Page 29 - Handbook Of Integral Equations

1.1-5. Kernels Containing Rational Functions

x y(t) dt
26. = f(x).
0 x + t
n
1 . For a polynomial right-hand side, f(x)= N A n x , the solution has the form
◦
n=0
N n k
A n n n
(–1)
y(x)= x , B n =(–1) ln 2 + .
B n k
n=0 k=1
n
2 .For f(x)= x λ N A n x , where λ is an arbitrary number (λ > –1), the solution has the
◦
n=0
form
N 1 t λ+n dt
n
y(x)= x λ A n x , B n = .
B n 0 1+ t
n=0

N
3 .For f(x)=ln x A n x n , the solution has the form
◦
n=0
N N
A n n A n I n n

y(x)=ln x x + 2 x ,
B n B n
n=0 n=0
n n
k 2 k
n (–1) n π (–1)
B n =(–1) ln 2 + , I n =(–1) + .
k 12 k 2
k=1 k=1
n
4 .For f(x)= A n ln x) , the solution of the equation has the form
◦
N
n=0
N

y(x)= A n Y n (x),
n=0
where the functions Y n = Y n (x) are given by
d x z dz
n λ 1 λ
Y n (x)= , I(λ)= .
dλ n I(λ) 1+ z
λ=0 0
5 .For f(x)= N A n cos(λ n ln x)+ N B n sin(λ n ln x), the solution of the equation has the
◦
n=1 n=1
form
N N

y(x)= C n cos(λ n ln x)+ D n sin(λ n ln x),
n=1 n=1
where the constants C n and D n are found by the method of undetermined coefficients.
6 . For arbitrary f(x), the transformation
◦
1 2z
–τ
–z
1 2τ
x = e , t = e , y(t)= e w(τ), f(x)= e g(z)
2 2
leads to an integral equation with difference kernel of the form 1.9.26:
z
w(τ) dτ
= g(z).
cosh(z – τ)
–∞
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 7