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 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 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 © 1998 by CRC Press LLC © 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 © 1998 by CRC Press LLC © 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 © 1998 by CRC Press LLC © 1998 by CRC Press LLC [...]... Bessel functions as follows: if Aλ > 0, then √ √ u1 (x) = ξ 1/2 J1 /3 2 Aλ ξ 3/ 2 , u2 (x) = ξ 1/2 Y1 /3 2 Aλ ξ 3/ 2 , 3 3 W = 3/ π, ξ = x + (λ/A); if Aλ < 0, then u1 (x) = ξ 1/2 I1 /3 2 3 √ –Aλ ξ 3/ 2 , W = 3, 2 u2 (x) = ξ 1/2 K1 /3 2 3 √ –Aλ ξ 3/ 2 , ξ = x + (λ/A) x 24 y(x) + A x sin[λ(x – t)]y(t) dt = f (x) a This is a special case of equation 2.9 .37 with g(x) = Ax and h(t) = 1 Solution: Aλ x y(x) = f (x) + x... in terms of Bessel functions or modified Bessel functions as follows: if Aλ > 0, then u1 (x) = ξ 1/2 J1 /3 √ Aλ ξ 3/ 2 , u2 (x) = ξ 1/2 Y1 /3 W = 3/ π, ξ = x + (λ/A); 2 3 2 3 √ Aλ ξ 3/ 2 , if Aλ < 0, then u1 (x) = ξ 1/2 I1 /3 2 3 √ –Aλ ξ 3/ 2 , W = 3, 2 u2 (x) = ξ 1/2 K1 /3 2 3 √ –Aλ ξ 3/ 2 , ξ = x + (λ/A) x 23 xeµ(x–t) sin[λ(x – t)]y(t) dt = f (x) y(x) + A a Solution: y(x) = f (x) + Aλ W x xeµ(x–t) u1 (x)u2... case of equation 2.9 .36 with g(t) = A 1◦ Solution with λ(A + λ) > 0: y(x) = f (x) – x Aλ k sin[k(x – t)]f (t) dt, where k= λ(A + λ) sinh[k(x – t)]f (t) dt, where k= –λ(λ + A) a 2◦ Solution with λ(A + λ) < 0: Aλ k y(x) = f (x) – x a 3 Solution with A = –λ: x y(x) = f (x) + λ2 (x – t)f (t) dt a x 17 sin3 [λ(x – t)]y(t) dt = f (x) y(x) + A a Using the formula sin3 β = – 1 sin 3 + 4 3 4 sin β, we arrive... and Y2 (x) can be expressed via the hypergeometric function x 33 y(x) – A a tan(λx) tan(λt) y(t) dt = f (x) Solution: x eA(x–t) y(x) = f (x) + A a tan(λx) f (t) dt tan(λt) © 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 154 x 34 y(x) – A tan(λt) tan(λx) a y(t) dt = f (x) Solution: x eA(x–t) y(x) = f (x) + A a tan(λt) f (t) dt tan(λx) x 35 tank (λx) tanm (µt)y(t) dt = f (x) y(x) – A a This is a special... tanm (µt) x 36 tk tanm (λx)y(t) dt = f (x) y(x) + A a This is a special case of equation 2.9.2 with g(x) = –A tanm (λx) and h(t) = tk x 37 xk tanm (λt)y(t) dt = f (x) y(x) + A a This is a special case of equation 2.9.2 with g(x) = –Axk and h(t) = tanm (λt) x 38 y(x) – A tan(kx) + B – AB(x – t) tan(kx) y(t) dt = f (x) a This is a special case of equation 2.9.7 with λ = B and g(x) = A tan(kx) x 39 y(x)... Press LLC © 1998 by CRC Press LLC Page 1 43 2 .3- 5 Kernels Containing Combinations of Hyperbolic Functions x 53 coshk (λx) sinhm (µt)y(t) dt = f (x) y(x) – A a This is a special case of equation 2.9.2 with g(x) = A coshk (λx) and h(t) = sinhm (µt) x 54 y(x) – A + B cosh(λx) + B(x – t)[λ sinh(λx) – A cosh(λx)] y(t) dt = f (x) a This is a special case of equation 2.9 .32 with b = B and g(x) = A x 55 y(x) –... CRC Press LLC © 1998 by CRC Press LLC Page 164 x 32 eµt cot(λx)y(t) dt = f (x) y(x) – A a This is a special case of equation 2.9.2 with g(x) = A cot(λx) and h(t) = eµt x 33 eµx cot(λt)y(t) dt = f (x) y(x) – A a This is a special case of equation 2.9.2 with g(x) = Aeµx and h(t) = cot(λt) 2.7-4 Kernels Containing Hyperbolic and Logarithmic Functions x 34 coshk (λx) lnm (µt)y(t) dt = f (x) y(x) – A a... Containing Tangent x 30 y(x) – A tan(λx)y(t) dt = f (x) a This is a special case of equation 2.9.2 with g(x) = A tan(λx) and h(t) = 1 Solution: x cos(λt) A/λ y(x) = f (x) + A tan(λx) f (t) dt cos(λx) a x 31 y(x) – A tan(λt)y(t) dt = f (x) a This is a special case of equation 2.9.2 with g(x) = A and h(t) = tan(λt) Solution: x cos(λt) A/λ y(x) = f (x) + A tanh(λt) f (t) dt cos(λx) a x 32 y(x) + A tan(λx)... y(x) – A + B sinh(λx) + B(x – t)[λ cosh(λx) – A sinh(λx)] y(t) dt = f (x) a This is a special case of equation 2.9 .33 with b = B and g(x) = A x 56 tanhk (λx) cothm (µt)y(t) dt = f (x) y(x) – A a This is a special case of equation 2.9.2 with g(x) = A tanhk (λx) and h(t) = cothm (µt) 2.4 Equations Whose Kernels Contain Logarithmic Functions 2.4-1 Kernels Containing Logarithmic Functions x 1 y(x) – A ln(λx)y(t)... formula sin3 β = – 1 sin 3 + 4 3 4 sin β, we arrive at an equation of the form 2.5.18: x y(x) + a – 1 A sin [3 (x – t)] + 3 A sin[λ(x – t)] y(t) dt = f (x) 4 4 x 18 y(x) + A1 sin[λ1 (x – t)] + A2 sin[λ2 (x – t)] y(t) dt = f (x) a 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 . 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. 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 +. 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 3 – 3 4 sinh β, we arrive