Page 21 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes or rearranging terms we have
x
x
x J n ( ) n J n ( ) x J n 1 ( )
x
x
x
x
III. 2n J n ( ) x [J n 1 ( ) J n 1 ( )]
To prove this we just make use of the above two recurrence relations I and II, here we have
x J (x) = n J (x) x J (x)
n n n+1
and x J (x) + n J (x) = x J (x)
n n n 1
Substracting we get

n J (x) = n J (x) x J (x) x J (x)
n n n+1 n 1
or 2n J (x) = x [J (x) + J (x)].
n n 1 n+1
Again rearranging terms we have

2n J (x) = x J (x) + x J (x)
n n+1 n 1
or 2n J (x) = x [J (x) + J (x)]
n n 1 n+1
You can see that relation III is not independent. It depends upon I and II recurring relations.
IV. 2 J (x) = J (x) J (x)
n n 1 n+1
Hint: Add recurrence relations I and II and simplify the result.

From recurrence relation I, we can show that
J (x) = J (x)
0 1
d n n
x
x
V. x J n ( ) x J n 1 ( )
dx
Now, the left hand side is
d n
x J n ( ) = n x n 1 J ( ) x n J
x
x
dx n n
x
x
= x n 1 [ n J n ( ) x J n ( )]
x
x
= x n 1 [ n J ( ) n J ( ) x J ( )] {From recurrence relation I}
x
n n n 1
= x n 1 [ x J ( )]
x
n 1
= x n J n 1 ( ) R . .S
H
x
Self Assessment
2. Prove
d n n
x J n ( ) x J n 1 ( )
x
x
dx

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