Page 126 - Handbook Of Integral Equations

x

50. g 1 (x)h 1 (t)+ g 2 (x)h 2 (t)+ g 3 (x)h 3 (t) y(t) dt = f(x),
a
where g 1 (x)h 1 (x)+ g 2 (x)h 2 (x)+ g 3 (x)h 3 (x) ≡ 0.
x

The substitution Y (x)= h 3 (t)y(t) dt followed by integration by parts leads to an integral
a
equation of the form 1.9.15:
x

h 1 (t) h 2 (t)
g 1 (x) + g 2 (x) Y (t) dt = –f(x).
a h 3 (t) t h 3 (t) t
x
αt
βt
51. Q(x – t)e y(ξ) dt = Ae px , ξ = e g(x – t).
–∞
Solution:
A p–α ∞ p–α –pz
y(ξ)= ξ β , q = Q(z)[g(z)] β e dz.
q 0
1.10. Some Formulas and Transformations

1. Let the solution of the integral equation
x
K(x, t)y(t) dt = f(x) (1)
a

have the form

y(x)= F f(x) , (2)
where F is some linear integro-differential operator. Then the solution of the more complicated
integral equation
x
K(x, t)g(x)h(t)y(t) dt = f(x) (3)
a
has the form
1 f(x)
y(x)= F . (4)
h(x) g(x)
Below are formulas for the solutions of integral equations of the form (3) for some specific
functions g(x) and h(t). In all cases, it is assumed that the solution of equation (1) is known and is
determined by formula (2).
(a) The solution of the equation

x
λ
K(x, t)(x/t) y(t) dt = f(x)
a
has the form
λ
–λ

y(x)= x F x f(x) .
(b) The solution of the equation
x
K(x, t)e λ(x–t) y(t) dt = f(x)
a
has the form
λx

y(x)= e F e –λx f(x) .

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 104