Page 348 - Handbook Of Integral Equations

b
46. y(x)+ f(t)y(x – t) dt = g(x).
a
n
◦
1 .For g(x)= A k exp(λ k x), the equation has a solution
k=1
n b
A k
y(x)= exp(λ k x), B k =1 + f(t) exp(–λ k t) dt.
B k a
k=1
n
k
2 . For polynomial right-hand side of the equation, g(x)= A k x , a solution has the form
◦
k=0
n

k
y(x)= B k x ,
k=0
where the constants B k can be found by the method of undetermined coefficients.
n
k
3 .For g(x)= e λx 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 can be found by the method of undetermined coefficients.
n
4 .For g(x)= A k cos(λ k x), a solution of the equation has the form
◦
k=1
n n

y(x)= B k cos(λ k x)+ C k sin(λ k x),
k=1 k=1
where the constants B k and C k can be found by the method of undetermined coefficients.
n
5 .For g(x)= A k sin(λ k x), a solution of the equation has the form
◦
k=1
n n

y(x)= B k cos(λ k x)+ C k sin(λ k x),
k=1 k=1
where the constants B k and C k can be found by the method of undetermined coefficients.
n
k
6 .For g(x) = cos(λx) 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 can be found by the method of undetermined coefficients.
n
k
7 .For g(x) = sin(λx) 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 can be found by the method of undetermined coefficients.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 327