Page 79 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 2n
(i) Equating the coefficients of t , on both sides, we have
1 1
H 2n (0) = ( 1) n
n
(2 )! ! n
n
(2 )!
n
or H (0) = ( 1)
2n
! n
(ii) Again equating the coefficients of t 2n 1 , on both sides of (i), we have
1
H 2n 1 (0) = 0 [Since R.H.S. of (i) does not involve
(2n 1)!
odd powers of t]

Hence H 2n 1 (0) = 0.

Example 5: Prove that
2 n t 2
dt
xt
x
P n ( ) = t e H n ( ) .
! n 0
Solution: We have
( /2)
n
x
H n ( ) = ( 1) r ! n (2 ) n 2r
x
r
r 0 r !(n 2 )
( /2)
n
t
H n ( ) = ( 1) r ! n (2 ) n 2r
x
xt
r
r 0 r !(n 2 )
n
2 n t 2 2 2 ( /2) ! n
xt
t e H n ( )dt = t e t ( 1) r (2 ) n 2r .dt
n
xt
! n 0 . ! 0 r !(n 2 )
r
n
r 0
( /2) n 2r 1 r n 2r 1
n
2 ( 1) x t 2 2 n r 1
= e t 2 dt
r
r !(n 2 )! 0
r 0
n
( /2) 2 n 2r 1 ( 1) x n 2r 1 1
r
= n r
r
r !(n 2 )! 2 2
r 0
n
Since 2 e t 2 t (2n 1) dt ( )
0
n
( /2) n 2r r n 2r
2 ( 1) x 2(n r ) !
=
r
r !(n 2 )!2 2(n r ) (n r )!
r 0
1 (2 )!
x
Since x 2x
2 2 ! x
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