Equations Whose Kernels Contain Special Functions

Linear Equations of the First Kind With Variable Limit of Integration

1.8. Equations Whose Kernels Contain Special Functions

96.

x a

Atanh^β(λx) +Bcos^γ(µt) +C

y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =Atanh^β(λx) andh(t) =Bcos^γ(µt) +C.

97.

x a

Atanh^β(λx) +Bsin^γ(µt) +C

y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =Atanh^β(λx) andh(t) =Bsin^γ(µt) +C.

1.7-6. Kernels Containing Logarithmic and Trigonometric Functions

98.

x a

Acos^β(λx) +Bln^γ(µt) +C

y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =Acos^β(λx) andh(t) =Bln^γ(µt) +C.

99.

x a

Acos^β(λt) +Bln^γ(µx) +C

y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =Bln^γ(µx) +Candh(t) =Acos^β(λt).

100.

_x

a

Asin^β(λx) +Bln^γ(µt) +C

y(t)dt =f(x).

This is a special case of equation 1.9.6 withg(x) =Asin^β(λx) andh(t) =Bln^γ(µt) +C.

101.

x a

Asin^β(λt) +Bln^γ(µx) +C

y(t)dt =f(x).

This is a special case of equation 1.9.6 withg(x) =Bln^γ(µx) andh(t) =Asin^β(λt) +C.

1.8. Equations Whose Kernels Contain Special

3.

x a

[AJ0(λx) +BJ0(λt)]y(t)dt=f(x).

ForB =–A, see equation 1.8.2. We consider the interval [a,x] in whichJ₀(λx) does not change its sign.

Solution withB≠–A:

y(x) =± 1 A+B

d dx

J₀(λx)–_A+B^A _x

a

J₀(λt)–_A+B^B f_t(t)dt

. Here the sign ofJ₀(λx) should be taken.

4.

_x

a

(x–t)J0[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = _x

a

g(t)dt, where

g(t) = 1 λ

d² dt² +λ²

3 t a

(t–τ)J1[λ(t–τ)]f(τ)dτ.

5.

_x

a

(x–t)J1[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = 1 3λ³

d² dx² +λ²

4 x a

(x–t)²J2[λ(x–t)]f(t)dt.

6.

_x

a

x–t2

J1[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = _x

a

g(t)dt, where

g(t) = 1 9λ³

d² dt² +λ²

5 t a

t–τ2

J2[λ(t–τ)]f(τ)dτ.

7.

_x

a

x–tn

Jn[λ(x–t)]y(t)dt=f(x), n= 0, 1, 2,. . .

Solution:

y(x) =A d²

dx² +λ²

2n+2 x a

(x–t)ⁿ⁺¹Jn+1[λ(x–t)]f(t)dt, A= 2

λ

2n+1 n! (n+ 1)!

(2n)! (2n+ 2)!.

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x⁽²ⁿ⁺¹⁾(a) = 0 are satisfied, then the solution of the integral equation can be written in the form

y(x) =A x

a

(x–t)²ⁿ⁺¹J2n+1[λ(x–t)]F(t)dt, F(t) = d²

dt² +λ² 2n+2

f(t)dt.

8.

x a

x–tn+1

Jn[λ(x–t)]y(t)dt=f(x), n= 0, 1, 2,. . . Solution:

y(x) = x

a

g(t)dt,

where

g(t) =A d²

dt² +λ²

2n+3 t a

(t–τ)ⁿ⁺¹J_n+1[λ(t–τ)]f(τ)dτ, A= 2

λ

2n+1 n! (n+ 1)!

(2n+ 1)! (2n+ 2)!.

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x⁽²ⁿ⁺²⁾(a) = 0 are satisfied, then the functiong(t) defining the solution can be written in the form

g(t) =A t

a

(t–τ)^n–ν–2J_n–ν–2[λ(t–τ)]F(τ)dτ, F(τ) = d²

dτ² +λ² 2n+2

f(τ).

9.

_x

a

(x–t)^1/2J_1/2[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = π 4λ²

d² dx² +λ²

3 x a

(x–t)^3/2J_3/2[λ(x–t)]f(t)dt.

10.

_x

a

(x–t)^3/2J_1/2[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = _x

a

g(t)dt, where

g(t) = π 8λ²

d² dt² +λ²

⁴ t a

(t–τ)^3/2J_3/2[λ(t–τ)]f(τ)dτ.

11.

x a

(x–t)^3/2J_3/2[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) =

√π 2^3/2λ^5/2

d² dx² +λ²

3 x a

sin[λ(x–t)]f(t)dt.

12.

_x

a

(x–t)^5/2J_3/2[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = x

a

g(t)dt, where

g(t) = π 128λ⁴

d² dt² +λ²

6 t a

(t–τ)^5/2J_5/2[λ(t–τ)]f(τ)dτ.

13.

x a

(x–t)

2n–1 2 J2n–1

2

[λ(x–t)]y(t)dt=f(x), n= 2, 3,. . .

Solution:

y(x) =

√π

√2λ²ⁿ⁺¹² (2n–2)!!

d² dx² +λ²

n _x

a

sin[λ(x–t)]f(t)dt.

14.

_x

a

[J_ν(λx)–J_ν(λt)]y(t)dt=f(x).

This is a special case of equation 1.9.2 withg(x) =J_ν(λx).

Solution: y(x) = d dx

xf_x(x) νJν(λx)–λxJν+1(λx)

.

15.

x a

[AJν(λx) +BJν(λt)]y(t)dt=f(x).

ForB =–A, see equation 1.8.14. We consider the interval [a,x] in whichJν(λx) does not change its sign.

Solution withB≠–A:

y(x) =± 1 A+B

d dx

J_ν(λx)–_A+B^A _x

a

J_ν(λt)–_A+B^B f_t(t)dt

. Here the sign ofJν(λx) should be taken.

16.

x a

[AJ_ν(λx) +BJ_µ(βt)]y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =AJ_ν(λx) andh(t) =BJ_µ(βt).

17.

_x

a

(x–t)^νJν[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) =A d²

dx² +λ² n x

a

(x–t)^n–ν–1J_n–ν–1[λ(x–t)]f(t)dt, A= 2

λ

n–1 Γ(ν+ 1)Γ(n–ν) Γ(2ν+ 1)Γ(2n–2ν–1), where–¹₂ <ν< ^n–1₂ andn= 1, 2,. . .

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x^(n–1)(a) = 0 are satisfied, then the solution of the integral equation can be written in the form

y(x) =A x

a

(x–t)^n–ν–1Jn–ν–1[λ(x–t)]F(t)dt, F(t) = d²

dt² +λ² n

f(t).

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

18.

x a

(x–t)^ν+1Jν[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = x

a

g(t)dt,

where

g(t) =A d²

dt² +λ² n _t

a

(t–τ)^n–ν–2J_n–ν–2[λ(t–τ)]f(τ)dτ, A= 2

λ

n–2 Γ(ν+ 1)Γ(n–ν–1) Γ(2ν+ 2)Γ(2n–2ν–3), where–1 <ν < ⁿ₂ –1 andn= 1, 2,. . .

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x^(n–1)(a) = 0 are satisfied, then the functiong(t) defining the solution can be written in the form

g(t) =A t

a

(t–τ)^n–ν–2Jn–ν–2[λ(t–τ)]F(τ)dτ, F(τ) = d²

dτ² +λ² n

f(τ).

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

19.

_x

a

J0

λ√ x–t

y(t)dt=f(x).

Solution:

y(x) = d² dx²

x a

I0

λ√ x–t

f(t)dt.

20.

x a

AJν

λ√ x

+BJν

λ√ t

y(t)dt=f(x).

We consider the interval [a,x] in whichJ_ν λ√

x

does not change its sign.

Solution withB≠–A:

y(x) =± 1 A+B

d dx

Jν

λ√

x–_A+B^A _x

a

Jν

λ√

t–_A+B^B f_t(t)dt

. Here the signJν

λ√ x

should be taken.

21.

x a

AJν

λ√ x

+BJµ

β√ t

y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =AJν

λ√ x

andh(t) =BJµ

β√ t

. 22.

x a

√x–t J1

λ√ x–t

y(t)dt=f(x).

Solution:

y(x) = 2 λ

d³ dx³

_x

a

I₀ λ√

x–t f(t)dt.

23.

x a

(x–t)^1/4J1/2

λ√ x–t

y(t)dt=f(x).

Solution:

y(x) = 2

πλ d² dx²

x a

cosh λ√

x–t

√x–t f(t)dt.

24.

_x

a

(x–t)^3/4J_3/2 λ√

x–t

y(t)dt=f(x).

Solution:

y(x) = 2^3/2

√π λ^3/2 d³ dx³

_x

a

cosh λ√

x–t

√x–t f(t)dt.

25.

x a

(x–t)^n/2Jn

λ√ x–t

y(t)dt=f(x), n= 0, 1, 2,. . . Solution:

y(x) = 2 λ

n dⁿ⁺² dxⁿ⁺²

x a

I0

λ√ x–t

f(t)dt.

26.

_x

a

x–t^2n–3

4 J2n–3 2

λ√ x–t

y(t)dt=f(x), n= 1, 2,. . . Solution:

y(x) = 1

√π 2

λ ^2n–3

2 dⁿ dxⁿ

_x

a

cosh λ√

x–t

√x–t f(t)dt.

27.

_x

a

(x–t)^–1/4J_–1/2 λ√

x–t

y(t)dt=f(x).

Solution:

y(x) = λ

2π d dx

x a

cosh λ√

x–t

√x–t f(t)dt.

28.

_x

a

(x–t)^ν/2Jν

λ√ x–t

y(t)dt=f(x).

Solution:

y(x) = 2 λ

n–2 dⁿ dxⁿ

_x

a

x–t^n–ν–2

2 I_n–ν–2 λ√

x–t f(t)dt, where–1 <ν <n–1,n= 1, 2,. . .

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x^(n–1)(a) = 0 are satisfied, then the solution of the integral equation can be written in the form

y(x) = 2 λ

n–2 _x

a

x–t^n–ν–2

2 I_n–ν–2 λ√

x–t

f_t⁽ⁿ⁾(t)dt.

29.

x 0

x²–t²–1/4

J_–1/2 λ√

x²–t²

y(t)dt=f(x).

Solution:

y(x) = 2λ

π d dx

x 0

tcosh λ√

x²–t²

√x²–t² f(t)dt.

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

30.

_∞

x

t²–x²–1/4

J–1/2

λ√

t²–x²

y(t)dt=f(x).

Solution:

y(x) =– 2λ

π d dx

_∞

x

t cosh λ√

t²–x²

√t²–x² f(t)dt.

31.

_x

0

x²–t²ν/2

J_ν λ√

x²–t²

y(t)dt=f(x), –1 <ν < 0.

Solution:

y(x) =λ d dx

_x

0

t

x²–t²–(ν+1)/2

I_–ν–1 λ√

x²–t² f(t)dt.

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

32.

_∞

x

t²–x²ν/2

Jν

λ√

t²–x²

y(t)dt=f(x), –1 <ν< 0.

Solution:

y(x) =–λ d dx

_∞

x

t

t²–x²–(ν+1)/2

I_–ν–1 λ√

t²–x² f(t)dt.

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

33.

x a

[At^kJν(λx) +Bx^mJµ(λt)]y(t)dt=f(x).

This is a special case of equation 1.9.15 withg1(x) =AJν(λx),h1(t) =t^k,g2(x) =Bx^m, and h₂(t) =J_µ(λt).

34.

x a

[AJ_ν²(λx) +BJ_ν²(λt)]y(t)dt=f(x).

Solution withB≠–A:

y(x) = 1 A+B

d dx

J_ν(λx)–_A+B^2A _x

a

J_ν(λt)–_A+B^2B f_t(t)dt

.

35.

_x

a

AJ_ν^k(λx) +BJ_µ^m(βt)

y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =AJ_ν^k(λx) andh(t) =BJ_µ^m(βt).

36.

x a

[Y0(λx)–Y0(λt)]y(t)dt=f(x).

Solution: y(x) =– d dx

f_x(x) λY₁(λx)

.

37.

_x

a

[Y_ν(λx)–Y_ν(λt)]y(t)dt=f(x).

Solution: y(x) = d dx

xf_x(x) νYν(λx)–λxYν+1(λx)

.

38.

x a

[AYν(λx) +BYν(λt)]y(t)dt=f(x).

ForB =–A, see equation 1.8.37. We consider the interval [a,x] in whichY_ν(λx) does not change its sign.

Solution withB≠–A:

y(x) =± 1 A+B

d dx

Y_ν(λx)–_A+B^A _x

a

Y_ν(λt)–_A+B^B f_t(t)dt

. Here the sign ofY_ν(λx) should be taken.

39.

x a

[At^kYν(λx) +Bx^mYµ(λt)]y(t)dt=f(x).

This is a special case of equation 1.9.15 withg₁(x) =AY_ν(λx),h₁(t) =t^k,g₂(x) =Bx^m, and h₂(t) =Y_µ(λt).

40.

x a

[AJν(λx)Yµ(βt) +BJν(λt)Yµ(βx)]y(t)dt=f(x).

This is a special case of equation 1.9.15 with g₁(x) = AY_ν(λx), h₁(t) = Y_µ(βt), g₂(x) = BY_µ(βx), andh₂(t) =J_ν(λt).

1.8-2. Kernels Containing Modified Bessel Functions 41.

x a

I0[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = 1 λ

d² dx² –λ²

2 x a

(x–t)I₁[λ(x–t)]f(t)dt.

42.

x a

[I0(λx) –I0(λt)]y(t)dt=f(x), f(a) =f_x(a) = 0.

Solution: y(x) = d dx

f_x(x) λI1(λx)

.

43.

x a

[AI0(λx) +BI0(λt)]y(t)dt =f(x).

ForB =–A, see equation 1.8.42. Solution withB≠–A:

y(x) =± 1 A+B

d dx

I0(λx)–_A+B^A x a

I0(λt)–_A+B^B f_t(t)dt

. Here the sign ofI_ν(λx) should be taken.

44.

_x

a

(x–t)I0[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = _x

a

g(t)dt, where

g(t) = 1 λ

d² dt² –λ²

3 t a

(t–τ)I1[λ(t–τ)]f(τ)dτ.

45.

x a

(x–t)I1[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = 1 3λ³

d² dx² –λ²

4 _x

a

(x–t)²I₂[λ(x–t)]f(t)dt.

46.

_x

a

(x–t)²I1[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = x

a

g(t)dt, where

g(t) = 1 9λ³

d² dt² –λ²

5 t a

t–τ2

I₂[λ(t–τ)]f(τ)dτ.

47.

x a

(x–t)ⁿIn[λ(x–t)]y(t)dt=f(x), n= 0, 1, 2,. . .

Solution:

y(x) =A d²

dx² –λ²

2n+2 _x

a

(x–t)ⁿ⁺¹I_n+1[λ(x–t)]f(t)dt, A= 2

λ

2n+1 n! (n+ 1)!

(2n)! (2n+ 2)!.

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x⁽²ⁿ⁺¹⁾(a) = 0 are satisfied, then the solution of the integral equation can be written in the form

y(x) =A _x

a

(x–t)²ⁿ⁺¹I_2n+1[λ(x–t)]F(t)dt, F(t) = d²

dt² –λ² 2n+2

f(t).

48.

x a

(x–t)ⁿ⁺¹I_n[λ(x–t)]y(t)dt=f(x), n= 0, 1, 2,. . .

Solution:

y(x) = _x

a

g(t)dt,

where

g(t) =A d²

dt² –λ²

2n+3 t a

(t–τ)ⁿ⁺¹In+1[λ(t–τ)]f(τ)dτ, A= 2

λ

2n+1 n! (n+ 1)!

(2n+ 1)! (2n+ 2)!.

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x⁽²ⁿ⁺²⁾(a) = 0 are satisfied, then the functiong(t) defining the solution can be written in the form

g(t) =A t

a

(t–τ)^n–ν–2In–ν–2[λ(t–τ)]F(τ)dτ, F(τ) = d²

dτ² –λ² 2n+2

f(τ).

49.

x a

(x–t)^1/2I1/2[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = π 4λ²

d² dx² –λ²

3 x a

(x–t)^3/2I_3/2[λ(x–t)]f(t)dt.

50.

x a

(x–t)^3/2I_1/2[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = x

a

g(t)dt, where

g(t) = π 8λ²

d² dt² –λ²

4 t a

(t–τ)^3/2I_3/2[λ(t–τ)]f(τ)dτ.

51.

x a

(x–t)^3/2I_3/2[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) =

√π 2^3/2λ^5/2

d² dx² –λ²

3 x a

sinh[λ(x–t)]f(t)dt.

52.

x a

(x–t)^5/2I3/2[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = x

a

g(t)dt, where

g(t) = π 128λ⁴

d² dt² –λ²

6 _t

a

(t–τ)^5/2I_5/2[λ(t–τ)]f(τ)dτ.

53.

_x

a

x–t^2n–1₂ I2n–1

2

[λ(x–t)]y(t)dt=f(x), n= 2, 3,. . .

Solution:

y(x) =

√π

√2λ²ⁿ⁺¹² (2n–2)!!

d² dx² –λ²

n x a

sinh[λ(x–t)]f(t)dt.

54.

_x

a

[Iν(λx)–Iν(λt)]y(t)dt=f(x).

This is a special case of equation 1.9.2 withg(x) =I_ν(λx).

55.

x a

[AIν(λx) +BIν(λt)]y(t)dt=f(x).

Solution withB≠–A:

y(x) = 1 A+B

d dx

Iν(λx)–_A+B^A x a

Iν(λt)–_A+B^B f_t(t)dt

.

56.

x a

[AIν(λx) +BIµ(βt)]y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =AIν(λx) andh(t) =BIµ(βt).

57.

_x

a

(x–t)^νI_ν[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) =A d²

dx² –λ² n _x

a

(x–t)^n–ν–1In–ν–1[λ(x–t)]f(t)dt, A= 2

λ

n–1 Γ(ν+ 1)Γ(n–ν) Γ(2ν+ 1)Γ(2n–2ν–1), where–¹₂ <ν< ^n–1₂ andn= 1, 2,. . .

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x^(n–1)(a) = 0 are satisfied, then the solution of the integral equation can be written in the form

y(x) =A x

a

(x–t)^n–ν–1In–ν–1[λ(x–t)]F(t)dt, F(t) = d²

dt² –λ² n

f(t).

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

58.

_x

a

(x–t)^ν+1I_ν[λ(x–t)]y(t)dt=f(x).

Solution:

y(x) = _x

a

g(t)dt,

where

g(t) =A d²

dt² –λ² n _t

a

(t–τ)^n–ν–2In–ν–2[λ(t–τ)]f(τ)dτ, A= 2

λ

n–2 Γ(ν+ 1)Γ(n–ν–1) Γ(2ν+ 2)Γ(2n–2ν–3), where–1 <ν < ⁿ₂ –1 andn= 1, 2,. . .

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x^(n–1)(a) = 0 are satisfied, then the functiong(t) defining the solution can be written in the form

g(t) =A t

a

(t–τ)^n–ν–2In–ν–2[λ(t–τ)]F(τ)dτ, F(τ) = d²

dτ² –λ² n

f(τ).

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

59.

_x

a

I0

λ√ x–t

y(t)dt=f(x).

Solution:

y(x) = d² dx²

_x

a

J₀ λ√

x–t f(t)dt.

60.

x a

AIν

λ√ x

+BIν

λ√ t

y(t)dt =f(x).

Solution withB≠–A:

y(x) = 1 A+B

d dx

Iν

λ√

x–_A+B^A x a

Iν

λ√

t–_A+B^B f_t(t)dt

.

61.

_x

a

AI_ν λ√

x

+BI_µ β√

t

y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =AIν

λ√ x

andh(t) =BIµ

β√ t

. 62.

x a

√x–t I1

λ√ x–t

y(t)dt=f(x).

Solution:

y(x) = 2 λ

d³ dx³

x a

J₀ λ√

x–t f(t)dt.

63.

_x

a

(x–t)^1/4I_1/2 λ√

x–t

y(t)dt=f(x).

Solution:

y(x) = 2

πλ d² dx²

x a

cos λ√

x–t

√x–t f(t)dt.

64.

x a

(x–t)^3/4I_3/2 λ√

x–t

y(t)dt=f(x).

Solution:

y(x) = 2^3/2

√π λ^3/2 d³ dx³

_x

a

cos λ√

x–t

√x–t f(t)dt.

65.

x a

(x–t)^n/2I_n λ√

x–t

y(t)dt=f(x), n= 0, 1, 2,. . .

Solution:

y(x) = 2 λ

n dⁿ⁺² dxⁿ⁺²

x a

J₀ λ√

x–t f(t)dt.

66.

x a

x–t^2n–3

4 I2n–3 2

λ√ x–t

y(t)dt=f(x), n= 1, 2,. . .

Solution:

y(x) = 1

√π 2

λ ^2n–3

2 dⁿ dxⁿ

x a

cos λ√

x–t

√x–t f(t)dt.

67.

_x

a

(x–t)^–1/4I_–1/2 λ√

x–t

y(t)dt=f(x).

Solution:

y(x) = λ

2π d dx

_x

a

cos λ√

x–t

√x–t f(t)dt.

68.

x a

(x–t)^ν/2Iν

λ√ x–t

y(t)dt=f(x).

Solution:

y(x) = 2 λ

n–2 dⁿ dxⁿ

x a

x–t^n–ν–2

2 Jn–ν–2

λ√ x–t

f(t)dt, where–1 <ν <n–1,n= 1, 2,. . .

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(a) =f_x(a) =· · ·=f_x^(n–1)(a) = 0 are satisfied, then the solution of the integral equation can be written in the form

y(x) = 2 λ

n–2 _x

a

x–t^n–ν–2

2 J_n–ν–2 λ√

x–t

f_t⁽ⁿ⁾(t)dt.

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

69.

_x

0

x²–t²–1/4

I_–1/2 λ√

x²–t²

y(t)dt =f(x).

Solution:

y(x) = 2λ

π d dx

x 0

tcos λ√

x²–t²

√x²–t² f(t)dt.

70.

_∞

x

t²–x²–1/4

I_–1/2 λ√

t²–x²

y(t)dt=f(x).

Solution:

y(x) =– 2λ

π d dx

_∞

x

tcos λ√

t²–x²

√t²–x² f(t)dt.

71.

x 0

x²–t²ν/2

Iν

λ√

x²–t²

y(t)dt=f(x), –1 <ν < 0.

Solution:

y(x) =λ d dx

x 0

t

x²–t²–(ν+1)/2

J–ν–1

λ√ x²–t²

f(t)dt.

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

72.

_∞

x

(t²–x²)^ν/2Iν

λ√

t²–x²

y(t)dt=f(x), –1 <ν < 0.

Solution:

y(x) =–λ d dx

_∞

x

t(t²–x²)^–(ν+1)/2J–ν–1

λ√ t²–x²

f(t)dt.

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

73.

x a

[At^kIν(λx) +Bx^sIµ(λt)]y(t)dt=f(x).

This is a special case of equation 1.9.15 withg₁(x) =AIν(λx),h₁(t) =t^k,g₂(x) =Bx^s, and h2(t) =Iµ(λt).

74.

x a

[AI_ν²(λx) +BI_ν²(λt)]y(t)dt=f(x).

Solution withB≠–A:

y(x) = 1 A+B

d dx

Iν(λx)–_A+B^2A _x

a

Iν(λt)–_A+B^2B f_t(t)dt

.

75.

_x

a

[AI_ν^k(λx) +BI_µ^s(βt)]y(t)dt=f(x).

This is a special case of equation 1.9.6 withg(x) =AI_ν^k(λx) andh(t) =BI_µ^s(βt).

76.

_x

a

[K0(λx)–K0(λt)]y(t)dt =f(x).

Solution: y(x) =– d dx

f_x(x) λK1(λx)

.

77.

x a

[K_ν(λx)–K_ν(λt)]y(t)dt=f(x).

This is a special case of equation 1.9.2 withg(x) =K_ν(λx).

78.

x a

[AKν(λx) +BKν(λt)]y(t)dt=f(x).

Solution withB≠–A:

y(x) = 1 A+B

d dx

Kν(λx)–_A+B^A _x

a

Kν(λt)–_A+B^B f_t(t)dt

.

79.

_x

a

[At^kKν(λx) +Bx^sKµ(λt)]y(t)dt=f(x).

This is a special case of equation 1.9.15 withg₁(x) =AK_ν(λx),h₁(t) =t^k,g₂(x) =Bx^s, and h₂(t) =K_µ(λt).

80.

x a

[AI_ν(λx)K_µ(βt) +BI_ν(λt)K_µ(βx)]y(t)dt=f(x).

This is a special case of equation 1.9.15 with g1(x) = AIν(λx), h1(t) = Kµ(βt), g2(x) = BKµ(βx), andh2(t) =Iν(λt).

1.8-3. Kernels Containing Associated Legendre Functions

81.

x a

(x²–t²)^–µ/2P_ν^µ x t

y(t)dt=f(x), 0 <a<∞.

HereP_ν^µ(x) is the associated Legendre function (see Supplement 10).

Solution:

y(x) =x^n+µ–1 dⁿ dxⁿ

x^1–µ

x a

(x²–t²)

n+µ–2

2 t^–nP_ν^2–n–µ t

x

f(t)dt

, whereµ< 1,ν ≥–¹₂, andn= 1, 2,. . .

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

82.

x a

(x²–t²)^–µ/2P_ν^µ t x

y(t)dt=f(x), 0 <a<∞.

HereP_ν^µ(x) is the associated Legendre function (see Supplement 10).

Solution:

y(x) = dⁿ dxⁿ

_x

a

(x²–t²)^n+µ–22 P_ν^2–n–µ x t

f(t)dt, whereµ< 1,ν ≥–¹₂, andn= 1, 2,. . .

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

83.

b x

(t²–x²)^–µ/2P_ν^µ x t

y(t)dt=f(x), 0 <b<∞.

HereP_ν^µ(x) is the associated Legendre function (see Supplement 10).

Solution:

y(x) = (–1)ⁿx^n+µ–1 dⁿ dxⁿ

x^1–µ

b x

(t²–x²)

n+µ–2

2 t^–nP_ν^2–n–µ t

x

f(t)dt

, whereµ< 1,ν ≥–¹₂, andn= 1, 2,. . .

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

84.

_b

x

(t²–x²)^–µ/2P_ν^µ t x

y(t)dt=f(x), 0 <b<∞.

HereP_ν^µ(x) is the associated Legendre function (see Supplement 10).

Solution:

y(x) = (–1)ⁿ dⁿ dxⁿ

_b

x

(t²–x²)^n+µ–22 P_ν^2–n–µ x t

f(t)dt, whereµ< 1,ν ≥–¹₂, andn= 1, 2,. . .

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

1.8-4. Kernels Containing Hypergeometric Functions

85.

x s

(x–t)^b–1Φ

a,b;λ(x–t)

y(t)dt=f(x).

HereΦ(a,b;z) is the degenerate hypergeometric function (see Supplement 10).

Solution:

y(x) = dⁿ dxⁿ

_x

s

(x–t)^n–b–1 Γ(b)Γ(n–b)Φ

–a,n–b;λ(x–t) f(t)dt, where 0 <b<nandn= 1, 2,. . .

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(s) =f_x(s) =· · · =f_x^(n–1)(s) = 0 are satisfied, then the solution of the integral equation can be written in the form

y(x) = x

s

(x–t)^n–b–1 Γ(b)Γ(n–b)Φ

–a,n–b;λ(x–t)

f_t⁽ⁿ⁾(t)dt.

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

86.

_x

s

(x–t)^c–1F

a,b,c; 1– x t

y(t)dt=f(x).

HereΦ(a,b,c;z) is the Gaussian hypergeometric function (see Supplement 10).

Solution:

y(x) =x^–a dⁿ dxⁿ

x^a

x s

(x–t)^n–c–1

Γ(c)Γ(n–c)F –a,n–b,n–c; 1– t x

f(t)dt

, where 0 <c<nandn= 1, 2,. . .

If the right-hand side of the equation is differentiable sufficiently many times and the conditionsf(s) =f_x(s) =· · · =f_x^(n–1)(s) = 0 are satisfied, then the solution of the integral equation can be written in the form

y(x) = _x

s

(x–t)^n–c–1

Γ(c)Γ(n–c)F –a,–b,n–c; 1– t x

f_t⁽ⁿ⁾(t)dt.

• Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

1.9. Equations Whose Kernels Contain Arbitrary