Page 869 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 849

15.3 Bessel Functions
c
a
1. With y = x J ν (bx ), compute
c
c
a
y =ax a−1 J ν (bx ) + x bcx c−1 J (bx )

ν
c
a
c

y =a(a − 1)x a−2 J ν (bx ) + (2ax a−1 bcx c−1 + x bc(c − 1)x c−2 )J (bx )

ν
a 2
c
+ x c x 2c−2 J (bx ).

ν
c
Substitute these into the differential equation to verify that y = J ν (bx ) is a solution.
2
2
3. y = c 1 J 1/3 (x ) + c 2 J −1/3 (x )
2
2
5. y = c 1 x −1 J 3/4 (2x ) + c 2 x −1 J −3/4 (2x )
4
3
4
3
7. y = c 1 x J 3/4 (2x ) + c 2 x J −3/4 (2x )
9. y = c 1 x −2 J 1/2 (3x ) + cx −2 J −1/2 (3x )
3
3
11. The differential equation transforms to
2
2
z y + zy + (z − 9)y = 0

with general solution
y(z) = c 1 J 3 (z) + c 2 Y 3 (z)
so
√ √
y(x) = c 1 J 3 ( x) + c 2 Y 3 ( x).
13. The differential equation transforms to
2
2
z y + zy + (z − 16)y = 0

giving y(x) = c 1 J 4 (2x 1/3 ) + c 2 Y 4 (2x 1/3 ).
15. The transformed differential equation is
2
2

x u + xu + (x − 1/4)u = 0
leading to y = c 1 x 2/3 J 1/2 (x) + c 2 x 2/3 Y 1/2 (x).
17. It is routine to check from the infinite series expansions that J (s) =−J 1 (s).Then

0
α
α
J 1 (s)ds =−J 0 (s)] = J 0 (α) = J 0 (0) − J 0 (α) = 1 − 0 = 1.
0
0
Now let s = αx to complete the solution.
n
n
19. From the infinite series, it is easy to check that (x J n (x)) = x J n−1 (x). Integrating this yields the first conclusion.

Next, (x −n J n (x)) =−x −n J n+1 (x). Integrating this gives the second expression.

21. Begin with the observation that x n+1 J n (αx)dx = (1/α)x n+1 J n+1 (αx).Then
1 J n+1 (α)

I n,0 = x n+1 J n (αx) = .
α
0
giving part (a). Part (b) follows by using the quoted identity again. The other parts follow by using the given hints.
23. Put t = x sin(θ) in Hankel’s integral, which is given in Problem 22.
25. f (x) = x
2.213145642J 1 (3.831705970x) − 0.5170987826J 1 (7.015586670x)
+ 1.104611216J 1 (10.17346814x) − 0.4549641786J 1 (13.32369194x)
+ 0.8113206562J 1 (16.47063005x)
27. f (x) = xe −x
1.256395517J 1 (3.831705970x) + 0.08237394412J 1 (7.015586670x)
+ 0.5976577270J 1 (10.173468144x) − 0.01994105804J 1 (13.32369194x)
+ 0.4181324338J 1 (16.47063005x)
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October 14, 2010 17:50 THM/NEIL Page-849 27410_25_Ans_p801-866