Question 0821

Maclaurin Series
2021 Paper 1 Question 7 Variant

Question

It is given that y=e2sin1x,{y=\mathrm{e}^{2 \sin^{-1} x},} for 1<x<1.{-1 < x < 1.}
(a)
Show that
(1x2)d2ydx2=xdydx+4y.(1-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=x\frac{\mathrm{d}y}{\mathrm{d}x}+4 y.
[4]
(b)
Find the first 4 terms of the Maclaurin expansion of e2sin1x.{\mathrm{e}^{2 \sin^{-1} x}.}
[5]

Answer