Page 54 - Handbook Of Integral Equations

x
y(t) dt
61. = f(x), 0 < λ <1.
a (sinh x – sinh t) λ
Solution: x
sin(πλ) d cosh tf(t) dt
y(x)= .
π dx a (sinh x – sinh t) 1–λ
x
62. (x – t) sinh[λ(x – t)]y(t) dt = f(x), f(a)= f (a)= f (a)=0.

x xx
a
Double differentiation yields
x x

2λ cosh[λ(x – t)]y(t) dt + λ 2 (x – t) sinh[λ(x – t)]y(t) dt = f (x).
xx
a a
Eliminating the second term on the left-hand side with the aid of the original equation, we
arrive at an equation of the form 1.3.1:
x

1 2
cosh[λ(x – t)]y(t) dt = f (x) – λ f(x) .

xx
a 2λ
Solution: x
1 1 3

y(x)= f xxx (x) – λf (x)+ λ f(t) dt.
x
2
2λ a
x

β γ
63. Ax + B sinh (λt)+ C]y(t) dt = f(x).
a
γ
β
This is a special case of equation 1.9.6 with g(x)= Ax and h(t)= B sinh (λt)+ C.
x
γ β
64. A sinh (λx)+ Bt + C]y(t) dt = f(x).
a
β
γ
This is a special case of equation 1.9.6 with g(x)= A sinh (λx) and h(t)= Bt + C.
x
λ µ β γ
65. Ax sinh t + Bt sinh x y(t) dt = f(x).
a
λ
γ
µ
This is a special case of equation 1.9.15 with g 1 (x)= Ax , h 1 (t) = sinh t, g 2 (x)= B sinh x,
β
and h 2 (t)= t .
1.3-3. Kernels Containing Hyperbolic Tangent
x

66. tanh(λx) – tanh(λt) y(t) dt = f(x).
a
This is a special case of equation 1.9.2 with g(x) = tanh(λx).
1 2
Solution: y(x)= cosh (λx)f (x) .

x
λ x
x

67. A tanh(λx)+ B tanh(λt) y(t) dt = f(x).
a
For B = –A, see equation 1.3.66. This is a special case of equation 1.9.4 with g(x) = tanh(λx).
1 d – A x – B

Solution: y(x)= tanh(λx) A+B tanh(λt) A+B f (t) dt .
t
A + B dx a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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