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x
x
86. (x – t) c–1 F a, b, c;1 – y(t) dt = f(x).
t
s
Here Φ(a, b, c; z) is the Gaussian hypergeometric function (see Supplement 10).
Solution:
d n a x (x – t) n–c–1 t
–a
y(x)= x x F –a, n – b, n – c;1 – f(t) dt ,
dx n s Γ(c)Γ(n – c) x
where 0 < c < n and n =1, 2, ...
If the right-hand side of the equation is differentiable sufficiently many times and the
conditions f(s)= f (s)= ··· = f (n–1) (s) = 0 are satisfied, then the solution of the integral

x x
equation can be written in the form
x (x – t) n–c–1 t
y(x)= F –a, –b, n – c;1 – f t (n) (t) dt.
s Γ(c)Γ(n – c) x
•
Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

1.9. Equations Whose Kernels Contain Arbitrary
Functions

1.9-1. Equations With Degenerate Kernel: K(x, t)= g 1 (x)h 1 (t)+ g 2 (x)h 2 (t)
x

1. g(x)h(t)y(t) dt = f(x).
a
1 d f(x) 1 g (x)

x

Solution: y = = f (x) – f(x).
x
2
h(x) dx g(x) g(x)h(x) g (x)h(x)
x

2. [g(x) – g(t)]y(t) dt = f(x).
a

It is assumed that f(a)= f (a)=0 and f /g ≠ const.
x
x
d x f (x)

x
Solution: y(x)= .
dx g (x)

x
x

3. [g(x) – g(t)+ b]y(t) dt = f(x).
a
Differentiation with respect to x yields an equation of the form 2.9.2:
1 x 1

y(x)+ g (x) y(t) dt = f (x).
x
x
b b
a
Solution: x
1 1 g(t) – g(x)

y(x)= f (x) – g (x) exp f (t) dt.

t
x
x
b b 2 a b
x

4. [Ag(x)+ Bg(t)]y(t) dt = f(x).
a
For B = –A, see equation 1.9.2.
Solution with B ≠ –A:
sign g(x) d – A x – B
y(x)= g(x) A+B g(t) A+B f (t) dt .

t
A + B dx a
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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