Page 44 - Handbook Of Integral Equations

x

3. cosh[λ(x – t)] + b y(t) dt = f(x).
a
For b = 0, see equation 1.3.1. For b = –1, see equation 1.3.2. For λ = 0, see equation 1.1.1.
Differentiating the equation with respect to x, we arrive at an equation of the form 2.3.16:
λ x f (x)

x
y(x)+ sinh[λ(x – t)]y(t) dt = .
b +1 a b +1
1 . Solution with b(b + 1)<0:
◦
f (x) λ 2 x –b

x

y(x)= – sin[k(x – t)]f (t) dt, where k = λ .
t
b +1 k(b +1) 2 a b +1
2 . Solution with b(b + 1)>0:
◦
f (x) λ 2 x b

x
y(x)= – sinh[k(x – t)]f (t) dt, where k = λ .

t
b +1 k(b +1) 2 a b +1
x
4. cosh(λx + βt)y(t) dt = f(x).
a
For β = –λ, see equation 1.3.1.
Differentiating the equation with respect to x twice, we obtain
x

cosh[(λ+β)x]y(x)+λ sinh(λx+βt)y(t) dt = f (x), (1)
x
a
x
2
cosh[(λ+β)x]y(x) +λ sinh[(λ+β)x]y(x)+λ cosh(λx+βt)y(t) dt = f (x). (2)
xx
x
a
Eliminating the integral term from (2) with the aid of the original equation, we arrive at
the first-order linear ordinary differential equation
2

w + λ tanh[(λ + β)x]w = f (x) – λ f(x), w = cosh[(λ + β)x]y(x). (3)

xx
x

Setting x = a in (1) yields the initial condition w(a)= f (a). On solving equation (3) with this
x
condition, after some manipulations we obtain the solution of the original integral equation
in the form
1 λ sinh[(λ + β)x]
y(x)= f (x) – f(x)

x
2
cosh[(λ + β)x] cosh [(λ + β)x]
λβ x k–2 λ
+ k+1 f(t) cosh [(λ + β)t] dt, k = .
cosh [(λ + β)x] a λ + β
x

5. [cosh(λx) – cosh(λt)]y(t) dt = f(x).
a
This is a special case of equation 1.9.2 with g(x) = cosh(λx).
1 d f (x)

x
Solution: y(x)= .
λ dx sinh(λx)
x

6. [A cosh(λx)+ B cosh(λt)]y(t) dt = f(x).
a
For B = –A, see equation 1.3.5. This is a special case of equation 1.9.4 with g(x) = cosh(λx).
1 d – A x – B

Solution: y(x)= cosh(λx) A+B cosh(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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