Page 38 - Handbook Of Integral Equations

x

λx λt
8. Ae + Be y(t) dt = f(x).
a
λx
For B = –A, see equation 1.2.6. This is a special case of equation 1.9.4 with g(x)= e .
1 d Aλ
x Bλ

Solution: y(x)= exp – x exp – t f (t) dt .
t
A + B dx A + B a A + B
x
λx λt
9. Ae + Be + C y(t) dt = f(x).
a
λx
This is a special case of equation 1.9.5 with g(x)= e .
x

λx µt
10. Ae + Be y(t) dt = f(x).
a
For λ = µ, see equation 1.2.8. This is a special case of equation 1.9.6 with g(x)= Ae λx and
µt
h(t)= Be .
x

λ(x–t) µ(x–t)

11. e – e y(t) dt = f(x), f(a)= f (a)=0.
x
a
Solution:
1
y(x)= f xx – (λ + µ)f + λµf , f = f(x).
x
λ – µ
x
λ(x–t) µ(x–t)
12. Ae + Be y(t) dt = f(x).
a
λx
For B = –A, see equation 1.2.11. This is a special case of equation 1.9.15 with g 1 (x)= Ae ,
µx
h 1 (t)= e –λt , g 2 (x)= Be , and h 2 (t)= e –µt .
Solution:
x
e λx d (µ–λ)x f(t) dt B(λ – µ)
y(x)= e Φ(x) , Φ(x)=exp x .
A + B dx a e µt t Φ(t) A + B
x

λ(x–t) µ(x–t)
13. Ae + Be + C y(t) dt = f(x).
a
This is a special case of equation 1.2.14 with β =0.
x
λ(x–t) µ(x–t) β(x–t)
14. Ae + Be + Ce y(t) dt = f(x).
a
Differentiating the equation with respect to x yields
x

λ(x–t) µ(x–t) β(x–t)

(A + B + C)y(x)+ Aλe + Bµe + Cβe y(t) dt = f (x).
x
a
Eliminating the term with e β(x–t) with the aid of the original equation, we arrive at an equation
of the form 2.2.10:
x

λ(x–t) µ(x–t)

(A + B + C)y(x)+ A(λ – β)e + B(µ – β)e y(t) dt = f (x) – βf(x).
x
a
In the special case A + B + C = 0, this is an equation of the form 1.2.12.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 16