Page 105 - Handbook Of Integral Equations

∞ √

2 2 –1/4 2 2

30. t – x J –1/2 λ t – x y(t) dt = f(x).
x
Solution:
√ 2 2
2λ d ∞ cosh λ t – x
y(x)= – t √ f(t) dt.
π dx x t – x 2
2
x
√
2 2 ν/2

2
31. x – t J ν λ x – t 2 y(t) dt = f(x), –1< ν <0.
0
Solution: x
d 2 –(ν+1)/2 √
2

y(x)= λ t x – t I –ν–1 λ x – t 2 f(t) dt.
2
dx
0
•
Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).
∞ √

2 2 ν/2
32. t – x J ν λ t – x 2 y(t) dt = f(x), –1< ν <0.
2
x
Solution:
d ∞ 2 2 –(ν+1)/2 √

2
y(x)= –λ t t – x I –ν–1 λ t – x 2 f(t) dt.
dx x
•
Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).
x

m
k
33. [At J ν (λx)+ Bx J µ (λt)]y(t) dt = f(x).
a
m
k
This is a special case of equation 1.9.15 with g 1 (x)= AJ ν (λx), h 1 (t)= t , g 2 (x)= Bx , and
h 2 (t)= J µ (λt).
x
2
2
34. [AJ (λx)+ BJ (λt)]y(t) dt = f(x).
ν ν
a
Solution with B ≠ –A:
1 d – 2A x – 2B

y(x)= J ν (λx) A+B J ν (λt) A+B f (t) dt .
t
A + B dx
a
x
k m
35. AJ (λx)+ BJ (βt) y(t) dt = f(x).
ν µ
a
m
k
This is a special case of equation 1.9.6 with g(x)= AJ (λx) and h(t)= BJ (βt).
ν µ
x
36. [Y 0 (λx) – Y 0 (λt)]y(t) dt = f(x).
a

d f (x)
x
Solution: y(x)= – .
dx λY 1 (λx)
x

37. [Y ν (λx) – Y ν (λt)]y(t) dt = f(x).
a
d xf (x)

x
Solution: y(x)= .
dx νY ν (λx) – λxY ν+1 (λx)
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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