Page 96 - Handbook Of Integral Equations

x

λ
68. e µ(x–t) (sin x – sin t) y(t) dt = f(x), 0 < λ <1.
a
Solution:
x –µt

1 d
2 e cos tf(t) dt sin(πλ)
y(x)= ke µx cos x , k = .
cos x dx (sin x – sin t) λ πλ
a
x

λ
λ
69. e µ(x–t) (sin x – sin t)y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.5.59:
x

λ
λ
(sin x – sin t)w(t) dt = e –µx f(x).
a
x
λ
λ

70. e µ(x–t) A sin x + B sin t y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.5.60:
x
λ λ
–µx
A sin x + B sin t w(t) dt = e f(x).
a
x µ(x–t)
e y(t) dt
71. = f(x), 0 < λ <1.
a (sin x – sin t) λ
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.5.61:
x

w(t) dt –µx
= e f(x).
(sin x – sin t) λ
a
x

µ(x–t) ν
72. Ae + B sin (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 sin (λx),
and h 2 (t)=1.
x
µ(x–t) ν
73. Ae + B sin (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) = sin (λt).
x

λ
λ

74. e µ(x–t) A tan x + B tan t y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.5.77:
x

λ λ
–µx
A tan x + B tan t w(t) dt = e f(x).
a
x
λ
β

75. e µ(x–t) A tan x + B tan t + C y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.9.6:
x
λ β
–µx
A tan x + B tan t + C w(t) dt = e f(x).
a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 74