Page 49 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 2 1/2 2 1/2
Now, (1 2xh h ) = {1 2 (x h ) (– ) }
h
n
= ( h P n ( )
)
x
n 0
n
n
x
= ( 1) h P n ( ) ...(i)
n 0
Again (1 2xh h 2 ) 1/2 = {1 2( x ) h h 2 } 1/2

n
= h P n ( x ) ...(ii)
n 0
From (i) and (ii) we have

n
n n
= h P n ( x ) ( 1) h P n ( )
x
n 0 n 0
n
Equating the coefficients of h from both sides, we get
n
P ( x ) = ( 1) P ( ).
x
n n
Deduction: Putting x = 1, we have
n
P ( 1) = ( 1) P (1)
n n
n
= ( 1) [ P n (1) 1].

1
Example 5: Prove that ( )a P n (1) ( n n 1)
2
( )b P n ( 1) ( 1) n 1 1 ( n n 1)
2

x
Solution: P n ( ) satisfies Legendre’s equation
2
2 d y dy
(1 x ) 2x ( n n 1)y 0, putting y P ( )
x
dx 2 dx n
 (1 x 2 )P n ( ) 2xP n ( ) n (n 1) ( ) = 0 ...(i)
P
x
x
x
n
(a) Putting x = 1, in (i) we have
2P n (1) n (n 1) (1) = 0
P
n
1
P
P (1) = ( n n 1) (1)
n
n 2
1

= ( n n 1) [ P n (1) 1].
2
(b) Putting x 1 in (i), we get

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