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 $y = f(x)$ .

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]