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]