Page 399 - Handbook Of Integral Equations

◦
3 . Solutions:
y 7 (x)= p(J 0 ln x – J 1 ), y 8 (x)= –p(J 0 ln x – J 1 ),
1/2

b
A m
p = , J m = (ln t) f(t) dt.
2
J J 2 – J 0 J 2
0 1 a
k
The equation has more complicated solutions of the form y(x)= n E k (ln x) , where the
k=0
constants E k can be found from the corresponding system of algebraic equations.
b
β
3. f(t)y(t)y(xt) dt = Ax .
a
1 . Solutions:
◦

β
β
y 1 (x)= A/I 0 x , y 2 (x)= – A/I 0 x ,
β
β
y 3 (x)= q(I 1 x – I 2 ) x , y 4 (x)= –q(I 1 x – I 2 ) x ,
where

b
A
I m = t 2β+m f(t) dt, q = , m =0, 1, 2.
2
I 2 (I 0 I 2 – I )
a 1
β
2 . The substitution y(x)= x w(x) leads to an equation of the form 6.2.2:
◦
b

2β
g(t)w(t)w(xt) dt = A, g(x)= f(x)x .
a
Therefore, the integral equation in question has more complicated solutions.
b
4. f(t)y(t)y(xt) dt = A ln x + B.
a
This equation has solutions of the form y(x)= p ln x+q. The constants p and q are determined
from the following system of two second-order algebraic equations:
2
2
2
I 1 p + I 0 pq = A, I 2 p +2I 1 pq + I 0 q = B,
where
b

m
I m = f(t)(ln t) dt, m =0, 1, 2.
a
b

λ λ
5. f(t)y(t)y(xt) dt = Ax ln x + Bx .
a
λ
The substitution y(x)= x w(x) leads to an equation of the form 6.2.4:
b
2λ
g(t)w(t)w(xt) dt = A ln x + B, g(t)= f(t)t .
a
∞ x

λ
6. f(t)y(t)y dt = Ax .
t
0
Solutions:
∞
A λ A λ
y 1 (x)= x , y 2 (x)= – x , I = f(t) dt.
I I
0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 379