Page 55 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes or

(x h )(1 2hx h 2 ) 3/2 = nh n 1 P n ( ) ...(B)
x
n 0
Differentiating (A) again by x, we have

n
x
h
( 2 )( 1/2)(1 2hx h 2 ) 3/2 = h P n ( )
n 0
or
n
x
h (1 2hx h 2 ) 3/2 = h P n ( ) ...(C)
n 0
Multiplying (B) by h and (C) by (x h), and subtracting we get

n
x
h n h n 1 P n ( ) = (x h ) h P n ( ) ...(D)
x
n 0 n 0
n
Now comparing the coefficients of h on both sides we have
x
n
( )P n ( ) = xP n ( ) P n 1 ( )
x
x
which is the recurrence relation II
III. Prove that
x
(2n 1)P n ( ) = P n 1 ( ) P n 1 ( )
x
x
Proof:
From recurrence relation I

x
x
(2n 1)xP n ( ) = (n 1)P n 1 ( ) nP n 1 ( )
x
Differentiating w.r.t. x we have
(2n 1)xP n ( ) (2n 1) ( ) = (n 1)P n 1 ( ) nP n ( ) ...(A)
x
x
x
x
P
n
From recurrence formula II
x
x
x
xP n ( ) = nP n ( ) P n 1 ( ) ...(B)
Substituting in (A) we have
x
(2n 1) nP n ( ) P n 1 ( ) (2n 1) P n ( ) = (n 1)P n 1 ( ) nP n ( )
x
x
x
x
(2n 1) (n 1) ( ) P n 1 ( ) = (n 1)P n 1 ( ) np n 1 ( )
P
x
x
x
x
n
or rearranging
x
(2n 1)(n 1) ( ) = (n 1)P ( ) (n 1)P ( )
P
x
x
n n 1 n 1
Removing common factor we have
(2n 1)P ( ) = P ( ) P ( )
x
x
x
n n 1 n 1
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