Page 92 - Handbook Of Integral Equations

x

37. e λ(x–t) (A ln x + B ln t)y(t) dt = f(x).
a
The substitution w(x)= e –λx y(x) leads to an equation of the form 1.4.4:

x
(A ln x + B ln t)w(t) dt = e –λx f(x).
a
x

2
2

38. e µ(x–t) A ln (λx)+ B ln (λt) y(t) dt = f(x).
a
The substitution w(x)= e –λx y(x) leads to an equation of the form 1.4.7:
x

2 2 –λx
A ln (λx)+ B ln (λt) w(t) dt = e f(x).
a
x n
39. e λ(x–t) ln(x/t) y(t) dt = f(x), n =1, 2, ...
a
Solution:
n+1
1 λx d –λx

y(x)= e x F λ (x), F λ (x)= e f(x).
n! x dx
x
40. e λ(x–t) ln(x/t) y(t) dt = f(x).
a
Solution:
2
2e λx d x e –λt f(t) dt
y(x)= x .
πx dx a t ln(x/t)
x
e
λ(x–t)
41. y(t) dt = f(x).
a ln(x/t)
Solution: x
1 λx d e –λt f(t) dt
y(x)= e .
π dx a t ln(x/t)
x

µ(x–t) ν
42. Ae + B ln (λx) y(t) dt = f(x).
a
ν
µx
This is a special case of equation 1.9.15 with g 1 (x)= Ae , h 1 (t)= e –µt , g 2 (x)= B ln (λx),
and h 2 (t)=1.
x
µ(x–t) ν
43. Ae + B ln (λt) y(t) dt = f(x).
a
µx
This is a special case of equation 1.9.15 with g 1 (x)= Ae , h 1 (t)= e –µt , g 2 (x)= B, and
ν
h 2 (t)=ln (λt).
x
λ
44. e µ(x–t) [ln(x/t)] y(t) dt = f(x), 0 < λ <1.
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.4.16:
x
λ
[ln(x/t)] w(t) dt = e –µx f(x).
a

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 70