Handbook of mathematics for engineers and scienteists part 181 pps

M., Ordinary Differential Equations and Their Solutions, D.. F., Handbook of Exact Solutions for Ordinary Differential Equations, 2nd Edition, Chapman & Hall/CRC Press, Boca Raton, 2003.

1228 ORDINARYDIFFERENTIALEQUATIONS

28. y 

xx + α(y  x) 2 = 

e βx f (y) + β

y 

x.

Solution:



e αy dy

F (y) + C1 = C2+

1

β e

βx, where F (y) =

e αy f (y) dy.

29. y 

xx + f (y)(y  x) 2+ g(y) = 0.

The substitution w(y) = (y  x)2 leads to a first-order linear equation: w y +2f (y)w+2 g (y) =0

30. y 

xx + f (y)(y  x) 2 – 1 2y 

x = e x g(y).

The substitution w(y) = e–x (y x )2 leads to a first-order linear equation: w  y+2f (y)w =2g (y).

31. y 

xx = xf (y)(y  x) 3

Taking y to be the independent variable, we obtain a linear equation with respect to x = x(y):

x 

yy = –f (y)x.

32. y 

xx = f (y)(y  x) 2+ g(x)y x .

Dividing by y  x, we obtain an exact differential equation Its solution follows from the equation:

ln|y 

x|=



f (y) dy +



g (x) dx + C.

Solving the latter for y  x , we arrive at a separable equation In addition, y = C1 is a singular

solution, with C1being an arbitrary constant

33. y 

xx = f (x)g(xy  x – y).

The substitution w = xy  x – y leads to a first-order separable equation: w  x = xf (x)g(w).

34. y 

xx =

y

x2f

 xy  x y



.

The substitution w(x) = xy  x /y leads to a first-order separable equation: xw  x = f (w)+w–w2

35. gy 

xx+ 1 2g 

x y x  = f (y)h y x  √

g

, g = g(x).

The substitution w(y) = y  x √ g leads to a first-order separable equation: ww 

y = f (y)h(w).

36. y 

xx = f y x 2 + ay

.

The substitution w(y) = (y  x)2+ay leads to a first-order separable equation: w y  =2f (w)+a.

References for Chapter T5

Kamke, E., Differentialgleichungen: L¨osungsmethoden und L ¨osungen, I, Gew¨ohnliche Differentialgleichungen,

B G Teubner, Leipzig, 1977.

Murphy, G M., Ordinary Differential Equations and Their Solutions, D Van Nostrand, New York, 1960 Polyanin, A D and Zaitsev, V F., Handbook of Exact Solutions for Ordinary Differential Equations, 2nd

Edition, Chapman & Hall/CRC Press, Boca Raton, 2003.

Zaitsev, V F and Polyanin, A D., Discrete-Group Methods for Integrating Equations of Nonlinear Mechanics,

CRC Press, Boca Raton, 1994.

Systems of Ordinary

Differential Equations

T6.1 Linear Systems of Two Equations

T6.1.1 Systems of First-Order Equations

1. x 

t = ax + by, y t  = cx + dy.

System of two constant-coefficient first-order linear homogeneous differential equations.

Let us write out the characteristic equation

and find its discriminant

1◦ Case ad – bc≠ 0 The origin of coordinates x = y =0is the only one stationary point;

it is

a node if D =0;

a node if D >0 and ad – bc >0;

a saddle if D >0 and ad – bc <0;

a focus if D <0 and a + d≠ 0;

a center if D <0 and a + d =0

1.1 Suppose D >0 The characteristic equation (1) has two distinct real roots, λ1

and λ2 The general solution of the original system of differential equations is expressed as

x = C1be λ1 t + C2be λ2t,

y = C1(λ1– a)e λ1t + C2(λ2– a)e λ2 t,

where C1and C2are arbitrary constants

1.2 Suppose D <0 The characteristic equation (1) has two complex conjugate roots,

λ1 , 2 = σ iβ The general solution of the original system of differential equations is given by

x = be σt

C1sin(βt) + C2cos(βt)

,

y = e σt5

[(σ – a)C1– βC2] sin(βt) + [βC1+ (σ – a)C2cos(βt)

,

where C1and C2are arbitrary constants

1.3 Suppose D =0and a≠d The characteristic equation (1) has two equal real roots,

λ1 = λ2 The general solution of the original system of differential equations is

x=2b



C1+ a C – d2 + C2t

 exp



a + d

2 t

 ,

y = [(d – a)C1+ C2+ (d – a)C2t] exp



a + d

2 t

 ,

1229

1230 SYSTEMS OFORDINARYDIFFERENTIALEQUATIONS

where C1and C2are arbitrary constants

1.4 Suppose a = d≠ 0and b =0 Solution:

x = C1e at, y = (cC

1t + C2)e at

1.5 Suppose a = d≠ 0and c =0 Solution:

x = (bC1t + C2)e at, y = C1e at.

2◦ Case ad – bc =0and a2+ b2>0 The whole of the line ax + by =0consists of singular points The system in question may be rewritten in the form

x 

t = ax + by, y t  = k(ax + by).

2.1 Suppose a + bk≠ 0 Solution:

x = bC1+ C2e(a+bk)t, y = –aC

1+ kC2e(a+bk)t.

2.2 Suppose a + bk =0 Solution:

x = C1(bkt –1) + bC2t, y = k2bC1t + (bk2t+1)C2

2. x 

t = a1x + b1y + c1 , y 

t = a2x + b2y + c2

The general solution of this system is given by the sum of its any particular solution and the general solution of the corresponding homogeneous system (see system T6.1.1.1)

1◦ Suppose a1b2– a2b1≠ 0 A particular solution:

x = x0, y = y0,

where the constants x0 and y0 are determined by solving the linear algebraic system of equations

a1x0+ b1y0+ c1=0, a2x0+ b2y0+ c2 =0

2◦ Suppose a1b2– a2b1=0and a21+ b21>0 Then the original system can be rewritten as

x 

t = ax + by + c1, y 

t = k(ax + by) + c2.

2.1 If σ = a + bk≠ 0, the original system has a particular solution of the form

x = bσ–1(c1k – c2)t – σ–2(ac1+ bc2), y = kx + (c2– c1k )t.

2.2 If σ = a + bk =0, the original system has a particular solution of the form

x= 12b (c2– c1k )t2+ c1t, y = kx + (c2– c1k )t.

3. x 

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

Solution:

x = e F (C1e G + C

2e–G), y = e F (C

1e G – C

2e–G),

where C1and C2are arbitrary constants, and

F =



f (t) dt, G=



g (t) dt.

4. x 

t = f (t)x + g(t)y, y  t = –g(t)x + f (t)y.

Solution:

x = F (C1cos G + C2sin G), y = F (–C1sin G + C2cos G),

where C1and C2are arbitrary constants, and

F = exp



f (t) dt

 , G=



g (t) dt.

5. x 

t = f (t)x + g(t)y, y  t = ag(t)x + [f (t) + bg(t)]y.

The transformation

x= exp



f (t) dt



u, y= exp



f (t) dt



v, τ =



g (t) dt

leads to a system of constant coefficient linear differential equations of the form T6.1.1.1:

u 

τ = v, v τ  = au + bv.

6. x 

t = f (t)x + g(t)y, y  t = a[f (t) + ah(t)]x + a[g(t) – h(t)]y.

Let us multiply the first equation by –a and add it to the second equation to obtain

y 

t – ax  t = –ah(t)(y – ax).

By setting U = y – ax and then integrating, one obtains

y – ax = C1exp



–a



h (t) dt



where C1 is an arbitrary constant On solving (∗) for y and on substituting the resulting

expression into the first equation of the system, one arrives at a first-order linear differential

equation for x.

7. x 

t = f (t)x + g(t)y, y  t = h(t)x + p(t)y.

1◦ Let us express y from the first equation and substitute into the second one to obtain a

second-order linear equation:

gx 

tt – (f g + gp + g t  )x  t + (f gp – g2h + f g  t – f t  g )x =0 (1) This equation is easy to integrate if, for example, the following conditions are met:

1) f gp – g2h + f g  t – f t  g=0;

2) f gp – g2h + f g  t – f t  g = ag, f g + gp + g  t = bg.

In the first case, equation (1) has a particular solution u = C = const In the second case, it

is a constant-coefficient equation

A considerable number of other solvable cases of equation (1) can be found in the handbooks by Kamke (1977) and Polyanin and Zaitsev (2003)

1232 SYSTEMS OFORDINARYDIFFERENTIALEQUATIONS

2◦ Suppose a particular solution of the system in question is known,

x = x0(t), y = y0(t).

Then the general solution can be written out in the form

x (t) = C1x0(t) + C2x0(t)



g (t)F (t)P (t)

x2

0(t)

dt,

y (t) = C1y0(t) + C2



F (t)P (t)

x0(t) + y0(t)



g (t)F (t)P (t)

x2

0(t)

dt

 ,

where C1and C2are arbitrary constants, and

F (t) = exp



f (t) dt

 , P (t) = exp



p (t) dt



T6.1.2 Systems of Second-Order Equations

1. x 

tt = ax + by, y  tt = cx + dy.

System of two constant-coefficient second-order linear homogeneous differential equations.

The characteristic equation has the form

λ4– (a + d)λ2+ ad – bc =0

1◦ Case ad – bc≠ 0

1.1 Suppose (a – d)2 +4bc ≠ 0 The characteristic equation has four distinct roots

λ1, , λ4 The general solution of the system in question is written as

x = C1be λ1 t + C2be λ2t + C3be λ3t + C4be λ4 t,

y = C1(λ21– a)e λ1t + C2(λ22– a)e λ2t + C3(λ23– a)e λ3 t + C4(λ24– a)e λ4 t,

where C1, , C4are arbitrary constants

1.2 Solution with (a – d)2+4bc=0and a≠d:

x=2C1



bt+ 2bk

a – d



e kt/2+2C2



bt– 2bk

a – d



e–kt/2+2bC3te kt/2+2bC4te–kt/2,

y = C1(d – a)te kt/2+ C2(d – a)te–kt/2+ C3[(d – a)t +2k ]e kt/2+ C4[(d – a)t –2k ]e–kt/2,

where C1, , C4are arbitrary constants and k = √

2(a + d)

1.3 Solution with a = d≠ 0and b =0:

x=2√ a C1e √

a t+2√ a C2e–√

a t,

y = cC1te √

a t – cC2te–√

a t + C3e √ a t + C4e–√

a t. 1.4 Solution with a = d≠ 0and c =0:

x = bC1te √

a t – bC2te–√

a t + C3e √ a t + C4e–√

a t,

y=2√ a C1e √ a t

+2√ a C2e–√ a t

2◦ Case ad – bc =0and a2+ b2 >0 The original system can be rewritten in the form

x 

tt = ax + by, y tt  = k(ax + by).

2.1 Solution with a + bk≠ 0:

x = C1exp t √

a + bk

+ C2exp –t √

a + bk

+ C3bt + C4b,

y = C1kexp t √

a + bk

+ C2kexp –t √

a + bk

– C3at – C4a

2.2 Solution with a + bk =0:

x = C1bt3+ C

2bt2+ C

3t + C4,

y = kx +6C1t+2C2.

2. x 

tt = a1x + b1y + c1 , y 

tt = a2x + b2y + c2

The general solution of this system is expressed as the sum of its any particular solution and the general solution of the corresponding homogeneous system (see system T6.1.2.1)

1◦ Suppose a1b2– a2b1≠ 0 A particular solution:

x = x0, y = y0,

where the constants x0 and y0 are determined by solving the linear algebraic system of equations

a1x0+ b1y0+ c1=0, a2x0+ b2y0+ c2 =0

2◦ Suppose a1b2– a2b1=0and a21+ b21>0 Then the system can be rewritten as

x 

tt = ax + by + c1, y 

tt = k(ax + by) + c2.

2.1 If σ = a + bk≠ 0, the original system has a particular solution

x= 12bσ–1(c

1k – c2)t2– σ–2(ac1+ bc2), y = kx + 12(c2– c1k )t2

2.2 If σ = a + bk =0, the system has a particular solution

x= 241 b (c2– c1k )t4+ 12c1t2, y = kx + 1

2(c2– c1k )t2.

3. x 

tt – ay t  + bx = 0, y  tt + ax  t + by = 0.

This system is used to describe the horizontal motion of a pendulum taking into account the rotation of the earth

Solution with a2+4b>0:

x = C1cos(αt) + C2sin(αt) + C3cos(βt) + C4sin(βt),

y = –C1sin(αt) + C2cos(αt) – C3sin(βt) + C4cos(βt), where C1, , C4are arbitrary constants and

α= 12a+ 12

a2+4b, β = 12a– 12

a2+4b

1234 SYSTEMS OFORDINARYDIFFERENTIALEQUATIONS

4. x 

tt + a1x 

t + b1y 

t + c1x + d1y = k1e iωt, y 

tt + a2x 

t + b2y 

t + c2x + d2y = k2e iωt.

Systems of this type often arise in oscillation theory (e.g., oscillations of a ship and a ship gyroscope) The general solution of this constant-coefficient linear nonhomogeneous system of differential equations is expressed as the sum of its any particular solution and

the general solution of the corresponding homogeneous system (with k1= k2=0)

1◦ A particular solution is sought by the method of undetermined coefficients in the form

x = A ∗ e iωt, y = B

∗ e iωt.

On substituting these expressions into the system of differential equations in question, one

arrives at a linear nonhomogeneous system of algebraic equations for the coefficients A ∗ and B ∗

2◦ The general solution of a homogeneous system of differential equations is determined

by a linear combination of its linearly independent particular solutions, which are sought using the method of undetermined coefficients in the form of exponential functions,

x = Ae λt, y = Be λt

On substituting these expressions into the system and on collecting the coefficients of the

unknowns A and B, one obtains

(λ2+ a1λ + c1)A + (b1λ + d1)B =0,

(a2λ + c2)A + (λ2+ b2λ + d2)B =0

For a nontrivial solution to exist, the determinant of this system must vanish This require-ment results in the characteristic equation

(λ2+ a1λ + c1)(λ2+ b2λ + d2) – (b1λ + d1)(a2λ + c2) =0,

which is used to determine λ If the roots of this equation, k1, , k4, are all distinct, then the general solution of the original system of differential equations has the form

x = –C1(b1λ1+ d1)e λ1t – C2(b1λ2+ d1)e λ2t – C3(b1λ1+ d1)e λ3 t – C4(b1λ4+ d1)e λ4 t,

y = C1(λ21+ a1λ1+ c1)e λ1t + C2(λ22+ a1λ2+ c1)e λ2t

+ C3(λ23+ a1λ3+ c1)e λ3 t + C4(λ24+ a1λ4+ c1)e λ4 t,

where C1, , C4are arbitrary constants

5. x 

tt = a(ty  t – y), y tt  = b(tx  t – x).

The transformation

leads to a first-order system:

u 

t = atv, v t  = btu.

The general solution of this system is expressed as

with ab >0:



u (t) = C1aexp 12√

ab t2

+ C2aexp –12√

ab t2 ,

v (t) = C1√

abexp 12√

ab t2

– C2√

abexp –12√

ab t2

;

with ab <0:



u (t) = C1acos 12√|

ab|t2

+ C2asin 12√|

ab|t2 ,

v (t) = –C1√

|ab| sin 12√

|ab|t2

+ C2√

|ab| cos 12√

|ab|t2 , (2)

Systems of this type often arise in oscillation theory (e.g., oscillations of a ship and a ship gyroscope) The general solution of this constant-coefficient linear nonhomogeneous system of differential... solution of a homogeneous system of differential equations is determined

by a linear combination of its linearly independent particular solutions, which are sought using the method of. .. expressions into the system of differential equations in question, one

arrives at a linear nonhomogeneous system of algebraic equations for the coefficients A ∗ and B ∗