Page 101 - Handbook Of Integral Equations

x

n+1
8. x – t J n [λ(x – t)]y(t) dt = f(x), n =0, 1, 2, ...
a
Solution:
x
y(x)= g(t) dt,
a
where
2 2n+3 t
d 2 n+1
g(t)= A + λ (t – τ) J n+1 [λ(t – τ)] f(τ) dτ,
dt 2 a
2 n!(n + 1)!

2n+1
A = .
λ (2n + 1)! (2n + 2)!
If the right-hand side of the equation is differentiable sufficiently many times and the
conditions f(a)= f (a)= ··· = f (2n+2) (a) = 0 are satisfied, then the function g(t)defining the

x x
solution can be written in the form
t 2n+2
d 2
g(t)= A (t – τ) n–ν–2 J n–ν–2 [λ(t – τ)]F(τ) dτ, F(τ)= + λ 2 f(τ).
dτ 2
a
x
9. (x – t) 1/2 J 1/2 [λ(x – t)]y(t) dt = f(x).
a
Solution:
π d 2 2 3 x 3/2
y(x)= + λ (x – t) J 3/2 [λ(x – t)] f(t) dt.
4λ 2 dx 2 a
x

10. (x – t) 3/2 J 1/2 [λ(x – t)]y(t) dt = f(x).
a
Solution:
x

y(x)= g(t) dt,
a
where
π d 2 2
4 t 3/2
g(t)= + λ (t – τ) J 3/2 [λ(t – τ)] f(τ) dτ.
8λ 2 dt 2 a

x
11. (x – t) 3/2 J 3/2 [λ(x – t)]y(t) dt = f(x).
a
Solution:
√ 2 3 x
π d 2
y(x)= + λ sin[λ(x – t)] f(t) dt.
2 3/2 5/2 dx 2 a
λ
x

5/2
12. (x – t) J 3/2 [λ(x – t)]y(t) dt = f(x).
a
Solution:
x

y(x)= g(t) dt,
a
where
π d 2 2 6 t 5/2
g(t)= + λ (t – τ) J 5/2 [λ(t – τ)] f(τ) dτ.
128λ 4 dt 2 a

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 79