Question 0803b

Maclaurin Series
Maclaurin Expansion

Question

A function f{f} is such that y=f(x),{y=f(x),} f(0)=3,{f(0)=3,} f(0)=1{f'(0)=1} and
(1+x2)d2ydx2+3dydx2y=8.(1+x^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3 \frac{\mathrm{d}y}{\mathrm{d}x} - 2 y = 8.
Find the Maclaurin series for y{y} up to and including the term in x3{x^3} in the form
a0+a1x+a2x2+a3x3+a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots

Attempt

a0={a_0=}
a1={a_1=}
a2={a_2=}
a3={a_3=}

Answer

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