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]