Page 289 - Handbook Of Integral Equations

b

β(x–t)
4. y(x) – λ Ae + B y(t) dt = f(x).
a
βx
This is a special case of equation 4.9.18 with g 1 (x)= e , h 1 (t)= Ae –βt , g 2 (x)=1, and
h 2 (t)= B.
Solution:
y(x)= f(x)+ λ(A 1 e βx + A 2 ),
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.18.

b

βx+µt (β+µ)t
5. y(x) – λ Ae + Be y(t) dt = f(x).
a
µt
This is a special case of equation 4.9.6 with g(x)= e βx and h(t)= e .
Solution:
y(x)= f(x)+ λ(A 1 e βx + A 2 ),
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.6.

b

α(x+t) β(x+t)
6. y(x) – λ Ae + Be y(t) dt = f(x).
a
βt
This is a special case of equation 4.9.14 with g(x)= e αx and h(t)= e .
Solution:
βx
y(x)= f(x)+ λ(A 1 e αx + A 2 e ),
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.14.

b
αx+βt βx+αt
7. y(x) – λ Ae + Be y(t) dt = f(x).
a
βt
This is a special case of equation 4.9.17 with g(x)= e αx and h(t)= e .
Solution:
βx
y(x)= f(x)+ λ(A 1 e αx + A 2 e ),
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.17.

b

(γ+µ)x νt+µx
8. y(x) – λ De + Ee y(t) dt = f(x).
a
µx
This is a special case of equation 4.9.18 with g 1 (x)= e (γ+µ)x , h 1 (t)= D, g 2 (x)= e , and
νt
h 2 (t)= Ee .
Solution:
µx
y(x)= f(x)+ λ[A 1 e (γ+µ)x + A 2 e ],
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.18.
b
9. y(x) – λ (Ae αx+βt + Be γx+δt )y(t) dt = f(x).
a
γx
βt
This is a special case of equation 4.9.18 with g 1 (x)= e αx , h 1 (t)= Ae , g 2 (x)= e , and
δt
h 2 (t)= Be .
Solution:
γx
y(x)= f(x)+ λ(A 1 e αx + A 2 e ),
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.18.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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