Page 193 - Handbook Of Integral Equations

x

5. y(x)+ g(x) – g(t) y(t) dt = f(x).
a
◦
1 . Differentiating the equation with respect to x yields
x

y (x)+ g (x) y(t) dt = f (x). (1)

x x x
a
x
Introducing the new variable Y (x)= y(t) dt, we obtain the second-order linear ordinary
a
differential equation

Y xx + g (x)Y = f (x), (2)

x
x
which must be supplemented by the initial conditions

Y (a)=0, Y (a)= f(a). (3)
x
Conditions (3) follow from the original equation and the definition of Y (x).
For exact solutions of second-order linear ordinary differential equations (2) with vari-
ous f(x), see E. Kamke (1977), G. M. Murphy (1960), and A. D. Polyanin and V. F. Zaitsev
(1995, 1996).
2 . Let Y 1 = Y 1 (x) and Y 2 = Y 2 (x) be two linearly independent solutions (Y 1 /Y 2 /≡ const) of
◦
the second-order linear homogeneous differential equation Y xx + g (x)Y = 0, which follows

x
from (2) for f(x) ≡ 0. In this case, the Wronskian is a constant:

W = Y 1 (Y 2 ) – Y 2 (Y 1 ) ≡ const .

x
x
Solving the nonhomogeneous equation (2) under the initial conditions (3) with arbitrary
f = f(x) and taking into account y(x)= Y (x), we obtain the solution of the original integral

x
equation in the form
1 x

y(x)= f(x)+ Y (x)Y (t) – Y (x)Y (t) f(t) dt, (4)
2
2
1
1
W a
where the primes stand for the differentiation with respect to the argument specified in the
parentheses.
◦
3 . Given only one nontrivial solution Y 1 = Y 1 (x) of the linear homogeneous differential
equation Y xx + g (x)Y = 0, one can obtain the solution of the nonhomogeneous equation (2)

x
under the initial conditions (3) by formula (4) with
x
dξ
W =1, Y 2 (x)= Y 1 (x) 2 ,
b Y (ξ)
1
where b is an arbitrary number.
x

6. y(x)+ g(x)+ h(t) y(t) dt = f(x).
a
◦
1 . Differentiating the equation with respect to x yields
x

y (x)+ g(x)+ h(x) y(x)+ g (x) y(t) dt = f (x).
x
x
x
a
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Page 172