Page 119 - Handbook Of Integral Equations

x

23. K(x – t)y(t) dt = A sin(λx).
–∞
Solution:

A
y(x)= B c sin(λx)+ B s cos(λx) ,
2
B + B 2 s
c
∞ ∞

B c = K(z) cos(λz) dz, B s = K(z) sin(λz) dz.
0 0
x
24. K(x – t)y(t) dt = Ae µx cos(λx).
–∞
Solution:

A µx
y(x)= e B c cos(λx) – B s sin(λx) ,
B + B 2
2
c s

∞ ∞
B c = K(z)e –µz cos(λz) dz, B s = K(z)e –µz sin(λz) dz.
0 0
x
25. K(x – t)y(t) dt = Ae µx sin(λx).
–∞
Solution:
A µx
y(x)= e B c sin(λx)+ B s cos(λx) ,
2
B + B s 2
c
∞ ∞

B c = K(z)e –µz cos(λz) dz, B s = K(z)e –µz sin(λz) dz.
0 0
x

26. K(x – t)y(t) dt = f(x).
–∞
k
◦
1 . For a polynomial right-hand side of the equation, f(x)= n A k x , the solution has the
k=0
form
n
k
y(x)= B k x ,
k=0
where the constants B k are found by the method of undetermined coefficients. The solution
◦
can also be obtained by the formula given in 1.9.17 (item 4 ).
k
2 .For f(x)= e λx n A k x , the solution 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. The solution
◦
can also be obtained by the formula given in 1.9.19 (item 3 ).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 97