T12.1. LINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1417 T12.1.1-5. Linear equations involving unknown function with rational argument. 34. y(x) – y a – x 1+bx =0. Solution: y(x)=Φ x, a – x 1 + bx , where Φ(x, z)=Φ(z, x) is any symmetric function with two arguments. 35. y(x) + y a – x 1+bx =0. Solution: y(x)=Φ x, a – x 1 + bx , where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 36. y(x) + y a – x 1+bx = f(x). The function f(x) must satisfy the condition f(x)=f a – x 1 + bx . Solution: y(x)= 1 2 f(x)+Φ x, a – x 1 + bx , where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 37. y(x) – y a – x 1+bx = f(x). Here, the function f(x) must satisfy the condition f(x)=–f a – x 1 + bx . Solution: y(x)= 1 2 f(x)+Φ x, a – x 1 + bx , where Φ(x, z)=Φ(z, x) is any symmetric function with two arguments. 38. y(x) – cy a – x 1+bx = f(x), c ≠ 1. Solution: y(x)= 1 1 – c 2 f(x)+ c 1 – c 2 f a – x 1 + bx . 39. y(x) + g(x)y a – x 1+bx = f(x). Solution: y(x)= f(x)–g(x)f(z) 1 – g(x)g(z) , z = a – x 1 + bx . 40. y(x) + cy ax – β x + b = f(x), β = a 2 + ab + b 2 . A special case of equation T12.1.2.12. 1418 FUNCTIONAL EQUATIONS 41. y(x) + cy bx + β a – x = f(x), β = a 2 + ab + b 2 . A special case of equation T12.1.2.12. 42. y(x) + g(x)y ax – β x + b = f(x), β = a 2 + ab + b 2 . A special case of equation T12.1.2.13. 43. y(x) + g(x)y bx + β a – x = f(x), β = a 2 + ab + b 2 . A special case of equation T12.1.2.13. T12.1.1-6. Linear functional equations involving y(x)andy √ a 2 – x 2 . 44. y(x) – y a 2 – x 2 =0, 0 ≤ x ≤ a. Solution: y(x)=Φ x, a 2 – x 2 , where Φ(x, z)=Φ(z, x) is any symmetric function with two arguments. 45. y(x) + y a 2 – x 2 =0, 0 ≤ x ≤ a. Solution: y(x)=Φ x, a 2 – x 2 , where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 46. y(x) + y a 2 – x 2 = b, 0 ≤ x ≤ a. Solution: y(x)= 1 2 b + Φ x, a 2 – x 2 , where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 47. y(x) + y a 2 – x 2 = f (x), 0 ≤ x ≤ a. Here, the function f(x) must satisfy the condition f(x)=f √ a 2 – x 2 . Solution: y(x)= 1 2 f(x)+Φ x, a 2 – x 2 , where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 48. y(x) – y a 2 – x 2 = f(x), 0 ≤ x ≤ a. Here, the function f(x) must satisfy the condition f(x)=–f √ a 2 – x 2 . Solution: y(x)= 1 2 f(x)+Φ x, a 2 – x 2 , where Φ(x, z)=Φ(z, x) is any symmetric function with two arguments. T12.1. LINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1419 49. y(x) + g(x)y a 2 – x 2 = f(x), 0 ≤ x ≤ a. Solution: y(x)= f(x)–g(x)f a 2 – x 2 1 – g(x)g √ a 2 – x 2 . T12.1.1-7. Linear functional equations involving y(sin x)andy(cos x). 50. y(sin x) – y(cos x) =0. Solution in implicit form: y(sin x)=Φ(sin x,cosx), where Φ(x, z)=Φ(z, x) is any symmetric function with two arguments. 51. y(sin x) + y(cos x) =0. Solution in implicit form: y(sin x)=Φ(sin x,cosx), where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 52. y(sin x) + y(cos x) = a. Solution in implicit form: y(sin x)= 1 2 a + Φ(sin x,cosx), where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 53. y(sin x) + y(cos x) = f (x). Here, the function f(x) must satisfy the condition f(x)=f π 2 – x . Solution in implicit form: y(sin x)= 1 2 f(x)+Φ(sin x,cosx), where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 54. y(sin x) – y(cos x) = f(x). Here, the function f(x) must satisfy the condition f(x)=–f π 2 – x . Solution in implicit form: y(sin x)= 1 2 f(x)+Φ(sin x,cosx), where Φ(x, z)=Φ(z, x) is any symmetric function with two arguments. 55. y(sin x) + g(x)y(cos x) = f (x). Solution in implicit form: y(sin x)= f(x)–g(x)f π 2 – x 1 – g(x)g π 2 – x . 1420 FUNCTIONAL EQUATIONS T12.1.1-8. Other equations involving unknown function with two different arguments. 56. y(x a ) – by(x) =0, a, b >0. Solution: y(x)=| ln x| p Θ ln | ln x| ln a , p = ln b ln a , where Θ(z)=Θ(z + 1) is an arbitrary periodic function with period 1. For Θ(z) ≡ const, we have a particular solution y(x)=C| ln x| p ,whereC is an arbitrary constant. 57. y(x) – y ω(x) =0, where ω ω(x) = x. Solution: y(x)=Φ x, ω(x) , where Φ(x, z)=Φ(z, x) is any symmetric function with two arguments. 58. y(x) + y ω(x) =0, where ω ω(x) = x. Solution: y(x)=Φ x, ω(x) , where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 59. y(x) + y ω(x) = b, where ω ω(x) = x. Solution: y(x)= 1 2 b + Φ x, ω(x) , where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 60. y(x) + y ω(x) = f(x), where ω ω(x) = x. Here, the function f(x) must satisfy the condition f(x)=f ω(x) . Solution: y(x)= 1 2 f(x)+Φ x, ω(x) , where Φ(x, z)=–Φ(z, x) is any antisymmetric function with two arguments. 61. y(x) – y ω(x) = f(x), where ω ω(x) = x. Here, the function f(x) must satisfy the condition f(x)=–f ω(x) . Solution: y(x)= 1 2 f(x)+Φ x, ω(x) , where Φ(x, z)=Φ(z, x) is any symmetric function with two arguments. 62. y(x) + g(x)y ω(x) = f(x), where ω ω(x) = x. Solution: y(x)= f(x)–g(x)f ω(x) 1 – g(x)g ω(x) . T12.1. LINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1421 T12.1.2. Other Linear Functional Equations T12.1.2-1. Second-order linear difference equations. 1. y n+2 + ay n+1 + by n =0. A homogeneous second-order linear difference equation defined on a discrete set of points x = 0, 1, 2, Notation adopted: y n = y(n). Let λ 1 and λ 2 be roots of the characteristic equation λ 2 + aλ + b = 0. 1 ◦ .Ifλ 1 ≠ λ 2 , the general solution of the difference equation has the form y n = y 1 λ n 1 – λ n 2 λ 1 – λ 2 – y 0 b λ n–1 1 – λ n–1 2 λ 1 – λ 2 , where y 1 and y 0 are arbitrary constants, equal to the values of y at the first two points. In the case of complex conjugate roots, one should separate the real and imaginary parts in the above solution. 2 ◦ .Ifλ 1 = λ 2 , the general solution of the difference equation is given by y n = y 1 nλ n–1 1 – y 0 b(n – 1)λ n–2 1 . 2. y n+2 + ay n+1 + by n = f n . A nonhomogeneous second-order linear difference equation definedonadiscretesetof points x = 0, 1, 2, Notation adopted: y n = y(n). Let λ 1 and λ 2 be roots of the characteristic equation λ 2 + aλ + b = 0. 1 ◦ .Ifλ 1 ≠ λ 2 , the general solution of the difference equation has the form y n = y 1 λ n 1 – λ n 2 λ 1 – λ 2 – y 0 b λ n–1 1 – λ n–1 2 λ 1 – λ 2 + n k=2 f n–k λ k–1 1 – λ k–1 2 λ 1 – λ 2 , where y 1 and y 0 are arbitrary constants, equal to the values of y at the first two points. In the case of complex conjugate roots, one should separate the real and imaginary parts in the above solution. 2 ◦ .Ifλ 1 = λ 2 , the general solution of the difference equation is given by y n = y 1 nλ n–1 1 – y 0 b(n – 1)λ n–2 1 + n k=2 f n–k (k – 1)λ k–2 1 . 3 ◦ . In boundary value problems, a finite set of points x = 0, 1, , N is often taken and the initial and final values of the unknown function, y 0 and y N , are prescribed. It is required to find the y n ≡ y(x)| x=n for 1 ≤ n ≤ N – 1. If λ 1 ≠ λ 2 , the solution is given by y n = y 0 λ N 1 λ n 2 – λ n 1 λ N 2 λ N 1 – λ N 2 + y N λ n 1 – λ n 2 λ N 1 – λ N 2 + n k=2 f n–k λ k–1 1 – λ k–1 2 λ 1 – λ 2 – λ n 1 – λ n 2 λ N 1 – λ N 2 N k=2 f N–k λ k–1 1 – λ k–1 2 λ 1 – λ 2 . For n = 1,thefirst sum is zero. 1422 FUNCTIONAL EQUATIONS 3. y(x +2) + ay(x +1) + by(x) =0. A homogeneous second-order constant-coefficient linear difference equation. Let us write out the characteristic equation: λ 2 + aλ + b = 0.(1) Consider the following cases. 1 ◦ . The roots λ 1 and λ 2 of the quadratic equation (1) are real and distinct. Then the general solution of the original finite-difference equation has the form y(x)=Θ 1 (x)λ x 1 + Θ 2 (x)λ x 2 ,(2) where Θ 1 (x)andΘ 2 (x) are arbitrary periodic functions with period 1, that is, Θ k (x)= Θ k (x + 1), k = 1, 2. For Θ k ≡ const, formula (2) gives particular solutions y(x)=C 1 λ x 1 + C 2 λ x 2 , where C 1 and C 2 are arbitrary constants. 2 ◦ . The quadratic equation (1) has equal roots: λ = λ 1 = λ 2 . In this case, the general solution of the original functional equation is given by y = Θ 1 (x)+xΘ 2 (x) λ x . 3 ◦ . In the case of complex conjugate roots, λ = ρ(cos β i sin β), the general solution of the original functional equation is expressed as y = Θ 1 (x)ρ x cos(βx)+Θ 2 (x)ρ x sin(βx), where Θ 1 (x)andΘ 2 (x) are arbitrary periodic functions with period 1. 4. y(x +2) + ay(x +1) + by(x) = f(x). A nonhomogeneous second-order constant-coefficient linear difference equation. 1 ◦ . Solution: y(x)=Y (x)+¯y(x), where Y (x) is the general solution of the corresponding homogeneous equation Y (x + 2)+ aY (x + 1)+bY (x)=0 (see the previous equation), and ¯y(x) is any particular solution of the nonhomogeneous equation. 2 ◦ .Forf (x)= n k=0 A k x n and a + b + 1 ≠ 1, the nonhomogeneous equation has a partic- ular solution of the form ¯y(x)= n k=0 B k x n ; the constants B k are found by the method of undetermined coefficients. 3 ◦ .Forf(x)= n k=1 A k exp(λ k x), the nonhomogeneous equation has a particular solution of the form ¯y(x)= n k=1 B k exp(λ k x); the constants B k are found by the method of undetermined coefficients. T12.1. LINEAR FUNCTIONAL EQUATIONS IN ONE INDEPENDENT VARIABLE 1423 4 ◦ .Forf(x)= n k=1 A k cos(λ k x), the nonhomogeneous equation has a particular solution of the form ¯y(x)= n k=1 B k cos(λ k x)+ n k=1 D k sin(λ k x); the constants B k and D k are found by the method of undetermined coefficients. 5 ◦ .Forf(x)= n k=1 A k sin(λ k x), the nonhomogeneous equation has a particular solution of the form ¯y(x)= n k=1 B k cos(λ k x)+ n k=1 D k sin(λ k x); the constants B k and D k are found by the method of undetermined coefficients. 5. y(x +2) + a(x +1)y(x +1) + bx(x +1)y(x) =0. This functional equation has particular solutions of the form y(x; λ)= ∞ 0 t x–1 e –t/λ dt,(1) where λ is a root of the square equation λ 2 + aλ + b = 0.(2) For the integral on the right-hand side of (1) to converge, the roots of equation (2) that satisfy the condition Re λ > 0 should be selected. If both roots, λ 1 and λ 2 , meet this condition, the general solution of the original functional equation is expressed as y(x)=Θ 1 (x)y(x, λ 1 )+Θ 2 (x)y(x, λ 2 ), where Θ 1 (x)andΘ 2 (x) are arbitrary periodic functions with period 1. T12.1.2-2. Linear equations involving composite functions y y(x) or y y(y(x)) . 6. y y(x) =0. Solution: y(x)= ϕ 1 (x)forx ≤ a, 0 for a ≤ x ≤ b, ϕ 2 (x)forb ≤ x, where a ≤ 0 and b ≥ 0 are arbitrary numbers; ϕ 1 (x)andϕ 2 (x) are arbitrary continuous functions satisfying the conditions ϕ 1 (a)=0, a ≤ ϕ 1 (x) ≤ b if x ≤ a; ϕ 2 (b)=0, a ≤ ϕ 2 (x) ≤ b if b ≤ x. 7. y y(x) – x =0. Babbage equation or the equation of involutory functions. It is a special case of equa- tion T12.1.2.21. 1 ◦ . Particular solutions: y 1 (x)=x, y 2 (x)=C – x, y 3 (x)= C x , y 4 (x)= C 1 – x 1 + C 2 x , where C, C 1 , C 2 are arbitrary constants. . a finite set of points x = 0, 1, , N is often taken and the initial and final values of the unknown function, y 0 and y N , are prescribed. It is required to find the y n ≡ y(x)| x=n for 1 ≤ n ≤. (x)= n k=0 A k x n and a + b + 1 ≠ 1, the nonhomogeneous equation has a partic- ular solution of the form ¯y(x)= n k=0 B k x n ; the constants B k are found by the method of undetermined coefficients. 3 ◦ .Forf(x)= n k=1 A k exp(λ k x),. solution of the corresponding homogeneous equation Y (x + 2)+ aY (x + 1)+bY (x)=0 (see the previous equation), and ¯y(x) is any particular solution of the nonhomogeneous equation. 2 ◦ .Forf (x)= n k=0 A k x n and