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b

3. ln |x – t| + β y(t) dt = f(x).
a
By setting
–β
–β
–β
x = e z, t = e τ, y(t)= Y (τ), f(x)= e g(z),
we arrive at an equation of the form 3.4.2:
B

β
β
ln |z – τ| Y (τ) dτ = g(z), A = ae , B = be .
A

a A
4. ln y(t) dt = f(x), –a ≤ x ≤ a.
–a |x – t|
This is a special case of equation 3.4.3 with b = –a. Solution with 0 < a <2A:
1 d a
y(x)= w(t, a)f(t) dt w(x, a)
2M (a) da –a

a ξ
1 d 1 d

– w(x, ξ) w(t, ξ)f(t) dt dξ
2 |x| dξ M (ξ) dξ –ξ

1 d a w(x, ξ) ξ
– w(t, ξ) df(t) dξ,
2 dx |x| M (ξ) –ξ

where
–1

2A M(ξ)
M(ξ)= ln , w(x, ξ)= ,
ξ π ξ – x 2
2
and the prime stands for the derivative.
•
Reference: I. C. Gohberg and M. G. Krein (1967).
a

x + t
5. ln y(t) dt = f(x).
x – t
0
Solution: a t
2 d F(t) dt d sf(s) ds
y(x)= – √ , F(t)= √ .
π dx 2 2 dt 2 2
2
x t – x 0 t – s
•
Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).
b

1+ λx
6. ln y(t) dt = f(x).
1+ λt
a
This is a special case of equation 3.8.3 with g(x) = ln(1 + λx).
Solution:
1 d
y(x)= (1 + λx)f (x) .

x
2λ dx
The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2 ◦
of equation 3.8.3).
b
β
7. ln x – ln t y(t) dt = f(x), 0 < β <1.
β
a
β
This is a special case of equation 3.8.3 with g(x)=ln x.
b
y(t)
8. dt = f(x), 0 < β <1.
a |ln(x/t)| β
This is a special case of equation 3.8.7 with g(x)=ln x + A, where A is an arbitrary number.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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