Page 495 - Handbook Of Integral Equations

9.5-3. Power-Law Generating Function

Consider the linear equation with power-law right-hand side
λ
L [y]= x . (10)
Suppose that the solution is known and is given by formula (2). In Table 5, solutions of the equation
L [y]= f(x) with various right-hand sides are presented which can be expressed via the solution of
Eq. (10).

TABLE 5
Solutions of the equation L [y]= f(x) with generating function of power-law form

No Right-Hand Side f(x) Solution y Solution Method
1 x λ y(x, λ) Original Equation

2 n A k x k n A k y(x, k) Follows from linearity
k=0 k=0
∂ Follows from linearity and
3 A ln x + B A y(x, λ) + By(x,0)
∂λ λ=0 from the results of row No 4
n
A ln x, ∂ n Follows from the results
4 A y(x, λ)
n = 0,1,2, ... ∂λ n of row No 5 for λ =0
λ=0
n λ
Ax x , ∂ n Differentiation
5 A y(x, λ)
n = 0,1,2, ... ∂λ n with respect to the parameter λ
Selection of the real

6 A cos(β ln x) A Re y(x, iβ)
part for λ = iβ
Selection of the imaginary

7 A sin(β ln x) A Im y(x, iβ)
part for λ = iβ
Selection of the real
µ

8 Ax cos(β ln x) A Re y(x, µ + iβ)
part for λ = µ + iβ
Selection of the imaginary
µ

9 Ax sin(β ln x) A Im y(x, µ + iβ)
part for λ = µ + iβ
Example 2. We seek a solution of the equation with power-law right-hand side
x 1 t
y(x)+ K y(t) dt = x λ
0 x x
λ
in the form y(x, λ)= kx by the method of indeterminate coefficients. We finally obtain
1 1
λ
λ
y(x, λ)= x , B(λ)= K(t)t dt.
1+ B(λ) 0
It follows from row 3 of Table 5 that the solution of the equation with logarithmic right-hand side
x 1 t
y(x)+ K y(t) dt = A ln x
0 x x
has the form
A AI 1
y(x)= ln x – ,
1+ I 0 (1 + I 0 ) 2
1 1

I 0 = K(t) dt, I 1 = K(t)ln tdt.
0 0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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