Page 77 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
x
Since H n ( ) is the solution of (i), hence, we have
H n ( ) 2x H n ( ) 2nH n ( ) = 0
x
x
x
Illustrative Examples

Example 1: Evaluate

xe x 2 H n ( )H m ( )dx
x
x
Solution: From recurrence formula II, we have

1
x
x
xH n ( ) = nH n 1 ( ) H n 1 ( )
x
2
x
xe x 2 H H ( )dx = e x 2 nH ( ) 1 H ( ) H ( )dx
x
x
x
n m n 1 n 1 m
2
2 1 x 2
x
= n e H n 1 ( )H m ( )dx e H n 1 ( )H m ( )dx
x
x
x
x
2
1 n 1
n
1
= n 2 (n 1)! n 1,m 2 (n 1)! n 1,m
2
= 2 n 1 ! n n 1,m 2 n n 1 ! n 1,m
where is Kronecker delta.
n
Example 2: Prove that H n 4 (n 1)H n 1
Solution: From recurrence formula I, we have
H n = 2nH n 1 ...(i)
Differentiating with respect to x, we have

H n = 2nH n 1 ...(ii)
Replacing n by (n 1) in (i), we have

H = 2(n 1)H ...(iii)
n 1 n 2
From (ii) and (iii), we have

H n = 4 (n n 1)H n 1

Example 3: Prove that, if m < n

m
d n 2 n !
x
x
H n ( ) = H n m ( )
dx m (n m )!

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