Question 0803b

Maclaurin Series

Maclaurin Expansion

Question

A function

{f}

is such that

{y=f(x),}

{f(0)=-2,}

{f'(0)=1}

and

(1+x^3) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 3 \frac{\mathrm{d}y}{\mathrm{d}x} + 2 y = 2.

Find the Maclaurin series for

{y}

up to and including the term in

{x^3}

in the form

a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots

{a_0=}

{a_1=}

{a_2=}

{a_3=}