Handbook of mathematics for engineers and scienteists part 208 ppt

A homogeneous second-order linear difference equation defined on a discrete set of points x=0,1, 2,.. Let λ1and λ2be roots of the characteristic equation λ2+ aλ + b =0.. In the case of c

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T12.1.1-5 Linear equations involving unknown function with rational argument.

34. y(x) – y a – x

1 + bx

= 0.

Solution:

y (x) =Φx, a – x

1+ bx

, whereΦ(x, z) = Φ(z, x) is any symmetric function with two arguments.

35. y(x) + y a – x

1 + bx

= 0.

Solution:

y (x) =Φx, a – x

1+ bx

, whereΦ(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) = 12f (x) +Φx, a – x

1+ bx

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) = 12f (x) +Φx, a – x

1+ bx

38. y(x) – cy a – x

1 + bx

= f (x), c ≠ 1.

Solution:

y (x) = 1

1– c2f (x) +

c

1– c2f

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 = a2+ ab + b2

A special case of equation T12.1.2.12

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41. y(x) + cy bx + β

a – x

= f (x), β = a2+ ab + b2

42. y(x) + g(x)y ax – β

x + b

= f (x), β = a2+ ab + b2

43. y(x) + g(x)y bx + β

a – x

= f (x), β = a2+ ab + b2

T12.1.1-6 Linear functional equations involving y(x) and y a2– x2

44. y(x) – y a2– x2

= 0, 0≤x≤ a.

Solution:

y (x) =Φ x,

a2– x2

45. y(x) + y a2– x2

= 0, 0≤ x≤a.

Solution:

y (x) =Φ x,

a2– x2

46. y(x) + y a2– x2

= b, 0 ≤x≤a.

Solution:

y (x) = 12b+Φ x,

a2– x2

47. y(x) + y a2– x2

= f (x), 0 ≤x≤a.

Here, the function f (x) must satisfy the condition f (x) = f a2– x2

Solution:

y (x) = 12f (x) +Φ x,

a2– x2

48. y(x) – y a2– x2

= f (x), 0 ≤x≤a.

Here, the function f (x) must satisfy the condition f (x) = –f a2– x2

Solution:

y (x) = 12f (x) +Φ x,

a2– x2

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49. y(x) + g(x)y a2– x2

= f (x), 0 ≤x≤a.

Solution:

y (x) = f (x) – g(x)f a

2– x2

1– g(x)g a2– x2

T12.1.1-7 Linear functional equations involving y(sin x) and y(cos x).

50. y(sin x) – y(cos x) = 0.

Solution in implicit form:

y (sin x) = Φ(sin x, cos x),

whereΦ(x, z) = Φ(z, x) is any symmetric function with two arguments.

51. y(sin x) + y(cos x) = 0.

y (sin x) = Φ(sin x, cos x),

whereΦ(x, z) = –Φ(z, x) is any antisymmetric function with two arguments.

52. y(sin x) + y(cos x) = a.

y (sin x) = 12a+Φ(sin x, cos x),

53. y(sin x) + y(cos x) = f (x).

Here, the function f (x) must satisfy the condition f (x) = f π2 – x

y (sin x) = 12f (x) + Φ(sin x, cos x),

54. y(sin x) – y(cos x) = f (x).

Here, the function f (x) must satisfy the condition f (x) = –f π2 – x

y (sin x) = 12f (x) + Φ(sin x, cos x),

whereΦ(x, z) = Φ(z, x) is any symmetric function with two arguments.

55. y(sin x) + g(x)y(cos x) = f (x).

y (sin x) = f (x) – g(x)f

π

2 – x

1– g(x)g π2 – x

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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|pΘ ln|ln x|

ln a

, p= ln b

ln a,

whereΘ(z) = Θ(z +1) is an arbitrary periodic function with period 1

ForΘ(z)≡const, we have a particular solution y(x) = C|ln x|p , where C is an arbitrary

constant

57. y(x) – y ω(x)

= 0, where ω ω(x)

= x.

Solution:

y (x) =Φ x , ω(x)

58. y(x) + y ω(x)

= 0, where ω ω(x)

= x.

Solution:

y (x) =Φ x , ω(x)

59. y(x) + y ω(x)

= b, where ω ω(x)

= x.

Solution:

y (x) = 12b+Φ x , ω(x)

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) = 12f (x) +Φ x , ω(x)

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) = 12f (x) +Φ x , ω(x)

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)

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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 λ1and λ2be roots of the characteristic equation

λ2+ aλ + b =0

1◦ If λ1≠λ2, the general solution of the difference equation has the form

y n = y1λ

n

1 – λ n2

λ1– λ2 – y0b

λ n–1

1 – λ n–2 1

λ1– λ2 ,

where y1and y0are 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◦ If λ1= λ2, the general solution of the difference equation is given by

y n = y1nλ n–1

1 – y0b (n –1)λ n–1 2

2. y n+2 + ay n+1 + by n = f n.

A nonhomogeneous second-order linear difference equation defined on a discrete set of

points x =0,1, 2, Notation adopted: y n = y(n).

Let λ1and λ2be roots of the characteristic equation

λ2+ aλ + b =0

1◦ If λ1≠λ2, the general solution of the difference equation has the form

y n = y1λ

n

1 – λ n2

λ1– λ2 – y0b

λ n–1

1 – λ n–2 1

λ1– λ2 +

n

k=2

f n–k

λ k–1

1 – λ k–2 1

λ1– λ2 , where y1and y0are 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◦ If λ1= λ2, the general solution of the difference equation is given by

y n = y1nλ n–1

1 – y0b (n –1)λ n–1 2+

n

k=2

f n–k (k –1)λ k–1 2

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, y0and y N, are prescribed It is required to

find the y n≡y (x)|x=nfor1 ≤n≤N–1

If λ1 ≠λ2, the solution is given by

y n = y0λ

N

1 λ n2 – λ n1λ N2

λ N

1 – λ N2

+ y N λ

n

1 – λ n2

λ N

1 – λ N2 +

n

k=2

f n–k

λ k–1

1 – λ k–2 1

λ1– λ2 –

λ n

1 – λ n2

λ N

1 – λ N2

N

k=2

f N–k

λ k–1

1 – λ k–2 1

λ1– λ2 .

For n =1, the first sum is zero

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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:

Consider the following cases

1◦ The roots λ1and λ2of the quadratic equation (1) are real and distinct Then the general

solution of the original finite-difference equation has the form

y (x) =Θ1(x)λ x1 +Θ2(x)λ x2, (2) where Θ1(x) and Θ2(x) are arbitrary periodic functions with period 1, that is, Θk (x) =

Θk (x +1), k =1, 2

ForΘk≡const, formula (2) gives particular solutions

y (x) = C1λ x

1 + C2λ x

2,

where C1and C2are arbitrary constants

2◦ The quadratic equation (1) has equal roots: λ = λ1 = λ2 In this case, the general

solution of the original functional equation is given by

y= Θ1(x) + xΘ2(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 =Θ1(x)ρ x cos(βx) +Θ2(x)ρ x sin(βx),

whereΘ1(x) andΘ2(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◦ For f (x) = n

k=0A k x

n and a + b +1 ≠ 1, the nonhomogeneous equation has a

partic-ular solution of the form ¯y(x) = n

k=0B k x

n ; the constants B

k are found by the method of

undetermined coefficients

3◦ For f (x) =n

k=1A k exp(λ k x), the nonhomogeneous equation has a particular solution of the form¯y(x) =n

k=1B k exp(λ k x ); the constants B kare found by the method of undetermined coefficients

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4◦ For f (x) = n

k=1A k cos(λ k x), the nonhomogeneous equation has a particular solution of the form¯y(x) = n

k=1B k cos(λ k x) +

n

k=1D k sin(λ k x ); the constants B k and D k are found by the method of undetermined coefficients

5◦ For f (x) = n

k=1A k sin(λ k x), the nonhomogeneous equation has a particular solution of the form¯y(x) = n

k=1B k cos(λ k x) +

n

k=1D 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

where λ is a root of the square equation

For the integral on the right-hand side of (1) to converge, the roots of equation (2) that satisfy

the condition Re λ >0should be selected If both roots, λ1and λ2, meet this condition, the general solution of the original functional equation is expressed as

y (x) =Θ1(x)y(x, λ1) +Θ2(x)y(x, λ2), whereΘ1(x) andΘ2(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) for x≤a,

0 for a≤x≤b,

ϕ2(x) for b≤x,

where a ≤ 0 and b≥ 0 are arbitrary numbers; ϕ1(x) and ϕ2(x) are arbitrary continuous

functions satisfying the conditions

ϕ1(a) =0, a≤ϕ1(x)≤b if x≤a;

ϕ2(b) =0, a≤ϕ2(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:

y1(x) = x, y2(x) = C – x, y3(x) = C

x, y4(x) = C1– x

1+ C2x,

where C, C1, C2are arbitrary constants

, , N is often taken and the initial and final values of the unknown function, y

and y N

and λ

solution of the original finite-difference equation has the form

y... y1and y0are 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

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