Page 116 - Handbook Of Integral Equations

x

12. [Ag(x)h(t)+ Bg(t)h(x)]y(t) dt = f(x).
a
For B = –A, see equation 1.9.11.
Solution with B ≠ –A:

A
B
x

1 d h(x) A+B h(t) A+B d f(t)
y(x)= dt .
(A + B)h(x) dx g(x) a g(t) dt h(t)
x

13. 1+[g(t) – g(x)]h(x) y(t) dt = f(x).
a
This is a special case of equation 1.9.15 with g 1 (x)=1 – g(x)h(x), h 1 (t)=1, g 2 (x)= h(x),
and h 2 (t)= g(t).
Solution:

x
d f(t) dt x

y(x)= h(x)Φ(x) , Φ(x)=exp g (t)h(t) dt .
t
dx a h(t) t Φ(t) a
x

–λ(x–t) λx λt
14. e + e g(t) – e g(x) h(x) y(t) dt = f(x).
a
λx
This is a special case of equation 1.9.15 with g 1 (x)= e h(x), h 1 (t)= g(t), g 2 (x)= e –λx –
λt
g(x)h(x), and h 2 (t)= e .
x
15. [g 1 (x)h 1 (t)+ g 2 (x)h 2 (t)]y(t) dt = f(x).
a
For g 2 /g 1 = const or h 2 /h 1 = const, see equation 1.9.1.

1 . Solution with g 1 (x)h 1 (x)+ g 2 (x)h 2 (x) /≡ 0 and f(x) /≡ const g 2 (x):
◦
x
1 d g 2 (x)h 1 (x)Φ(x) f(t) dt
y(x)= , (1)
h 1 (x) dx g 1 (x)h 1 (x)+ g 2 (x)h 2 (x) a g 2 (t) t Φ(t)

where
x
h 2 (t) g 2 (t)h 1 (t) dt
Φ(x)=exp . (2)
a h 1 (t) t g 1 (t)h 1 (t)+ g 2 (t)h 2 (t)
If f(x) ≡ const g 2 (x), the solution is given by formulas (1) and (2) in which the subscript 1
must be changed by 2 and vice versa.

2 . Solution with g 1 (x)h 1 (x)+ g 2 (x)h 2 (x) ≡ 0:
◦
1 d (f/g 2 ) x 1 d (f/g 2 ) x
y(x)= = – ,
h 1 dx (g 1 /g 2 ) h 1 dx (h 2 /h 1 )
x x
where f = f(x), g 2 = g 2 (x), h 1 = h 1 (x), and h 2 = h 2 (x).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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