Handbook of mathematics for engineers and scienteists part 207 pps

The simplest particular solution corresponds toΘx≡ 1.. Γx – μm, whereΓx is the gamma function, Θx is an arbitrary periodic function with period 1... Solution: y x = Θxb x/a+¯yx, where Θx

TABLE T12.1

Particular solutions of the nonhomogeneous functional equation y(x +1) – y(x) = f (x)

No. Right-hand side of equation, f (x) Particular solution,¯y(x)

4 x n , n =0 , 1 , 2,

1

n+1 B n (x), where the B n (x) are Bernoulli polynomials.

The generating function: te

xt

e t– 1 =

∞



n=0

B n (x) t

n

n!

x

ψ (x) = – C +

 1

0

1– t x–1

1– t dt is the logarithmic derivative of

the gamma function, C =0 5772 . is the Euler constant

1

x

a – 1a λx

2sinh b

2sinh b

a– 1a x



x– a

a– 1



 ∞

0 t

x–1

e–t dtis the gamma function

2sin a

2sin a

2 –

sin[a(2x – 1 )]

4sin a

2 +

sin[a(2x – 1 )]

4sin a

4 sin2a – x cos[a(2x – 1 )]

2sin a

4 sin 2a + x sin[a(2x – 1 )]

2sin a

24 a x sin(bx), a >0, a≠ 1 a x a sin[b(x –1)] – sin(bx)

a2– 2acos b +1

25 a x cos(bx), a >0, a≠ 1 a x a cos[b(x –1)] – cos(bx)

a2– 2acos b +1

2◦ Solution for a <0:

y (x) =Θ1(x)|a|x sin(πx) +Θ2(x)|a|x cos(πx),

where theΘk (x) =Θk (x +1) are arbitrary periodic functions with period 1 (k =1, 2)

Remark See also Subsection 17.2.1, Example 1.

4. y(x + 1) – ay(x) = f (x).

A nonhomogeneous first-order constant-coefficient linear difference equation.

1◦ Solution:

y (x) =



Θ1(x)|a|x sin(πx) +Θ2(x)|a|x cos(πx) + ¯y(x) if a <0, whereΘ(x), Θ1(x), and Θ2(x) are arbitrary periodic functions with period 1, and ¯y(x) is

any particular solution of the nonhomogeneous equation

2◦ For f (x) = n

k=0b k x

n and a≠ 1, the nonhomogeneous equation has a particular solution

of the form ¯y(x) = n

k=0A k x

n ; the constants B k are found by the method of undetermined

coefficients

3◦ For f (x) = n

k=1b k e

λkx, the nonhomogeneous equation has a particular solution of the

form

¯y(x) =

⎧

⎪

⎪

⎨

⎪

⎪

⎩

n



k=1

b k

b m xe λm(x–1 )+

n



k=1 ,k m

b k

e λk – a e λkx if a = e λm,

where m =1, , n.

4◦ For f (x) = n

k=1b k cos(β k x), the nonhomogeneous equation has a particular solution of

the form

¯y(x) =n

k=1

b k

a2+1–2a cos β k



(cos β k – a) cos(β k x ) + sin β k sin(β k x)

5◦ For f (x) = n

k=1b k sin(β k x), the nonhomogeneous equation has a particular solution of

the form

¯y(x) =n

k=1

b k

a2+1–2a cos β k



(cos β k – a) sin(β k x ) – sin β k cos(β k x)

5. y(x + 1) – xy(x) = 0.

Solution:

y (x) = Θ(x)Γ(x), Γ(x) =

 ∞

0 t

x–1e–t dt,

whereΓ(x) is the gamma function, Θ(x) = Θ(x +1) is an arbitrary periodic function with period 1

The simplest particular solution corresponds toΘ(x)≡ 1

6. y(x + 1) – a(x – b)(x – c)y(x) = 0.

Solution:

y (x) = Θ(x)a x Γ(x – b)Γ(x – c),

whereΓ(x) is the gamma function, Θ(x) is an arbitrary periodic function with period 1.

7. y(x + 1) – R(x)y(x) = 0, R(x) = a (x – λ1)(x – λ2) (x – λ n)

(x – μ1)(x – μ2) (x – μ m) .

Solution:

y (x) = Θ(x)a x Γ(x – λ1)Γ(x – λ2) Γ(x – λ n)

Γ(x – μ1)Γ(x – μ2) Γ(x – μm), whereΓ(x) is the gamma function, Θ(x) is an arbitrary periodic function with period 1.

8. y(x + 1) – ae λx y(x)= 0.

Solution:

y (x) = Θ(x)a xexp 12λx2– 1

2λx

, whereΘ(x) is an arbitrary periodic function with period 1.

9. y(x + 1) – ae μx2 +λx y(x)= 0.

Solution:

y (x) = Θ(x)a xexp1

3μx3+ 12(λ – μ)x2+ 16(μ –3λ )x

, whereΘ(x) is an arbitrary periodic function with period 1.

10. y(x + 1) – f (x)y(x) = 0.

Here, f (x) = f (x +1) is a given periodic function with period 1

Solution:

y (x) = Θ(x)f (x)x

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

ForΘ(x)≡const, we have a particular solution y(x) = C

f (x)x

, where C is an arbitrary

constant

11. y(x + a) – by(x) = 0.

Solution:

y (x) = Θ(x)b x/a, whereΘ(x) = Θ(x + a) is an arbitrary periodic function with period a.

ForΘ(x) ≡ const, we have particular solution y(x) = Cb x/a , where C is an arbitrary

constant

12. y(x + a) – by(x) = f (x).

1◦ Solution:

y (x) = Θ(x)b x/a+¯y(x),

where Θ(x) = Θ(x + a) is an arbitrary periodic function with period a, and ¯y(x) is any

particular solution of the nonhomogeneous equation

2◦ For f (x) =n

k=0A k x

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

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

13. y(x + a) – bxy(x) = 0, a, b> 0.

Solution:

y (x) = Θ(x)

 ∞

0 t

(x/a)–1e–t/(ab) dt,

whereΘ(x) = Θ(x + a) is an arbitrary periodic function with period a.

14. y(x + a) – f (x)y(x) = 0.

Here, f (x) = f (x + a) is a given periodic function with period a.

Solution:

y (x) = Θ(x)f (x)x/a

, whereΘ(x) = Θ(x + a) is an arbitrary periodic function with period a.

ForΘ(x) ≡ const, we have a particular solution y(x) = C

f (x)x/a

, where C is an

arbitrary constant

T12.1.1-2 Linear functional equations involving y(x) and y(ax).

Solution for x >0:

y (x) = x λΘ ln x

ln a

 , λ= ln b

ln a,

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

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

constant

1◦ Solution:

y (x) = Y (x) + ¯y(x), where Y (x) is the general solution of the homogeneous equation Y (ax) – 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, the nonhomogeneous equation has a particular solution of the

form

¯y(x) =n k=0

A k

a k – b x k, a k – b≠ 0

3◦ For f (x) = ln xn

k=0A k x

k, the nonhomogeneous equation has a particular solution of the

form

¯y(x) =n

k=1

x k B

k ln x + C k

, B k = A k

a k – b, C k = –

A k a k ln a (a k – b)2 .

17. y( 2x) – a cos x y(x) = 0.

Solution for a >0and x >0:

y (x) = xlnlna2 –1sin x Θln x

ln2

 ,

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

T12.1.1-3 Linear functional equations involving y(x) and y(a – x).

18. y(x) = y(–x).

This functional equation may be treated as a definition of even functions.

Solution:

y (x) = ϕ (x) + ϕ(–x)

where ϕ(x) is an arbitrary function.

19. y(x) = –y(–x).

This functional equation may be treated as a definition of odd functions.

Solution:

y (x) = ϕ (x) – ϕ(–x)

where ϕ(x) is an arbitrary function.

20. y(x) – y(a – x) = 0.

1◦ Solution:

y (x) = Φ(x, a – x),

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

2◦ Specific particular solutions may be obtained using the formula

y (x) =Ψ ϕ (x) + ϕ(a – x)

by specifying the functionsΨ(z) and ϕ(x).

21. y(x) + y(a – x) = 0.

1◦ Solution:

y (x) = Φ(x, a – x),

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

2◦ Specific particular solutions may be obtained using the formula

y (x) = (2x – a)Ψ ϕ (x) + ϕ(a – x)

by specifyingΨ(z) and ϕ(x).

22. y(x) + y(a – x) = b.

Solution:

y (x) = 12b+Φ(x, a – x),

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

Particular solutions: y(x) = b sin2

πx

2a



and y(x) = b cos2

πx

2a



23. y(x) + y(a – x) = f (x).

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

Solution:

y (x) = 12f (x) + Φ(x, a – x),

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

24. y(x) – y(a – x) = f (x).

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

Solution:

y (x) = 12f (x) + Φ(x, a – x),

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

25. y(x) + g(x)y(a – x) = f (x).

Solution:

y (x) = f (x) – g(x)f (a – x)

1– g(x)g(a – x) [if g(x)g(a – x)1]

T12.1.1-4 Linear functional equations involving y(x) and y(a/x).

26. y(x) – y(a/x) = 0.

Babbage equation.

Solution:

y (x) = Φ(x, a/x),

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

27. y(x) + y(a/x) = 0.

Solution:

y (x) = Φ(x, a/x),

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

28. y(x) + y(a/x) = b.

Solution:

y (x) = 12b+Φ(x, a/x),

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

29. y(x) + y(a/x) = f (x).

The right-hand side must satisfy the condition f (x) = f (a/x).

Solution:

y (x) = 12f (x) + Φ(x, a/x),

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

30. y(x) – y(a/x) = f (x).

Here, the function f (x) must satisfy the condition f (x) = –f (a/x).

Solution:

y (x) = 12f (x) + Φ(x, a/x),

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

31. y(x) + x a y( 1/x) = 0.

Solution:

y (x) = (1– x a)Φ(x,1/x), whereΦ(x, z) = Φ(z, x) is any symmetric function with two arguments.

32. y(x) – x a y( 1/x) = 0.

Solution:

y (x) = (1+ x a)Φ(x,1/x), whereΦ(x, z) = Φ(z, x) is any symmetric function with two arguments.

33. y(x) + g(x)y(a/x) = f (x).

Solution:

y (x) = f (x) – g(x)f (a/x)

1– g(x)g(a/x) [if g(x)g(a/x)1]