Page 154 - Handbook Of Integral Equations

x

2
2
30. y(x)+ A (x – t )e λ(x–t) y(t) dt = f(x).
0
The substitution u(x)= e –λx y(x) leads to an equation of the form 2.1.11:
x
2
2
u(x)+ A (x – t )u(t) dt = f(x)e –λx .
0
x
n λ(x–t)
31. y(x)+ A (x – t) e y(t) dt = f(x), n =1, 2, ...
a
Solution:
x
y(x)= f(x)+ R(x – t)f(t) dt,
a
n
1 λx
R(x)= e exp(σ k x) σ k cos(β k x) – β k sin(β k x) ,
n +1
k=0
where
1
2πk 1
2πk
σ k = |An!| n+1 cos , β k = |An!| n+1 sin for A <0,
n +1 n +1
1
2πk + π 1
2πk + π
σ k = |An!| n+1 cos , β k = |An!| n+1 sin for A >0.
n +1 n +1
x
exp[λ(x – t)]
32. y(x)+ b √ y(t) dt = f(x).
a x – t
Solution:
x
2
y(x)= e λx F(x)+ πb 2 exp[πb (x – t)]F(t) dt ,
a
where
x
e –λt f(t)
F(x)= e –λx f(x) – b √ dt.
a x – t
x

k λ(x–t)
33. y(x)+ A (x – t)t e y(t) dt = f(x).
a
The substitution u(x)= e –λx y(x) leads to an equation of the form 2.1.49:
x

k
u(x)+ A (x – t)t u(t) dt = f(x)e –λx .
a
x

k
k
34. y(x)+ A (x – t )e λ(x–t) y(t) dt = f(x).
a
The substitution u(x)= e –λx y(x) leads to an equation of the form 2.1.52:
x
k
k
u(x)+ A (x – t )u(t) dt = f(x)e –λx .
a
x
e µ(x–t)
35. y(x) – λ α y(t) dt = f(x), 0 < α <1.
0 (x – t)
Solution:
x ∞ 1–α n
λΓ(1 – α)x
µx
y(x)= f(x)+ R(x – t)f(t) dt, where R(x)= e .
0 xΓ n(1 – α)
n=1
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 133