Page 106 - Handbook Of Integral Equations

x

38. [AY ν (λx)+ BY ν (λt)]y(t) dt = f(x).
a
For B = –A, see equation 1.8.37. We consider the interval [a, x] in which Y ν (λx) does not
change its sign.
Solution with B ≠ –A:
1 d – A x – B

y(x)= ± Y ν (λx) A+B Y ν (λt) A+B f (t) dt .
t
A + B dx a
Here the sign of Y ν (λx) should be taken.
x

m
k
39. [At Y ν (λx)+ Bx Y µ (λt)]y(t) dt = f(x).
a
m
k
This is a special case of equation 1.9.15 with g 1 (x)= AY ν (λx), h 1 (t)= t , g 2 (x)= Bx , and
h 2 (t)= Y µ (λt).
x
40. [AJ ν (λx)Y µ (βt)+ BJ ν (λt)Y µ (βx)]y(t) dt = f(x).
a
This is a special case of equation 1.9.15 with g 1 (x)= AY ν (λx), h 1 (t)= Y µ (βt), g 2 (x)=
BY µ (βx), and h 2 (t)= J ν (λt).

1.8-2. Kernels Containing Modified Bessel Functions

x
41. I 0 [λ(x – t)]y(t) dt = f(x).
a
Solution:
1 d 2 2 2 x
y(x)= – λ (x – t) I 1 [λ(x – t)] f(t) dt.
λ dx 2 a
x

42. [I 0 (λx)– I 0 (λt)]y(t) dt = f(x), f(a)= f (a)=0.
x
a
d f (x)

x
Solution: y(x)= .
dx λI 1 (λx)
x
43. [AI 0 (λx)+ BI 0 (λt)]y(t) dt = f(x).
a
For B = –A, see equation 1.8.42. Solution with B ≠ –A:
1 d – A x – B

y(x)= ± I 0 (λx) A+B I 0 (λt) A+B f (t) dt .
t
A + B dx
a
Here the sign of I ν (λx) should be taken.
x
44. (x – t)I 0 [λ(x – t)]y(t) dt = f(x).
a
Solution:
x
y(x)= g(t) dt,
a
where
2 3 t
1 d 2
g(t)= – λ (t – τ) I 1 [λ(t – τ)] f(τ) dτ.
λ dt 2 a

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 84