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Question 0803b
Maclaurin Series
Maclaurin Expansion
Question
Generate new
A function
f
{f}
f
is such that
y
=
f
(
x
)
,
{y=f(x),}
y
=
f
(
x
)
,
f
(
0
)
=
4
,
{f(0)=4,}
f
(
0
)
=
4
,
f
′
(
0
)
=
−
5
{f'(0)=-5}
f
′
(
0
)
=
−
5
and
(
1
+
x
3
)
d
2
y
d
x
2
−
d
y
d
x
−
y
=
−
7.
(1+x^3) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \frac{\mathrm{d}y}{\mathrm{d}x} - y = -7.
(
1
+
x
3
)
d
x
2
d
2
y
−
d
x
d
y
−
y
=
−
7.
Find the Maclaurin series for
y
{y}
y
up to and including the term in
x
3
{x^3}
x
3
in the form
a
0
+
a
1
x
+
a
2
x
2
+
a
3
x
3
+
…
a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots
a
0
+
a
1
x
+
a
2
x
2
+
a
3
x
3
+
…
Attempt
a
0
=
{a_0=}
a
0
=
a
1
=
{a_1=}
a
1
=
a
2
=
{a_2=}
a
2
=
a
3
=
{a_3=}
a
3
=
Answer
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