Page 107 - Handbook Of Integral Equations

x

45. (x – t)I 1 [λ(x – t)]y(t) dt = f(x).
a
Solution:
1 d 2 2 4 x 2
y(x)= – λ (x – t) I 2 [λ(x – t)] f(t) dt.
3λ 3 dx 2 a
x

2
46. (x – t) I 1 [λ(x – t)]y(t) dt = f(x).
a
Solution:
x
y(x)= g(t) dt,
a
where
1 d 2 2 5 t
2
g(t)= – λ t – τ I 2 [λ(t – τ)] f(τ) dτ.
9λ 3 dt 2 a
x
n
47. (x – t) I n [λ(x – t)]y(t) dt = f(x), n =0, 1, 2, ...
a
Solution:

2 2n+2 x
d
y(x)= A – λ 2 (x – t) n+1 I n+1 [λ(x – t)] f(t) dt,
dx 2
a
2 n!(n + 1)!

2n+1
A = .
λ (2n)! (2n + 2)!
If the right-hand side of the equation is differentiable sufficiently many times and the

conditions f(a)= f (a)= ··· = f x (2n+1) (a) = 0 are satisfied, then the solution of the integral
x
equation can be written in the form
x 2n+2
d 2
y(x)= A (x – t) 2n+1 I 2n+1 [λ(x – t)]F(t) dt, F(t)= 2 – λ 2 f(t).
a dt
x

48. (x – t) n+1 I 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
g(t)= A – λ 2 (t – τ) n+1 I 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 x (2n+2) (a) = 0 are satisfied, then the function g(t)defining the
x
solution can be written in the form
t 2n+2
d 2
g(t)= A (t – τ) n–ν–2 I n–ν–2 [λ(t – τ)]F(τ) dτ, F(τ)= 2 – λ 2 f(τ).
a dτ

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 85