Page 212 - Handbook Of Integral Equations

∞

62. y(x)+ K(x – t)y(t) dt = f(x).
x
k
◦
1 . For a polynomial right-hand side, f(x)= n A k x , a solution has the form
k=0
n
k
y(x)= B k x ,
k=0
where the constants B k are found by the method of undetermined coefficients. One can also
make use of the formula given in item 4 of equation 2.9.53 to construct the solution.
◦
k
2 .For f(x)= e λx n A k x , a solution of the equation has the form
◦
k=0
n
k
y(x)= e λx B k x ,
k=0
where the constants B k are found by the method of undetermined coefficients. One can also
make use of the formula given in item 3 of equation 2.9.55 to construct the solution.
◦
3 .For f(x)= n A k exp(λ k x), a solution of the equation has the form
◦
k=0
n
A k
∞
y(x)= exp(λ k x), B k =1 + K(–z) exp(λ k z) dz.
B k 0
k=0
k
4 .For f(x) = cos(λx) n A k x a solution of the equation 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 , a solution of the equation 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 B k and C k are found by the method of undetermined coefficients.

6 .For f(x)= n A k cos(λ k x), a solution of the equation 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 =1 + K(–z) cos(λ k z) dz, B sk = K(–z) sin(λ k z) dz.
0 0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 191