Handbook of mathematics for engineers and scienteists part 178 pot

The general solution can be obtained by formulas given in Item1◦of equation T5.1.23.. The general solution can be obtained by formulas given in Item1◦of equation T5.1.23.. The general so

Ordinary Differential Equations

T5.1 First-Order Equations

 In this Section we shall often use the term “solution” to mean “general solution.”

1. y 

x = f (y).

Autonomous equation.

Solution: x =



dy

f (y) + C.

Particular solutions: y = A k , where the A k are roots of the algebraic (transcendental)

equation f (A k) =0

2. y 

x = f (x)g(y).

Separable equation.

Solution:



dy

g (y) =



f (x) dx + C.

Particular solutions: y = A k , where the A k are roots of the algebraic (transcendental)

equation g(A k) =0

3. g(x)y 

x = f1(x)y + f0(x).

Linear equation.

Solution:

y = Ce F + e F



e–F f0(x)

g (x) dx, where F (x) =



f1(x)

g (x) dx.

4. g(x)y 

x = f1(x)y + f0(x)y k.

Bernoulli equation Here, k is an arbitrary number For k≠ 1, the substitution w(x) = y1–k leads to a linear equation: g(x)w  x= (1– k)f1(x)w + (1– k)f0(x).

Solution:

y1 –k = Ce F + (1– k)e F



e–F f0(x)

g (x) dx, where F (x) = (1– k)



f1(x)

g (x) dx.

5. y 

x = f (y/x).

Homogeneous equation The substitution u(x) = y/x leads to a separable equation: xu  x=

f (u) – u.

1207

6. y 

x = ay2+ bx n.

Special Riccati equation, n is an arbitrary number.

1◦ Solution for n≠–2:

y = –1

a

w  x

w , w (x) = √

x

*

C1J1

2k

1

k

√

ab x k

+ C2Y1

2k

1

k

√

ab x k+

,

where k = 12(n +2); J m (z) and Y m (z) are Bessel functions (see Subsection 18.6).

2◦ Solution for n = –2:

y= λ

x – x2aλ

 ax

2aλ+1x2aλ + C

–1

,

where λ is a root of the quadratic equation aλ2+ λ + b =0

7. y 

x = y2+ f (x)y – a2– af (x).

Particular solution: y0 = a The general solution can be obtained by formulas given in

Item1◦of equation T5.1.23.

8. y 

x = f (x)y2+ ay – ab – b2f (x).

Particular solution: y0 = b The general solution can be obtained by formulas given in

Item1◦of equation T5.1.23.

9. y 

x = y2+ xf (x)y + f (x).

Particular solution: y0= –1/x The general solution can be obtained by formulas given in Item1◦of equation T5.1.23.

10. y 

x = f (x)y2– ax n f (x)y + anx n–1.

Particular solution: y0 = ax n The general solution can be obtained by formulas given in Item1◦of equation T5.1.23.

11. y 

x = f (x)y2+ anx n–1 – a2x2n f (x).

Particular solution: y0 = ax n The general solution can be obtained by formulas given in Item1◦of equation T5.1.23.

12. y 

x = –(n + 1)x n y2+ x n+1 f (x)y – f (x).

Particular solution: y0= x–n–1 The general solution can be obtained by formulas given in Item1◦of equation T5.1.23.

13. xy 

x = f (x)y2+ ny + ax2n f (x).

Solution: y =

⎧

⎪

⎪

√

a x ntan*√

a



x n–1f (x) dx + C+

if a >0,

|a|x ntanh*

–

|a|



x n–1f (x) dx + C+

if a <0

14. xy 

x = x2n f (x)y2+ [ax n f (x) – n]y + bf (x).

The substitution z = x n y leads to a separable equation: z  x = x n–1f (x)(z2+ az + b).

15. y 

x = f (x)y2+ g(x)y – a2f (x) – ag(x).

Particular solution: y0 = a The general solution can be obtained by formulas given in

Item1◦of equation T5.1.23.

16. y 

x = f (x)y2+ g(x)y + anx n–1 – a2x2n f (x) – ax n g(x).

Particular solution: y0 = ax n The general solution can be obtained by formulas given in Item1◦of equation T5.1.23.

17. y 

x = ae λx y2+ ae λx f (x)y + λf (x).

Particular solution: y0= –λ

a e

–λx The general solution can be obtained by formulas given

in Item1◦of equation T5.1.23.

18. y 

x = f (x)y2– ae λx f (x)y + aλe λx.

Particular solution: y0= ae λx The general solution can be obtained by formulas given in Item1◦of equation T5.1.23.

19. y 

x = f (x)y2+ aλe λx – a2e2λx f (x).

Particular solution: y0= ae λx The general solution can be obtained by formulas given in Item1◦of equation T5.1.23.

20. y 

x = f (x)y2+ λy + ae2λx f (x).

Solution: y =

⎧

⎪

⎪

√

a e λxtan*√

a



e λx f (x) dx + C+

if a >0,

|a|e λxtanh*

–

|a|



e λx f (x) dx + C+

if a <0

21. y 

x = y2– f2(x) + f 

x (x).

Particular solution: y0 = f (x) The general solution can be obtained by formulas given in

Item1◦of equation T5.1.23.

22. y 

x = f (x)y2– f (x)g(x)y + g  x (x).

Particular solution: y0= g(x) The general solution can be obtained by formulas given in

Item1◦of equation T5.1.23.

23. y 

x = f (x)y2+ g(x)y + h(x).

General Riccati equation.

1◦ Given a particular solution y0= y0(x) of the Riccati equation, the general solution can

be written as:

y = y0(x) + Φ(x)*C– Φ(x)f2(x) dx

+– 1 , where

Φ(x) = exp 2f2(x)y0(x) + f1(x)

dx

4

To the particular solution y0(x) there corresponds C = ∞.

2◦ The substitution

u (x) = exp



–



f2y dx



reduces the general Riccati equation to a second-order linear equation:

f2u 

xx–



(f2) x + f1f2

u 

x + f0f22u=0, which often may be easier to solve than the original Riccati equation

3◦ For more details about the Riccati equation, see Subsection 12.1.4 Many solvable

equations of this form can be found in the books by Kamke (1977) and Polyanin and Zaitsev (2003)

24. yy 

x = y + f (x).

Abel equation of the second kind in the canonical form Many solvable equations of this

form can be found in the books by Zaitsev and Polyanin (1994) and Polyanin and Zaitsev (2003)

25. yy 

x = f (x)y + g(x).

Abel equation of the second kind Many solvable equations of this form can be found in the

books by Zaitsev and Polyanin (1994) and Polyanin and Zaitsev (2003)

26. yy 

x = f (x)y2+ g(x)y + h(x).

Abel equation of the second kind Many solvable equations of this form can be found in the

books by Zaitsev and Polyanin (1994) and Polyanin and Zaitsev (2003)

 In equations T5.1.27–T5.1.48, the functions f, g, and h are arbitrary composite functions

whose arguments can depend on both x and y.

27. y 

x = f (ax + by + c).

If b≠ 0, the substitution u(x) = ax + by + c leads to a separable equation: u  x = bf (u) + a.

28. y 

x = f (y + ax n + b) – anx n–1.

The substitution u = y + ax n + b leads to a separable equation: u  x = f (u).

29. y 

x =

y

x f (x

n y m).

Generalized homogeneous equation The substitution z = x n y m leads to a separable

equation: xz  x = nz + mzf (z).

30. y 

x = –

n

m

y

x + y k f (x)g(x n y m).

The substitution z = x n y m leads to a separable equation: z 

x = mx

n–nk

m f (x)z k+m– m 1g (z).

31. y 

x = f

 ax + by + c

αx + βy + γ



.

See Paragraph 12.1.2-3, Item2◦.

32. y 

x = x n–1 y1–m f (ax n + by m).

The substitution w = ax n + by m leads to a separable equation: w  x = x n–1[an + bmf (w)].

33. [x n f (y) + xg(y)]y 

x = h(y).

This is a Bernoulli equation with respect to x = x(y) (see equation T5.1.4).

34. x[f (x n y m ) + mx k g(x n y m )]y 

x = y[h(x n y m ) – nx k g(x n y m)].

The transformation t = x n y m , z = x–k leads to a linear equation with respect to z = z(t):

t [nf (t) + mh(t)]z  t = –kf (t)z – kmg(t).

35. x[f (x n y m ) + my k g(x n y m )]y  x = y[h(x n y m ) – ny k g(x n y m)].

The transformation t = x n y m , z = y–k leads to a linear equation with respect to z = z(t):

t [nf (t) + mh(t)]z  t = –kh(t)z + kng(t).

36. x[sf (x n y m ) – mg(x k y s )]y 

x = y[ng(x k y s ) – kf(x n y m)].

The transformation t = x n y m , w = x k y s leads to a separable equation: tf (t)w 

t = wg(w).

37. [f(y) + amx n y m–1 ]y 

x + g(x) + anx n–1 y m= 0.

Solution:



f (y) dy +



g (x) dx + ax n y m = C.

38. y 

x = e–λx f (e λx y).

The substitution u = e λx y leads to a separable equation: u  x = f (u) + λu.

39. y 

x = e λy f (e λy x).

The substitution u = e λy x leads to a separable equation: xu  x = λu2f (u) + u.

40. y 

x = yf (e αx y m).

The substitution z = e αx y m leads to a separable equation: z 

x = αz + mzf (z).

41. y 

x =

1

x f (x

n e αy).

The substitution z = x n e αy leads to a separable equation: xz 

x = nz + αzf (z).

42. y 

x = f (x)e λy + g(x).

The substitution u = e–λy leads to a linear equation: u  x = –λg(x)u–λf (x).

43. y 

x = –

n

x + f (x)g(x n e y).

The substitution z = x n e y leads to a separable equation: z 

x = f (x)zg(z).

44. y 

x = –

α

m y + y k f (x)g(e αx y m).

The substitution z = e αx y m leads to a separable equation:

z 

x = m exp

*α

m(1– k)x

+

f (x)z k+m– m 1g (z).

45. y 

x = e αx–βy f (ae αx + be βy).

The substitution w = ae αx + be βy leads to a separable equation: w  x = e αx [aα + bβf (w)].

46. [e αx f (y) + aβ]y 

x + e βy g(x) + aα = 0.

Solution:



e–βy f (y) dy +

e–αx g (x) dx – ae–αx–βy = C.

47. x[f (x n e αy ) + αyg(x n e αy )]y 

x = h(x n e αy ) – nyg(x n e αy).

The substitution t = x n e αy leads to a linear equation with respect to y = y(t):

t [nf (t) + αh(t)]y t  = –ng(t)y + h(t).

48. [f(e αx y m ) + mxg(e αx y m )]y 

x = y[h(e αx y m ) – αxg(e αx y m)].

The substitution t = e αx y m leads to a linear equation with respect to x = x(t):

t [αf (t) + mh(t)]x  t = mg(t)x + f (t).

T5.2 Second-Order Linear Equations

Preliminary remarks A homogeneous linear equation of the second order has the general

form

f2(x)y  xx + f1(x)y x  + f0(x)y =0 (1)

Let y0 = y0(x) be a nontrivial particular solution (y0  0) of this equation Then the general solution of equation (1) can be found from the formula

y = y0

C1+ C2

 e–F

y2 0

dx



, where F = f1

f2 dx. (2) For specific equations described below, often only particular solutions are given, while the general solutions can be obtained with formula (2)

T5.2.1 Equations Involving Power Functions

1. y 

xx + ay = 0.

Equation of free oscillations.

Solution: y =



C1sinh(x √

|a|) + C2cosh(x √

|a|) if a <0,

C1sin(x √

a ) + C2cos(x √

a) if a >0

2. y 

xx – ax n y = 0.

1◦ For n = –2, this is the Euler equation T5.2.1.12 (the solution is expressed in terms of elementary function)

2◦ Assume2/ (n +2) =2m+1, where m is an integer Then the solution is

y=

⎧

⎪

⎪

⎩

x (x1–2 D)m+1



C1exp

 √

a

q x

q

+ C2exp



–

√ a

q x

q

if m≥ 0,

(x1–2 D)–m



C1exp

 √

a

q x

q

+ C2exp



–

√ a

q x

q

if m <0,

where D = d

dx , q = n+2

2 =

1

2m+1.

3◦ For any n, the solution is expressed in terms of Bessel functions and modified Bessel

functions:

y=

⎧

⎪

⎪

⎪

⎪

C1√

x J 1 2

 √

–a

q x

q

+ C2√

x Y1 2

 √

–a

q x

q

if a <0,

C1√

x I 1 2

 √

a

q x

q

+ C2√

x K 1 2

 √

a

q x

q

if a >0,

where q = 12(n +2) The functions J ν (z), Y ν (z) and I ν (z), K ν (z) are described in

Sec-tions 18.6 and 18.7 in detail; see also equaSec-tions T5.2.1.13 and T5.2.1.14

3. y 

xx + ay x  + by = 0.

Second-order constant coefficient linear equation In physics this equation is called an equation of damped vibrations.

Solution: y =

⎧

⎪

⎪

exp –12ax

C1exp 12λx

+ C2exp –12λx

if λ2 = a2–4b>0, exp –12ax

C1sin 12λx

+ C2cos 12λx

if λ2 =4b – a2>0, exp –12ax C1x + C2

if a2 =4b

4. y 

xx + ay x  + (bx + c)y = 0.

1◦ Solution with b≠ 0:

y = exp –12ax

ξ

C1J1 3 23

√

b ξ3 2

+ C2Y1 3 23

√

b ξ3 2

, ξ = x + 4c – a2

4b ,

where J1 3(z) and Y1 3(z) are Bessel functions.

2◦ For b =0, see equation T5.2.1.3

Abel equation of the second kind in the canonical form Many solvable equations of this

form can be found in the books by Zaitsev and Polyanin (1994) and Polyanin and Zaitsev (2003)... g(x).

Abel equation of the second kind Many solvable equations of this form can be found in the

books by Zaitsev and Polyanin (1994) and Polyanin and Zaitsev (2003)

26.... h(x).

Abel equation of the second kind Many solvable equations of this form can be found in the

books by Zaitsev and Polyanin (1994) and Polyanin and Zaitsev (2003)