The variables x and y are related by \(\frac{{{\text{d}}y}}{{{\text{d}}x}} - y\tan x = \cos x\) .

(a)     Find the Maclaurin series for y up to and including the term in \({x^2}\) given that

\(y = - \frac{\pi }{2}\) when x = 0 .

(b)     Solve the differential equation given that y = 0 when \(x = \pi \) . Give the solution in the form \(y = f(x)\) .

(a)     from \(\frac{{{\text{d}}y}}{{{\text{d}}x}} - y\tan x + \cos x\) , \(f'(0) = 1\)     A1

now \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = y{\sec ^2}x + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x - \sin x\)     M1A1A1A1

Note: Award A1 for each term on RHS.

\( \Rightarrow f''(0) = - \frac{\pi }{2}\)     A1

\( \Rightarrow y = - \frac{\pi }{2} + x - \frac{{\pi {x^2}}}{4}\)     A1

[7 marks]

(b)     recognition of integrating factor     (M1)

integrating factor is \({{\text{e}}^{\int { - \tan x{\text{d}}x} }}\)

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

\( = \cos x\)     (A1)

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

\( \Rightarrow y\cos x = \frac{1}{2}\int {(1 + \cos 2x){\text{d}}x} \)     A1

\( \Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} + k\)     A1

when \(x = \pi ,{\text{ }}y = 0 \Rightarrow k = - \frac{\pi }{2}\)     M1A1

\( \Rightarrow y\cos x = \frac{x}{2} + \frac{{\sin 2x}}{4} - \frac{\pi }{2}\)     (A1)

\( \Rightarrow y = \sec x\left( {\frac{x}{2} + \frac{{\sin 2x}}{4} - \frac{\pi }{2}} \right)\)     A1

[10 marks]

Total [17 marks]

Part (a) of the question was set up in an unusual way, which caused a problem for a number of candidates as they tried to do part (b) first and then find the Maclaurin series by a standard method. Few were successful as they were usually weaker candidates and made errors in finding the solution \(y = f(x)\) . The majority of candidates knew how to start part (b) and recognised the need to use an integrating factor, but a number failed because they missed out the negative sign on the integrating factor, did not realise that \({{\text{e}}^{\ln \cos x}} = \cos x\) or were unable to integrate \({{{\cos }^2}x}\) . Having said this, a number of candidates succeeded in gaining full marks on this question.