HANDBOOK OFINTEGRAL EQUATIONS phần 5 ppsx

2
◦
.Forλ(2A + λ)=–k
2
< 0, the general solution of equation (1) is given by
y(x)=C
1
cosh(kx)+C
2
sinh(kx)+f(x) –
2Aλ
k

x
a
sinh[k(x – t)] f(t) dt, (3)
where C
1
and C
2
are arbitrary constants.
For λ(2A + λ)=k
2
> 0, the general solution of equation (1) is given by
y(x)=C
1
cos(kx)+C
2
sin(kx)+f(x) –
2Aλ
k


x
a
sin[k(x – t)] f(t) dt. (4)
For λ =2A, the general solution of equation (1) is given by
y(x)=C
1
+ C
2
x + f(x)+4A
2

x
a
(x – t)f(t) dt. (5)
The constants C
1
and C
2
in solutions (3)–(5) are determined by conditions (2).
30. y(x)+A

b
a
t sin(λ|x – t|)y(t) dt = f(x).
This is a special case of equation 4.9.39 with g(t)=At. The solution of the integral equation
can be written via the Bessel functions (or modified Bessel functions) of order 1/3.
31. y(x)+A

b

a
sin
3
(λ|x – t|)y(t) dt = f(x).
Using the formula sin
3
β = –
1
4
sin 3β +
3
4
sin β, we arrive at an equation of the form 4.5.32
with n =2:
y(x)+

b
a

–
1
4
A sin(3λ|x – t|)+
3
4
A sin(λ|x – t|)

y(t) dt = f(x).
32. y(x)+


b
a

n

k=1
A
k
sin(λ
k
|x – t|)

y(t) dt = f (x), –∞ < a < b < ∞.
1
◦
. Let us remove the modulus in the kth summand of the integrand:
I
k
(x)=

b
a
sin(λ
k
|x – t|)y(t) dt =

x
a
sin[λ
k

(x – t)]y(t) dt +

b
x
sin[λ
k
(t – x)]y(t) dt. (1)
Differentiating (1) with respect to x twice yields
I

k
= λ
k

x
a
cos[λ
k
(x – t)]y(t) dt – λ
k

b
x
cos[λ
k
(t – x)]y(t) dt,
I

k
=2λ

k
y(x) – λ
2
k

x
a
sin[λ
k
(x – t)]y(t) dt – λ
2
k

b
x
sin[λ
k
(t – x)]y(t) dt,
(2)
where the primes denote the derivatives with respect to x. By comparing formulas (1) and (2),
we find the relation between I

k
and I
k
:
I

k
=2λ

k
y(x) – λ
2
k
I
k
, I
k
= I
k
(x). (3)
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2
◦
. With the aid of (1), the integral equation can be rewritten in the form
y(x)+
n

k=1
A
k
I
k
= f(x). (4)
Differentiating (4) with respect to x twice and taking into account (3), we find that
y

xx

(x)+σ
n
y(x) –
n

k=1
A
k
λ
2
k
I
k
= f

xx
(x), σ
n
=2
n

k=1
A
k
λ
k
. (5)
Eliminating the integral I
n
from (4) and (5) yields

y

xx
(x)+(σ
n
+ λ
2
n
)y(x)+
n–1

k=1
A
k
(λ
2
n
– λ
2
k
)I
k
= f

xx
(x)+λ
2
n
f(x). (6)
Differentiating (6) with respect to x twice and eliminating I

n–1
from the resulting equation
with the aid of (6), we obtain a similar equation whose left-hand side is a second-order linear
differential operator (acting on y) with constant coefficients plus the sum
n–2

k=1
B
k
I
k
.Ifwe
successively eliminate I
n–2
, I
n–3
, , with the aid of double differentiation, then we finally
arrive at a linear nonhomogeneous ordinary differential equation of order 2n with constant
coefficients.
3
◦
. The boundary conditions for y(x) can be found by setting x = a in the integral equation
and all its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b
in the integral equation and all its derivatives obtained by means of double differentiation.)
33. y(x) – λ

∞
–∞
sin(x – t)
x – t

y(t) dt = f (x).
Solution:
y(x)=f(x)+
λ
√
2π – πλ

∞
–∞
sin(x – t)
x – t
f(t) dt, λ ≠

2
π
.
•
Reference: F. D. Gakhov and Yu. I. Cherskii (1978).
4.5-3. Kernels Containing Tangent
34. y(x) – λ

b
a
tan(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tan(βx) and h(t)=1.
35. y(x) – λ

b
a
tan(βt)y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = tan(βt).
36. y(x) – λ

b
a
[A tan(βx)+B tan(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.4 with g(x) = tan(βx).
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37. y(x) – λ

b
a
tan(βx)
tan(βt)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tan(βx) and h(t)=
1
tan(βt)
.
38. y(x) – λ

b
a
tan(βt)
tan(βx)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=
1

tan(βx)
and h(t) = tan(βt).
39. y(x) – λ

b
a
tan
k
(βx) tan
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tan
k
(βx) and h(t) = tan
m
(µt).
40. y(x) – λ

b
a
t
k
tan
m
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tan
m
(βx) and h(t)=t
k
.

41. y(x) – λ

b
a
x
k
tan
m
(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=x
k
and h(t) = tan
m
(βt).
42. y(x) – λ

b
a
[A + B(x – t) tan(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.8 with h(t) = tan(βt).
43. y(x) – λ

b
a
[A + B(x – t) tan(βx)]y(t) dt = f (x).
This is a special case of equation 4.9.10 with h(x) = tan(βx).
4.5-4. Kernels Containing Cotangent
44. y(x) – λ

b

a
cot(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cot(βx) and h(t)=1.
45. y(x) – λ

b
a
cot(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = cot(βt).
46. y(x) – λ

b
a
[A cot(βx)+B cot(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.4 with g(x) = cot(βx).
Page 288
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47. y(x) – λ

b
a
cot(βx)
cot(βt)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cot(βx) and h(t)=
1
cot(βt)
.
48. y(x) – λ


b
a
cot(βt)
cot(βx)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=
1
cot(βx)
and h(t) = cot(βt).
49. y(x) – λ

b
a
cot
k
(βx) cot
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cot
k
(βx) and h(t) = cot
m
(µt).
50. y(x) – λ

b
a
t
k

cot
m
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cot
m
(βx) and h(t)=t
k
.
51. y(x) – λ

b
a
x
k
cot
m
(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=x
k
and h(t) = cot
m
(βt).
52. y(x) – λ

b
a
[A + B(x – t) cot(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.8 with h(t) = cot(βt).
53. y(x) – λ


b
a
[A + B(x – t) cot(βx)]y(t) dt = f (x).
This is a special case of equation 4.9.10 with h(x) = cot(βx).
4.5-5. Kernels Containing Combinations of Trigonometric Functions
54. y(x) – λ

b
a
cos
k
(βx) sin
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cos
k
(βx) and h(t) = sin
m
(µt).
55. y(x) – λ

b
a
[A sin(αx) cos(βt)+B sin(γx) cos(δt)]y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x) = sin(αx), h
1
(t)=A cos(βt), g
2

(x) = sin(γx),
and h
2
(t)=B cos(δt).
56. y(x) – λ

b
a
tan
k
(γx) cot
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tan
k
(γx) and h(t) = cot
m
(µt).
57. y(x) – λ

b
a
[A tan(αx) cot(βt)+B tan(γx) cot(δt)]y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x) = tan(αx), h
1
(t)=A cot(βt), g
2
(x) = tan(γx),

and h
2
(t)=B cot(δt).
Page 289
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4.5-6. A Singular Equation
58. Ay(x) –
B
2π

2π
0
cot

t – x
2

y(t) dt = f (x), 0 ≤ x ≤ 2π.
Here the integral is understood in the sense of the Cauchy principal value. Without loss of
generality we may assume that A
2
+ B
2
=1.
Solution:
y(x)=Af(x)+
B
2π


2π
0
cot

t – x
2

f(t) dt +
B
2
2πA

2π
0
f(t) dt.
•
Reference: I. K. Lifanov (1996).
4.6. Equations Whose Kernels Contain Inverse
Trigonometric Functions
4.6-1. Kernels Containing Arccosine
1. y(x) – λ

b
a
arccos(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arccos(βx) and h(t)=1.
2. y(x) – λ

b
a

arccos(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = arccos(βt).
3. y(x) – λ

b
a
arccos(βx)
arccos(βt)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arccos(βx) and h(t)=
1
arccos(βt)
.
4. y(x) – λ

b
a
arccos(βt)
arccos(βx)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=
1
arccos(βx)
and h(t) = arccos(βt).
5. y(x) – λ

b
a
arccos
k

(βx) arccos
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arccos
k
(βx) and h(t) = arccos
m
(µt).
6. y(x) – λ

b
a
t
k
arccos
m
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arccos
m
(βx) and h(t)=t
k
.
7. y(x) – λ

b
a
x
k
arccos
m

(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=x
k
and h(t) = arccos
m
(βt).
Page 290
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8. y(x) – λ

b
a
[A + B(x – t) arccos(βx)]y(t) dt = f(x).
This is a special case of equation 4.9.10 with h(x) = arccos(βx).
9. y(x) – λ

b
a
[A + B(x – t) arccos(βt)]y(t) dt = f(x).
This is a special case of equation 4.9.8 with h(t) = arccos(βt).
4.6-2. Kernels Containing Arcsine
10. y(x) – λ

b
a
arcsin(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arcsin(βx) and h(t)=1.
11. y(x) – λ


b
a
arcsin(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = arcsin(βt).
12. y(x) – λ

b
a
arcsin(βx)
arcsin(βt)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arcsin(βx) and h(t)=
1
arcsin(βt)
.
13. y(x) – λ

b
a
arcsin(βt)
arcsin(βx)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=
1
arcsin(βx)
and h(t) = arcsin (βt).
14. y(x) – λ

b
a

arcsin
k
(βx) arcsin
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arcsin
k
(βx) and h(t) = arcsin
m
(µt).
15. y(x) – λ

b
a
t
k
arcsin
m
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arcsin
m
(βx) and h(t)=t
k
.
16. y(x) – λ

b
a
x
k

arcsin
m
(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=x
k
and h(t) = arcsin
m
(βt).
17. y(x) – λ

b
a
[A + B(x – t) arcsin(βt)]y(t) dt = f(x).
This is a special case of equation 4.9.8 with h(t) = arcsin(βt).
18. y(x) – λ

b
a
[A + B(x – t) arcsin(βx)]y(t) dt = f(x).
This is a special case of equation 4.9.10 with h(x) = arcsin(βx).
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4.6-3. Kernels Containing Arctangent
19. y(x) – λ

b
a
arctan(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arctan(βx) and h(t)=1.

20. y(x) – λ

b
a
arctan(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = arctan(βt).
21. y(x) – λ

b
a
[A arctan(βx)+B arctan(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.4 with g(x) = arctan(βx).
22. y(x) – λ

b
a
arctan(βx)
arctan(βt)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arctan(βx) and h(t)=
1
arctan(βt)
.
23. y(x) – λ

b
a
arctan(βt)
arctan(βx)
y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x)=
1
arctan(βx)
and h(t) = arctan(βt).
24. y(x) – λ

b
a
arctan
k
(βx) arctan
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arctan
k
(βx) and h(t) = arctan
m
(µt).
25. y(x) – λ

b
a
t
k
arctan
m
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arctan
m
(βx) and h(t)=t

k
.
26. y(x) – λ

b
a
x
k
arctan
m
(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=x
k
and h(t) = arctan
m
(βt).
27. y(x) – λ

b
a
[A + B(x – t) arctan(βt)]y(t) dt = f(x).
This is a special case of equation 4.9.8 with h(t) = arctan(βt).
28. y(x) – λ

b
a
[A + B(x – t) arctan(βx)]y(t) dt = f(x).
This is a special case of equation 4.9.10 with h(x) = arctan(βx).
Page 292
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4.6-4. Kernels Containing Arccotangent
29. y(x) – λ

b
a
arccot(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arccot(βx) and h(t)=1.
30. y(x) – λ

b
a
arccot(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = 1 and h(t) = arccot(βt).
31. y(x) – λ

b
a
[A arccot(βx)+B arccot(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.4 with g(x) = arccot(βx).
32. y(x) – λ

b
a
arccot(βx)
arccot(βt)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arccot(βx) and h(t)=
1
arccot(βt)

.
33. y(x) – λ

b
a
arccot(βt)
arccot(βx)
y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=
1
arccot(βx)
and h(t) = arccot(βt).
34. y(x) – λ

b
a
arccot
k
(βx) arccot
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arccot
k
(βx) and h(t) = arccot
m
(µt).
35. y(x) – λ

b
a

t
k
arccot
m
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = arccot
m
(βx) and h(t)=t
k
.
36. y(x) – λ

b
a
x
k
arccot
m
(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=x
k
and h(t) = arccot
m
(βt).
37. y(x) – λ

b
a
[A + B(x – t) arccot(βt)]y(t) dt = f(x).
This is a special case of equation 4.9.8 with h(t) = arccot(βt).

38. y(x) – λ

b
a
[A + B(x – t) arccot(βx)]y(t) dt = f(x).
This is a special case of equation 4.9.10 with h(x) = arccot(βx).
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4.7. Equations Whose Kernels Contain Combinations of
Elementary Functions
4.7-1. Kernels Containing Exponential and Hyperbolic Functions
1. y(x) – λ

b
a
e
µ(x–t)
cosh[β(x – t)]y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x)=e
µx
cosh(βx), h
1
(t)=e
–µt
cosh(βt),
g
2

(x)=e
µx
sinh(βx), and h
2
(t)=–e
–µt
sinh(βt).
2. y(x) – λ

b
a
e
µ(x–t)
sinh[β(x – t)]y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x)=e
µx
sinh(βx), h
1
(t)=e
–µt
cosh(βt),
g
2
(x)=e
µx
cosh(βx), and h
2
(t)=–e

–µt
sinh(βt).
3. y(x) – λ

b
a
te
µ(x–t)
sinh[β(x – t)]y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x)=e
µx
sinh(βx), h
1
(t)=te
–µt
cosh(βt),
g
2
(x)=e
µx
cosh(βx), and h
2
(t)=–te
–µt
sinh(βt).
4.7-2. Kernels Containing Exponential and Logarithmic Functions
4. y(x) – λ


b
a
e
µt
ln(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = ln(βx) and h(t)=e
µt
.
5. y(x) – λ

b
a
e
µx
ln(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=e
µx
and h(t) = ln(βt).
6. y(x) – λ

b
a
e
µ(x–t)
ln(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=e
µx
ln(βx) and h(t)=e
–µt
.

7. y(x) – λ

b
a
e
µ(x–t)
ln(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=e
µx
and h(t)=e
–µt
ln(βt).
8. y(x) – λ

b
a
e
µ(x–t)
(ln x – ln t)y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x)=e
µx
ln x, h
1
(t)=e
–µt
, g
2
(x)=e

µx
, and
h
2
(t)=–e
–µt
ln t.
Page 294
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© 1998 by CRC Press LLC
9. y(x)+
b
2
– a
2
2a

∞
0
1
t
exp

–a



ln
x
t





y(t) dt = f (x).
Solution with a >0,b > 0, and x >0:
y(x)=f(x)+
a
2
– b
2
2b

∞
0
1
t
exp

–b



ln
x
t





f(t) dt.
•
Reference: F. D. Gakhov and Yu. I. Cherskii (1978).
4.7-3. Kernels Containing Exponential and Trigonometric Functions
10. y(x) – λ

b
a
e
µt
cos(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cos(βx) and h(t)=e
µt
.
11. y(x) – λ

b
a
e
µx
cos(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=e
µx
and h(t) = cos(βt).
12. y(x) – λ

∞
0
e
µ(x–t)

cos(xt)y(t) dt = f (x).
Solution:
y(x)=
f(x)
1 –
π
2
λ
2
+
λ
1 –
π
2
λ
2

∞
0
e
µ(x–t)
cos(xt)f(t) dt, λ ≠ ±

2/π.
13. y(x) – λ

b
a
e
µ(x–t)

cos[β(x – t)]y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x)=e
µx
cos(βx), h
1
(t)=e
–µt
cos(βt),
g
2
(x)=e
µx
sin(βx), and h
2
(t)=e
–µt
sin(βt).
14. y(x) – λ

b
a
e
µt
sin(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = sin(βx) and h(t)=e
µt
.
15. y(x) – λ


b
a
e
µx
sin(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=e
µx
and h(t) = sin(βt).
16. y(x) – λ

∞
0
e
µ(x–t)
sin(xt)y(t) dt = f (x).
Solution:
y(x)=
f(x)
1 –
π
2
λ
2
+
λ
1 –
π
2
λ

2

∞
0
e
µ(x–t)
sin(xt)f(t) dt, λ ≠ ±

2/π.
Page 295
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© 1998 by CRC Press LLC
17. y(x) – λ

b
a
e
µ(x–t)
sin[β(x – t)]y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x)=e
µx
sin(βx), h
1
(t)=e
–µt
cos(βt),
g
2

(x)=e
µx
cos(βx), and h
2
(t)=–e
–µt
sin(βt).
18. y(x) – λ

b
a
e
µ(x–t)

n

k=1
A
k
sin[β
k
(x – t)]

y(t) dt = f (x), n =1,2,
This is a special case of equation 4.9.20.
19. y(x) – λ

b
a
te

µ(x–t)
sin[β(x – t)]y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x)=e
µx
sin(βx), h
1
(t)=te
–µt
cos(βt),
g
2
(x)=e
µx
cos(βx), and h
2
(t)=–te
–µt
sin(βt).
20. y(x) – λ

b
a
xe
µ(x–t)
sin[β(x – t)]y(t) dt = f (x).
This is a special case of equation 4.9.18 with g
1
(x)=xe

µx
sin(βx), h
1
(t)=e
–µt
cos(βt),
g
2
(x)=xe
µx
cos(βx), and h
2
(t)=–e
–µt
sin(βt).
21. y(x) – λ

b
a
e
µt
tan(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tan(βx) and h(t)=e
µt
.
22. y(x) – λ

b
a
e

µx
tan(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=e
µx
and h(t) = tan(βt).
23. y(x) – λ

b
a
e
µ(x–t)
[tan(βx) – tan(βt)]y(t) dt = f(x).
This is a special case of equation 4.9.18 with g
1
(x)=e
µx
tan(βx), h
1
(t)=e
–µt
, g
2
(x)=e
µx
,
and h
2
(t)=–e
–µt
tan(βt).

24. y(x) – λ

b
a
e
µt
cot(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cot(βx) and h(t)=e
µt
.
25. y(x) – λ

b
a
e
µx
cot(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=e
µx
and h(t) = cot(βt).
4.7-4. Kernels Containing Hyperbolic and Logarithmic Functions
26. y(x) – λ

b
a
cosh
k
(βx)ln
m
(µt)y(t) dt = f (x).

This is a special case of equation 4.9.1 with g(x) = cosh
k
(βx) and h(t)=ln
m
(µt).
Page 296
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© 1998 by CRC Press LLC
27. y(x) – λ

b
a
cosh
k
(βt)ln
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=ln
m
(µx) and h(t) = cosh
k
(βt).
28. y(x) – λ

b
a
sinh
k
(βx)ln
m

(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = sinh
k
(βx) and h(t)=ln
m
(µt).
29. y(x) – λ

b
a
sinh
k
(βt)ln
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=ln
m
(µx) and h(t) = sinh
k
(βt).
30. y(x) – λ

b
a
tanh
k
(βx)ln
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tanh

k
(βx) and h(t)=ln
m
(µt).
31. y(x) – λ

b
a
tanh
k
(βt)ln
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=ln
m
(µx) and h(t) = tanh
k
(βt).
32. y(x) – λ

b
a
coth
k
(βx)ln
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = coth
k
(βx) and h(t)=ln

m
(µt).
33. y(x) – λ

b
a
coth
k
(βt)ln
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=ln
m
(µx) and h(t) = coth
k
(βt).
4.7-5. Kernels Containing Hyperbolic and Trigonometric Functions
34. y(x) – λ

b
a
cosh
k
(βx) cos
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cosh
k
(βx) and h(t) = cos
m

(µt).
35. y(x) – λ

b
a
cosh
k
(βt) cos
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cos
m
(µx) and h(t) = cosh
k
(βt).
36. y(x) – λ

b
a
cosh
k
(βx) sin
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cosh
k
(βx) and h(t) = sin
m
(µt).
Page 297

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© 1998 by CRC Press LLC
37. y(x) – λ

b
a
cosh
k
(βt) sin
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = sin
m
(µx) and h(t) = cosh
k
(βt).
38. y(x) – λ

b
a
sinh
k
(βx) cos
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = sinh
k
(βx) and h(t) = cos
m
(µt).

39. y(x) – λ

b
a
sinh
k
(βt) cos
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cos
m
(µx) and h(t) = sinh
k
(βt).
40. y(x) – λ

b
a
sinh
k
(βx) sin
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = sinh
k
(βx) and h(t) = sin
m
(µt).
41. y(x) – λ


b
a
sinh
k
(βt) sin
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = sin
m
(µx) and h(t) = sinh
k
(βt).
42. y(x) – λ

b
a
tanh
k
(βx) cos
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tanh
k
(βx) and h(t) = cos
m
(µt).
43. y(x) – λ

b
a

tanh
k
(βt) cos
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cos
m
(µx) and h(t) = tanh
k
(βt).
44. y(x) – λ

b
a
tanh
k
(βx) sin
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tanh
k
(βx) and h(t) = sin
m
(µt).
45. y(x) – λ

b
a
tanh
k

(βt) sin
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = sin
m
(µx) and h(t) = tanh
k
(βt).
4.7-6. Kernels Containing Logarithmic and Trigonometric Functions
46. y(x) – λ

b
a
cos
k
(βx)ln
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cos
k
(βx) and h(t)=ln
m
(µt).
Page 298
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© 1998 by CRC Press LLC
47. y(x) – λ

b
a

cos
k
(βt)ln
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=ln
m
(µx) and h(t) = cos
k
(βt).
48. y(x) – λ

b
a
sin
k
(βx)ln
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = sin
k
(βx) and h(t)=ln
m
(µt).
49. y(x) – λ

b
a
sin
k

(βt)ln
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=ln
m
(µx) and h(t) = sin
k
(βt).
50. y(x) – λ

b
a
tan
k
(βx)ln
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = tan
k
(βx) and h(t)=ln
m
(µt).
51. y(x) – λ

b
a
tan
k
(βt)ln
m

(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=ln
m
(µx) and h(t) = tan
k
(βt).
52. y(x) – λ

b
a
cot
k
(βx)ln
m
(µt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = cot
k
(βx) and h(t)=ln
m
(µt).
53. y(x) – λ

b
a
cot
k
(βt)ln
m
(µx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=ln

m
(µx) and h(t) = cot
k
(βt).
4.8. Equations Whose Kernels Contain Special
Functions
4.8-1. Kernels Containing Bessel Functions
1. y(x) – λ

b
a
J
ν
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=J
ν
(βx) and h(t)=1.
2. y(x) – λ

b
a
J
ν
(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = 1 and h(t)=J
ν
(βt).
3. y(x)+λ

∞

0
tJ
ν
(xt)y(t) dt = f (x), ν > –
1
2
.
Solution:
y(x)=
f(x)
1 – λ
2
–
λ
1 – λ
2

∞
0
tJ
ν
(xt)f(t) dt, λ ≠ ±1.
Page 299
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© 1998 by CRC Press LLC
4. y(x)+λ

∞
0
J

ν

2
√
xt

y(t) dt = f (x).
By setting x =
1
2
z
2
, t =
1
2
τ
2
, y(x)=Y (z), and f(x)=F (z), we arrive at an equation of the
form 4.8.3:
Y (z)+λ

∞
0
τJ
ν
(zτ)Y (τ) dτ = F (z).
5. y(x) – λ

b
a

[A + B(x – t)J
ν
(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.8 with h(t)=J
ν
(βt).
6. y(x) – λ

b
a
[A + B(x – t)J
ν
(βx)]y(t) dt = f (x).
This is a special case of equation 4.9.10 with h(x)=J
ν
(βx).
7. y(x) – λ

b
a
[AJ
µ
(αx)+BJ
ν
(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.5 with g(x)=AJ
µ
(αx) and h(t)=BJ
ν
(βt).

8. y(x) – λ

b
a
[AJ
µ
(x)J
ν
(t)+BJ
ν
(x)J
µ
(t)]y(t) dt = f (x).
This is a special case of equation 4.9.17 with g(x)=J
µ
(x) and h(t)=J
ν
(t).
9. y(x) – λ

b
a
Y
ν
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=Y
ν
(βx) and h(t)=1.
10. y(x) – λ


b
a
Y
ν
(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = 1 and h(t)=Y
ν
(βt).
11. y(x) – λ

b
a
[A + B(x – t)Y
ν
(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.8 with h(t)=Y
ν
(βt).
12. y(x) – λ

b
a
[A + B(x – t)Y
ν
(βx)]y(t) dt = f (x).
This is a special case of equation 4.9.10 with h(x)=Y
ν
(βx).
13. y(x) – λ


b
a
[AY
µ
(αx)+BY
ν
(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.5 with g(x)=AY
µ
(αx) and h(t)=BY
ν
(βt).
14. y(x) – λ

b
a
[AY
µ
(x)Y
µ
(t)+BY
ν
(x)Y
ν
(t)]y(t) dt = f (x).
This is a special case of equation 4.9.14 with g(x)=Y
µ
(x) and h(t)=Y
ν
(t).

15. y(x) – λ

b
a
[AY
µ
(x)Y
ν
(t)+BY
ν
(x)Y
µ
(t)]y(t) dt = f (x).
This is a special case of equation 4.9.17 with g(x)=Y
µ
(x) and h(t)=Y
ν
(t).
Page 300
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© 1998 by CRC Press LLC
4.8-2. Kernels Containing Modified Bessel Functions
16. y(x)–λ

b
a
I
ν
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=I

ν
(βx) and h(t)=1.
17. y(x)–λ

b
a
I
ν
(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = 1 and h(t)=I
ν
(βt).
18. y(x)–λ

b
a
[A + B(x – t)I
ν
(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.8 with h(t)=I
ν
(βt).
19. y(x)–λ

b
a
[A + B(x – t)I
ν
(βx)]y(t) dt = f (x).
This is a special case of equation 4.9.10 with h(x)=I

ν
(βx).
20. y(x)–λ

b
a
[AI
µ
(αx)+BI
ν
(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.5 with g(x)=AI
µ
(αx) and h(t)=BI
ν
(βt).
21. y(x)–λ

b
a
[AI
µ
(x)I
µ
(t)+BI
ν
(x)I
ν
(t)]y(t) dt = f (x).
This is a special case of equation 4.9.14 with g(x)=I

µ
(x) and h(t)=I
ν
(t).
22. y(x)–λ

b
a
[AI
µ
(x)I
ν
(t)+BI
ν
(x)I
µ
(t)]y(t) dt = f (x).
This is a special case of equation 4.9.17 with g(x)=I
µ
(x) and h(t)=I
ν
(t).
23. y(x)–λ

b
a
K
ν
(βx)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x)=K

ν
(βx) and h(t)=1.
24. y(x)–λ

b
a
K
ν
(βt)y(t) dt = f (x).
This is a special case of equation 4.9.1 with g(x) = 1 and h(t)=K
ν
(βt).
25. y(x)–λ

b
a
[A + B(x – t)K
ν
(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.8 with h(t)=K
ν
(βt).
26. y(x)–λ

b
a
[A + B(x – t)K
ν
(βx)]y(t) dt = f (x).
This is a special case of equation 4.9.10 with h(x)=K

ν
(βx).
Page 301
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
27. y(x) – λ

b
a
[AK
µ
(αx)+BK
ν
(βt)]y(t) dt = f (x).
This is a special case of equation 4.9.5 with g(x)=AK
µ
(αx) and h(t)=BK
ν
(βt).
28. y(x) – λ

b
a
[AK
µ
(x)K
µ
(t)+BK
ν
(x)K

ν
(t)]y(t) dt = f (x).
This is a special case of equation 4.9.14 with g(x)=K
µ
(x) and h(t)=K
ν
(t).
29. y(x) – λ

b
a
[AK
µ
(x)K
ν
(t)+BK
ν
(x)K
µ
(t)]y(t) dt = f (x).
This is a special case of equation 4.9.17 with g(x)=K
µ
(x) and h(t)=K
ν
(t).
4.9. Equations Whose Kernels Contain Arbitrary
Functions
4.9-1. Equations With Degenerate Kernel: K(x, t)=g
1
(x)h

1
(t)+···+ g
n
(x)h
n
(t)
1. y(x) – λ

b
a
g(x)h(t)y(t) dt = f (x).
1
◦
. Assume that λ ≠


b
a
g(t)h(t) dt

–1
.
Solution:
y(x)=f(x)+λkg(x), where k =

1 – λ

b
a
g(t)h(t) dt


–1

b
a
h(t)f(t) dt.
2
◦
. Assume that λ =


b
a
g(t)h(t) dt

–1
.
For

b
a
h(t)f(t) dt = 0, the solution has the form
y = f(x)+Cg(x),
where C is an arbitrary constant.
For

b
a
h(t)f(t) dt ≠ 0, there is no solution.
The limits of integration may take the values a = –∞ and/or b = ∞, provided that the

corresponding improper integral converges.
2. y(x) – λ

b
a
[g(x)+g(t)]y(t) dt = f (x).
The characteristic values of the equation:
λ
1
=
1
g
1
+
√
(b – a)g
2
, λ
2
=
1
g
1
–
√
(b – a)g
2
,
where
g

1
=

b
a
g(x) dx, g
2
=

b
a
g
2
(x) dx.
Page 302
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
1
◦
. Solution with λ ≠ λ
1,2
:
y(x)=f(x)+λ[A
1
g(x)+A
2
],
where the constants A
1
and A

2
are given by
A
1
=
f
1
– λ[f
1
g
1
– (b – a)f
2
]
[g
2
1
– (b – a)g
2
]λ
2
– 2g
1
λ +1
, A
2
=
f
2
– λ(f

2
g
1
– f
1
g
2
)
[g
2
1
– (b – a)g
2
]λ
2
– 2g
1
λ +1
,
f
1
=

b
a
f(x) dx, f
2
=

b

a
f(x)g(x) dx.
2
◦
. Solution with λ = λ
1
≠ λ
2
and f
1
= f
2
=0:
y(x)=f(x)+Cy
1
(x), y
1
(x)=g(x)+

g
2
b – a
,
where C is an arbitrary constant and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
3

◦
. Solution with λ = λ
2
≠ λ
1
and f
1
= f
2
=0:
y(x)=f(x)+Cy
2
(x), y
2
(x)=g(x) –

g
2
b – a
,
where C is an arbitrary constant and y
2
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
2
.
4
◦
. The equation has no multiple characteristic values.
3. y(x) – λ


b
a
[g(x) – g(t)]y(t) dt = f (x).
The characteristic values of the equation:
λ
1
=
1

g
2
1
– (b – a)g
2
, λ
2
= –
1

g
2
1
– (b – a)g
2
,
where
g
1
=


b
a
g(x) dx, g
2
=

b
a
g
2
(x) dx.
1
◦
. Solution with λ ≠ λ
1,2
:
y(x)=f(x)+λ[A
1
g(x)+A
2
],
where the constants A
1
and A
2
are given by
A
1
=

f
1
+ λ[f
1
g
1
– (b – a)f
2
]
[(b – a)g
2
– g
2
1
]λ
2
+1
, A
2
=
–f
2
+ λ(f
2
g
1
– f
1
g
2

)
[(b – a)g
2
– g
2
1
]λ
2
+1
,
f
1
=

b
a
f(x) dx, f
2
=

b
a
f(x)g(x) dx.
2
◦
. Solution with λ = λ
1
≠ λ
2
and f

1
= f
2
=0:
y(x)=f(x)+Cy
1
(x), y
1
(x)=g(x)+
1 – λ
1
g
1
λ
1
(b – a)
,
where C is an arbitrary constant and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
3
◦
. The solution with λ = λ
2
≠ λ
1
and f

1
= f
2
= 0 is given by the formulas of item 2
◦
in
which one must replace λ
1
and y
1
(x)byλ
2
and y
2
(x), respectively.
4
◦
. The equation has no multiple characteristic values.
Page 303
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
4. y(x) – λ

b
a
[Ag(x)+Bg(t)]y(t) dt = f (x).
The characteristic values of the equation:
λ
1,2
=

(A + B)g
1
±

(A – B)
2
g
2
1
+4AB(b – a)g
2
2AB[g
2
1
– (b – a)g
2
]
,
where
g
1
=

b
a
g(x) dx, g
2
=

b

a
g
2
(x) dx.
1
◦
. Solution with λ ≠ λ
1,2
:
y(x)=f(x)+λ[A
1
g(x)+A
2
],
where the constants A
1
and A
2
are given by
A
1
=
Af
1
–λAB[f
1
g
1
–(b–a)f
2

]
AB[g
2
1
–(b–a)g
2
]λ
2
–(A+B)g
1
λ+1
, A
2
=
Bf
2
–λAB(f
2
g
1
–f
1
g
2
)
AB[g
2
1
–(b–a)g
2

]λ
2
–(A+B)g
1
λ+1
,
f
1
=

b
a
f(x) dx, f
2
=

b
a
f(x)g(x) dx.
2
◦
. Solution with λ = λ
1
≠ λ
2
and f
1
= f
2
=0:

y(x)=f(x)+Cy
1
(x), y
1
(x)=g(x)+
1 – λ
1
Ag
1
λ
1
A(b – a)
,
where C is an arbitrary constant and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
3
◦
. The solution with λ = λ
2
≠ λ
1
and f
1
= f
2
= 0 is given by the formulas of item 2

◦
in
which one must replace λ
1
and y
1
(x)byλ
2
and y
2
(x), respectively.
4
◦
. Solution with λ = λ
1,2
= λ
∗
and f
1
= f
2
= 0, where the characteristic value λ
∗
=
2
(A + B)g
1
is double:
y(x)=f(x)+Cy
∗

(x), y
∗
(x)=g(x) –
(A – B)g
1
2A(b – a)
.
Here C is an arbitrary constant and y
∗
(x) is an eigenfunction of the equation corresponding
to λ
∗
.
5. y(x) – λ

b
a
[g(x)+h(t)]y(t) dt = f (x).
The characteristic values of the equation:
λ
1,2
=
s
1
+ s
3
±

(s
1

– s
3
)
2
+4(b – a)s
2
2[s
1
s
3
– (b – a)s
2
]
,
where
s
1
=

b
a
g(x) dx, s
2
=

b
a
g(x)h(x) dx, s
3
=


b
a
h(x) dx.
Page 304
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
1
◦
. Solution with λ ≠ λ
1,2
:
y(x)=f(x)+λ[A
1
g(x)+A
2
],
where the constants A
1
and A
2
are given by
A
1
=
f
1
– λ[f
1
s

3
– (b – a)f
2
]
[s
1
s
3
– (b – a)s
2
]λ
2
– (s
1
+ s
3
)λ +1
, A
2
=
f
2
– λ(f
2
s
1
– f
1
s
2

)
[s
1
s
3
– (b – a)s
2
]λ
2
– (s
1
+ s
3
)λ +1
,
f
1
=

b
a
f(x) dx, f
2
=

b
a
f(x)h(x) dx.
2
◦

. Solution with λ = λ
1
≠ λ
2
and f
1
= f
2
=0:
y(x)=f(x)+Cy
1
(x), y
1
(x)=g(x)+
1 – λ
1
s
1
λ
1
(b – a)
,
where C is an arbitrary constant and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
3
◦

. The solution with λ = λ
2
≠ λ
1
and f
1
= f
2
= 0 is given by the formulas of item 2
◦
in
which one must replace λ
1
and y
1
(x)byλ
2
and y
2
(x), respectively.
4
◦
. Solution with λ = λ
1,2
= λ
∗
and f
1
= f
2

= 0, where the characteristic value λ
∗
=
2
s
1
+ s
3
is double:
y(x)=f(x)+Cy
∗
(x), y
∗
(x)=g(x) –
s
1
– s
3
2(b – a)
.
Here C is an arbitrary constant and y
∗
(x) is an eigenfunction of the equation corresponding
to λ
∗
.
6. y(x) – λ

b
a

[Ag(x)+Bg(t)]h(t) y(t) dt = f(x).
The characteristic values of the equation:
λ
1,2
=
(A + B)s
1
±

(A – B)
2
s
2
1
+4ABs
0
s
2
2AB(s
2
1
– s
0
s
2
)
,
where
s
0

=

b
a
h(x) dx, s
1
=

b
a
g(x)h(x) dx, s
2
=

b
a
g
2
(x)h(x) dx.
1
◦
. Solution with λ ≠ λ
1,2
:
y(x)=f(x)+λ[A
1
g(x)+A
2
],
where the constants A

1
and A
2
are given by
A
1
=
Af
1
– ABλ(f
1
s
1
– f
2
s
0
)
AB(s
2
1
– s
0
s
2
)λ
2
– (A + B)s
1
λ +1

, A
2
=
Bf
2
– ABλ(f
2
s
1
– f
1
s
2
)
AB(s
2
1
– s
0
s
2
)λ
2
– (A + B)s
1
λ +1
,
f
1
=


b
a
f(x)h(x) dx, f
2
=

b
a
f(x)g(x)h(x) dx.
2
◦
. Solution with λ = λ
1
≠ λ
2
and f
1
= f
2
=0:
y(x)=f(x)+Cy
1
(x), y
1
(x)=g(x)+
1 – λ
1
As
1

λ
1
As
0
,
where C is an arbitrary constant and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
Page 305
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
3
◦
. The solution with λ = λ
2
≠ λ
1
and f
1
= f
2
= 0 is given by the formulas of item 2
◦
in
which one must replace λ
1
and y

1
(x)byλ
2
and y
2
(x), respectively.
4
◦
. Solution with λ = λ
1,2
= λ
∗
and f
1
= f
2
= 0, where the characteristic value λ
∗
=
2
(A + B)s
1
is double:
y(x)=f(x)+Cy
∗
(x),
where C is an arbitrary constant and y
∗
(x) is an eigenfunction of the equation corresponding
to λ

∗
. Two cases are possible.
(a) If A ≠ B, then 4ABs
0
s
2
= –(A – B)
2
s
2
1
and
y
∗
(x)=g(x) –
(A – B)s
1
2As
0
.
(b) If A = B, then, in view of 4ABs
0
s
2
= –(A – B)
2
s
2
1
=0,wehave

y
∗
(x)=

g(x) for s
0
≠ 0 and s
2
=0,
1 for s
0
= 0 and s
2
≠ 0,
C
1
g(x)+C
2
for s
0
= s
2
=0,
where C
1
and C
2
are arbitrary constants.
7. y(x) – λ


b
a
[Ag(x)+Bg(t)+C]h(t) y(t) dt = f(x).
The characteristic values of the equation:
λ
1,2
=
(A + B)s
1
+ Cs
0
±

(A – B)
2
s
2
1
+2(A + B)Cs
1
s
0
+ C
2
s
2
0
+4ABs
0
s

2
2AB(s
2
1
– s
0
s
2
)
,
where
s
0
=

b
a
h(x) dx, s
1
=

b
a
g(x)h(x) dx, s
2
=

b
a
g

2
(x)h(x) dx.
1
◦
. Solution with λ ≠ λ
1,2
:
y(x)=f(x)+λ[A
1
g(x)+A
2
],
where the constants A
1
and A
2
are given by
A
1
=
Af
1
– ABλ(f
1
s
1
– f
2
s
0

)
AB(s
2
1
– s
0
s
2
)λ
2
– [(A + B)s
1
+ Cs
0
]λ +1
,
A
2
=
C
1
f
1
+ Bf
2
– ABλ(f
2
s
1
– f

1
s
2
)
AB(s
2
1
– s
0
s
2
)λ
2
– [(A + B)s
1
+ Cs
0
]λ +1
,
f
1
=

b
a
f(x)h(x) dx, f
2
=

b

a
f(x)g(x)h(x) dx.
2
◦
. Solution with λ = λ
1
≠ λ
2
and f
1
= f
2
=0:
y(x)=f(x)+

Cy
1
(x), y
1
(x)=g(x)+
1 – λ
1
As
1
λ
1
As
0
,
where


C is an arbitrary constant and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
Page 306
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
3
◦
. The solution with λ = λ
2
≠ λ
1
and f
1
= f
2
= 0 is given by the formulas of item 2
◦
in
which one must replace λ
1
and y
1
(x)byλ
2
and y

2
(x), respectively.
4
◦
. Solution with λ = λ
1,2
= λ
∗
and f
1
= f
2
= 0, where the characteristic value λ
∗
=
2
(A + B)s
1
+ Cs
0
is double:
y(x)=f(x)+

Cy
∗
(x),
where

C is an arbitrary constant and y
∗

(x) is an eigenfunction of the equation corresponding
to λ
∗
. Two cases are possible.
(a) If As
1
– (Bs
1
+ Cs
0
) ≠ 0, then 4As
0
(Bs
2
+ Cs
1
)=–[(A – B)s
1
– Cs
0
]
2
and
y
∗
(x)=g(x) –
(A – B)s
1
– Cs
0

2As
0
.
(b) If (A – B)s
1
= Cs
0
, then, in view of 4As
0
(Bs
2
+ Cs
1
)=–[(A – B)s
1
– Cs
0
]
2
=0,we
have
y
∗
(x)=

g(x) for s
0
≠ 0 and Bs
2
= –Cs

1
,
1 for s
0
= 0 and Bs
2
≠ –Cs
1
,

C
1
g(x)+

C
2
for s
0
= 0 and Bs
2
= –Cs
1
,
where

C
1
and

C

2
are arbitrary constants.
8. y(x) – λ

b
a
[A + B(x – t)h(t)]y(t) dt = f (x).
The characteristic values of the equation:
λ
1,2
=
A(b – a) ±

[A(b – a) – 2Bh
1
]
2
+2Bh
0
[A(b
2
– a
2
) – 2Bh
2
]
B

A(b – a)[2h
1

– (b + a)h
0
] – 2B(h
2
1
– h
0
h
2
)

,
where
h
0
=

b
a
h(x) dx, h
1
=

b
a
xh(x) dx, h
2
=

b

a
x
2
h(x) dx.
1
◦
. Solution with λ ≠ λ
1,2
:
y(x)=f(x)+λ(A
1
+ A
2
x),
where the constants A
1
and A
2
are given by
A
1
=
f
1
– λ

B(f
1
h
1

+ f
2
h
2
) –
1
2
Af
2
(b
2
– a
2
)

B

A(b – a)

h
1
–
1
2
(b + a)h
0

– B(h
2
1

– h
0
h
2
)

λ
2
+ A(b – a)λ +1
,
A
2
=
f
2
– λ[A(b – a)f
2
– B(f
1
h
0
+ f
2
h
1
)]
B

A(b – a)


h
1
–
1
2
(b + a)h
0

– B(h
2
1
– h
0
h
2
)

λ
2
+ A(b – a)λ +1
,
f
1
= A

b
a
f(x) dx – B

b

a
xf(x)h(x) dx, f
2
= B

b
a
f(x)h(x) dx.
2
◦
. Solution with λ = λ
1
≠ λ
2
and f
1
= f
2
=0:
y(x)=f(x)+Cy
1
(x), y
1
(x)=1+
2 – 2λ
1
[A(b – a) – Bh
1
]
λ

1
[A(b
2
– a
2
) – 2Bh
2
]
x,
where C is an arbitrary constant, and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
Page 307
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
3
◦
. The solution with λ = λ
2
≠ λ
1
and f
1
= f
2
= 0 is given by the formulas of item 2
◦

in
which one must replace λ
1
and y
1
(x)byλ
2
and y
2
(x), respectively.
4
◦
. Solution with λ = λ
1,2
= λ
∗
and f
1
= f
2
= 0, where the characteristic value λ
∗
=
2
A(b – a)
(A ≠ 0) is double:
y(x)=f(x)+Cy
∗
(x),
where C is an arbitrary constant, and y

∗
(x) is an eigenfunction of the equation corresponding
to λ
∗
. Two cases are possible.
(a) If A(b – a) – 2Bh
1
≠ 0, then
y
∗
(x)=1–
A(b – a) – 2Bh
1
A(b
2
– a
2
) – 2Bh
2
x.
(b) If A(b – a) – 2Bh
1
= 0, then, in view of h
0
[A(b
2
– a
2
) – 2Bh
2

] = 0, we have
y
∗
(x)=



1 for h
0
≠ 0 and A(b
2
– a
2
)=2Bh
2
,
x for h
0
= 0 and A(b
2
– a
2
) ≠ 2Bh
2
,
C
1
+ C
2
x for h

0
= 0 and A(b
2
– a
2
)=2Bh
2
,
where C
1
and C
2
are arbitrary constants.
9. y(x) – λ

b
a
[A +(Bx + Ct)h(t)]y(t) dt = f (x).
The characteristic values of the equation:
λ
1,2
=
A(b – a)+(C + B)h
1
±
√
D
B

A(b – a)[2h

1
– (b + a)h
0
]+2C(h
2
1
– h
0
h
2
)

,
D =[A(b – a)+(C – B)h
1
]
2
+2Bh
0
[A(b
2
– a
2
)+2Ch
2
],
where
h
0
=


b
a
h(x) dx, h
1
=

b
a
xh(x) dx, h
2
=

b
a
x
2
h(x) dx.
1
◦
. Solution with λ ≠ λ
1,2
:
y(x)=f(x)+λ(A
1
+ A
2
x),
where the constants A
1

and A
2
are given by
A
1
= ∆
–1

f
1
– λ

Bf
1
h
1
– Cf
2
h
2
–
1
2
A(b
2
– a
2
)f
2


,
A
2
= ∆
–1

f
2
– λ

A(b – a)f
2
– Bf
1
h
0
+ Cf
2
h
1

,
∆ = B

A(b – a)

h
1
–
1

2
(b + a)h
0

+ C(h
2
1
– h
0
h
2
)

λ
2
+[A(b – a)+(B + C)h
1
]λ +1,
f
1
= A

b
a
f(x) dx + C

b
a
xf(x)h(x) dx, f
2

= B

b
a
f(x)h(x) dx.
2
◦
. Solution with λ = λ
1
≠ λ
2
and f
1
= f
2
=0:
y(x)=f(x)+

Cy
1
(x), y
1
(x)=1+
2 – 2λ
1
[A(b – a)+Ch
1
]
λ
1

[A(b
2
– a
2
)+2Ch
2
]
x,
where

C is an arbitrary constant and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
Page 308
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
3
◦
. The solution with λ = λ
2
≠ λ
1
and f
1
= f
2
= 0 is given by the formulas of item 2

◦
in
which one must replace λ
1
and y
1
(x)byλ
2
and y
2
(x), respectively.
4
◦
. Solution with λ = λ
1,2
= λ
∗
and f
1
= f
2
= 0, where the characteristic value λ
∗
=
2
A(b – a)+(B + C)h
1
is double:
y(x)=f(x)+


Cy
∗
(x),
where

C is an arbitrary constant and y
∗
(x) is an eigenfunction of the equation corresponding
to λ
∗
. Two cases are possible.
(a) If A(b – a)+(C – B)h
1
≠ 0, then
y
∗
(x)=1–
A(b – a)+(C – B)h
1
A(b
2
– a
2
)+2Ch
2
x.
(b) If A(b – a)+(C – B)h
1
= 0, then, in view of h
0

[A(b
2
– a
2
)+2Ch
2
]=0,wehave
y
∗
(x)=



1 for h
0
≠ 0 and A(b
2
– a
2
)=–2Ch
2
,
x for h
0
= 0 and A(b
2
– a
2
) ≠ –2Ch
2

,

C
1
+

C
2
x for h
0
= 0 and A(b
2
– a
2
)=–2Ch
2
,
where

C
1
and

C
2
are arbitrary constants.
10. y(x) – λ

b
a

[A + B(x – t)h(x)]y(t) dt = f (x).
The characteristic values of the equation:
λ
1,2
=
A(b – a) ±

[A(b – a)+2Bh
1
]
2
– 2Bh
0
[A(b
2
– a
2
)+2Bh
2
]
B{h
0
[A(b
2
– a
2
)+2Bh
2
] – 2h
1

[A(b – a)+Bh
1
]}
,
where
h
0
=

b
a
h(x) dx, h
1
=

b
a
xh(x) dx, h
2
=

b
a
x
2
h(x) dx.
1
◦
. Solution with λ ≠ λ
1,2

:
y(x)=f(x)+λ

AE
1
+(BE
1
x + E
2
)h(x)

,
where the constants E
1
and E
2
are given by
E
1
= ∆
–1

f
1
+ λB(f
1
h
1
– f
2

h
0
)

,
E
2
= B∆
–1

–f
2
+ λf
2

A(b – a)+Bh
1

+ λf
1

1
2
A(b
2
– a
2
)+Bh
2


,
∆ = B

h
0

1
2
A(b
2
– a
2
)+Bh
2

– h
1
[A(b – a)+Bh
1
]

λ
2
– A(b – a)λ +1,
f
1
=

b
a

f(x) dx, f
2
=

b
a
xf(x) dx.
2
◦
. Solution with λ = λ
1
≠ λ
2
and f
1
= f
2
=0:
y(x)=f(x)+Cy
1
(x), y
1
(x)=A + Bxh(x)+
1 – λ
1
[A(b – a)+Bh
1
]
λ
1

h
0
h(x),
where C is an arbitrary constant and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
Page 309
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
3
◦
. The solution with λ = λ
2
≠ λ
1
and f
1
= f
2
= 0 is given by the formulas of item 2
◦
in
which one must replace λ
1
and y
1
(x)byλ

2
and y
2
(x), respectively.
4
◦
. Solution with λ = λ
1,2
= λ
∗
and f
1
= f
2
= 0, where the characteristic value λ
∗
=
2
A(b – a)
(A ≠ 0) is double:
y(x)=f(x)+Cy
∗
(x),
where C is an arbitrary constant and y
∗
(x) is an eigenfunction of the equation corresponding
to λ
∗
. Two cases are possible.
(a) If A(b – a) ≠ –2Bh

1
, then
y
∗
(x)=A + Bxh(x) –
A(b – a)+2Bh
1
2h
0
h(x).
(b) If A(b – a)=–2Bh
1
, then, in view of h
0
[A(b
2
– a
2
)+2Bh
2
]=0,wehave
y
∗
(x)=



A + Bxh(x) for h
0
≠ 0 and A(b

2
– a
2
)=–2Bh
2
,
h(x) for h
0
= 0 and A(b
2
– a
2
) ≠ –2Bh
2
,
C
1
[A + Bxh(x)] + C
2
h(x) for h
0
= 0 and A(b
2
– a
2
)=–2Bh
2
,
where C
1

and C
2
are arbitrary constants.
11. y(x) – λ

b
a
[A +(Bx + Ct)h(x)]y(t) dt = f (x).
The characteristic values of the equation:
λ
1,2
=
A(b – a)+(B + C)h
1
±
√
D
C{2h
1
[A(b – a)+Bh
1
] – h
0
[A(b
2
– a
2
)+2Bh
2
]}

,
D =[A(b – a)+(B – C)h
1
]
2
+2Ch
0
[A(b
2
– a
2
)+2Bh
2
],
where
h
0
=

b
a
h(x) dx, h
1
=

b
a
xh(x) dx, h
2
=


b
a
x
2
h(x) dx.
1
◦
. Solution with λ ≠ λ
1,2
:
y(x)=f(x)+λ

AE
1
+(BE
1
x + E
2
)h(x)

,
where the constants E
1
and E
2
are given by
E
1
= ∆

–1
[f
1
– λC(f
1
h
1
– f
2
h
0
)],
E
2
= C∆
–1

f
2
– λf
2

A(b – a)+Bh
1

– λf
1

1
2

A(b
2
– a
2
)+Bh
2

,
∆ = C

h
1
[A(b – a)+Bh
1
] – h
0

1
2
A(b
2
– a
2
)+Bh
2

λ
2
– [A(b – a)+(B + C)h
1

]λ +1,
f
1
=

b
a
f(x) dx, f
2
=

b
a
xf(x) dx.
2
◦
. Solution with λ = λ
1
≠ λ
2
and f
1
= f
2
=0:
y(x)=f(x)+

Cy
1
(x), y

1
(x)=A + Bxh(x)+
1 – λ
1
[A(b – a)+Bh
1
]
λ
1
h
0
h(x),
where

C is an arbitrary constant and y
1
(x) is an eigenfunction of the equation corresponding
to the characteristic value λ
1
.
Page 310
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC