Page 71 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
1 d 1 d 2tx
2 2tx
e = t e
2 dx 2 dx
2
1 d 2tx 2 2tx
or e = t e
2 dx
1 d n 2tx
n
e = t e 2tx
2 dx
Hence

1 d 2 2tx 1 1 d 2 n 2tx
exp 2 e = 2 e
4 dx ! n 4 dx
n 0
( 1) n 1 d 2n
= e 2tx
! n 2 dx
n 0
( 1) n
2n
= t e 2tx [from (ii)]
! n
n 0
( 1) n
= e 2tx t 2n
! n
n 0
2 (2tx t 2 )
t
2tx
= e . e e
1 d 2 1 t n
x
or exp. (2 ) n = H n ( )
tx
4 dx 2 ! n ! n
n 0 n 0
n
Equating the coefficient of t from the two sides, we have
1 d 2 1 1
n
exp. 2 x n = H ( )
x
4 dx 2 ! n ! n n
1 d 2
or H n ( ) = 2 n exp. x n .
x
4 dx 2
Self Assessment

5. Obtain the expression for H 2 ( ) from generating function e 2xt t 2 .
x
d
x
6. Obtain the expression for H n ( ) from the generating function e 2xt t 2 .
dx
7. Show that for odd n.
H n (0) 0

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