HANDBOOK OFINTEGRAL EQUATIONS phần 3 potx

30. y(x)+A

x
0
(x
2
– t
2
)e
λ(x–t)
y(t) dt = f(x).
The substitution u(x)=e
–λx
y(x) leads to an equation of the form 2.1.11:
u(x)+A

x
0
(x
2
– t
2
)u(t) dt = f (x)e
–λx
.
31. y(x)+A

x
a
(x – t)
n

e
λ(x–t)
y(t) dt = f(x), n =1,2,
Solution:
y(x)=f(x)+

x
a
R(x – t)f(t) dt,
R(x)=
1
n +1
e
λx
n

k=0
exp(σ
k
x)

σ
k
cos(β
k
x) – β
k
sin(β
k
x)


,
where
σ
k
= |An!|
1
n+1
cos

2πk
n +1

, β
k
= |An!|
1
n+1
sin

2πk
n +1

for A <0,
σ
k
= |An!|
1
n+1
cos


2πk + π
n +1

, β
k
= |An!|
1
n+1
sin

2πk + π
n +1

for A >0.
32. y(x)+b

x
a
exp[λ(x – t)]
√
x – t
y(t) dt = f(x).
Solution:
y(x)=e
λx

F (x)+πb
2


x
a
exp[πb
2
(x – t)]F (t) dt

,
where
F (x)=e
–λx
f(x) – b

x
a
e
–λt
f(t)
√
x – t
dt.
33. y(x)+A

x
a
(x – t)t
k
e
λ(x–t)
y(t) dt = f(x).
The substitution u(x)=e

–λx
y(x) leads to an equation of the form 2.1.49:
u(x)+A

x
a
(x – t)t
k
u(t) dt = f (x)e
–λx
.
34. y(x)+A

x
a
(x
k
– t
k
)e
λ(x–t)
y(t) dt = f(x).
The substitution u(x)=e
–λx
y(x) leads to an equation of the form 2.1.52:
u(x)+A

x
a
(x

k
– t
k
)u(t) dt = f (x)e
–λx
.
35. y(x) – λ

x
0
e
µ(x–t)
(x – t)
α
y(t) dt = f(x), 0 < α <1.
Solution:
y(x)=f(x)+

x
0
R(x – t)f(t) dt, where R(x)=e
µx
∞

n=1

λΓ(1 – α)x
1–α

n

xΓ

n(1 – α)

.
Page 133
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
36. y(x)+A

x
a
exp

λ(x
2
– t
2
)

y(t) dt = f(x).
Solution:
y(x)=f(x) – A

x
a
exp

λ(x
2

– t
2
) – A(x – t)

f(t) dt.
37. y(x)+A

x
a
exp

λx
2
+ βt
2

y(t) dt = f(x).
In the case β = –λ, see equation 2.2.36. This is a special case of equation 2.9.2 with
g(x)=–A exp

λx
2
) and h(t)=exp

βt
2

.
38. y(x)+A


∞
x
exp

–λ
√
t – x

y(t) dt = f(x).
This is a special case of equation 2.9.62 with K(x)=A exp

–λ
√
–x

.
39. y(x)+A

x
a
exp

λ(x
µ
– t
µ
)

y(t) dt = f(x), µ >0.
This is a special case of equation 2.9.2 with g(x)=–A exp


λx
µ

and h(t)=exp

–λt
µ

.
Solution:
y(x)=f(x) – A

x
a
exp

λ(x
µ
– t
µ
) – A(x – t)

f(t) dt.
40. y(x)+k

x
0
1
x

exp

–λ
t
x

y(t) dt = g(x).
This is a special case of equation 2.9.71 with f(z)=ke
–λz
.
For a polynomial right-hand side, g(x)=
N

n=0
A
n
x
n
, a solution is given by
y(x)=
N

n=0
A
n
1+kB
n
x
n
, B

n
=
n!
λ
n+1
– e
–λ
n

k=0
n!
k!
1
λ
n–k+1
.
2.3. Equations Whose Kernels Contain Hyperbolic
Functions
2.3-1. Kernels Containing Hyperbolic Cosine
1. y(x) – A

x
a
cosh(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A cosh(λx) and h(t)=1.
Solution:
y(x)=f(x)+A

x
a

cosh(λx)exp

A
λ

sinh(λx) – sinh(λt)


f(t) dt.
Page 134
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
2. y(x) – A

x
a
cosh(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = cosh(λt).
Solution:
y(x)=f(x)+A

x
a
cosh(λt)exp

A
λ

sinh(λx) – sinh(λt)



f(t) dt.
3. y(x)+A

x
a
cosh[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.28 with g(t)=A. Therefore, solving the original integral
equation is reduced to solving the second-order linear nonhomogeneous ordinary differential
equation with constant coefficients
y

xx
+ Ay

x
– λ
2
y = f

xx
– λ
2
f, f = f (x),
under the initial conditions
y(a)=f(a), y

x
(a)=f


x
(a) – Af(a).
Solution:
y(x)=f(x)+

x
a
R(x – t)f(t) dt,
R(x)=exp

–
1
2
Ax


A
2
2k
sinh(kx) – A cosh(kx)

, k =

λ
2
+
1
4
A
2

.
4. y(x)+

x
a

n

k=1
A
k
cosh[λ
k
(x – t)]

y(t) dt = f(x).
This equation can be reduced to an equation of the form 2.2.19 by using the identity
cosh z ≡
1
2

e
z
+ e
–z

. Therefore, the integral equation in question can be reduced to a
linear nonhomogeneous ordinary differential equation of order 2n with constant coefficients.
5. y(x) – A


x
a
cosh(λx)
cosh(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
cosh(λx)
cosh(λt)
f(t) dt.
6. y(x) – A

x
a
cosh(λt)
cosh(λx)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
cosh(λt)

cosh(λx)
f(t) dt.
7. y(x) – A

x
a
cosh
k
(λx) cosh
m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A cosh
k
(λx) and h(t) = cosh
m
(µt).
Page 135
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
8. y(x)+A

x
a
t cosh[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.28 with g(t)=At.
9. y(x)+A

x
a
t

k
cosh
m
(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–A cosh
m
(λx) and h(t)=t
k
.
10. y(x)+A

x
a
x
k
cosh
m
(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–Ax
k
and h(t) = cosh
m
(λt).
11. y(x) –

x
a

A cosh(kx)+B – AB(x – t) cosh(kx)


y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A cosh(kx).
Solution:
y(x)=f(x)+

x
a
R(x, t)f(t) dt,
R(x, t)=[A cosh(kx)+B]
G(x)
G(t)
+
B
2
G(t)

x
t
e
B(x–s)
G(s) ds, G(x)=exp

A
k
sinh(kx)

.
12. y(x)+

x

a

A cosh(kt)+B + AB(x – t) cosh(kt)

y(t) dt = f(x).
This is a special case of equation 2.9.8 with λ = B and g(t)=A cosh(kt).
Solution:
y(x)=f(x)+

x
a
R(x, t)f(t) dt,
R(x, t)=–[A cosh(kt)+B]
G(t)
G(x)
+
B
2
G(x)

x
t
e
B(t–s )
G(s) ds, G(x)=exp

A
k
sinh(kx)


.
13. y(x)+A

∞
x
cosh

λ
√
t – x

y(t) dt = f(x).
This is a special case of equation 2.9.62 with K(x)=A cosh

λ
√
–x

.
2.3-2. Kernels Containing Hyperbolic Sine
14. y(x) – A

x
a
sinh(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A sinh(λx) and h(t)=1.
Solution:
y(x)=f(x)+A

x

a
sinh(λx)exp

A
λ

cosh(λx) – cosh(λt)


f(t) dt.
Page 136
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
15. y(x) – A

x
a
sinh(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = sinh(λt).
Solution:
y(x)=f(x)+A

x
a
sinh(λt)exp

A
λ

cosh(λx) – cosh(λt)



f(t) dt.
16. y(x)+A

x
a
sinh[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.30 with g(x)=A.
1
◦
. Solution with λ(A – λ)>0:
y(x)=f(x) –
Aλ
k

x
a
sin[k(x – t)]f(t) dt, where k =

λ(A – λ).
2
◦
. Solution with λ(A – λ)<0:
y(x)=f(x) –
Aλ
k

x
a

sinh[k(x – t)]f(t) dt, where k =

λ(λ – A).
3
◦
. Solution with A = λ:
y(x)=f(x) – λ
2

x
a
(x – t)f(t) dt.
17. y(x)+A

x
a
sinh
3
[λ(x – t)]y(t) dt = f (x).
Using the formula sinh
3
β =
1
4
sinh 3β –
3
4
sinh β, we arrive at an equation of the form 2.3.18:
y(x)+


x
a

1
4
A sinh

3λ(x – t)

–
3
4
A sinh[λ(x – t)]

y(t) dt = f (x).
18. y(x)+

x
a

A
1
sinh[λ
1
(x – t)] + A
2
sinh[λ
2
(x – t)]


y(t) dt = f(x).
1
◦
. Introduce the notation
I
1
=

x
a
sinh[λ
1
(x – t)]y(t) dt, I
2
=

x
a
sinh[λ
2
(x – t)]y(t) dt,
J
1
=

x
a
cosh[λ
1
(x – t)]y(t) dt, J

2
=

x
a
cosh[λ
2
(x – t)]y(t) dt.
Successively differentiating the integral equation four times yields (the first line is the original
equation)
y + A
1
I
1
+ A
2
I
2
= f , f = f(x), (1)
y

x
+ A
1
λ
1
J
1
+ A
2

λ
2
J
2
= f

x
, (2)
y

xx
+(A
1
λ
1
+ A
2
λ
2
)y + A
1
λ
2
1
I
1
+ A
2
λ
2

2
I
2
= f

xx
, (3)
y

xxx
+(A
1
λ
1
+ A
2
λ
2
)y

x
+ A
1
λ
3
1
J
1
+ A
2

λ
3
2
J
2
= f

xxx
, (4)
y

xxxx
+(A
1
λ
1
+ A
2
λ
2
)y

xx
+(A
1
λ
3
1
+ A
2

λ
3
2
)y + A
1
λ
4
1
I
1
+ A
2
λ
4
2
I
2
= f

xxxx
. (5)
Page 137
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Eliminating I
1
and I
2
from (1), (3), and (5), we arrive at a fourth-order linear ordinary
differential equation with constant coefficients:

y

xxxx
– (λ
2
1
+ λ
2
2
– A
1
λ
1
– A
2
λ
2
)y

xx
+(λ
2
1
λ
2
2
– A
1
λ
1

λ
2
2
– A
2
λ
2
1
λ
2
)y =
f

xxxx
– (λ
2
1
+ λ
2
2
)f

xx
+ λ
2
1
λ
2
2
f.

(6)
The initial conditions can be obtained by setting x = a in (1)–(4):
y(a)=f(a), y

x
(a)=f

x
(a),
y

xx
(a)=f

xx
(a) – (A
1
λ
1
+ A
2
λ
2
)f(a),
y

xxx
(a)=f

xxx

(a) – (A
1
λ
1
+ A
2
λ
2
)f

x
(a).
(7)
On solving the differential equation (6) under conditions (7), we thus find the solution of the
integral equation.
2
◦
. Consider the characteristic equation
z
2
– (λ
2
1
+ λ
2
2
– A
1
λ
1

– A
2
λ
2
)z + λ
2
1
λ
2
2
– A
1
λ
1
λ
2
2
– A
2
λ
2
1
λ
2
= 0, (8)
whose roots, z
1
and z
2
, determine the solution structure of the integral equation.

Assume that the discriminant of equation (8) is positive:
D ≡ (A
1
λ
1
– A
2
λ
2
– λ
2
1
+ λ
2
2
)
2
+4A
1
A
2
λ
1
λ
2
>0.
In this case, the quadratic equation (8) has the real (different) roots
z
1
=

1
2
(λ
2
1
+ λ
2
2
– A
1
λ
1
– A
2
λ
2
)+
1
2
√
D, z
2
=
1
2
(λ
2
1
+ λ
2

2
– A
1
λ
1
– A
2
λ
2
) –
1
2
√
D.
Depending on the signs of z
1
and z
2
the following three cases are possible.
Case 1.Ifz
1
> 0 and z
2
> 0, then the solution of the integral equation has the form
(i = 1, 2):
y(x)=f(x)+

x
a
{B

1
sinh[µ
1
(x – t)] + B
2
sinh

µ
2
(x – t)

f(t) dt, µ
i
=
√
z
i
,
where
B
1
= A
1
λ
1
(µ
2
1
– λ
2

2
)
µ
1
(µ
2
2
– µ
2
1
)
+ A
2
λ
2
(µ
2
1
– λ
2
1
)
µ
1
(µ
2
2
– µ
2
1

)
, B
2
= A
1
λ
1
(µ
2
2
– λ
2
2
)
µ
2
(µ
2
1
– µ
2
2
)
+ A
2
λ
2
(µ
2
2

– λ
2
1
)
µ
2
(µ
2
1
– µ
2
2
)
.
Case 2.Ifz
1
< 0 and z
2
< 0, then the solution of the integral equation has the form
y(x)=f(x)+

x
a
{B
1
sin[µ
1
(x – t)] + B
2
sin


µ
2
(x – t)

f(t) dt, µ
i
=

|z
i
|,
where the coefficients B
1
and B
2
are found by solving the following system of linear algebraic
equations:
B
1
µ
1
λ
2
1
+ µ
2
1
+
B

2
µ
2
λ
2
1
+ µ
2
2
+1=0,
B
1
µ
1
λ
2
2
+ µ
2
1
+
B
2
µ
2
λ
2
2
+ µ
2

2
+1=0.
Case 3.Ifz
1
> 0 and z
2
< 0, then the solution of the integral equation has the form
y(x)=f(x)+

x
a
{B
1
sinh[µ
1
(x – t)] + B
2
sin

µ
2
(x – t)

f(t) dt, µ
i
=

|z
i
|,

where B
1
and B
2
are determined from the following system of linear algebraic equations:
B
1
µ
1
λ
2
1
– µ
2
1
+
B
2
µ
2
λ
2
1
+ µ
2
2
+1=0,
B
1
µ

1
λ
2
2
– µ
2
1
+
B
2
µ
2
λ
2
2
+ µ
2
2
+1=0.
Page 138
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
19. y(x)+

x
a

n

k=1

A
k
sinh[λ
k
(x – t)]

y(t) dt = f(x).
1
◦
. This equation can be reduced to an equation of the form 2.2.19 with the aid of the formula
sinh z =
1
2

e
z
– e
–z

. Therefore, the original integral equation can be reduced to a linear
nonhomogeneous ordinary differential equation of order 2n with constant coefficients.
2
◦
. Let us find the roots z
k
of the algebraic equation
n

k=1
λ

k
A
k
z – λ
2
k
+ 1 = 0. (1)
By reducing it to a common denominator, we arrive at the problem of determining the roots
of an nth-degree characteristic polynomial.
Assume that all z
k
are real, different, and nonzero. Let us divide the roots into two groups
z
1
>0, z
2
>0, , z
s
> 0 (positive roots);
z
s+1
<0, z
s+2
<0, , z
n
< 0 (negative roots).
Then the solution of the integral equation can be written in the form
y(x)=f(x)+

x

a

s

k=1
B
k
sinh

µ
k
(x–t)

+
n

k=s +1
C
k
sin

µ
k
(x–t)


f(t) dt, µ
k
=


|z
k
|. (2)
The coefficients B
k
and C
k
are determined from the following system of linear algebraic
equations:
s

k=0
B
k
µ
k
λ
2
m
– µ
2
k
+
n

k=s +1
C
k
µ
k

λ
2
m
+ µ
2
k
+1=0, µ
k
=

|z
k
|, m =1, , n. (3)
In the case of a nonzero root z
s
= 0, we can introduce the new constant D = B
s
µ
s
and
proceed to the limit µ
s
→ 0. As a result, the term D(x – t) appears in solution (2) instead of
B
s
sinh

µ
s
(x – t)


and the corresponding terms Dλ
–2
m
appear in system (3).
20. y(x) – A

x
a
sinh(λx)
sinh(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
sinh(λx)
sinh(λt)
f(t) dt.
21. y(x) – A

x
a
sinh(λt)
sinh(λx)
y(t) dt = f(x).
Solution:

y(x)=f(x)+A

x
a
e
A(x–t)
sinh(λt)
sinh(λx)
f(t) dt.
22. y(x) – A

x
a
sinh
k
(λx) sinh
m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A sinh
k
(λx) and h(t) = sinh
m
(µt).
Page 139
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23. y(x)+A

x
a

t sinh[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.30 with g(t)=At.
Solution:
y(x)=f(x)+
Aλ
W

x
a
t

u
1
(x)u
2
(t) – u
2
(x)u
1
(t)

f(t) dt,
where u
1
(x), u
2
(x) is a fundamental system of solutions of the second-order linear ordinary
differential equation u

xx

+ λ(Ax – λ)u = 0, and W is the Wronskian.
The functions u
1
(x) and u
2
(x) are expressed in terms of Bessel functions or modified
Bessel functions, depending on the sign of Aλ, as follows:
if Aλ > 0, then
u
1
(x)=ξ
1/2
J
1/3

2
3
√
Aλ ξ
3/2

, u
2
(x)=ξ
1/2
Y
1/3

2
3

√
Aλ ξ
3/2

,
W =3/π, ξ = x – (λ/A);
if Aλ < 0, then
u
1
(x)=ξ
1/2
I
1/3

2
3
√
–Aλ ξ
3/2

, u
2
(x)=ξ
1/2
K
1/3

2
3
√

–Aλ ξ
3/2

,
W = –
3
2
, ξ = x – (λ/A).
24. y(x)+A

x
a
x sinh[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.31 with g(x)=Ax and h(t)=1.
Solution:
y(x)=f(x)+
Aλ
W

x
a
x

u
1
(x)u
2
(t) – u
2
(x)u

1
(t)

f(t) dt,
where u
1
(x), u
2
(x) is a fundamental system of solutions of the second-order linear ordinary
differential equation u

xx
+ λ(Ax – λ)u = 0, and W is the Wronskian.
The functions u
1
(x), u
2
(x), and W are specified in 2.3.23.
25. y(x)+A

x
a
t
k
sinh
m
(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–A sinh
m
(λx) and h(t)=t

k
.
26. y(x)+A

x
a
x
k
sinh
m
(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–Ax
k
and h(t) = sinh
m
(λt).
27. y(x) –

x
a

A sinh(kx)+B – AB(x – t) sinh(kx)

y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A sinh(kx).
Solution:
y(x)=f(x)+

x
a

R(x, t)f(t) dt,
R(x, t)=[A sinh(kx)+B]
G(x)
G(t)
+
B
2
G(t)

x
t
e
B(x–s)
G(s) ds, G(x)=exp

A
k
cosh(kx)

.
Page 140
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© 1998 by CRC Press LLC
28. y(x)+

x
a

A sinh(kt)+B + AB(x – t) sinh(kt)


y(t) dt = f(x).
This is a special case of equation 2.9.8 with λ = B and g(t)=A sinh(kt).
Solution:
y(x)=f(x)+

x
a
R(x, t)f(t) dt,
R(x, t)=–[sinh(kt)+B]
G(t)
G(x)
+
B
2
G(x)

x
t
e
B(t–s)
G(s) ds, G(x)=exp

A
k
cosh(kx)

.
29. y(x)+A

∞

x
sinh

λ
√
t – x

y(t) dt = f(x).
This is a special case of equation 2.9.62 with K(x)=A sinh

λ
√
–x

.
2.3-3. Kernels Containing Hyperbolic Tangent
30. y(x) – A

x
a
tanh(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A tanh(λx) and h(t)=1.
Solution:
y(x)=f(x)+A

x
a
tanh(λx)

cosh(λx)

cosh(λt)

A/λ
f(t) dt.
31. y(x) – A

x
a
tanh(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = tanh(λt).
Solution:
y(x)=f(x)+A

x
a
tanh(λt)

cosh(λx)
cosh(λt)

A/λ
f(t) dt.
32. y(x)+A

x
a

tanh(λx) – tanh(λt)

y(t) dt = f(x).

This is a special case of equation 2.9.5 with g(x)=A tanh(λx).
Solution:
y(x)=f(x)+
1
W

x
a

Y

1
(x)Y

2
(t) – Y

2
(x)Y

1
(t)

f(t) dt,
where Y
1
(x), Y
2
(x) is a fundamental system of solutions of the second-order linear ordinary
differential equation cosh

2
(λx)Y

xx
+ AλY =0,W is the Wronskian, and the primes stand for
the differentiation with respect to the argument specified in the parentheses.
As shown in A. D. Polyanin and V. F. Zaitsev (1996), the functions Y
1
(x) and Y
2
(x) can
be represented in the form
Y
1
(x)=F

α, β,1;
e
λx
1+e
λx

, Y
2
(x)=Y
1
(x)

x
a

dξ
Y
2
1
(ξ)
, W =1,
where F (α, β, γ; z) is the hypergeometric function, in which α and β are determined from
the algebraic system α + β =1, αβ = –A/λ.
Page 141
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© 1998 by CRC Press LLC
33. y(x) – A

x
a
tanh(λx)
tanh(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
tanh(λx)
tanh(λt)
f(t) dt.
34. y(x) – A


x
a
tanh(λt)
tanh(λx)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
tanh(λt)
tanh(λx)
f(t) dt.
35. y(x) – A

x
a
tanh
k
(λx) tanh
m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A tanh
k
(λx) and h(t) = tanh
m
(µt).
36. y(x)+A


x
a
t
k
tanh
m
(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–A tanh
m
(λx) and h(t)=t
k
.
37. y(x)+A

x
a
x
k
tanh
m
(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–Ax
k
and h(t) = tanh
m
(λt).
38. y(x)+A

∞

x
tanh[λ(t – x)]y(t) dt = f (x).
This is a special case of equation 2.9.62 with K(z)=A tanh(–λz).
39. y(x)+A

∞
x
tanh

λ
√
t – x

y(t) dt = f(x).
This is a special case of equation 2.9.62 with K(z)=A tanh

λ
√
–z

.
40. y(x) –

x
a

A tanh(kx)+B – AB(x – t) tanh(kx)

y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A tanh(kx).

41. y(x)+

x
a

A tanh(kt)+B + AB(x – t) tanh(kt)

y(t) dt = f(x).
This is a special case of equation 2.9.8 with λ = B and g(t)=A tanh(kt).
2.3-4. Kernels Containing Hyperbolic Cotangent
42. y(x) – A

x
a
coth(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A coth(λx) and h(t)=1.
Solution:
y(x)=f(x)+A

x
a
coth(λx)

sinh(λx)
sinh(λt)

A/λ
f(t) dt.
Page 142
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© 1998 by CRC Press LLC
43. y(x) – A

x
a
coth(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = coth(λt).
Solution:
y(x)=f(x)+A

x
a
coth(λt)

sinh(λx)
sinh(λt)

A/λ
f(t) dt.
44. y(x) – A

x
a
coth(λt)
coth(λx)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x

a
e
A(x–t)
coth(λt)
coth(λx)
f(t) dt.
45. y(x) – A

x
a
coth(λx)
coth(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
coth(λx)
coth(λt)
f(t) dt.
46. y(x) – A

x
a
coth
k
(λx) coth

m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A coth
k
(λx) and h(t) = coth
m
(µt).
47. y(x)+A

x
a
t
k
coth
m
(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–A coth
m
(λx) and h(t)=t
k
.
48. y(x)+A

x
a
x
k
coth
m
(λt)y(t) dt = f(x).

This is a special case of equation 2.9.2 with g(x)=–Ax
k
and h(t) = coth
m
(λt).
49. y(x)+A

∞
x
coth[λ(t – x)]y(t) dt = f (x).
This is a special case of equation 2.9.62 with K(z)=A coth(–λz).
50. y(x)+A

∞
x
coth

λ
√
t – x

y(t) dt = f(x).
This is a special case of equation 2.9.62 with K(z)=A coth

λ
√
–z

.
51. y(x) –


x
a

A coth(kx)+B – AB(x – t) coth(kx)

y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A coth(kx).
52. y(x)+

x
a

A coth(kt)+B + AB(x – t) coth(kt)

y(t) dt = f(x).
This is a special case of equation 2.9.8 with λ = B and g(t)=A coth(kt).
Page 143
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© 1998 by CRC Press LLC
2.3-5. Kernels Containing Combinations of Hyperbolic Functions
53. y(x) – A

x
a
cosh
k
(λx) sinh
m
(µt)y(t) dt = f(x).

This is a special case of equation 2.9.2 with g(x)=A cosh
k
(λx) and h(t) = sinh
m
(µt).
54. y(x) –

x
a

A + B cosh(λx)+B(x – t)[λ sinh(λx) – A cosh(λx)]

y(t) dt = f(x).
This is a special case of equation 2.9.32 with b = B and g(x)=A.
55. y(x) –

x
a

A + B sinh(λx)+B(x – t)[λ cosh(λx) – A sinh(λx)]

y(t) dt = f(x).
This is a special case of equation 2.9.33 with b = B and g(x)=A.
56. y(x) – A

x
a
tanh
k
(λx) coth

m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A tanh
k
(λx) and h(t) = coth
m
(µt).
2.4. Equations Whose Kernels Contain Logarithmic
Functions
2.4-1. Kernels Containing Logarithmic Functions
1. y(x) – A

x
a
ln(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A ln(λx) and h(t)=1.
Solution:
y(x)=f(x)+A

x
a
ln(λx)e
–A(x–t)
(λx)
Ax
(λt)
At
f(t) dt.
2. y(x) – A


x
a
ln(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = ln(λt).
Solution:
y(x)=f(x)+A

x
a
ln(λt)e
–A(x–t)
(λx)
Ax
(λt)
At
f(t) dt.
3. y(x)+A

x
a
(ln x – ln t)y(t) dt = f (x).
This is a special case of equation 2.9.5 with g(x)=A ln x.
Solution:
y(x)=f(x)+
1
W

x
a


u

1
(x)u

2
(t) – u

2
(x)u

1
(t)

f(t) dt,
where the primes denote differentiation with respect to the argument specified in the paren-
theses; and u
1
(x), u
2
(x) is a fundamental system of solutions of the second-order linear
homogeneous ordinary differential equation u

xx
+ Ax
–1
u = 0, with u
1
(x) and u
2

(x) expressed
in terms of Bessel functions or modified Bessel functions, depending on the sign of A:
W =
1
π
, u
1
(x)=
√
xJ
1

2
√
Ax

, u
2
(x)=
√
xY
1

2
√
Ax

for A >0,
W = –
1

2
, u
1
(x)=
√
xI
1

2
√
–Ax

, u
2
(x)=
√
xK
1

2
√
–Ax

for A <0.
Page 144
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© 1998 by CRC Press LLC
4. y(x) – A

x

a
ln(λx)
ln(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
ln(λx)
ln(λt)
f(t) dt.
5. y(x) – A

x
a
ln(λt)
ln(λx)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
ln(λt)
ln(λx)

f(t) dt.
6. y(x) – A

x
a
ln
k
(λx)ln
m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A ln
k
(λx) and h(t)=ln
m
(µt).
7. y(x)+a

∞
x
ln(t – x)y(t) dt = f (x).
This is a special case of equation 2.9.62 with K(x)=a ln(–x).
For f(x)=
m

k=1
A
k
exp(–λ
k
x), where λ

k
> 0, a solution of the equation has the form
y(x)=
m

k=1
A
k
B
k
exp(–λ
k
x), B
k
=1–
a
λ
k
(ln λ
k
+ C),
where C = 0.5772 is the Euler constant.
8. y(x)+a

∞
x
ln
2
(t – x)y(t) dt = f (x).
This is a special case of equation 2.9.62 with K(x)=a ln

2
(–x).
For f(x)=
m

k=1
A
k
exp(–λ
k
x), where λ
k
> 0, a solution of the equation has the form
y(x)=
m

k=1
A
k
B
k
exp(–λ
k
x), B
k
=1+
a
λ
k


1
6
π
2
+ (ln λ
k
+ C)
2

,
where C = 0.5772 is the Euler constant.
2.4-2. Kernels Containing Power-Law and Logarithmic Functions
9. y(x) – A

x
a
x
k
ln
m
(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=Ax
k
and h(t)=ln
m
(λt).
10. y(x) – A

x
a

t
k
ln
m
(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A ln
m
(λx) and h(t)=t
k
.
Page 145
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© 1998 by CRC Press LLC
11. y(x) –

x
a

A ln(kx)+B – AB(x – t) ln(kx)

y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A ln(kx).
12. y(x)+

x
a

A ln(kt)+B + AB(x – t) ln(kt)

y(t) dt = f(x).

This is a special case of equation 2.9.8 with λ = B and g(t)=A ln(kt).
13. y(x)+a

∞
x
(t – x)
n
ln(t – x)y(t) dt = f (x), n =1,2,
For f(x)=
m

k=1
A
k
exp(–λ
k
x), where λ
k
> 0, a solution of the equation has the form
y(x)=
m

k=1
A
k
B
k
exp(–λ
k
x), B

k
=1+
an!
λ
n+1
k

1+
1
2
+
1
3
+ ···+
1
n
– ln λ
k
– C

,
where C = 0.5772 is the Euler constant.
14. y(x)+a

∞
x
ln(t – x)
√
t – x
y(t) dt = f(x).

This is a special case of equation 2.9.62 with K(–x)=ax
–1/2
ln x.
For f(x)=
m

k=1
A
k
exp(–λ
k
x), where λ
k
> 0, a solution of the equation has the form
y(x)=
m

k=1
A
k
B
k
exp(–λ
k
x), B
k
=1– a

π
λ

k

ln(4λ
k
)+C

,
where C = 0.5772 is the Euler constant.
2.5. Equations Whose Kernels Contain Trigonometric
Functions
2.5-1. Kernels Containing Cosine
1. y(x) – A

x
a
cos(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A cos(λx) and h(t)=1.
Solution:
y(x)=f(x)+A

x
a
cos(λx)exp

A
λ

sin(λx) – sin(λt)



f(t) dt.
2. y(x) – A

x
a
cos(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = cos(λt).
Solution:
y(x)=f(x)+A

x
a
cos(λt)exp

A
λ

sin(λx) – sin(λt)


f(t) dt.
Page 146
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
3. y(x)+A

x
a
cos[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.34 with g(t)=A. Therefore, solving this integral

equation is reduced to solving the following second-order linear nonhomogeneous ordinary
differential equation with constant coefficients:
y

xx
+ Ay

x
+ λ
2
y = f

xx
+ λ
2
f, f = f (x),
with the initial conditions
y(a)=f(a), y

x
(a)=f

x
(a) – Af(a).
1
◦
. Solution with |A| >2|λ|:
y(x)=f(x)+

x

a
R(x – t)f(t) dt,
R(x)=exp

–
1
2
Ax


A
2
2k
sinh(kx) – A cosh(kx)

, k =

1
4
A
2
– λ
2
.
2
◦
. Solution with |A| <2|λ|:
y(x)=f(x)+

x

a
R(x – t)f(t) dt,
R(x)=exp

–
1
2
Ax


A
2
2k
sin(kx) – A cos(kx)

, k =

λ
2
–
1
4
A
2
.
3
◦
. Solution with λ = ±
1
2

A:
y(x)=f(x)+

x
a
R(x – t)f(t) dt, R(x)=exp

–
1
2
Ax

1
2
A
2
x – A

.
4. y(x)+

x
a

n

k=1
A
k
cos[λ

k
(x – t)]

y(t) dt = f(x).
This integral equation is reduced to a linear nonhomogeneous ordinary differential equation
of order 2n with constant coefficients. Set
I
k
(x)=

x
a
cos[λ
k
(x – t)]y(t) dt. (1)
Differentiating (1) with respect to x twice yields
I

k
= y(x) – λ
k

x
a
sin[λ
k
(x – t)]y(t) dt,
I

k

= y

x
(x) – λ
2
k

x
a
cos[λ
k
(x – t)]y(t) dt,
(2)
where the primes stand for differentiation with respect to x. Comparing (1) and (2), we see
that
I

k
= y

x
(x) – λ
2
k
I
k
, I
k
= I
k

(x). (3)
Page 147
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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 taking into account (3) yields
y

xx
(x)+σ
n
y

x
(x) –
n

k=1
A
k
λ

2
k
I
k
= f

xx
(x), σ
n
=
n

k=1
A
k
. (5)
Eliminating the integral I
n
from (4) and (5), we obtain
y

xx
(x)+σ
n
y

x
(x)+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 followed by eliminating I
n–1
from the resulting
expression with the aid of (6) yields a similar equation whose left-hand side is a fourth-
order differential operator (acting on y) with constant coefficients plus the sum
n–2

k=1
B

k
I
k
.
Successively eliminating the terms I
n–2
, I
n–3
, using double differentiation and formula (3),
we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2n with
constant coefficients.
The initial conditions for y(x) can be obtained by setting x = a in the integral equation
and all its derivative equations.
5. y(x) – A

x
a
cos(λx)
cos(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
cos(λx)
cos(λt)
f(t) dt.

6. y(x) – A

x
a
cos(λt)
cos(λx)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
cos(λt)
cos(λx)
f(t) dt.
7. y(x) – A

x
a
cos
k
(λx) cos
m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A cos
k
(λx) and h(t) = cos
m

(µt).
8. y(x)+A

x
a
t cos[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.34 with g(t)=At.
9. y(x)+A

x
a
t
k
cos
m
(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–A cos
m
(λx) and h(t)=t
k
.
Page 148
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© 1998 by CRC Press LLC
10. y(x)+A

x
a
x
k

cos
m
(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–Ax
k
and h(t) = cos
m
(λt).
11. y(x) –

x
a

A cos(kx)+B – AB(x – t) cos(kx)

y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A cos(kx).
Solution:
y(x)=f(x)+

x
a
R(x, t)f(t) dt,
R(x, t)=[A cos(kx)+B]
G(x)
G(t)
+
B
2
G(t)


x
t
e
B(x–s)
G(s) ds, G(x)=exp

A
k
sin(kx)

.
12. y(x)+

x
a

A cos(kt)+B + AB(x – t) cos(kt)

y(t) dt = f(x).
This is a special case of equation 2.9.8 with λ = B and g(t)=A cos(kt).
Solution:
y(x)=f(x)+

x
a
R(x, t)f(t) dt,
R(x, t)=–[A cos(kt)+B]
G(t)
G(x)

+
B
2
G(x)

x
t
e
B(t–s )
G(s) ds, G(x)=exp

A
k
sin(kx)

.
13. y(x)+A

∞
x
cos

λ
√
t – x

y(t) dt = f(x).
This is a special case of equation 2.9.62 with K(x)=A cos

λ

√
–x

.
2.5-2. Kernels Containing Sine
14. y(x) – A

x
a
sin(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A sin(λx) and h(t)=1.
Solution:
y(x)=f(x)+A

x
a
sin(λx)exp

A
λ

cos(λt) – cos(λx)


f(t) dt.
15. y(x) – A

x
a
sin(λt)y(t) dt = f(x).

This is a special case of equation 2.9.2 with g(x)=A and h(t) = sin(λt).
Solution:
y(x)=f(x)+A

x
a
sin(λt)exp

A
λ

cos(λt) – cos(λx)


f(t) dt.
Page 149
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
16. y(x)+A

x
a
sin[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.36 with g(t)=A.
1
◦
. Solution with λ(A + λ)>0:
y(x)=f(x) –
Aλ
k


x
a
sin[k(x – t)]f(t) dt, where k =

λ(A + λ).
2
◦
. Solution with λ(A + λ)<0:
y(x)=f(x) –
Aλ
k

x
a
sinh[k(x – t)]f(t) dt, where k =

–λ(λ + A).
3
◦
. Solution with A = –λ:
y(x)=f(x)+λ
2

x
a
(x – t)f(t) dt.
17. y(x)+A

x

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 2.5.18:
y(x)+

x
a

–
1
4
A sin[3λ(x – t)] +
3
4
A sin[λ(x – t)]

y(t) dt = f (x).
18. y(x)+

x

a

A
1
sin[λ
1
(x – t)] + A
2
sin[λ
2
(x – t)]

y(t) dt = f(x).
This equation can be solved by the same method as equation 2.3.18, by reducing it to a
fourth-order linear ordinary differential equation with constant coefficients.
Consider the characteristic equation
z
2
+(λ
2
1
+ λ
2
2
+ A
1
λ
1
+ A
2

λ
2
)z + λ
2
1
λ
2
2
+ A
1
λ
1
λ
2
2
+ A
2
λ
2
1
λ
2
= 0, (1)
whose roots, z
1
and z
2
, determine the solution structure of the integral equation.
Assume that the discriminant of equation (1) is positive:
D ≡ (A

1
λ
1
– A
2
λ
2
+ λ
2
1
– λ
2
2
)
2
+4A
1
A
2
λ
1
λ
2
>0.
In this case, the quadratic equation (1) has the real (different) roots
z
1
= –
1
2

(λ
2
1
+ λ
2
2
+ A
1
λ
1
+ A
2
λ
2
)+
1
2
√
D, z
2
= –
1
2
(λ
2
1
+ λ
2
2
+ A

1
λ
1
+ A
2
λ
2
) –
1
2
√
D.
Depending on the signs of z
1
and z
2
the following three cases are possible.
Case 1.Ifz
1
> 0 and z
2
> 0, then the solution of the integral equation has the form
(i = 1, 2):
y(x)=f(x)+

x
a
{B
1
sinh[µ

1
(x – t)] + B
2
sinh

µ
2
(x – t)

f(t) dt, µ
i
=
√
z
i
,
where the coefficients B
1
and B
2
are determined from the following system of linear algebraic
equations:
B
1
µ
1
λ
2
1
+ µ

2
1
+
B
2
µ
2
λ
2
1
+ µ
2
2
– 1=0,
B
1
µ
1
λ
2
2
+ µ
2
1
+
B
2
µ
2
λ

2
2
+ µ
2
2
– 1=0.
Page 150
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© 1998 by CRC Press LLC
Case 2.Ifz
1
< 0 and z
2
< 0, then the solution of the integral equation has the form
y(x)=f(x)+

x
a
{B
1
sin[µ
1
(x – t)] + B
2
sin

µ
2
(x – t)


f(t) dt, µ
i
=

|z
i
|,
where B
1
and B
2
are determined from the system
B
1
µ
1
λ
2
1
– µ
2
1
+
B
2
µ
2
λ
2
1

– µ
2
2
– 1=0,
B
1
µ
1
λ
2
2
– µ
2
1
+
B
2
µ
2
λ
2
2
– µ
2
2
– 1=0.
Case 3.Ifz
1
> 0 and z
2

< 0, then the solution of the integral equation has the form
y(x)=f(x)+

x
a
{B
1
sinh[µ
1
(x – t)] + B
2
sin

µ
2
(x – t)

f(t) dt, µ
i
=

|z
i
|,
where B
1
and B
2
are determined from the system
B

1
µ
1
λ
2
1
+ µ
2
1
+
B
2
µ
2
λ
2
1
– µ
2
2
– 1=0,
B
1
µ
1
λ
2
2
+ µ
2

1
+
B
2
µ
2
λ
2
2
– µ
2
2
– 1=0.
Remark. The solution of the original integral equation can be obtained from the solution
of equation 2.3.18 by performing the following change of parameters:
λ
k
→ iλ
k
, µ
k
→ iµ
k
, A
k
→ –iA
k
, B
k
→ –iB

k
, i
2
= –1(k = 1, 2).
19. y(x)+

x
a

n

k=1
A
k
sin[λ
k
(x – t)]

y(t) dt = f(x).
1
◦
. This integral equation can be reduced to a linear nonhomogeneous ordinary differential
equation of order 2n with constant coefficients. Set
I
k
(x)=

x
a
sin[λ

k
(x – t)]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, I

k
= λ
k
y(x) – λ
2
k

x
a
sin[λ
k
(x – t)]y(t) dt, (2)
where the primes stand for differentiation with respect to x. Comparing (1) and (2), we see
that
I


k
= λ
k
y(x) – λ
2
k
I
k
, I
k
= I
k
(x). (3)
With 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 taking into account (3) yields
y

xx
(x)+σ
n

y(x) –
n

k=1
A
k
λ
2
k
I
k
= f

xx
(x), σ
n
=
n

k=1
A
k
λ
k
. (5)
Page 151
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Eliminating the integral I
n

from (4) and (5), we obtain
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 followed by eliminating I
n–1
from the resulting
expression with the aid of (6) yields a similar equation whose left-hand side is a fourth-
order differential operator (acting on y) with constant coefficients plus the sum
n–2

k=1
B
k
I
k
.
Successively eliminating the terms I
n–2
, I
n–3
, using double differentiation and formula (3),
we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2n with
constant coefficients.
The initial conditions for y(x) can be obtained by setting x = a in the integral equation
and all its derivative equations.
2
◦
. Let us find the roots z
k
of the algebraic equation
n

k=1

λ
k
A
k
z + λ
2
k
+ 1 = 0. (7)
By reducing it to a common denominator, we arrive at the problem of determining the roots
of an nth-degree characteristic polynomial.
Assume that all z
k
are real, different, and nonzero. Let us divide the roots into two groups
z
1
>0, z
2
>0, , z
s
> 0 (positive roots);
z
s+1
<0, z
s+2
<0, , z
n
< 0 (negative roots).
Then the solution of the integral equation can be written in the form
y(x)=f(x)+


x
a

s

k=1
B
k
sinh

µ
k
(x–t)

+
n

k=s +1
C
k
sin

µ
k
(x–t)


f(t) dt, µ
k
=


|z
k
|. (8)
The coefficients B
k
and C
k
are determined from the following system of linear algebraic
equations:
s

k=0
B
k
µ
k
λ
2
m
+ µ
2
k
+
n

k=s +1
C
k
µ

k
λ
2
m
– µ
2
k
– 1=0, µ
k
=

|z
k
| m =1,2, , n. (9)
In the case of a nonzero root z
s
= 0, we can introduce the new constant D = B
s
µ
s
and
proceed to the limit µ
s
→ 0. As a result, the term D(x – t) appears in solution (8) instead of
B
s
sinh

µ
s

(x – t)

and the corresponding terms Dλ
–2
m
appear in system (9).
20. y(x) – A

x
a
sin(λx)
sin(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
sin(λx)
sin(λt)
f(t) dt.
21. y(x) – A

x
a
sin(λt)
sin(λx)
y(t) dt = f(x).

Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
sin(λt)
sin(λx)
f(t) dt.
Page 152
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© 1998 by CRC Press LLC
22. y(x) – A

x
a
sin
k
(λx) sin
m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A sin
k
(λx) and h(t) = sin
m
(µt).
23. y(x)+A

x

a
t sin[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.36 with g(t)=At.
Solution:
y(x)=f(x)+
Aλ
W

x
a
t

u
1
(x)u
2
(t) – u
2
(x)u
1
(t)

f(t) dt,
where u
1
(x), u
2
(x) is a fundamental system of solutions of the second-order linear ordinary
differential equation u


xx
+ λ(Ax + λ)u = 0, and W is the Wronskian.
Depending on the sign of Aλ, the functions u
1
(x) and u
2
(x) are expressed in terms of
Bessel functions or modified Bessel functions as follows:
if Aλ > 0, then
u
1
(x)=ξ
1/2
J
1/3

2
3
√
Aλ ξ
3/2

, u
2
(x)=ξ
1/2
Y
1/3

2

3
√
Aλ ξ
3/2

,
W =3/π, ξ = x +(λ/A);
if Aλ < 0, then
u
1
(x)=ξ
1/2
I
1/3

2
3
√
–Aλ ξ
3/2

, u
2
(x)=ξ
1/2
K
1/3

2
3

√
–Aλ ξ
3/2

,
W = –
3
2
, ξ = x +(λ/A).
24. y(x)+A

x
a
x sin[λ(x – t)]y(t) dt = f (x).
This is a special case of equation 2.9.37 with g(x)=Ax and h(t)=1.
Solution:
y(x)=f(x)+
Aλ
W

x
a
x

u
1
(x)u
2
(t) – u
2

(x)u
1
(t)

f(t) dt,
where u
1
(x), u
2
(x) is a fundamental system of solutions of the second-order linear ordinary
differential equation u

xx
+ λ(Ax + λ)u = 0, and W is the Wronskian.
The functions u
1
(x), u
2
(x), and W are specified in 2.5.23.
25. y(x)+A

x
a
t
k
sin
m
(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–A sin
m

(λx) and h(t)=t
k
.
26. y(x)+A

x
a
x
k
sin
m
(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–Ax
k
and h(t) = sin
m
(λt).
27. y(x) –

x
a

A sin(kx)+B – AB(x – t) sin(kx)

y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A sin(kx).
Solution:
y(x)=f(x)+

x

a
R(x, t)f(t) dt,
R(x, t)=[A sin(kx)+B]
G(x)
G(t)
+
B
2
G(t)

x
t
e
B(x–s)
G(s) ds, G(x)=exp

–
A
k
cos(kx)

.
Page 153
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© 1998 by CRC Press LLC
28. y(x)+

x
a


A sin(kt)+B + AB(x – t) sin(kt)

y(t) dt = f(x).
This is a special case of equation 2.9.8 with λ = B and g(t)=A sin(kt).
Solution:
y(x)=f(x)+

x
a
R(x, t)f(t) dt,
R(x, t)=–[A sin(kt)+B]
G(t)
G(x)
+
B
2
G(x)

x
t
e
B(t–s )
G(s) ds, G(x)=exp

–
A
k
cos(kx)

.

29. y(x)+A

∞
x
sin

λ
√
t – x

y(t) dt = f(x).
This is a special case of equation 2.9.62 with K(x)=A sin

λ
√
–x

.
2.5-3. Kernels Containing Tangent
30. y(x) – A

x
a
tan(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A tan(λx) and h(t)=1.
Solution:
y(x)=f(x)+A

x
a

tan(λx)



cos(λt)
cos(λx)



A/λ
f(t) dt.
31. y(x) – A

x
a
tan(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = tan(λt).
Solution:
y(x)=f(x)+A

x
a
tanh(λt)



cos(λt)
cos(λx)




A/λ
f(t) dt.
32. y(x)+A

x
a

tan(λx) – tan(λt)

y(t) dt = f(x).
This is a special case of equation 2.9.5 with g(x)=A tan(λx).
Solution:
y(x)=f(x)+
1
W

x
a

Y

1
(x)Y

2
(t) – Y

2
(x)Y


1
(t)

f(t) dt,
where Y
1
(x), Y
2
(x) is a fundamental system of solutions of the second-order linear ordinary
differential equation cos
2
(λx)Y

xx
+ AλY =0,W is the Wronskian, and the primes stand for
the differentiation with respect to the argument specified in the parentheses.
As shown in A. D. Polyanin and V. F. Zaitsev (1995, 1996), the functions Y
1
(x) and Y
2
(x)
can be expressed via the hypergeometric function.
33. y(x) – A

x
a
tan(λx)
tan(λt)
y(t) dt = f(x).

Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
tan(λx)
tan(λt)
f(t) dt.
Page 154
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© 1998 by CRC Press LLC
34. y(x) – A

x
a
tan(λt)
tan(λx)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
tan(λt)
tan(λx)
f(t) dt.

35. y(x) – A

x
a
tan
k
(λx) tan
m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A tan
k
(λx) and h(t) = tan
m
(µt).
36. y(x)+A

x
a
t
k
tan
m
(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–A tan
m
(λx) and h(t)=t
k
.
37. y(x)+A


x
a
x
k
tan
m
(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–Ax
k
and h(t) = tan
m
(λt).
38. y(x) –

x
a

A tan(kx)+B – AB(x – t) tan(kx)

y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A tan(kx).
39. y(x)+

x
a

A tan(kt)+B + AB(x – t) tan(kt)

y(t) dt = f(x).
This is a special case of equation 2.9.8 with λ = B and g(t)=A tan(kt).

2.5-4. Kernels Containing Cotangent
40. y(x) – A

x
a
cot(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A cot(λx) and h(t)=1.
Solution:
y(x)=f(x)+A

x
a
cot(λx)



sin(λx)
sin(λt)



A/λ
f(t) dt.
41. y(x) – A

x
a
cot(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = cot(λt).
Solution:

y(x)=f(x)+A

x
a
coth(λt)



sin(λx)
sin(λt)



A/λ
f(t) dt.
42. y(x) – A

x
a
cot(λx)
cot(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
cot(λx)

cot(λt)
f(t) dt.
Page 155
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© 1998 by CRC Press LLC
43. y(x) – A

x
a
cot(λt)
cot(λx)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
cot(λt)
cot(λx)
f(t) dt.
44. y(x)+A

x
a
t
k
cot
m

(λx)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–A cot
m
(λx) and h(t)=t
k
.
45. y(x)+A

x
a
x
k
cot
m
(λt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=–Ax
k
and h(t) = cot
m
(λt).
46. y(x) –

x
a

A cot(kx)+B – AB(x – t) cot(kx)

y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A cot(kx).
47. y(x)+


x
a

A cot(kt)+B + AB(x – t) cot(kt)

y(t) dt = f(x).
This is a special case of equation 2.9.8 with λ = B and g(t)=A cot(kt).
2.5-5. Kernels Containing Combinations of Trigonometric Functions
48. y(x) – A

x
a
cos
k
(λx) sin
m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A cos
k
(λx) and h(t) = sin
m
(µt).
49. y(x) –

x
a

A + B cos(λx) – B(x – t)[λ sin(λx)+A cos(λx)]


y(t) dt = f(x).
This is a special case of equation 2.9.38 with b = B and g(x)=A.
50. y(x) –

x
a

A + B sin(λx)+B(x – t)[λ cos(λx) – A sin(λx)]

y(t) dt = f(x).
This is a special case of equation 2.9.39 with b = B and g(x)=A.
51. y(x) – A

x
a
tan
k
(λx) cot
m
(µt)y(t) dt = f(x).
This is a special case of equation 2.9.2 with g(x)=A tan
k
(λx) and h(t) = cot
m
(µt).
2.6. Equations Whose Kernels Contain Inverse
Trigonometric Functions
2.6-1. Kernels Containing Arccosine
1. y(x) – A


x
a
arccos(λx)y(t) dt = f (x).
This is a special case of equation 2.9.2 with g(x)=A arccos(λx) and h(t)=1.
Page 156
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© 1998 by CRC Press LLC
2. y(x) – A

x
a
arccos(λt)y(t) dt = f (x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = arccos(λt).
3. y(x) – A

x
a
arccos(λx)
arccos(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
arccos(λx)
arccos(λt)
f(t) dt.

4. y(x) – A

x
a
arccos(λt)
arccos(λx)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
arccos(λt)
arccos(λx)
f(t) dt.
5. y(x) –

x
a

A arccos(kx)+B – AB(x – t) arccos(kx)

y(t) dt = f(x).
This is a special case of equation 2.9.7 with λ = B and g(x)=A arccos(kx).
6. y(x)+

x
a


A arccos(kt)+B + AB(x – t) arccos(kt)

y(t) dt = f(x).
This is a special case of equation 2.9.8 with λ = B and g(t)=A arccos(kt).
2.6-2. Kernels Containing Arcsine
7. y(x) – A

x
a
arcsin(λx)y(t) dt = f (x).
This is a special case of equation 2.9.2 with g(x)=A arcsin(λx) and h(t)=1.
8. y(x) – A

x
a
arcsin(λt)y(t) dt = f (x).
This is a special case of equation 2.9.2 with g(x)=A and h(t) = arcsin(λt).
9. y(x) – A

x
a
arcsin(λx)
arcsin(λt)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a

e
A(x–t)
arcsin(λx)
arcsin(λt)
f(t) dt.
10. y(x) – A

x
a
arcsin(λt)
arcsin(λx)
y(t) dt = f(x).
Solution:
y(x)=f(x)+A

x
a
e
A(x–t)
arcsin(λt)
arcsin(λx)
f(t) dt.
Page 157
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© 1998 by CRC Press LLC