Page 150 - Handbook Of Integral Equations

x

–λ(x–t) βx λx+µt µx+λt
16. y(x) – λe + Ae µe – λe y(t) dt = f(x).
a
βx
This is a special case of equation 2.9.24 with h(x)= Ae .
Assume that f(a) = 0. Solution:
x
x
d e (2λ+β)x f(t)
y(x)= w(t) dt, w(x)= e –λx (λ+β)t Φ(t) dt ,
a dx Φ(x) a e t
λ – µ (λ+µ+β)x

Φ(x)=exp A e .
λ + µ + β
x
(λ+1)x+t λx+2t λx+t t
17. y(x)+ ABe – ABe – Ae – Be y(t) dt = f(x).
a
t
x
The transformation z = e , τ = e , Y (z)= y(x) leads to an equation of the form 2.1.56:
z
λ+1 λ λ
Y (z)+ ABz – ABz τ – Az – B Y (τ) dτ = F(z),
b
a
where F(z)= f(x) and b = e .
x
x+λt (λ+1)t λt t
18. y(x)+ ABe – ABe + Ae + Be y(t) dt = f(x).
a
t
x
The transformation z = e , τ = e , Y (z)= y(x) leads to an equation of the form 2.1.57 (in
which λ is substituted by λ – 1):
z
λ–1 λ λ–1
Y (z)+ ABzτ – ABτ + Aτ + B Y (τ) dτ = F(z),
b
a
where F(z)= f(x) and b = e .
n
x
19. y(x)+ A k e λ k (x–t) y(t) dt = f(x).
a
k=1
1 . This integral equation can be reduced to an nth-order linear nonhomogeneous ordinary
◦
differential equation with constant coefficients. Set
x
I k (x)= e λ k (x–t) y(t) dt. (1)
a
Differentiating (1) with respect to x yields

x
λ k (x–t)
I = y(x)+ λ k e y(t) dt, (2)
k
a
where the prime stands for differentiation with respect to x. From the comparison of (1)
with (2) we see that

I = y(x)+ λ k I k , I k = I k (x). (3)
k
The integral equation can be written in terms of I k (x) as follows:
n

y(x)+ A k I k = f(x). (4)
k=1

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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