Page 136 - Advanced engineering mathematics

116 CHAPTER 3 The Laplace Transform

This is an infinite series if k is not a positive integer. Now set k =−1/2 and x = 1/s in the
2
binomial series. For s > 1, this gives us
−1/2
C 1

Y(s) = 1 +
s s 2
C 1 1 (1)(3) 1
= 1 − + + ···
s 2 s 2 2 2! s 4
2
∞ m
(−1) (2m)! 1

= C .
m
(2 m!) 2 s 2m+1
m=0
Then
∞ m
(−1) (2m)! 1

y(t) = C L −1
m
(2 m!) 2 s 2m+1
m=0
∞ m 2m
(−1) (2m)! t
= C
m
(2 m!) 2 (2m)!
m=0
∞ m
(−1)
2m
= C t .
m
(2 m!) 2
m=0
If we impose the condition y(0)=1, then C =1, and we have the solution called the Bessel
function of the first kind of order zero:
∞ m
(−1)
2m
J 0 (t) = t .
m
(2 m!) 2
m=0
We will now solve Bessel’s equation of any positive integer order n. Bessel’s equation is
2
2
2

t y + ty + (t − n )y = 0.
Change variables by setting
−n
y(t) = t w(t).
Compute y and y , substitute into Bessel’s equation, and carry out some routine algebra to obtain

tw + (1 − 2n)w + tw = 0.

Now apply the Laplace transform to obtain
d
d 2

− s W − sw(0) − w (0) + (1 − 2n)(sW − w(0)) − W = 0.
ds ds
After carrying out these differentiations, we obtain
2

(−1 − s )W + (−2s + (1 − 2n)s)W + w(0) − (1 − 2n)w(0) = 0.
We will seek a solution satisfying w(0) = 0, so this equation becomes
2
(1 + s )W + (1 + 2n)sW = 0.

This is separable. Write
W (2n + 1)s
=− .
W 1 + s 2
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October 14, 2010 14:14 THM/NEIL Page-116 27410_03_ch03_p77-120