Section 5.5
Legendre’s Differential Equation
89
Another faster way to see the answer is to simply note that
P
3
is an odd function, so
its integral over any symmetric interval is 0. There is yet another more important
reason for this integral to equal 0. In fact,
1

1
P
n
(
x
)
dx
= 0
for all
n
= 0
.
This is a consequence of orthogonality that you will study in Section 5.6.
9.
This is Legendre’s equation with
n
(
n
+ 1) = 30 so
n
= 5. Its general solution
is of the form
y
=
c
1
P
5
(
x
) +
c
2
Q
5
(
x
)
=
c
1
1
8
(63
x
5

70
x
3
+ 15
x
) +
c
2
(
1

15
x
2
+ 30
x
4
+
· · ·
)
=
c
1
(63
x
5

70
x
3
+ 15
x
) +
c
2
(
1

15
x
2
+ 30
x
4
+
· · ·
)
In finding
P
5
(
x
), we used the given formulas in the text. In finding the first few
terms of
Q
5
(
x
), we used (3) with
n
= 5. (If you are comparing with the answers in
your textbook, just remember that
c
1
and
c
2
are arbitrary constants.)
13.
This is Legendre’s equation with
n
(
n
+ 1) = 6 or
n
= 2. Its general solution
is
y
=
c
1
P
2
(
x
) +
c
2
Q
2
(
x
).
The solution will be bounded on [

1
,
1] if and only
if
c
2
= 0; that’s because
P
2
is bounded in [

1
,
1] but
Q
2
is not.
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 Fall '11
 StuartChalk
 Calculus, Recurrence relation, Legendre’s equation

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