Question
The function \(f(x)\) is defined by the series \(f(x) = 1 + \frac{{(x + 2)}}{{3 \times 1}} + \frac{{{{(x + 2)}^2}}}{{{3^2} \times 2}} + \frac{{{{(x + 2)}^3}}}{{{3^3} \times 3}} + \ldots \) .
Write down the general term.
Find the interval of convergence.
Solve the differential equation \((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\) , giving your answer in the form \(u = f(v)\) .
Answer/Explanation
Markscheme
the general term is \(\frac{{{{(x + 2)}^n}}}{{{3^n}n}}\) A1
[1 mark]
\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}}(n + 1)}} \times \frac{{{3^n}n}}{{{{(x + 2)}^n}}}} \right]\) M1A1A1
\( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{(x + 2){n^{}}}}{{3(n + 1)}}} \right]\) A1
\( = \frac{{(x + 2)}}{3}\) since \( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^{}}}}{{n + 1}}} \right] = 1\) A1R1
the series is convergent if \(\left| {\frac{{(x + 2)}}{3}} \right| < 1\) R1
then \( – 3 < x + 2 < 3 \Rightarrow – 5 < x < 1\) A1
if \(x = – 5\) , series is \(1 – 1 + \frac{1}{2} – \frac{1}{3} + \ldots + \frac{{{{( – 1)}^n}}}{n} + \ldots \) which converges M1A1
if \(x = 1\) , series is \(1 + 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} + \ldots \) which diverges M1A1
the interval of convergence is \( – 5 \le x < 1\) A1
[13 marks]
\((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\)
\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} = \frac{{(u + 3{v^3})}}{{2v}} = \frac{u}{{2v}} + \frac{{3{v^2}}}{2}\) M1A1
\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} – \frac{u}{{2v}} = \frac{{3{v^2}}}{2}\) A1
IF is \({{\rm{e}}^{\int {\frac{1}{{2v}}} {\rm{d}}v}} = {{\rm{e}}^{\frac{1}{2}\ln v}}\) M1
\( = {v^{\frac{1}{2}}}\) A1
\(\frac{u}{{\sqrt v }} = \int {\frac{{3{v^{\frac{3}{2}}}}}{2}} {\rm{d}}v\) M1
\( = \frac{3}{5}{v^{\frac{5}{2}}} + c\) A1
\(u = \frac{3}{5}{v^3} + c\sqrt v \) A1
[8 marks]
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = 2{\cos ^4}x\) given that \(y = 1\) when \(x = 0\).
(a) Solve the differential equation, giving your answer in the form \(y = f(x)\).
(b) (i) By differentiating both sides of the differential equation, show that
\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = – 10\sin x{\cos ^3}x\]
(ii) Hence find the first four terms of the Maclaurin series for \(y\).
Answer/Explanation
Markscheme
(a) integrating factor \( = {e^{\int {\tan x{\text{d}}x} }}\) M1
\( = {{\text{e}}^{\ln \sec x}}\) A1
\( = \sec x\) A1
\(\sec x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x\tan x = 2{\cos ^3}x\) (M1)
integrating,
\(y\sec x = 2\int {{{\cos }^3}x{\text{d}}x} \) A1
\( = 2\int {\cos x(1 – {{\sin }^2}x){\text{d}}x} \) A1
\( = 2\left( {\sin x – \frac{{{{\sin }^3}x}}{3}} \right) + C\) A1
Note: Condone the absence of \(C\).
(substituting \(x = 0,{\text{ }}y = 1\))
\(1 = C\) M1
the solution is
\(y = 2\cos x\left( {\sin x – \frac{{{{\sin }^3}x}}{3}} \right) + \cos x\) A1
[9 marks]
(b) (i) differentiating the equation,
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 8{\cos ^3}x\sin x\) A1A1
Note: A1 for each side.
substituting for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\),
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\left( {2{{\cos }^4}x – y\tan x} \right) = – 8{\cos ^3}x\sin x\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y({\sec ^2}x – {\tan ^2}x) = – 8{\cos ^3}x\sin x – 2\tan x{\cos ^4}x\) (or equivalent) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = – 10\sin x{\cos ^3}x\) AG
(ii) differentiating again,
\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}} = – 10{\cos ^4}x + {\text{term involving }} \sin x\) A1
it follows that
\(y(0) = 1,{\text{ }}y'(0) = 2\) A1
\(y”(0) = – 1,{\text{ }}y”'(0) = – 12\) A1
attempting to use \(y = y(0) + xy'(0) + \frac{{{x^2}}}{2}y”(0) + \frac{{{x^3}}}{6}y”'(0) + \ldots \) (M1)
\(y = 1 + 2x – \frac{{{x^2}}}{2} – 2{x^3}\) A1
[9 marks]
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x + y – 1\) with boundary condition \(y = 1\) when \(x = 0\).
Using Euler’s method with increments of \(0.2\), find an approximate value for \(y\) when \(x = 1\).
Explain how Euler’s method could be improved to provide a better approximation.
Solve the differential equation to find an exact value for \(y\) when \(x = 1\).
(i) Find the first three non-zero terms of the Maclaurin series for \(y\).
(ii) Hence find an approximate value for \(y\) when \(x = 1\).
Answer/Explanation
Markscheme
(M1)(A1)(A1)(A1)
Note: Award M1 for equivalent of setting up first row of table, A1 for each of row 2, 3 and 5.
approximate solution \(y = 1.98\) A1
make the increments smaller or any specific correct instruction – for example change increment from \(0.2\) to \(0.1\) A1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} – y = 2x – 1\)
integrating factor is \({{\text{e}}^{\int { – 1{\text{d}}x} }} = {{\text{e}}^{ – x}}\) (M1)(A1)
\(\frac{{\text{d}}}{{{\text{d}}x}}(y{{\text{e}}^{ – x}}) = {{\text{e}}^{ – x}}(2x – 1)\) M1
attempt at integration by parts of \(\int {{{\text{e}}^{ – x}}(2x – 1){\text{d}}x} \) (M1)
\( = – (2x – 1){{\text{e}}^{ – x}} + \int {2{{\text{e}}^{ – x}}{\text{d}}x} \) A1
\( = – (2x – 1){{\text{e}}^{ – x}} – 2{{\text{e}}^{ – x}}( + c)\) A1
\(y{{\text{e}}^{ – x}} = – (1 + 2x){{\text{e}}^{ – x}} + c\)
\(y = – (1 + 2x) + c{{\text{e}}^x}\)
when \(x = 0,{\text{ }}y = 1 \Rightarrow c = 2\) M1
\(y = – (1 + 2x) + 2{{\text{e}}^{ – x}}\) A1
when \(x = 1,{\text{ }}y = – 3 + 2{\text{e}}\) A1
(i) METHOD 1
\(f(0) = 1,{\text{ }}f'(0) = 0\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 + \frac{{{\text{d}}y}}{{{\text{d}}x}} \Rightarrow {f^2}(0) = 2\) A1
\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} \Rightarrow {f^3}(0) = 2\) A1
hence \(y = 1 + {x^2} + \frac{{{x^3}}}{3} + \ldots \) A1
Note: Accuracy marks are independent of each other.
METHOD 2
using Maclaurin series for \({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots \) M1
\(y = – 1 – 2x + 2\left( {1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots } \right)\) M1A1
\(y = 1 + {x^2} + \frac{{{x^3}}}{3} + \ldots \) A1
(ii) when \(x = 1,{\text{ }}y = 1 + 1 + \frac{1}{3} = \frac{7}{3} = 2.33\) A1
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{x}{y}\), where \(y \ne 0\).
Find the general solution of the differential equation, expressing your answer in the form \(f(x,{\text{ }}y) = c\), where \(c\) is a constant.
(i) Hence find the particular solution passing through the points \((1,{\rm{ \pm }}\sqrt 2 )\).
(ii) Sketch the graph of your solution and name the type of curve represented.
(i) Write down the particular solution passing through the points \((1,{\text{ }} \pm 1)\).
(ii) Give a geometrical interpretation of this solution in relation to part (b).
(i) Find the general solution of the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{x}{y} + \frac{y}{x}\), where \(xy \ne 0\).
(ii) Find the particular solution passing through the point \((1,{\text{ }}\sqrt 2 )\).
(iii) Sketch the particular solution.
(iv) The graph of the solution only contains points with \(\left| x \right| > a\).
Find the exact value of \(a,{\text{ }}a > 0\).
Answer/Explanation
Markscheme
attempt to separate the variables M1
\(\int {y\frac{{{\text{d}}y}}{{{\text{d}}x}}{\text{d}}x = \int {x{\text{d}}x} } \) A1
Note: Accept \(\int {y{\text{d}}y = \int {x{\text{d}}x} } \).
obtain \(\frac{1}{2}{y^2} = \frac{1}{2}{x^2} + {\text{ constant }}( \Rightarrow {y^2} – {x^2} = c)\) A1
[3 marks]
(i) substitute the coordinates for both points M1
\({( \pm \sqrt 2 )^2} – {1^2} = 1\)
obtain \({y^2} – {x^2} = 1\) or equivalent A1
(ii) 
A1A1
Note: A1 for general shape including two branches and symmetry;
A1 for values of the intercepts.
(rectangular) hyperbola A1
[5 marks]
(i) \({y^2} – {x^2} = 0\) A1
(ii) the two straight lines \(y = \pm x\) A1
form the asymptotes to the hyperbola found above, or equivalent A1
[3 marks]
(i) the equation is homogeneous, so attempt to substitute \(y = vx\) M1
as a first step write \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x\frac{{{\text{d}}v}}{{{\text{d}}x}} + v\) (A1)
then \(x\frac{{{\text{d}}v}}{{{\text{d}}x}} + v = \frac{1}{v} + v\) A1
attempt to solve the resulting separable equation M1
\(\int {v{\text{d}}v = \int {\frac{1}{x}{\text{d}}x} } \) A1
obtain \(\frac{1}{2}{v^2} = \ln \left| x \right| + {\text{ constant}} \Rightarrow {y^2} = 2{x^2}\ln \left| x \right| + c{x^2}\) A1
(ii) substituting the coordinates (M1)
obtain \(c = 2 \Rightarrow {y^2} = 2{x^2}\ln \left| x \right| + 2{x^2}\) A1
(iii) 
A1
(iv) since \({y^2} > 0\) and \({x^2} \ne 0\) R1
\(\ln \left| x \right| > – 1 \Rightarrow \left| x \right| > {{\text{e}}^{ – 1}}\) A1
\(a = {{\text{e}}^{ – 1}}\) A1
Note: The R1 may be awarded for a correct reason leading to subsequent correct work.
[12 marks]
Question
Consider the differential equation
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = 2{\sec ^2}x,{\text{ }}0 \leqslant x < \frac{\pi }{2}\), given that \(y = 1\) when \(x = 0\).
By considering integration as the reverse of differentiation, show that for
\(0 \leqslant x < \frac{\pi }{2}\)
\[\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C.} \]
Hence, using integration by parts, show that
\[\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C.} \]
Find an integrating factor and hence solve the differential equation, giving your answer in the form \(y = f(x)\).
Starting with the differential equation, show that
\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x.\]
Hence, by using your calculator to draw two appropriate graphs or otherwise, find the \(x\)-coordinate of the point of inflection on the graph of \(y = f(x)\).
Answer/Explanation
Markscheme
\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\ln (\sec x + \tan x)} \right) = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}}\) M1
\( = \sec x\) A1
therefore \(\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C} \) AG
[4 marks]
\(\int {{{\sec }^3}x{\text{d}}x = \int {\sec x \times {{\sec }^2}x{\text{d}}x} } \) M1
\( = \sec x\tan x – \int {\sec x{{\tan }^2}x{\text{d}}x} \) A1A1
\( = \sec x\tan x – \int {\sec x({{\sec }^2}x – 1){\text{d}}x} \) A1
\( = \sec x\tan x – \int {{{\sec }^3}x{\text{d}}x + \int {\sec x{\text{d}}x} } \)
\( = \sec x\tan x – \int {{{\sec }^3}x{\text{d}}x + \ln (\sec x + \tan x)} \) A1
\(2\int {{{\sec }^3}x{\text{d}}x = \left( {\sec x\tan x + \ln (\sec x + \tan x)} \right)} \) A1
therefore
\(\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C} \) AG
[4 marks]
\({\text{int factor}} = {{\text{e}}^{\int {\tan x{\text{d}}x} }}\) (M1)
\( = {{\text{e}}^{\ln \sec x}}\) (A1)
\( = \sec x\) A1
the differential equation can be written as
\(\frac{{\text{d}}}{{{\text{d}}x}}(y\sec x) = 2{\sec ^3}x\) M1A1
integrating,
\(y\sec x = \sec x\tan x + \ln (\sec x + \tan x) + C\) A1
putting \(x = 0,{\text{ }}y = 1,\) M1
\(C = 1\) A1
the solution is \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\) A1
[??? marks]
differentiating the differential equation,
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\) A1A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + (2{\sec ^2}x – y\tan x)\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x\) AG
[??? marks]
at a point of inflection, \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 0\) so \(y = 2{\sec ^2}x\tan x\) (M1)
therefore the point of inflection can be found as the point of intersection of the graphs of \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\)
and \(y = 2{\sec ^2}x\tan x\) (M1)
drawing these graphs on the calculator, \(x = 0.605\) A2
[??? marks]
Question
It is given that \(\left( {5x + y} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = \left( {x + 5y} \right)\) and that when \(x = 0,\,\,y = 2\).
Use Euler’s method with step length 0.1 to find an approximate value of \(y\) when \(x = 0.4\).
Show that \(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 – {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\).
Show that \(\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = – 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} – 3\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\).
Find the Maclaurin expansion for \(y\) up to and including the term in \({{x^3}}\).
Answer/Explanation
Markscheme
Euler’s method with step length \(h = 0.1\) to find \(y\) when \(x = 0.4\)
Note: Accept 3 significant figures in the table.
first line of table (M1)(A1)
line 2 (A1)
line 3 (A1)
hence \(y\) = 3.65 A1
Note: Accept any answer that rounds to 3.65.
[5 marks]
\(\left( {5x + y} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + 5y\)
\(\left( {5 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 + 5\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1A1A1
Note: Award M1 for a valid attempt to differentiate, A1 for LHS, A1 for RHS.
\(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 + 5\frac{{{\text{d}}y}}{{{\text{d}}x}} – 5\frac{{{\text{d}}y}}{{{\text{d}}x}} – {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\)
\(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 – {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\) AG
[3 marks]
\(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}1 – {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\)
\(\left( {5 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + \left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = – 2\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\) M1A1A1A1
\(\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = – 2\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right) – 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} – \left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\)
\(\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = – 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} – 3\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\) AG
[4 marks]
when \(x = 0\,\,\,y = 2\)
when \(x = 0\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = 5\) A1
when \(x = 0\,\,\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = – 12\) A1
when \(x = 0\,\,\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = 120\) A1
Note: Allow follow through from incorrect values of derivatives.
\(y = 2 + 5x – 6{x^2} + 20{x^3}\) M1A1
[5 marks]
Question
Draw slope fields for the following cases for \( – 2 \leqslant x \leqslant 2,\,\, – 2 \leqslant y \leqslant 2\)
Explain what isoclines tell you about the slope field in the following case:
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2\).
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + 1\).
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x – 1\).
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \) constant.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( x \right)\).
The slope field for the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + y\) for \( – 4 \leqslant x \leqslant 4,\,\, – 4 \leqslant y \leqslant 4\) is shown in the following diagram.
Explain why the slope field indicates that the only linear solution is \(y = – x – 1\).
Given that all the isoclines from a slope field of a differential equation are straight lines through the origin, find two examples of the differential equation.
Answer/Explanation
Markscheme
A2
[2 marks]
A2
[2 marks]
A2
[2 marks]
the slope is the same everywhere A1
[1 mark]
all points that have the same \(x\) coordinate have the same slope A1
[1 mark]
this is where a straight line appears on the slope field A1
There is no other straight line, all the other solutions are curves A1
[2 marks]
given \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( {x,\,y} \right)\), the isoclines are \(f\left( {x,\,y} \right) = k\) (M1)
here the isoclines are \(y = kx\) (or \(x = ky\)) (A1)
any two differential equations of the correct form, for example
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ky}}{x},\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{kx}}{y},\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = {\text{sin}}\left( {\frac{y}{x}} \right),\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = {\text{sin}}\left( {\frac{x}{y}} \right)\) A1A1
[4 marks]
Question
Consider the differential equation\[\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x = x(\sec x – \tan x),{\text{ where }}y = 3{\text{ when }}x = 0.\]
Use Euler’s method with a step length of \(0.1\) to find an approximate value for \(y\) when \(x = 0.3\) .
(i) By differentiating the above differential equation, obtain an expression involving \(\frac{{{{\rm{d}}^{\rm{2}}}y}}{{{\rm{d}}{x^2}}}\) .
(ii) Hence determine the Maclaurin series for \(y\) up to the term in \({{x^2}}\) .
(iii) Use the result in part (ii) to obtain an approximate value for \(y\) when \(x = 0.3\) .
(i) Show that \(\sec x + \tan x\) is an integrating factor for solving this differential equation.
(ii) Solve the differential equation, giving your answer in the form \(y = f(x)\) .
(iii) Hence determine which of the two approximate values for y when \(x = 0.3\) , obtained in parts (a) and (b), is closer to the true value.
Answer/Explanation
Markscheme
Note: The A1 marks above are for correct entries in the \(y\) column.
\(y(0.3) \approx 2.21\) A1
[5 marks]
(i) use of product rule on either side M1
\(\frac{{{{\rm{d}}^{\rm{2}}}y}}{{{\rm{d}}{x^2}}} + \sec x\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + y\sec x\tan x = \sec x – \tan x + x(\sec x\tan x – {\sec ^2}x)\) A1A1
(ii) \(y(0) = 3\)
\(y'(0) = – 3\), \(y”(0) = 4\) A1A1
the quadratic approximation is
\(y = \left( {y(0) + xy'(0) + |\frac{{{x^2}y”(0)}}{2} = } \right)3 – 3x + 2{x^2}\) (M1)A1
(iii) using this approximation, \(y(0.3) \approx 2.28\) A1
[8 marks]
(i) EITHER
\(\frac{{\rm{d}}}{{{\rm{d}}x}}(\sec x + \tan x) = \sec x\tan x + {\sec ^2}x\) A1
\(\sec x(\sec x + \tan x) = {\sec ^2}x + \sec x\tan x\) A1
as these two expressions are the same, this is an integrating factor R1AG
OR
\((\sec x + \tan x)\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + y\sec x} \right) = (\sec x + \tan x)x(\sec x – \tan x)\) M1
Note: RHS does not need to be shown.
\({\rm{LHS}} = \frac{{{\rm{d}}y}}{{{\rm{d}}x}}(\sec x + \tan x) + y(\sec x|\tan x + {\sec ^2}x)\) A1
\( = \frac{{\rm{d}}}{{{\rm{d}}x}}y(\sec x + \tan x)\) A1
making LHS an exact derivative
OR
integrating factor \( = {{\rm{e}}^{\int {\sec x{\rm{d}}x} }}\) M1
since \(\frac{{\rm{d}}}{{{\rm{d}}x}}\ln (\sec x + \tan x) = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}} = \sec x\) M1A1
integrating factor \( = {{\rm{e}}^{\ln (\sec x + \tan x)}} = \sec x + \tan x\) AG
(ii) \(\frac{{\rm{d}}}{{{\rm{d}}x}}(y\left[ {\sec x + \tan x} \right]) = x({\sec ^2}x – {\tan ^2}x) = x\) M1A1
\(y(\sec x + \tan x) = \frac{{{x^2}}}{2} + c\) A1
\(x = 0,y = 3 \Rightarrow c = 3\) M1A1
\(y = \frac{{{x^2} + 6}}{{2(\sec x + \tan x)}}\) A1
(iii) when \(x = 0.3,y = 2.245 \ldots \) A1
the closer approximation is obtained by using the series in part (b) R1
[11 marks]
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