Page 195 - Handbook Of Integral Equations

x

10. y(x) – g 1 (t)+ g 2 (t)x y(t) dt = f(x).
a
This equation can be rewritten in the form of equation 2.9.12 with g 1 (t)= g(t)+ th(t) and
g 2 (t)= –h(t).
x

11. y(x) – g(x)+ h(x)(x – t) y(t) dt = f(x).
a
1 . The solution of the integral equation can be represented in the form y(x)= Y , where
◦

xx
Y = Y (x) is the solution of the second-order linear nonhomogeneous ordinary differential
equation

Y xx – g(x)Y – h(x)Y = f(x), (1)
x
under the initial conditions

Y (a)= Y (a)=0. (2)
x
2 . Let Y 1 = Y 1 (x) and Y 2 = Y 2 (x) be two nontrivial linearly independent solutions of the
◦

second-order linear homogeneous differential equation Y –g(x)Y –h(x)Y =0, which follows
x
xx
from (1) for f(x) ≡ 0. Then the solution of the nonhomogeneous differential equation (1)
under conditions (2) is given by
x f(t)

Y (x)= Y 2 (x)Y 1 (t) – Y 1 (x)Y 2 (t) dt, W(t)= Y 1 (t)Y (t) – Y 2 (t)Y (t), (3)

1
2
a W(t)
where W(t) is the Wronskian and the primes denote the derivatives.
Substituting (3) into (1), we obtain the solution of the original integral equation in the
form
1
x

y(x)= f(x)+ R(x, t)f(t) dt, R(x, t)= [Y (x)Y 1 (t) – Y (x)Y 2 (t)]. (4)
2
1
a W(t)
◦
3 . Let Y 1 = Y 1 (x) be a nontrivial particular solution of the homogeneous differential equa-
tion (1) (with f ≡ 0) satisfying the initial condition Y 1 (a) ≠ 0. Then the function
x W(t) x
Y 2 (x)= Y 1 (x) 2 dt, W(x)=exp g(s) ds (5)
a [Y 1 (t)] a
is another nontrivial solution of the homogeneous equation. Substituting (5) into (4) yields
the solution of the original integral equation in the form
x
y(x)= f(x)+ R(x, t)f(t) dt,
a
W(x) Y 1 (t) Y 1 (t) x W(s)

R(x, t)= g(x) +[g(x)Y (x)+ h(x)Y 1 (x)] ds,
1
Y 1 (x) W(t) W(t) t [Y 1 (s)] 2
x
where W(x)=exp g(s) ds .
a
x

12. y(x) – g(t)+ h(t)(t – x) y(t) dt = f(x).
a
Solution:
x

y(x)= f(x)+ R(x, t)f(t) dt,
a
Y (x)W(x) t ds
R(x, t)= g(t) + Y (x)W(x)[g(t)Y (t)+ h(t)Y (t)] ,

t
Y (t)W(t) x W(s)[Y (s)] 2
t

W(t)=exp g(t) dt ,
b
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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