Page 194 - Handbook Of Integral Equations

x

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

Y + g(x)+ h(x) Y + g (x)Y = f (x), (1)

xx x x x
which must be supplemented by the initial conditions
Y (a)=0, Y (a)= f(a). (2)

x
Conditions (3) follow from the original equation and the definition of Y (x).
For exact solutions of second-order linear ordinary differential equations (1) 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)+ h(x) Y + g (x)Y =0,

x
x
which follows from (1) for f(x) ≡ 0.
Solving the nonhomogeneous equation (1) under the initial conditions (2) 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
x

y(x)= f(x)+ R(x, t)f(t) dt,
a
∂ 2 Y 1 (x)Y 2 (t) – Y 2 (x)Y 1 (t)

R(x, t)= , W(x)= Y 1 (x)Y (x) – Y 2 (x)Y (x),
1
2
∂x∂t W(t)
where W(x) is the Wronskian and the primes stand for the differentiation with respect to the
argument specified in the parentheses.
x

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

8. y(x)+ g(t)+ λ + λ(x – t)g(t) y(t) dt = f(x).
a
Solution:
x

y(x)= f(x)+ R(x, t)f(t) dt,
a
G(t) λ 2 x λ(t–s) x
R(x, t)= –[g(t)+ λ] + e G(s) ds, G(x)=exp g(s) ds .
G(x) G(x) t a
x

9. y(x) – g 1 (x)+ g 2 (x)t y(t) dt = f(x).
a
This equation can be rewritten in the form of equation 2.9.11 with g 1 (x)= g(x)+ xh(x) and
g 2 (x)= –h(x).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 173