Page 144 - Handbook Of Integral Equations

3 . For arbitrary f(x), expandable into power series, the formulas of item 2 can be used, in
◦
◦
which one should set N = ∞. In this case, the radius of convergence of the solution y(x)is
equal to the radius of convergence of f(x).
4 .For
◦
N
n
f(x) = ln(kx) A n x ,
n=0
a solution has the form
N N
n n
y(x) = ln(kx) B n x + D n x ,
n=0 n=0
where the constants B n and D n are found by the method of undetermined coefficients. To
obtain the general solution we must add the solution (1) of the homogeneous equation.
In Mikhailov (1966), solvability conditions for the integral equation in question were
investigated for various classes of f(x).

λ x y(t) dt
62. y(x) – = f(x).
x µ 0 (ax + bt) 1–µ
Here a >0, b > 0, and µ is an arbitrary number.
◦
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
λI(β) = 1, where I(β)= z (a + bz) µ–1 dz. (2)
0
◦
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
1
1 n µ–1

λ n = , I(n)= z (a + bz) dz.
I(n) 0
Here C is an arbitrary constant, and β = β(λ) is determined by the transcendental equation (2).
◦
3 . 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
1 –1

¯
n
where λ n = z (a + bz) µ–1 ln zdz .
0
◦
◦
4 . For arbitrary f(x) expandable into power series, the formulas of item 2 can be used, in
which one should set N = ∞. In this case, the radius of convergence of the solution y(x)is
equal to the radius of convergence of f(x).
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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