Show that \(f''(x) = 2\left( {f'(x) - f(x)} \right)\).

By further differentiation of the result in part (a) , find the Maclaurin expansion of \(f(x)\), as far as the term in \({x^5}\).

\(f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x\)     M1A1

\(f''(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x - {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x = 2{{\text{e}}^x}\cos x\)     A1

\( = 2\left( {{{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x - {{\text{e}}^x}\sin x} \right)\)     M1

\( = 2\left( {f'(x) - f(x)} \right)\)     AG

[4 marks]

\(f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f''(0) = 2(1 - 0) = 2\)     (M1)A1

Note:     Award M1 for attempt to find \(f(0)\), \(f'(0)\) and \(f''(0)\).

\(f'''(x) = 2\left( {f''(x) - f'(x)} \right)\)     (M1)

\(f'''(0) = 2(2 - 1) = 2,{\text{ }}{f^{IV}}(0) = 2(2 - 2) = 0,{\text{ }}{f^V}(0) = 2(0 - 2) =  - 4\)     A1

so \(f(x) = x + \frac{2}{{2!}}{x^2} + \frac{2}{{3!}}{x^3} - \frac{4}{{5!}} +  \ldots \)     (M1)A1

\( = x + {x^2} + \frac{1}{3}{x^3} - \frac{1}{{30}}{x^5} +  \ldots \)

[6 marks]

Total [10 marks]