Page 120 - Handbook Of Integral Equations

3 .For f(x)= n A k exp(λ k x), the solution has the form
◦
k=0
n
∞
A k

y(x)= exp(λ k x), B k = K(z) exp(–λ k z) dz.
B k 0
k=0
k
◦
4 .For f(x) = cos(λx) n A k x , the solution has the form
k=0
n n
k k
y(x) = cos(λx) B k x + sin(λx) C k x ,
k=0 k=0
where the constants B k and C k are found by the method of undetermined coefficients.

k
5 .For f(x) = sin(λx) n A k x , the solution has the form
◦
k=0
n n
k k
y(x) = cos(λx) B k x + sin(λx) C k x ,
k=0 k=0
where the constants B k and C k are found by the method of undetermined coefficients.
◦
6 .For f(x)= n A k cos(λ k x), the solution has the form
k=0
n
A k
y(x)= B ck cos(λ k x) – B sk sin(λ k x) ,
B 2 + B 2
k=0 ck sk
∞ ∞

B ck = K(z) cos(λ k z) dz, B sk = K(z) sin(λ k z) dz.
0 0
◦
7 .For f(x)= n A k sin(λ k x), the solution has the form
k=0
n
A k
y(x)= B ck sin(λ k x)+ B sk cos(λ k x) ,
B 2 + B 2
k=0 ck sk
∞ ∞

B ck = K(z) cos(λ k z) dz, B sk = K(z) sin(λ k z) dz.
0 0
∞

n
27. K(x – t)y(t) dt = Ax , n =0, 1, 2, ...
x
This is a special case of equation 1.9.29 with λ =0.
1 . Solution with n =0:
◦
A ∞
y(x)= , B = K(–z) dz.
B
0
2 . Solution with n =1:
◦
A AC ∞ ∞
y(x)= x – , B = K(–z) dz, C = zK(–z) dz.
B B 2
0 0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 98