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Question 0821
Maclaurin Series
2021 Paper 1 Question 7 Variant
Question
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It is given that
y
=
e
3
sin
−
1
x
,
{y=\mathrm{e}^{3 \sin^{-1} x},}
y
=
e
3
s
i
n
−
1
x
,
for
−
1
<
x
<
1.
{-1 < x < 1.}
−
1
<
x
<
1.
(a)
Show that
(
1
−
x
2
)
d
2
y
d
x
2
=
x
d
y
d
x
+
9
y
.
(1-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=x\frac{\mathrm{d}y}{\mathrm{d}x}+9 y.
(
1
−
x
2
)
d
x
2
d
2
y
=
x
d
x
d
y
+
9
y
.
[4]
(b)
Find the first 4 terms of the Maclaurin expansion of
e
3
sin
−
1
x
.
{\mathrm{e}^{3 \sin^{-1} x}.}
e
3
s
i
n
−
1
x
.
[5]
Answer
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