Page 53 - Handbook Of Integral Equations

Eliminating the first term on the left-hand side with the aid of the original equation, we obtain

x 1
2 2

sinh n–2 [λ(x – t)]y(t) dt = 2 f (x) – λ n f(x) .

xx
a λ n(n – 1)
This equation has the same form as the original equation, but the exponent of the kernel has
been reduced by two.
By applying this technique sufficiently many times, we finally arrive at simple integral
equations of the form 1.1.1 (for even n) or 1.3.34 (for odd n).
x
√

55. sinh λ x – t y(t) dt = f(x).
a
Solution: √
2 d 2 x cos λ x – t
y(x)= √ f(t) dt.
πλ dx 2 a x – t
x
√

56. sinh x – sinh ty(t) dt = f(x).
a
Solution:
2 1 d
2 x cosh tf(t) dt
y(x)= cosh x √ .
π cosh x dx a sinh x – sinh t
x
y(t) dt
57. √ = f(x).
a sinh x – sinh t
Solution:
1 d x cosh tf(t) dt
y(x)= √ .
π dx a sinh x – sinh t
x

λ
58. (sinh x – sinh t) y(t) dt = f(x), 0 < λ <1.
a
Solution:
x
1 d cosh tf(t) dt sin(πλ)

2
y(x)= k cosh x , k = .
cosh x dx (sinh x – sinh t) λ πλ
a
x
µ µ
59. (sinh x – sinh t)y(t) dt = f(x).
a
µ
This is a special case of equation 1.9.2 with g(x) = sinh x.
1 d f (x)

x
Solution: y(x)= .
µ dx cosh x sinh µ–1 x
x
µ µ
60. A sinh (λx)+ B sinh (λt) y(t) dt = f(x).
a
µ
This is a special case of equation 1.9.4 with g(x) = sinh (λx).
Solution with B ≠ –A:
1 d – Aµ x – Bµ

y(x)= sinh(λx) A+B sinh(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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