Page 29 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes From (i) we have
cos (x sin ) = J + 2J cos 2 + 2J cos 4 ...
0 2 4

x
cos ( sin )d J d 2J cos2 d ...
0 2
0 0 0
J 0.
Deduction: We have to prove that

1
x
J 0 ( ) cos ( cos )d
x
0
1 x 2 cos 2 x 4 cos 4 x 6 cos 6
= 1 ... d ...(iii)
2! 4! 6!
0
1.3.5...(2r 1)
2r
Since cos d
r
2.4.6...(2 )
0
from definite integrals.
from (iii), we get
1 x 2 1 x 4 1.3 x 6 1.3.5
J (x) = ...
0
2! 2 4! 2.4 6! 2.4.6
= 1 x 2 x 4 x 6 ...
2
2 2 2 2 4 2 2 2 4 6 2
x 2 x 4 x 6
= 1 2 2 2 6 ...
2 2 (2!) 2 (3!)
r
( 1) x 2r
= r
r
(2 . !)
r 0
Self Assessment

3. Verify directly from the representation

1
J (x) = cos( sin )d
x
0
0
,
that J (x) satisfies Bessel s equation in which n = 0
0
Example 3: Prove
1
ax
e j (bx )dx ,a 0.
0 2 2
{(a b )}
0

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