Page 198 - Handbook Of Integral Equations

n
x
19. y(x) – g k (x)(x – t) k–1 y(t) dt = f(x).
a
k=1
The solution can be represented in the form
x
y(x)= f(x)+ R(x, t)f(t) dt. (1)
a
Here the resolvent R(x, t)isgiven by
n
d w
(n)
(n)
R(x, t)= w x , w x = , (2)
dx n
where w is the solution of the nth-order linear homogeneous ordinary differential equation
w (n) – g 1 (x)w (n–1) – g 2 (x)w x (n–2) – 2g 3 (x)w x (n–3) – ··· – (n – 1)! g n (x)w = 0 (3)
x
x
satisfying the following initial conditions at x = t:

w = w = ··· = w x (n–2) =0, w x (n–1) = 1. (4)
x=t x x=t x=t x=t
Note that the differential equation (3) implicitly depends on t via the initial conditions (4).
•
References: E. Goursat (1923), A. F. Verlan’ and V. S. Sizikov (1987).
n
x
20. y(x) – g k (t)(t – x) k–1 y(t) dt = f(x).
a
k=1
The solution can be represented in the form
x

y(x)= f(x)+ R(x, t)f(t) dt. (1)
a
Here the resolvent R(x, t)isgiven by
n
d u
(n)
(n)
R(x, t)= –u t , u t = , (2)
dt n
where u is the solution of the nth-order linear homogeneous ordinary differential equation
u (n) + g 1 (t)u (n–1) + g 2 (t)u (n–2) +2g 3 (t)u (n–3) + ... +(n – 1)! g n (t)u = 0, (3)
t t t t
satisfying the following initial conditions at t = x:
(n–2) (n–1)
u = u = ··· = u t =0, u t = 1. (4)
t=x t t=x t=x t=x
Note that the differential equation (3) implicitly depends on x via the initial conditions (4).
•
References: E. Goursat (1923), A. F. Verlan’ and V. S. Sizikov (1987).
x
λx+µt µx+λt
21. y(x)+ e – e g(t)y(t) dt = f(x).
a
Let us differentiate the equation twice and then eliminate the integral terms from the resulting
relations and the original equation. As a result, we arrive at the second-order linear ordinary
differential equation

y xx – (λ + µ)y + (λ – µ)e (λ+µ)x g(x)+ λµ y = f (x) – (λ + µ)f (x)+ λµf(x),

x
x
xx

which must be supplemented by the initial conditions y(a)= f(a), y (a)= f (a).
x
x
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 177