Page 91 - Handbook Of Integral Equations

x

µ(x–t) λ
29. Ae + B tanh x y(t) dt = f(x).
a
λ
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 tanh x,
and h 2 (t)=1.
x

µ(x–t) λ
30. Ae + B tanh t 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
λ
h 2 (t) = tanh t.
x
λ
λ

31. e µ(x–t) A coth x + B coth t y(t) dt = f(x).
a
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.3.90:
x

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

32. e µ(x–t) A coth x + B coth t + C y(t) dt = f(x).
a
λ
The substitution w(x)= e –µx y(x) leads to an equation of the form 1.9.6 with g(x)= A coth x,
β
h(t)= B coth t + C:
x

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

µ(x–t) λ
33. Ae + B coth x y(t) dt = f(x).
a
λ
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 coth x,
and h 2 (t)=1.
x
µ(x–t) λ
34. Ae + B coth t 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
λ
h 2 (t) = coth t.
1.7-2. Kernels Containing Exponential and Logarithmic Functions
x

35. e λ(x–t) (ln x – ln t)y(t) dt = f(x).
a
Solution:

y(x)= e λx xϕ (x)+ ϕ (x) , ϕ(x)= e –λx f(x).

xx x
x
36. e λ(x–t) ln(x – t)y(t) dt = f(x).
0
The substitution w(x)= e –λx y(x) leads to an equation of the form 1.4.2:
x

ln(x – t)w(t) dt = e –λx f(x).
0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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