Page 200 - Handbook Of Integral Equations

x

λx λx
25. y(x) – g(x)+ be + b(x – t)e [λ – g(x)] y(t) dt = f(x).
a
λx
This is a special case of equation 2.9.16 with h(x)= be .
Solution:
x
y(x)= f(x)+ R(x, t)f(t) dt,
a
G(x) 2 2λx λx H(x) x G(s)
λx
R(x, t)=[g(x)+ be ] +(b e + bλe ) ds,
G(t) G(t) t H(s)
x b

where G(x)=exp g(s) ds and H(x)=exp e λx .
a λ
x
λ(x–t) λx
λt
26. y(x)+ λe + e g (x) – λe g(t) h(t) y(t) dt = f(x).
x
a
λx
This is a special case of equation 2.9.17 with ϕ(x)= e .
x
–λ(x–t) λt
27. y(x) – λe + e λx
g (t) – λe g(x) h(x) y(t) dt = f(x).
t
a
λx
This is a special case of equation 2.9.18 with ϕ(x)= e .
x

28. y(x)+ cosh[λ(x – t)]g(t)y(t) dt = f(x).
a
Differentiating the equation with respect to x twice yields
x

y (x)+ g(x)y(x)+ λ sinh[λ(x – t)]g(t)y(t) dt = f (x), (1)

x x
a
x

2

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

xx x xx
a
Eliminating the integral term from (2) with the aid of the original equation, we arrive at
the second-order linear ordinary differential equation
2 2

y + g(x)y x – λ y = f (x) – λ f(x). (3)
xx
xx
By setting x = a in the original equation and (1), we obtain the initial conditions for y = y(x):

y(a)= f(a), y (a)= f (a) – f(a)g(a). (4)
x
x
Equation (3) under conditions (4) determines the solution of the original integral equation.
x
29. y(x)+ cosh[λ(x – t)]g(x)h(t)y(t) dt = f(x).
a
The substitution y(x)= g(x)u(x) leads to an equation of the form 2.9.28:
x
u(x)+ cosh[λ(x – t)]g(t)h(t)u(t) dt = f(x)/g(x).
a

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 179