Page 125 - Handbook Of Integral Equations

x n
t
m
44. K y(t) dt = P n (x), P n (x)= x λ A m x .
0 x m=0
Solution:
n 1
y(x)= x λ A m x m–1 , I m = K(z)z λ+m–1 dz.
I m 0
m=0
The integral I 0 is supposed to converge.

x

45. g 1 (x) h 1 (t) – h 1 (x) + g 2 (x) h 2 (t) – h 2 (x) y(t) dt = f(x).
a
This is a special case of equation 1.9.50 with g 3 (x)= –g 1 (x)h 1 (x) – g 2 (x)h 2 (x) and h 3 (t)=1.
x
The substitution Y (x)= y(t) dt followed by integration by parts leads to an integral
a
equation of the form 1.9.15:

x

g 1 (x) h 1 (t) + g 2 (x) h 2 (t) Y (t) dt = –f(x).
t t
a
x

λ(x–t) λ(x–t)
46. g 1 (x) h 1 (t) – e h 1 (x) + g 2 (x) h 2 (t) – e h 2 (x) y(t) dt = f(x).
a

This is a special case of equation 1.9.50 with g 3 (x)= –e λx g 1 (x)h 1 (x)+ g 2 (x)h 2 (x) , and
h 3 (t)= e –λt .
x

The substitution Y (x)= e –λt y(t) dt followed by integration by parts leads to an integral
a
equation of the form 1.9.15:

x
λt λt
g 1 (x) e h 1 (t) + g 2 (x) e h 2 (t) Y (t) dt = –f(x).
t t
a
x

λ µ λ+β µ–β λ+γ µ–γ
47. Ag (x)g (t)+ Bg (x)g (t) – (A + B)g (x)g (t) y(t) dt = f(x).
a
µ
λ
This is a special case of equation 1.9.50 with g 1 (x)= Ag (x), h 1 (t)= g (t), g 2 (x)= Bg λ+β (x),
h 2 (t)= g µ–β (t), g 3 (x)= –(A + B)g λ+γ (x), and h 3 (t)= g µ–γ (t).
x
λ µ λ+β µ–β
48. Ag (x)h(x)g (t)+ Bg (x)h(x)g (t)
a

– (A + B)g λ+γ (x)g µ–γ (t)h(t) y(t) dt = f(x).
µ
λ
This is a special case of equation 1.9.50 with g 1 (x)= Ag (x)h(x), h 1 (t)= g (t), g 2 (x)=
Bg λ+β (x)h(x), h 2 (t)= g µ–β (t), g 3 (x)= –(A + B)g λ+γ (x), and h 3 (t)= g µ–γ (t)h(t).
x

λ µ λ+β µ–β
49. Ag (x)h(x)g (t)+ Bg (x)h(t)g (t)
a

– (A + B)g λ+γ (x)g µ–γ (t)h(t) y(t) dt = f(x).
λ
µ
This is a special case of equation 1.9.50 with g 1 (x)= Ag (x)h(x), h 1 (t)= g (t), g 2 (x)=
Bg λ+β (x), h 2 (t)= g µ–β (t)h(t), g 3 (x)= –(A + B)g λ+γ (x), and h 3 (t)= g µ–γ (t)h(t).
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 103