Page 121 - Handbook Of Integral Equations

3 . 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 = K(–z) dz, C = zK(–z) dz, D = z K(–z) dz.
0 0 0
◦
4 . Solution with n =3, 4, ... is given by
n λx ∞
∂ e λz
y n (x)= A , B(λ)= K(–z)e dz.
∂λ n B(λ)
λ=0 0
∞

28. K(x – t)y(t) dt = Ae λx .
x
Solution:
A λx ∞ λz
y(x)= e , B = K(–z)e dz.
B 0
The expression for B is the Laplace transform of the function K(–z) with parameter p=–λ and
can be calculated with the aid of tables of Laplace transforms given (e.g., see Supplement 4).

∞

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

∞ ∞
B = K(–z)e λz dz, C = zK(–z)e λz dz.
0 0
It is convenient to calculate the coefficients B and C using tables of Laplace transforms with
parameter p = –λ.
2 . 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 = K(–z)e λz dz, C = zK(–z)e λz dz, D = z K(–z)e λz dz.
0 0 0
◦
3 . Solution with n =3, 4, ... is given by:
∂ ∂ n e λx ∞ λz
y n (x)= y n–1 (x)= A , B(λ)= K(–z)e dz.
∂λ ∂λ n B(λ) 0
∞

30. K(x – t)y(t) dt = A cosh(λx).
x
Solution:
A λx A –λx 1 A A
1 A A
y(x)= e + e = + cosh(λx)+ – sinh(λx),
2B + 2B – 2 B + B – 2 B + B –
∞ ∞

B + = K(–z)e λz dz, B – = K(–z)e –λz dz.
0 0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 99