Page 213 - Handbook Of Integral Equations

7 .For f(x)= n A k sin(λ k x), a solution of the equation has the form
◦
k=0
n
A k

y(x)= 2 2 B ck sin(λ k x) – B sk cos(λ k x) ,
B + B
k=0 ck sk
∞ ∞

B ck =1 + K(–z) cos(λ k z) dz, B sk = K(–z) sin(λ k z) dz.
0 0
8 .For f(x) = cos(λx) n A k exp(µ k x), a solution of the equation has the form
◦
k=0
n n
A k B ck A k B sk

y(x) = cos(λx) 2 2 exp(µ k x) + sin(λx) 2 2 exp(µ k x),
B + B B + B
k=0 ck sk k=0 ck sk
∞ ∞

B ck =1 + K(–z) exp(µ k z) cos(λz) dz, B sk = K(–z) exp(µ k z) sin(λz) dz.
0 0
9 .For f(x) = sin(λx) n A k exp(µ k x), a solution of the equation has the form
◦
k=0
n n
A k B ck A k B sk
y(x) = sin(λx) exp(µ k x) – cos(λx) exp(µ k x),
B 2 + B 2 B 2 + B 2
k=0 ck sk k=0 ck sk
∞ ∞

B ck =1 + K(–z) exp(µ k z) cos(λz) dz, B sk = K(–z) exp(µ k z) sin(λz) dz.
0 0
10 . In the general case of arbitrary right-hand side f = f(x), the solution of the integral
◦
equation can be represented in the form
˜
1 c+i∞ f(p) px
y(x)= e dp,
˜
2πi c–i∞ 1+ k(–p)
∞ ∞

˜
˜
f(p)= f(x)e –px dx, k(–p)= K(–z)e pz dz.
0 0
˜
˜
To calculate f(p) and k(–p), it is convenient to use tables of Laplace transforms, and to
determine y(x), tables of inverse Laplace transforms.
2.9-3. Other Equations
x 1 t
63. y(x)+ f y(t) dt =0.
0 x x
Eigenfunctions of this integral equation are determined by the roots of the following tran-
scendental (algebraic) equation for the parameter λ:

1

λ
f(z)z dz = –1. (1)
0
◦
1 . For a real simple root λ k of equation (1) there is a corresponding eigenfunction
λ k
y k (x)= x .

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 192