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.