Power series methods of solving ordinary differential equations 513
Hence, P 2 (x)=−
1
2(
1 − 3 x^2)
=1
2(3x^2 −1)Problem 14. Determine the Legendre poly-
nomialP 3 (x).Since inP 3 (x),n=k=3, then from the second part of
equation (47), i.e. the odd powers ofx:
y=a 1{
x−(k− 1 )(k+ 2 )
3!x^3+(k− 1 )(k− 3 )(k+ 2 )(k+ 4 )
5!x^5 − ···}i.e.y=a 1
{
x−
( 2 )( 5 )
3!x^3 +
( 2 )( 0 )( 5 )( 7 )
5!x^5}=a 1{
x−5
3x^3 + 0}a 1 is chosen to makey=1whenx=1.
i.e. 1=a 1
{
1 −5
3}
=a 1(
−2
3)
from which,a 1 =−3
2Hence,P 3 (x)=−
3
2(
x−5
3x^3)
orP 3 (x)=1
2(5x^3 − 3 x)Rodrigue’s formula
An alternative method of determining Legendre poly-
nomials is by usingRodrigue’s formula, which states:
Pn(x)=1
2 nn!dn(
x^2 − 1)ndxn(48)This is demonstrated in thefollowingworked problems.
Problem 15. Determine the Legendre polynomial
P 2 (x)using Rodrigue’s formula.In Rodrigue’s formula, Pn(x)=
1
2 nn!dn(
x^2 − 1)ndxn
and whenn=2,
P 2 (x)=1
22 2!d^2 (x^2 − 1 )^2
dx^2=1
23d^2 (x^4 − 2 x^2 + 1 )
dx^2
d
dx(x^4 − 2 x^2 + 1 )= 4 x^3 − 4 xandd^2(
x^4 − 2 x^2 + 1)dx^2=d( 4 x^3 − 4 x)
dx= 12 x^2 − 4Hence,P 2 (x)=1
23d^2(
x^4 − 2 x^2 + 1)dx^2=1
8(
12 x^2 − 4)i.e. P 2 (x)=1
2(
3 x^2 − 1)
the same as in Problem 13.Problem 16. Determine the Legendre polynomial
P 3 (x)using Rodrigue’s formula.In Rodrigue’s formula,Pn(x)=1
2 nn!dn(
x^2 − 1)ndxnand
whenn=3,P 3 (x)=1
23 3!d^3(
x^2 − 1) 3dx^3=1
23 ( 6 )d^3(
x^2 − 1)(
x^4 − 2 x^2 + 1)dx^3=1
( 8 )( 6 )d^3(
x^6 − 3 x^4 + 3 x^2 − 1)dx^3d(
x^6 − 3 x^4 + 3 x^2 − 1)dx= 6 x^5 − 12 x^3 + 6 xd(
6 x^5 − 12 x^3 + 6 x)dx= 30 x^4 − 36 x^2 + 6andd(
30 x^4 − 36 x^2 + 6)dx= 120 x^3 − 72 xHence,P 3 (x)=
1
( 8 )( 6 )d^3(
x^6 − 3 x^4 + 3 x^2 − 1)dx^3=1
( 8 )( 6 )(
120 x^3 − 72 x)
=1
8(
20 x^3 − 12 x)i.e. P 3 (x)=1
2(
5 x^3 − 3 x)
the same as in Problem 14.Now try the following exerciseExercise 198 Legendre’sequation and
Legendre polynomials- Determine the power series solution of
the Legendre equation:
(
1 −x^2
)
y′′− 2 xy′+k(k+ 1 )y=0when
(a)k=0(b)k=2, up to and including the