Page 206 - Handbook Of Integral Equations

For equations 2.9.42–2.9.51, only particular solutions are given. To obtain the general solu-
tion, one must add the general solution of the corresponding homogeneous equation 2.9.41 to the
particular solution.

x
n
42. y(x)+ K(x – t)y(t) dt = Ax , n =0, 1, 2, ...
–∞
This is a special case of equation 2.9.44 with λ =0.
◦
1 . A solution with n =0:
A ∞
y(x)= , B =1 + K(z) dz.
B 0
◦
2 . A solution with n =1:
A AC ∞ ∞
y(x)= x + , B =1 + K(z) dz, C = zK(z) dz.
B B 2 0 0
3 . A solution with n =2:
◦
A 2 AC AC 2 AD
y 2 (x)= x +2 x +2 – ,
B B 2 B 3 B 2

∞ ∞ ∞
2
B =1 + K(z) dz, C = zK(z) dz, D = z K(z) dz.
0 0 0
4 . A solution with n =3, 4, ... is given by:
◦
∂ n e λx
–λz
∞
y n (x)= A , B(λ)=1 + K(z)e dz.
∂λ n B(λ)
λ=0 0
x
43. y(x)+ K(x – t)y(t) dt = Ae λx .
–∞
A solution:
A λx ∞ –λz
y(x)= e , B =1 + K(z)e dz.
B
0
The integral term in the expression for B is the Laplace transform of K(z), which may be
calculated using tables of Laplace transforms (e.g., see Supplement 4).
x

n λx
44. y(x)+ K(x – t)y(t) dt = Ax e , n =1, 2, ...
–∞
1 . A solution with n =1:
◦
A λx AC λx
y 1 (x)= xe + e ,
B B 2
∞ ∞

B =1 + K(z)e –λz dz, C = zK(z)e –λz dz.
0 0
It is convenient to calculate B and C using tables of Laplace transforms.
2 . A solution with n =2:
◦
2
A 2 λx AC λx AC AD λx
y 2 (x)= x e +2 xe + 2 – e ,
B B 2 B 3 B 2
∞ ∞ ∞

2
B =1 + K(z)e –λz dz, C = zK(z)e –λz dz, D = z K(z)e –λz dz.
0 0 0
◦
3 . A solution with n =3, 4, ... is given by:
∂ ∂ n e λx ∞ –λz
y n (x)= y n–1 (x)= A , B(λ)=1 + K(z)e dz.
∂λ ∂λ n B(λ) 0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 185