Page 31 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
Example 4: Using generating function or otherwise, show that

J n x 1 n J n ( )
x
Solution: We have

x 1
z
x
J ( )z n e 2 z
n
n
Replacing x by x in (i), we get
x 1 x 1
z z
J n ( x )z n e 2 z e 2 z
n

x
J n ( ).( z ) n [by (i)]
n

n n
J ( x )z n J ( ).( 1) z
x
n n
n n
n
Equating the coefficient of z from both sides of (ii) gives
n
J x 1 J ( ) .
x
n n
Example 5: If n > 1, show that

x
n
x
x n 1 J n ( )dx = x J n 1 ( )
x
0
Solution: From recurrence formula I, we have
d n
n
x
x J n ( ) = x J ( ) ...(i)
x
dx n 1
Replacing n by (n + 1) in (i), we get
d n 1
x
x J n 1 ( ) = x n 1 J ( ) ...(ii)
x
dx n
, ,
Integrating (i) w.r.t. x between the limits 0 and x, we get
x
x
x n 1 J n 1 ( ) = x n 1 J n ( )dx
x
x
0
0
x
x
x
or x n 1 J n ( )dx = x n 1 J n 1 ( )
0

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