Page 137 - Handbook Of Integral Equations

2.1-5. Kernels Containing Rational Functions

x

40. y(x)+ x –3 t 2Ax +(1 – A)t y(t) dt = f(x).
a
This equation can be obtained by differentiating the equation
x x

2 2 3
Ax t +(1 – A)xt y(t) dt = F(x), F(x)= t f(t) dt,
a a
which has the form 1.1.17:
Solution:
1 d –A x A–1 1 x 3
y(x)= x t ϕ (t) dt , ϕ(x)= t f(t) dt.

t
x dx a x a
x y(t) dt
41. y(x) – λ = f(x).
0 x + t
Dixon’s equation. This is a special case of equation 2.1.62 with a = b = 1 and µ =0.
1 . The solution of the homogeneous equation (f ≡ 0) is
◦
y(x)= Cx β (β > –1, λ > 0). (1)
Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation

1
β
z dz
λI(β) = 1, where I(β)= . (2)
0 1+ z
◦
2 . For a polynomial right-hand side,
N
n
f(x)= A n x
n=0
the solution bounded at zero is given by

N
A n

 n
 x for λ < λ 0 ,
1 – (λ/λ n )



n=0
y(x)=
 N
A n n β


 x + Cx for λ > λ 0 and λ ≠ λ n ,

 1 – (λ/λ n )
n=0
n

1 n (–1) m
λ n = , I(n)=(–1) ln 2 + ,
I(n) m
m=1
where C is an arbitrary constant, and β =β(λ) is determined by the transcendental equation (2).
For special λ = λ n (n =1, 2, ... ), the solution differs in one term and has the form
n–1 N
¯
A m m A m m λ n n n
y(x)= x + x – A n x ln x + Cx ,
1 – (λ n /λ m ) 1 – (λ n /λ m ) λ n
m=0 m=n+1
π 2 n (–1) k –1
¯
where λ n =(–1) n+1 + .
12 k=1 k 2
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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