THÔNG TIN TÀI LIỆU
113. x a cot(λx) tan(µt) + cot(βx) tan(γt) y(t) dt = f(x). This is a special case of equation 1.9.15 with g 1 (x) = cot(λx), h 1 (t) = tan(µt), g 2 (x) = cot(βx), and h 2 (t) = tan(γt). 114. x a tan(λx) tan(µt) + cot(βx) cot(γt) y(t) dt = f(x). This is a special case of equation 1.9.15 with g 1 (x) = tan(λx), h 1 (t) = tan(µt), g 2 (x) = cot(βx), and h 2 (t) = cot(γt). 115. x a A tan β (λx)+B cot γ (µt) y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x)=A tan β (λx) and h(t)=B cot γ (µt). 116. x a A cot β (λx)+B tan γ (µt) y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x)=A cot β (λx) and h(t)=B tan γ (µt). 117. x a Ax λ tan µ t + Bt β cot γ x y(t) dt = f(x). This is a special case of equation 1.9.15 with g 1 (x)=Ax λ , h 1 (t) = tan µ t, g 2 (x)=B cot γ x, and h 2 (t)=t β . 118. x a Ax λ cot µ t + Bt β tan γ x y(t) dt = f(x). This is a special case of equation 1.9.15 with g 1 (x)=Ax λ , h 1 (t) = cot µ t, g 2 (x)=B tan γ x, and h 2 (t)=t β . 1.6. Equations Whose Kernels Contain Inverse Trigonometric Functions 1.6-1. Kernels Containing Arccosine 1. x a arccos(λx) – arccos(λt) y(t) dt = f(x). This is a special case of equation 1.9.2 with g(x) = arccos(λx). Solution: y(x)=– 1 λ d dx √ 1 – λ 2 x 2 f x (x) . 2. x a A arccos(λx)+B arccos(λt) y(t) dt = f(x). For B = –A, see equation 1.6.1. This is a special case of equation 1.9.4 with g(x) = arccos(λx). Solution: y(x)= 1 A + B d dx arccos(λx) – A A+B x a arccos(λt) – B A+B f t (t) dt . Page 59 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 3. x a A arccos(λx)+B arccos(µt)+C y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x) = arccos(λx) and h(t)=B arccos(µt)+C. 4. x a arccos(λx) – arccos(λt) n y(t) dt = f(x), n =1,2, The right-hand side of the equation is assumed to satisfy the conditions f (a)=f x (a)=···= f (n) x (a)=0. Solution: y(x)= (–1) n λ n n! √ 1 – λ 2 x 2 √ 1 – λ 2 x 2 d dx n+1 f(x). 5. x a arccos(λt) – arccos(λx) y(t) dt = f(x). This is a special case of equation 1.9.38 with g(x)=1– arccos(λx). Solution: y(x)= 2 π ϕ(x) 1 ϕ(x) d dx 2 x a ϕ(t)f(t) dt √ arccos(λt) – arccos(λx) , ϕ(x)= 1 √ 1 – λ 2 x 2 . 6. x a y(t) dt √ arccos(λt) – arccos(λx) = f(x). Solution: y(x)= λ π d dx x a ϕ(t)f(t) dt √ arccos(λt) – arccos(λx) , ϕ(x)= 1 √ 1 – λ 2 x 2 . 7. x a arccos(λt) – arccos(λx) µ y(t) dt = f(x), 0 < µ <1. Solution: y(x)=kϕ(x) 1 ϕ(x) d dx 2 x a ϕ(t)f(t) dt [arccos(λt) – arccos(λx)] µ , ϕ(x)= 1 √ 1 – λ 2 x 2 , k = sin(πµ) πµ . 8. x a arccos µ (λx) – arccos µ (λt) y(t) dt = f(x). This is a special case of equation 1.9.2 with g(x) = arccos µ (λx). Solution: y(x)=– 1 λµ d dx f x (x) √ 1 – λ 2 x 2 arccos µ–1 (λx) . 9. x a y(t) dt arccos(λt) – arccos(λx) µ = f(x), 0 < µ <1. Solution: y(x)= λ sin(πµ) π d dx x a ϕ(t)f(t) dt [arccos(λt) – arccos(λx)] 1–µ , ϕ(x)= 1 √ 1 – λ 2 x 2 . 10. x a A arccos β (λx)+B arccos γ (µt)+C y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x)=A arccos β (λx) and h(t)=B arccos γ (µt)+C. Page 60 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 1.6-2. Kernels Containing Arcsine 11. x a arcsin(λx) – arcsin(λt) y(t) dt = f(x). This is a special case of equation 1.9.2 with g(x) = arcsin(λx). Solution: y(x)= 1 λ d dx √ 1 – λ 2 x 2 f x (x) . 12. x a A arcsin(λx)+B arcsin(λt) y(t) dt = f(x). For B = –A, see equation 1.6.11. This is a special case of equation 1.9.4 with g(x) = arcsin(λx). Solution: y(x)= sign x A + B d dx arcsin(λx) – A A+B x a arcsin(λt) – B A+B f t (t) dt . 13. x a A arcsin(λx)+B arcsin(µt)+C y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x)=A arcsin(λx) and h(t)=B arcsin(µt)+C. 14. x a arcsin(λx) – arcsin(λt) n y(t) dt = f(x), n =1,2, The right-hand side of the equation is assumed to satisfy the conditions f (a)=f x (a)=···= f (n) x (a)=0. Solution: y(x)= 1 λ n n! √ 1 – λ 2 x 2 √ 1 – λ 2 x 2 d dx n+1 f(x). 15. x a arcsin(λx) – arcsin(λt) y(t) dt = f(x). Solution: y(x)= 2 π ϕ(x) 1 ϕ(x) d dx 2 x a ϕ(t)f(t) dt √ arcsin(λx) – arcsin(λt) , ϕ(x)= 1 √ 1 – λ 2 x 2 . 16. x a y(t) dt √ arcsin(λx) – arcsin(λt) = f(x). Solution: y(x)= λ π d dx x a ϕ(t)f(t) dt √ arcsin(λx) – arcsin(λt) , ϕ(x)= 1 √ 1 – λ 2 x 2 . 17. x a arcsin(λx) – arcsin(λt) µ y(t) dt = f(x), 0 < µ <1. Solution: y(x)=kϕ(x) 1 ϕ(x) d dx 2 x a ϕ(t)f(t) dt [arcsin(λx) – arcsin(λt)] µ , ϕ(x)= 1 √ 1 – λ 2 x 2 , k = sin(πµ) πµ . Page 61 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 18. x a arcsin µ (λx) – arcsin µ (λt) y(t) dt = f(x). This is a special case of equation 1.9.2 with g(x) = arcsin µ (λx). Solution: y(x)= 1 λµ d dx f x (x) √ 1 – λ 2 x 2 arcsin µ–1 (λx) . 19. x a y(t) dt arcsin(λx) – arcsin(λt) µ = f(x), 0 < µ <1. Solution: y(x)= λ sin(πµ) π d dx x a ϕ(t)f(t) dt [arcsin(λx) – arcsin(λt)] 1–µ , ϕ(x)= 1 √ 1 – λ 2 x 2 . 20. x a A arcsin β (λx)+B arcsin γ (µt)+C y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x)=A arcsin β (λx) and h(t)=B arcsin γ (µt)+C. 1.6-3. Kernels Containing Arctangent 21. x a arctan(λx) – arctan(λt) y(t) dt = f(x). This is a special case of equation 1.9.2 with g(x) = arctan(λx). Solution: y(x)= 1 λ d dx (1 + λ 2 x 2 ) f x (x) . 22. x a A arctan(λx)+B arctan(λt) y(t) dt = f(x). For B = –A, see equation 1.6.21. This is a special case of equation 1.9.4 with g(x) = arctan(λx). Solution: y(x)= sign x A + B d dx arctan(λx) – A A+B x a arctan(λt) – B A+B f t (t) dt . 23. x a A arctan(λx)+B arctan(µt)+C y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x)=A arctan(λx) and h(t)=B arctan(µt)+C. 24. x a arctan(λx) – arctan(λt) n y(t) dt = f(x), n =1,2, The right-hand side of the equation is assumed to satisfy the conditions f (a)=f x (a)=···= f (n) x (a)=0. Solution: y(x)= 1 λ n n!(1+λ 2 x 2 ) (1 + λ 2 x 2 ) d dx n+1 f(x). Page 62 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 25. x a arctan(λx) – arctan(λt) y(t) dt = f(x). Solution: y(x)= 2 π ϕ(x) 1 ϕ(x) d dx 2 x a ϕ(t)f(t) dt √ arctan(λx) – arctan(λt) , ϕ(x)= 1 1+λ 2 x 2 . 26. x a y(t) dt √ arctan(λx) – arctan(λt) = f(x). Solution: y(x)= λ π d dx x a ϕ(t)f(t) dt √ arctan(λx) – arctan(λt) , ϕ(x)= 1 1+λ 2 x 2 . 27. x a √ t arctan x – t t y(t) dt = f(x). The equation can be rewritten in terms of the Gaussian hypergeometric function in the form x a (x – t) γ–1 F α, β, γ;1– x t y(t) dt = f (x), where α = 1 2 , β =1, γ = 3 2 . See 1.8.86 for the solution of this equation. 28. x a arctan(λx) – arctan(λt) µ y(t) dt = f(x), 0 < µ <1. Solution: y(x)=kϕ(x) 1 ϕ(x) d dx 2 x a ϕ(t)f(t) dt [arctan(λx) – arctan(λt)] µ , ϕ(x)= 1 1+λ 2 x 2 , k = sin(πµ) πµ . 29. x a arctan µ (λx) – arctan µ (λt) y(t) dt = f(x). This is a special case of equation 1.9.2 with g(x) = arctan µ (λx). Solution: y(x)= 1 λµ d dx (1 + λ 2 x 2 )f x (x) arctan µ–1 (λx) . 30. x a y(t) dt arctan(λx) – arctan(λt) µ = f(x), 0 < µ <1. Solution: y(x)= λ sin(πµ) π d dx x a ϕ(t)f(t) dt [arctan(λx) – arctan(λt)] 1–µ , ϕ(x)= 1 1+λ 2 x 2 . 31. x a A arctan β (λx)+B arctan γ (µt)+C y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x)=A arctan β (λx) and h(t)=B arctan γ (µt)+C. Page 63 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 1.6-4. Kernels Containing Arccotangent 32. x a arccot(λx) – arccot(λt) y(t) dt = f(x). This is a special case of equation 1.9.2 with g(x) = arccot(λx). Solution: y(x)=– 1 λ d dx (1 + λ 2 x 2 ) f x (x) . 33. x a A arccot(λx)+B arccot(λt) y(t) dt = f(x). For B = –A, see equation 1.6.32. This is a special case of equation 1.9.4 with g(x) = arccot(λx). Solution: y(x)= 1 A + B d dx arccot(λx) – A A+B x a arccot(λt) – B A+B f t (t) dt . 34. x a A arccot(λx)+B arccot(µt)+C y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x)=A arccot(λx) and h(t)=B arccot(µt)+C. 35. x a arccot(λx) – arccot(λt) n y(t) dt = f(x), n =1,2, The right-hand side of the equation is assumed to satisfy the conditions f (a)=f x (a)=···= f (n) x (a)=0. Solution: y(x)= (–1) n λ n n!(1+λ 2 x 2 ) (1 + λ 2 x 2 ) d dx n+1 f(x). 36. x a arccot(λt) – arccot(λx) y(t) dt = f(x). Solution: y(x)= 2 π ϕ(x) 1 ϕ(x) d dx 2 x a ϕ(t)f(t) dt √ arccot(λt) – arccot(λx) , ϕ(x)= 1 1+λ 2 x 2 . 37. x a y(t) dt √ arccot(λt) – arccot(λx) = f(x). Solution: y(x)= λ π d dx x a ϕ(t)f(t) dt √ arccot(λt) – arccot(λx) , ϕ(x)= 1 1+λ 2 x 2 . 38. x a arccot(λt) – arccot(λx) µ y(t) dt = f(x), 0 < µ <1. Solution: y(x)=kϕ(x) 1 ϕ(x) d dx 2 x a ϕ(t)f(t) dt [arccot(λt) – arccot(λx)] µ , ϕ(x)= 1 1+λ 2 x 2 , k = sin(πµ) πµ . Page 64 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 39. x a arccot µ (λx) – arccot µ (λt) y(t) dt = f(x). This is a special case of equation 1.9.2 with g(x) = arccot µ (λx). Solution: y(x)=– 1 λµ d dx (1 + λ 2 x 2 )f x (x) arccot µ–1 (λx) . 40. x a y(t) dt arccot(λt) – arccot(λx) µ = f(x), 0 < µ <1. Solution: y(x)= λ sin(πµ) π d dx x a ϕ(t)f(t) dt [arccot(λt) – arccot(λx)] 1–µ , ϕ(x)= 1 1+λ 2 x 2 . 41. x a A arccot β (λx)+B arccot γ (µt)+C y(t) dt = f(x). This is a special case of equation 1.9.6 with g(x)=A arccot β (λx) and h(t)=B arccot γ (µt)+C. 1.7. Equations Whose Kernels Contain Combinations of Elementary Functions 1.7-1. Kernels Containing Exponential and Hyperbolic Functions 1. x a e µ(x–t) A 1 cosh[λ 1 (x – t)] + A 2 cosh[λ 2 (x – t)] y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.8: x a A 1 cosh[λ 1 (x – t)] + A 2 cosh[λ 2 (x – t)] w(t) dt = e –µx f(x). 2. x a e µ(x–t) cosh 2 [λ(x – t)]y(t) dt = f(x). Solution: y(x)=ϕ(x) – 2λ 2 k x a e µ(x–t) sinh[k(x – t)]ϕ(x) dt, k = λ √ 2, ϕ(x)=f x (x) – µf(x). 3. x a e µ(x–t) cosh 3 [λ(x – t)]y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.15: x a cosh 3 [λ(x – t)]w(t) dt = e –µx f(x). 4. x a e µ(x–t) cosh 4 [λ(x – t)]y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.19: x a cosh 4 [λ(x – t)]w(t) dt = e –µx f(x). Page 65 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 5. x a e µ(x–t) cosh(λx) – cosh(λt) n y(t) dt = f(x), n =1,2, Solution: y(x)= 1 λ n n! e µx sinh(λx) 1 sinh(λx) d dx n+1 F µ (x), F µ (x)=e –µx f(x). 6. x a e µ(x–t) √ cosh x – cosh ty(t) dt = f (x), f (a)=0. Solution: y(x)= 2 π e µx sinh x 1 sinh x d dx 2 x a e –µt sinh tf(t) dt √ cosh x – cosh t . 7. x a e µ(x–t) y(t) dt √ cosh x – cosh t = f(x). Solution: y(x)= 1 π e µx d dx x a e –µt sinh tf(t) dt √ cosh x – cosh t . 8. x a e µ(x–t) (cosh x – cosh t) λ y(t) dt = f(x), 0 < λ <1. The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.23: x a (cosh x – cosh t) λ w(t) dt = e –µx f(x). 9. x a Ae µ(x–t) + B cosh λ x y(t) dt = f(x). This is a special case of equation 1.9.15 with g 1 (x)=Ae µx , h 1 (t)=e –µt , g 2 (x)=B cosh λ x, and h 2 (t)=1. 10. x a Ae µ(x–t) + B cosh λ t y(t) dt = f(x). This is a special case of equation 1.9.15 with g 1 (x)=Ae µx , h 1 (t)=e –µt , g 2 (x)=B, and h 2 (t) = cosh λ t. 11. x a e µ(x–t) (cosh λ x – cosh λ t)y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.24: x a (cosh λ x – cosh λ t)w(t) dt = e –µx f(x). 12. x a e µ(x–t) A cosh λ x + B cosh λ t y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.25: x a A cosh λ x + B cosh λ t w(t) dt = e –µx f(x). Page 66 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 13. x a e µ(x–t) y(t) dt (cosh x – cosh t) λ = f(x), 0 < λ <1. Solution: y(x)= sin(πλ) π e µx d dx x a e –µt sinh tf(t) dt (cosh x – cosh t) 1–λ . 14. x a e µ(x–t) A 1 sinh[λ 1 (x – t)] + A 2 sinh[λ 2 (x – t)] y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.41: x a A 1 sinh[λ 1 (x – t)] + A 2 sinh[λ 2 (x – t)] w(t) dt = e –µx f(x). 15. x a e µ(x–t) sinh 2 [λ(x – t)]y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.43: x a sinh 2 [λ(x – t)]w(t) dt = e –µx f(x). 16. x a e µ(x–t) sinh 3 [λ(x – t)]y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.49: x a sinh 3 [λ(x – t)]w(t) dt = e –µx f(x). 17. x a e µ(x–t) sinh n [λ(x – t)]y(t) dt = f(x), n =2,3, The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.54: x a sinh n [λ(x – t)]w(t) dt = e –µx f(x). 18. x a e µ(x–t) sinh k √ x – t y(t) dt = f(x). Solution: y(x)= 2 πk e µx d 2 dx 2 x a e –µt cos k √ x – t √ x – t f(t) dt. 19. x a e µ(x–t) √ sinh x – sinh ty(t) dt = f (x). Solution: y(x)= 2 π e µx cosh x 1 cosh x d dx 2 x a e –µt cosh tf(t) dt √ sinh x – sinh t . 20. x a e µ(x–t) y(t) dt √ sinh x – sinh t = f(x). Solution: y(x)= 1 π e µx d dx x a e –µt cosh tf(t) dt √ sinh x – sinh t . Page 67 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 21. x a e µ(x–t) (sinh x – sinh t) λ y(t) dt = f(x), 0 < λ <1. The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.58: x a (sinh x – sinh t) λ w(t) dt = e –µx f(x). 22. x a e µ(x–t) (sinh λ x – sinh λ t)y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.59: x a (sinh λ x – sinh λ t)w(t) dt = e –µx f(x). 23. x a e µ(x–t) A sinh λ x + B sinh λ t y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.60: x a A sinh λ x + B sinh λ t w(t) dt = e –µx f(x). 24. x a Ae µ(x–t) + B sinh λ x y(t) dt = f(x). This is a special case of equation 1.9.15 with g 1 (x)=Ae µx , h 1 (t)=e –µt , g 2 (x)=B sinh λ x, and h 2 (t)=1. 25. x a Ae µ(x–t) + B sinh λ t y(t) dt = f(x). This is a special case of equation 1.9.15 with g 1 (x)=Ae µx , h 1 (t)=e –µt , g 2 (x)=B, and h 2 (t) = sinh λ t. 26. x a e µ(x–t) y(t) dt (sinh x – sinh t) λ = f(x), 0 < λ <1. Solution: y(x)= sin(πλ) π e µx d dx x a e –µt cosh tf(t) dt (sinh x – sinh t) 1–λ . 27. x a e µ(x–t) A tanh λ x + B tanh λ t y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.3.77: x a A tanh λ x + B tanh λ t w(t) dt = e –µx f(x). 28. x a e µ(x–t) A tanh λ x + B tanh β t + C y(t) dt = f(x). The substitution w(x)=e –µx y(x) leads to an equation of the form 1.9.6 with g(x)=A tanh λ x, g(t)=B tanh β t + C: x a A tanh λ x + B tanh β t + C w(t) dt = e –µx f(x). Page 68 © 1998 by CRC Press LLC © 1998 by CRC Press LLC [...]... Marichev (1993) x x2 – t 2 69 –1/4 √ I–1 /2 λ x2 – t2 y(t) dt = f (x) 0 Solution: 2 d π dx y(x) = ∞ 70 t 2 – x2 –1/4 x 0 √ cos λ x2 – t2 √ t f (t) dt x2 – t2 √ I–1 /2 λ t2 – x2 y(t) dt = f (x) x Solution: 2 d π dx y(x) = – x x2 – t 2 71 ν /2 ∞ x √ cos λ t2 – x2 √ t f (t) dt t2 – x2 √ Iν λ x2 – t2 y(t) dt = f (x), –1 < ν < 0 0 Solution: y(x) = λ d dx x t x2 – t2 –(ν+1) /2 √ J–ν–1 λ x2 – t2 f (t) dt 0 • Reference:... = x x2 – t 2 29 –1/4 2 λ x n 2 x–t n–ν 2 2 √ In–ν 2 λ x – t ft(n) (t) dt a √ J–1 /2 λ x2 – t2 y(t) dt = f (x) 0 Solution: y(x) = 2 d π dx x 0 √ cosh λ x2 – t2 √ t f (t) dt x2 – t2 • Reference: S G Samko, A A Kilbas, and O I Marichev (1993) © 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 82 ∞ 30 t 2 – x2 –1/4 √ J–1 /2 λ t2 – x2 y(t) dt = f (x) x Solution: 2 d π dx y(x) = – x x2 – t 2 31 ν /2 ∞ x... y(x) = π 4 2 3 d2 + 2 dx2 x (x – t)3 /2 J3 /2 [λ(x – t)] f (t) dt a x (x – t)3 /2 J1 /2 [λ(x – t)]y(t) dt = f (x) 10 a Solution: x y(x) = g(t) dt, a where g(t) = π d2 + 2 2 dt2 8λ 4 t (t – τ )3 /2 J3 /2 [λ(t – τ )] f (τ ) dτ a x (x – t)3 /2 J3 /2 [λ(x – t)]y(t) dt = f (x) 11 a Solution: √ π y(x) = 3 /2 5 /2 2 λ 3 d2 + 2 dx2 x sin[λ(x – t)] f (t) dt a x (x – t)5 /2 J3 /2 [λ(x – t)]y(t) dt = f (x) 12 a Solution:... Page 85 x (x – t)1 /2 I1 /2 [λ(x – t)]y(t) dt = f (x) 49 a Solution: y(x) = d2 – 2 dx2 π 4 2 3 x (x – t)3 /2 I3 /2 [λ(x – t)] f (t) dt a x (x – t)3 /2 I1 /2 [λ(x – t)]y(t) dt = f (x) 50 a Solution: x y(x) = g(t) dt, a where d2 – 2 dt2 π g(t) = 8 2 4 t (t – τ )3 /2 I3 /2 [λ(t – τ )] f (τ ) dτ a x (x – t)3 /2 I3 /2 [λ(x – t)]y(t) dt = f (x) 51 a Solution: √ y(x) = π 3 /2 λ5 /2 2 3 d2 – 2 dx2 x sinh[λ(x – t)]... x (x – t)5 /2 I3 /2 [λ(x – t)]y(t) dt = f (x) 52 a Solution: x y(x) = g(t) dt, a where g(t) = x x–t 53 2n–1 2 d2 – 2 dt2 π 128 λ4 6 t (t – τ )5 /2 I5 /2 [λ(t – τ )] f (τ ) dτ a I 2n–1 [λ(x – t)]y(t) dt = f (x), a n = 2, 3, 2 Solution: √ y(x) = √ 2n+1 2 2 π (2n – 2) !! n d2 – 2 dx2 x sinh[λ(x – t)] f (t) dt a x [Iν (λx) – Iν (λt)]y(t) dt = f (x) 54 a This is a special case of equation 1.9 .2 with g(x)... Solution: y(x) = x x–t 6 2 1 3λ3 4 d2 + 2 dx2 x (x – t )2 J2 [λ(x – t)] f (t) dt a J1 [λ(x – t)]y(t) dt = f (x) a Solution: x y(x) = g(t) dt, a where g(t) = x x–t 7 n 1 9λ3 5 d2 + 2 dt2 t t–τ 2 J2 [λ(t – τ )] f (τ ) dτ a Jn [λ(x – t)]y(t) dt = f (x), n = 0, 1, 2, a Solution: y(x) = A d2 + 2 dx2 A= 2n +2 x (x – t)n+1 Jn+1 [λ(x – t)] f (t) dt, 2 λ a 2n+1 n! (n + 1)! (2n)! (2n + 2) ! If the right-hand... – x x2 – t 2 31 ν /2 ∞ x √ cosh λ t2 – x2 √ t f (t) dt t2 – x2 √ Jν λ x2 – t2 y(t) dt = f (x), –1 < ν < 0 0 Solution: y(x) = λ d dx x –(ν+1) /2 t x2 – t2 √ I–ν–1 λ x2 – t2 f (t) dt 0 • Reference: S G Samko, A A Kilbas, and O I Marichev (1993) ∞ 32 t 2 – x2 ν /2 √ Jν λ t2 – x2 y(t) dt = f (x), –1 < ν < 0 x Solution: y(x) = –λ d dx ∞ t t2 – x2 –(ν+1) /2 √ I–ν–1 λ t2 – x2 f (t) dt x • Reference: S G Samko,... Solution: y(x) = 1 3λ3 4 d2 – 2 dx2 x (x – t )2 I2 [λ(x – t)] f (t) dt a x (x – t )2 I1 [λ(x – t)]y(t) dt = f (x) 46 a Solution: x y(x) = g(t) dt, a where g(t) = 1 9λ3 5 d2 – 2 dt2 t t–τ 2 I2 [λ(t – τ )] f (τ ) dτ a x (x – t)n In [λ(x – t)]y(t) dt = f (x), 47 n = 0, 1, 2, a Solution: y(x) = A 2n +2 d2 – 2 dx2 x (x – t)n+1 In+1 [λ(x – t)] f (t) dt, a 2n+1 2 λ A= n! (n + 1)! (2n)! (2n + 2) ! If the right-hand... dt, a where π g(t) = 128 λ4 d2 + 2 dt2 6 t (t – τ )5 /2 J5 /2 [λ(t – τ )] f (τ ) dτ a © 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 79 x (x – t) 13 2n–1 2 J 2n–1 [λ(x – t)]y(t) dt = f (x), a n = 2, 3, 2 Solution: √ y(x) = √ 2 2n+1 2 n d2 + 2 dx2 π (2n – 2) !! x sin[λ(x – t)] f (t) dt a x [Jν (λx) – Jν (λt)]y(t) dt = f (x) 14 a This is a special case of equation 1.9 .2 with g(x) = Jν (λx) d... t)3/4 J3 /2 λ x – t y(t) dt = f (x) a Solution: x 25 x 23 /2 d3 y(x) = √ 3 /2 3 dx πλ a √ cosh λ x – t √ f (t) dt x–t √ (x – t)n /2 Jn λ x – t y(t) dt = f (x), n = 0, 1, 2, a Solution: y(x) = x x–t 26 2n–3 4 n 2 λ a n = 1, 2, 2 Solution: 1 y(x) = √ π x √ I0 λ x – t f (t) dt √ J 2n–3 λ x – t y(t) dt = f (x), a 27 x dn +2 dxn +2 2n–3 2 2 λ dn dxn x a √ cosh λ x – t √ f (t) dt x–t √ (x – t)–1/4 J–1 /2 λ x . t n J n [λ(x – t)]y(t) dt = f(x), n =0,1 ,2, Solution: y(x)=A d 2 dx 2 + λ 2 2n +2 x a (x – t) n+1 J 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. dt, where g(t)= 1 λ d 2 dt 2 + λ 2 3 t a (t – τ) J 1 [λ(t – τ)] f(τ) dτ. 5. x a (x – t)J 1 [λ(x – t)]y(t) dt = f(x). Solution: y(x)= 1 3λ 3 d 2 dx 2 + λ 2 4 x a (x – t) 2 J 2 [λ(x – t)] f(t). f(a)=f x (a)=···= f (2n+1) x (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+1 J 2n+1 [λ(x – t)]F(t) dt, F (t)= d 2 dt 2 + λ 2 2n +2 f(t)
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