”(1)\frac{{{{(x – 1)}^2}}}{{2!}} + f”'(1)\frac{{{{(x – 1)}^3}}}{{3!}}\) (M1)
\( = – 1 + \frac{{{{(x – 1)}^2}}}{{2!}} – \frac{{{{(x – 1)}^3}}}{{3!}}\) A1A1A1
[7 marks]
Total: [12 marks]
Examiners report
Part(a) was well done by many candidates. In part(b)(i), however, it was disappointing to see so many candidates unable to differentiate \(x(\ln x – 1)\) correctly. Again, too many candidates were able to quote the general form of a Taylor series expansion, but not how to apply it to the given function.
Question
a.Given that \(y = \ln \left( {\frac{{1 + {{\text{e}}^{ – x}}}}{2}} \right)\), show that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\text{e}}^{ – y}}}}{2} – 1\).[5]
b.Hence, by repeated differentiation of the above differential equation, find the Maclaurin series for y as far as the term in \({x^3}\), showing that two of the terms are zero.[11]
▶️Answer/Explanation
Markscheme
METHOD 1
\(y = \ln \left( {\frac{{1 + {{\text{e}}^{ – x}}}}{2}} \right)\)
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – 2{{\text{e}}^{ – x}}}}{{2(1 + {{\text{e}}^{ – x}})}} = \frac{{ – {{\text{e}}^{ – x}}}}{{1 + {{\text{e}}^{ – x}}}}\) M1A1
now \(\frac{{1 + {{\text{e}}^{ – x}}}}{2} = {{\text{e}}^y}\) M1
\( \Rightarrow 1 + {{\text{e}}^{ – x}} = 2{{\text{e}}^y}\)
\( \Rightarrow {{\text{e}}^{ – x}} = 2{{\text{e}}^y} – 1\) (A1)
\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – 2{{\text{e}}^y} + 1}}{{2{{\text{e}}^y}}}\) (A1)
Note: Only one of the two above A1 marks may be implied.
\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\text{e}}^{ – y}}}}{2} = – 1\) AG
Note: Candidates may find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) as a function of x and then work backwards from the given answer. Award full marks if done correctly.
METHOD 2
\(y = \ln \left( {\frac{{1 + {{\text{e}}^{ – x}}}}{2}} \right)\)
\( \Rightarrow {{\text{e}}^y} = \frac{{1 + {{\text{e}}^{ – x}}}}{2}\) M1
\( \Rightarrow {{\text{e}}^{ – x}} = 2{{\text{e}}^y} – 1\)
\( \Rightarrow x = – \ln (2{{\text{e}}^y} – 1)\) A1
\( \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}y}} = – \frac{1}{{2{{\text{e}}^y} – 1}} \times 2{{\text{e}}^y}\) M1A1
\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2{{\text{e}}^y} – 1}}{{ – 2{{\text{e}}^y}}}\) A1
\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\text{e}}^{ – y}}}}{2} – 1\) AG
[5 marks]
METHOD 1
when \(x = 0,{\text{ }}y = \ln 1 = 0\) A1
when \(x = 0,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{2} – 1 = – \frac{1}{2}\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = – \frac{{{{\text{e}}^{ – y}}}}{2}\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1A1
when \(x = 0,{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\) A1
\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{{{{\text{e}}^{ – y}}}}{2}{\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2} – \frac{{{{\text{e}}^{ – y}}}}{2}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}\) M1A1A1
when \(x = 0,{\text{ }}\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{1}{2} \times \frac{1}{4} – \frac{1}{2} \times \frac{1}{4} = 0\) A1
\(y = f(0) + f'(0)x + \frac{{f”(0)}}{{2!}}{x^2} + \frac{{f”'(0)}}{{3!}}{x^3}\)
\( \Rightarrow y = 0 – \frac{1}{2}x + \frac{1}{8}{x^2} + 0{x^3} + \ldots \) (M1)A1
two of the above terms are zero AG
METHOD 2
when \(x = 0,{\text{ }}y = \ln 1 = 0\) A1
when \(x = 0,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{2} – 1 = – \frac{1}{2}\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{ – {{\text{e}}^{ – y}}}}{2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – {{\text{e}}^{ – y}}}}{2}\left( {\frac{{{{\text{e}}^{ – y}}}}{2} – 1} \right) = \frac{{ – {{\text{e}}^{2y}}}}{4} + \frac{{{{\text{e}}^{ – y}}}}{2}\) M1A1
when \(x = 0,{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = – \frac{1}{4} + \frac{1}{2} = \frac{1}{4}\) A1
\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \left( {\frac{{{{\text{e}}^{ – 2y}}}}{2} – \frac{{{{\text{e}}^{ – y}}}}{2}} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1A1A1
when \(x = 0,{\text{ }}\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = – \frac{1}{2} \times \left( {\frac{1}{2} – \frac{1}{2}} \right) = 0\) A1
\(y = f(0) + f'(0)x + \frac{{f”(0)}}{{2!}}{x^2} + \frac{{f”'(0)}}{{3!}}{x^3}\)
\( \Rightarrow y = 0 – \frac{1}{2}x + \frac{1}{8}{x^2} + 0{x^3} + \ldots \) (M1)A1
two of the above terms are zero AG
[11 marks]
Examiners report
Many candidates were successful in (a) with a variety of methods seen. In (b) the use of the chain rule was often omitted when differentiating \({{{\text{e}}^{ – y}}}\) with respect to x. A number of candidates tried to repeatedly differentiate the original expression, which was not what was asked for, although partial credit was given for this. In this case, they often found problems in simplifying the algebra.
Many candidates were successful in (a) with a variety of methods seen. In (b) the use of the chain rule was often omitted when differentiating \({{{\text{e}}^{ – y}}}\) with respect to x. A number of candidates tried to repeatedly differentiate the original expression, which was not what was asked for, although partial credit was given for this. In this case, they often found problems in simplifying the algebra.
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
Consider the differential equation
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2{{\text{e}}^x} + y\tan x\) , given that y = 1 when x = 0 .
The domain of the function y is \(\left[ {0,\frac{\pi }{2}} \right[\).
By finding the values of successive derivatives when x = 0 , find the Maclaurin series for y as far as the term in \({x^3}\) .
(i) Differentiate the function \({{\text{e}}^x}(\sin x + \cos x)\) and hence show that
\[\int {{{\text{e}}^x}\cos x{\text{d}}x = \frac{1}{2}{{\text{e}}^x}(\sin x + \cos x) + c} .\]
(ii) Find an integrating factor for the differential equation and hence find the solution in the form \(y = f(x)\) .
▶️Answer/Explanation
Markscheme
we note that \(y(0) = 1\) and \(y'(0) = 2\) A1
\(y” = 2{{\text{e}}^x} + y’\tan x + y{\sec ^2}x\) M1
\(y”(0) = 3\) A1
\(y”’ = 2{{\text{e}}^x} + y”\tan x + 2y'{\sec ^2}x + 2y{\sec ^2}x\tan x\) M1
\(y”'(0) = 6\) A1
the maclaurin series solution is therefore
\(y = 1 + 2x + \frac{{3{x^2}}}{2} + {x^3} + \ldots \) A1
[6 marks]
(i) \(\frac{{\text{d}}}{{{\text{d}}x}}\left( {{{\text{e}}^x}(\sin x + \cos x)} \right) = {{\text{e}}^x}(\sin x + \cos x) + {{\text{e}}^x}(\cos x – \sin x)\) M1
\( = 2{{\text{e}}^x}\cos x\) A1
it follows that
\(\int {{{\text{e}}^x}\cos x{\text{d}}x = \frac{1}{2}{{\text{e}}^x}(\sin x + \cos x) + c} \) AG
(ii) the differential equation can be written as
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\tan x = 2{{\text{e}}^x}\) M1
\({\text{IF}} = {{\text{e}}^{\int { – \tan x{\text{d}}x} }} = {{\text{e}}^{\ln \cos x}} = \cos x\) M1A1
\(\cos x\frac{{{\text{d}}y}}{{{\text{d}}x}} – y\sin x = 2{{\text{e}}^x}\cos x\) M1
integrating,
\(y\cos x = {{\text{e}}^x}(\sin x + \cos x) + C\) A1
y = 1 when x = 0 gives C = 0 M1
therefore
\(y = {{\text{e}}^x}(1 + \tan x)\) A1
[9 marks]
Examiners report
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