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.

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]