Page 39 - Handbook Of Integral Equations

x

λ(x–t) µ(x–t) β(x–t)
15. Ae + Be + Ce – A – B – C y(t) dt = f(x), f(a)= f (a)=0.

x
a
Differentiating with respect to x, we arrive at an equation of the form 1.2.14:
x

λ(x–t) µ(x–t) β(x–t)

Aλe + Bµe + Cβe y(t) dt = f (x).
x
a
x

λx+µt µx+λt

16. e – e y(t) dt = f(x), f(a)= f (a)=0.
x
a
µt
This is a special case of equation 1.9.11 with g(x)= e λx and h(t)= e .
Solution:

f xx – (λ + µ)f (x)+ λµf(x)

x
y(x)= .
(λ – µ) exp[(λ + µ)x]
x
λx+µt µx+λt
17. Ae + Be y(t) dt = f(x).
a
For B = –A, see equation 1.2.16. This is a special case of equation 1.9.12 with g(x)= e λx
µt
and h(t)= e .
Solution:
1 d A x B d f(t) µ – λ
y(x)= Φ (x) Φ (t) dt , Φ(x)=exp x .
(A + B)e µx dx a dt e µt A + B
x

λx+µt βx+γt
18. Ae + Be y(t) dt = f(x).
a
λx
µt
βx
This is a special case of equation 1.9.15 with g 1 (x)= Ae , h 1 (t)= e , g 2 (x)= Be , and
γt
h 2 (t)= e .
x

2λx 2βt λx βt
19. Ae + Be + Ce + De + E y(t) dt = f(x).
a
βt
This is a special case of equation 1.9.6 with g(x)= Ae 2λx +Ce λx and h(t)= Be 2βt +De +E.
x

λx+βt 2βt λx βt
20. Ae + Be + Ce + De + E y(t) dt = f(x).
a
λx
This is a special case of equation 1.9.15 with g 1 (x)= e , h 1 (t)= Ae βt + D, and g 2 (x)=1,
h 2 (t)= Be 2βt + De βt + E.
x

2λx λx+βt λx βt
21. Ae + Be + Ce + De + E y(t) dt = f(x).
a
βt
This is a special case of equation 1.9.15 with g 1 (x)= Be λx + D, h 1 (t)= e , and g 2 (x)=
Ae 2λx + Ce λx + E, h 2 (t)=1.
x

λx µt µx
22. 1+ Ae (e – e )y(t) dt = f(x).
a
λx
This is a special case of equation 1.9.13 with g(x)= e µx and h(x)= Ae .
Solution:
x
d λx f(t) dt Aµ (λ+µ)x
y(x)= e Φ(x) , Φ(x)=exp e .
dx a e λt t Φ(t) λ + µ
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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