Page 48 - Handbook Of Integral Equations

√
x cosh λ x – t
28. √ y(t) dt = f(x).
a x – t
Solution: √
x
1 d cos λ x – t

y(x)= √ f(t) dt.
π dx x – t
a
√
x
cosh λ x – t
2 2
29. √ y(t) dt = f(x).
2
0 x – t 2
Solution: √
2
2 d x cos λ x – t 2
y(x)= t √ f(t) dt.
π dx 0 x – t 2
2
√
∞ cosh λ t – x
2 2
30. √ y(t) dt = f(x).
2
x t – x 2
Solution: √
2
2 d ∞ cos λ t – x 2
y(x)= – t √ f(t) dt.
π dx 2 2
x t – x
x

β γ
31. Ax + B cosh (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 cosh (λt)+ C.
x

γ β
32. A cosh (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 cosh (λx) and h(t)= Bt + C.
x
λ µ β γ
33. Ax cosh t + Bt cosh x y(t) dt = f(x).
a
µ
γ
λ
This is a special case of equation 1.9.15 with g 1 (x)= Ax , h 1 (t) = cosh t, g 2 (x)= B cosh x,
β
and h 2 (t)= t .
1.3-2. Kernels Containing Hyperbolic Sine
x

34. sinh[λ(x – t)]y(t) dt = f(x), f(a)= f (a)=0.
x
a
1

Solution: y(x)= f (x) – λf(x).
xx
λ
x

35. sinh[λ(x – t)] + b y(t) dt = f(x).
a
Differentiating the equation with respect to x, we arrive at an equation of the form 2.3.3:
λ x 1

y(x)+ cosh[λ(x – t)]y(t) dt = f (x).
x
b a b
Solution:
1 x

y(x)= f (x)+ R(x – t)f (t) dt,

x
t
b a
√
λ λx λ λ 1+4b 2
R(x)= exp – sinh(kx) – cosh(kx) , k = .
b 2 2b 2bk 2b
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Page 26