Differential equations and boundary value problems computing and modeling 5th edition edwards soluti by john.albers185

Differential Equations and Boundary Value Problems Computing and Modeling 5th Edition

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CHAPTER 4 INTRODUCTION TO SYSTEMS OF DIFFERENTIAL EQUATIONS

This chapter bridges the gap between the treatment of a single differential equation in Chapters 1-3 and the comprehensive treatment of linear and nonlinear systems in Chapters 5-6. It also is designed to offer some flexibility in the treatment of linear systems, depending on the background in linear algebra that students are assumed to have Sections 4.1 and 4.2 can stand alone as a very brief introduction to linear systems without the use of linear algebra and matrices. The final Section 4.3 of this chapter extends to systems the numerical approximation techniques of Chapter 2.

SECTION 4.1

FIRST-ORDER SYSTEMS AND APPLICATIONS

1. Let x1 = x and x2 = x1  = x  , so that x  = x  = 7x 3x  + t2 Equivalent system: x  = x , x  = 7x 3x + t2 1 2 2 1 2 2. Let x = x and x = x  = x  , so that x  = x  = 4x + x3 Equivalent system: 1 2 1 2 x  = x , x  = 4x + x3 . 1 2 2 1 1 3. Let x1 = x and x2 = x1  = x  , so that x2  = x  = 26x 2x  + 82cos4t . Equivalent system: x1  = x2, x2  = 26x1 2x2 + 82cos4t .

Copyright © 2015 Pearson Education, Inc. 4 4 4. Let x = x , x = x  = x  , and x = x  = x  , so that x  = x  = 2x  x  +1+ tet . Equivalent 1 system: 2 1 3 2 3 x  = x , x  = x , x  = 2x x + 1 + tet . 1 2 2 3 3 3 2 5. Let x1 = x , x2 = x1  = x  , x3 = x2  = x  , and x4 = x3  = x  , so that x  = x(4) = 3x  x + et sin2t Equivalent system: x  = x , x  = x , x  = x , x  = 3x x + et sin2t . 1 2 2 3 3 4 4 2 1 6. Let x1 = x, x2 = x1  = x  , x3 = x2  = x  , and x4 = x3  = x  , so that x  = x(4) = x + 3x  6x  + cos3t Equivalent system: 241

Copyright © 2015 Pearson Education, Inc. 1 2 2 1 2 1 1 1 1 242 FIRST-ORDER SYSTEMS AND APPLICATIONS x1  = x2, x2  = x3, x3  = x4, x4  = x1 + 3x2 6x3 + cos3t . (1 t2 ) x tx 7. Let x1 = x and x2 = x1  = x  , so that x  = x , x2  = x  = t2x  = (1 t2 ) x t2 tx Equivalent system: 8. Let x = x , x = x  = x  , and x = x  = x  , so that x  = x  = 1 ( 5x 3tx 2t2x  + ln t). 1 2 1 3 2 3 t3 Equivalent system: x  = x , x  = x , t3x  = 5x 3tx + 2t2x + lnt . 1 2 2 3 3 1 2 3 9. Let x = x , x = x  = x  , and x = x  = x  , so that x  = x  = (x )2 + cos x . Equivalent 1 2 1 3 2 3 system: x  = x , x  = x , x  = x 2 + cos x . 1 2 2 3 3 2 1 10. Let x1 = x , x2 = x1  = x  , y1 = y , and y2 = y1  = y  , so that x2  = x  = 5x 4y and y2  = y  = 4x + 5y . Equivalent system: x1  = x2, x2  = 5x1 4y1, y1  = y2, y2  = 4x1 + 5y1 . 11. Let x1 = x , x2 = x1  = x  , y1 = y , and y2 = y1  = y  , so that x2  = x  = and y2  = y  = . Equivalent system: x  = x , x  = kx1 , y  = y , y  = ky1 . 1 2 2 (x 2 + y 2 )3/2 1 2 2 ( x 2 + y 2 )3/2 12. Let x = x , x = x  = x  , y = y , and y = y  = y  , so that x  = x  = 2 y  and 1 2 1 1 2 1 2 3 y  = y  = 2 x  Equivalent system: 2 3 x + y + y kx ky

Copyright © 2015 Pearson Education, Inc. x  = x , x  = 2 y , y  = y , y  = 2 x 1 2 2 3 2 1 2 2 3 2 13. Let x1 = x , x2 = x1  = x  , y1 = y , and y2 = y1  = y  , so that x2  = x  = 75x + 25y and y2  = y  = 50x 50y + 50cos5t . Equivalent system:

17. The computation x

= x yields the single linear second-order equation x  + x = 0 with characteristic equation r2 + 1 = 0 and general solution x (t) = Acost + B sin t Then the original first equation y = x  gives y (t) = B cost Asin t The figure shows a direction field and typical solution curves (obviously circles?) for the given system. Problem 17

= y

Copyright © 2015 Pearson Education, Inc. Section 4.1 243 x1  = x2, x2  = 75x1 + 25y1, y1  = y2, y2  = 50x1 50y1 + 50cos5t . 14. Let x1 = x , x2 = x1  = x  , y1 = y , and y2 = y1  = y  , so that x2  = x  = 4x + 2y 3x  and y2  = y  = 3x y 2 y  + cost . Equivalent system: x1  = x2, x2  = 4x1 + 2y1 3x2, y1  = y2, y2  = 3x1 y1 2y2 + cost . 15. Let x1 = x , x2 = x1  = x  , y1 = y , y2 = y1  = y  , z1 = z , and z2 = z1  = z  , so that x2  = x  = 3x y + 2z , y2  = y  = x + y 4z , and z2  = z  = 5x y z . Equivalent system: 16. Let x1 = x , x2 = x1  = x  , x1  = x2 , y1  = y2 , z1  = z2, y1 = y , and x2  = 3x1 y1 + 2z1 y2  = x1 + y1 4z1 z2  = 5x1 y1 z1 y2 = y1  = y  , so that x2  = x  = x(1 y) and y2  = y  = y (1 x) .

x1  = x2, x2  = x1 (1 y1), y1  = y2, y2  = y1 (1 x1) .

Equivalent system:

5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x





18. The computation x  = y  = x yields the single linear second-order equation x  x = 0

with characteristic equation r2 1 = 0 and general solution x (t) = Aet + Be t . Then the original first equation y = x  gives y (t) = Aet Be t . The figure shows a direction field and some typical solution curves of this system. It appears that the typical solution curve is a branch of a hyperbola.

19. Thecomputation x  = 2y  = 4x yields the single linear second-order equation

x  + 4x = 0 with characteristic equation r2 + 4 = 0 and general solution

x (t) = Acos2t + B sin2t Then the original first equation y = 1 x  gives 2

y (t) = B cos2t + Asin2t . Finally, the condition x (0) = 1 implies that A = 1, and then

the condition y (0) = 0 gives B = 0. Hence the desired particular solution is given by

x (t) = cos2t, y (t) = sin2t .

The figure shows a direction field and some typical circular solution curves for the given system.

Problem 19

20. Thecomputation x  = 10y  = 100x yields the single linear second-order equation

x  +100x = 0 with characteristic equation r2 +100 = 0 and general solution

244 FIRST-ORDER SYSTEMS AND APPLICATIONS

5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x Problem 20 5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x

x (t) = Acos10t + B sin10t . Then the original first equation y = 1 10 x  gives

y (t) = B cos10t Asin10t . Finally, the condition x (0) = 3 implies that A = 3, and then

the condition y (0) = 4 gives B = 4 Hence the desired particular solution is given by

x (t) = 3cos10t + 4sin10t, y (t) = 4cos10t 3sin10t .

The typical solution curve is a circle, as the figure suggests.

21. Thecomputation x  = 1 y  = 4x 2 yields the single linear second-order equation

x  + 4x = 0 with characteristic equation r2 + 4 = 0 and general solution

x (t) = Acos2t + B sin2t . Then the original first equation y = 2x  gives y (t) = 4B cos2t 4 Asin2t . The figure shows a direction field and some typical elliptical solution curves.

22. Thecomputation x  = 8y  = 16x yields the single linear second-order equation

x  +16x = 0 with characteristic equation r2 + 16 = 0 and general solution

x (t) = Acos4t + B sin4t . Then the original first equation y = 1 x  gives 8

y t = B cos 4t A sin 4t . The typical solution curve is an ellipse. The figure shows a 2 2 direction field and some typical solution curves.

23. Thecomputation x  = y  = 6x y = 6x x  yields the single linear second-order equation

x  + x  6x = 0 with characteristic equation r2 + r 6 = 0, characteristic roots r = 3 and 2, and general solution x (t) = Ae 3t + Be2t . Then the original first equation y = x  gives y (t) = 3Ae 3t + 2Be2t . Finally, the initial conditions

x (0) = A + B = 1, y (0) = 3A + 2B = 2 implythat A = 0 and B = 1, so the desired particular solution is given by

( ) Section 4.1 245

Problem

5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x Problem

5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x

The figure shows a direction field and some typical solution curves.

Copyright © 2015 Pearson Education, Inc. 246 FIRST-ORDER SYSTEMS AND APPLICATIONS Problem 23 5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x Problem 24 5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x 24. Thecomputation x  = y  = 10x + 7y = 10x 7x  yields the single linear second-order equation x  + 7x  + 10x = 0 with characteristic equation r2 + 7r +10 = 0, characteristic roots r = 2 and 5, and general solution x (t) = Ae 2t + Be 5t . Then the original first equation y = x  gives y (t) = 2 Ae 2t + 5Be 5t Finally, the initial conditions x (0) = A + B = 2, y (0) = 2A + 5B = 7 implythat A = 17 3 and B = 11 , so the desired particular solution is given by 3 x (t) = 1 (17e 2t 11e 5t ), y (t) = 1 (34e 2t 55e 5t ) . 3 3 It

tangent to

y = 2x The figure 25. Thecomputation x  = y  = 13x 4y = 13x + 4x  yields

single linear second-order equation x  4x  + 13x = 0 with characteristic equation r2 4r +13 = 0 and characteristic roots r = 2  3i ; hence the general solution is x (t) = e2t ( Acos3t + Bsin3t) The initial condition x (0) = 0 thengives A = 0 , so x (t) = Be2t sin3t . Then the original first

appears that the typical solution curve is

the straight line shows a direction field and some typical solution curves.

the

equation y = x  gives y (t) = e2t (3B cos3t + 2B sin3t) . Finally, the initial condition

y (0) = 3 gives B = 1, so the desired particular solution is given by

x (t) = e2t sin3t, y (t) = e2t (3cos3t + 2sin3t) .

The figure shows a direction field and some typical solution curves.

= 3x . The figure shows a direction field

26. Thecomputation x  = y  = 9x + 6y = 9x + 6x  yields the single linear second-order equation x  6x  + 9x = 0 with characteristic equation r2 6r + 9 = 0 and repeated characteristic root r = 3,3, so its general solution is given by x (t) = ( A + Bt)e3t Then the original first equation y = x  gives y (t) = (3A + B + 3Bt) e3t . It appears that the typical solution curve is tangent to the straight line and some typical solution curves.

27. (a) Substituting the general solution found in Problem 17 we get

x2 + y2 = C2 , the equation of a circle of radius C = .

Substituting the general solution found in Problem 18, we get

equation of a hyperbola.

28. (a) Substituting the general solution found in Problem 19 we get

Section 4.1 247 Problem

5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x Problem

5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x

x2 + y2 = ( Acost + Bsin t)2 + (B cost Asin t)2 = ( A2 + B2 )(cos2 t + sin2 t) = ( A2 + B2 ), or

(b)

x2 y2 = ( Aet + Be t )2 ( Aet Be t )2 = 4AB ,

the

B cos2t

Asin2t)2 = ( A2 + B2 )(cos2 2t + sin2 2t)

+ y2 = ( Acos2t +

sin

)

+ (

or x2 + y2 = C2 , the equation of a circle of radius C = .

(b) Substituting the general solution found in Problem 21 we get

or 16x2 + y2 = C2 , the equation of an ellipse with semi-axes 1 and 4.

29. When we solve Equations (20) and (21) in the text for

and x + y = 3Be2t Hence

and

we get

30. Looking at Fig. 4.1.11 in the text, we see that the first spring is stretched by x1, the second spring is stretched by x2 x1 , and the third spring is compressed by x2 . Hence New-

31. Looking at Fig. 4.1.12 in the text, we see that

We get the desired equations when we multiply each of these equations by L T and set

32. The concentration of salt in tank i is ci = xi 100 for i = 1,2,3,… and each inflow-outflow

2 =

( Acos2t + Bsin2t)2 + (4Bcos2t 4Asin2t)2 = 16( A2 + B2 )(cos2 2t + sin2 2t) = 16( A2 + B2 ),

x2 + y

e t

e2t

2x y = 3Ae t

(2x y)2 (x + y) = (3Ae t )2  3Be2t = 27A2B = C . Clearly y = 2x or y = x

4x3 3xy2 + y3 = C .

if C = 0, and expansion gives the equation

1x1  = k1 ( x1) + k2 (x2 x1) and m2x2  = k2 (x2 x1) k3 (x2 )

ton's second law gives m

my  = T sin + T sin  T tan + T tan = Ty1 + T ( y2 y1) and 1 1 2 1 2 L L my  = T sin T sin  T tan T tan = T ( y2 y1) Ty2 . 2 2 3 2 3 L L

k = mL T

33. We apply Kirchhoff's law to each loop in Figure 4.1.14 in the text, and immediately get the equations

34. First we apply Kirchhoff's law to each loop in Figure 4.1.14 in the text, denoting by Q the charge on the capacitor, and get the equations 50I1 +1000Q = 100,

. Then we differentiate each equation and substitute

35. If  is the polar angular coordinate of the point ( x, y) then Newton’s second law gives

36. If we write ( x  , y ) for the velocity vector and v = for the speed, then

, y

is a unit vector pointing in the direction of the velocity vector, and so the com-

ponents of the air resistance force Fr are given by

37. If r = ( x, y, z) is the particle's position vector, then Newton's law mr  = F gives

2(I1  I2  ) + 50I1 = 100sin60t, 2(I2  I1 ) + 25I2 = 0.

Q = 0

Q = I1 I2

I1  = 20(I1 I2 ), I2  = 40(I1 I2 )

1000

to get the system

and we write F = k x2 + y2 = k , r 2 mx  = F cos = k r 2  x = kx , r r3 my  = F sin = k r 2  y = ky . r r3





 

 

F = kv2 

 ,

  = ( kvx , kvy)  

i j k

SECTION 4.2

THE METHOD OF ELIMINATION

1. The second differential equation y  = 2y has the exponential solution

Substitution in the first differential equation gives the linear first-order equation

with integrating factor

this equation in the usual way

The figure shows a direction field and some typical solution curves.

2. From the first differential equation we get y = 1 (x x

of these expressions for y and y  into the second differential equation yields the secondorderequation x  + 2x  + x = 0 with general solution

y = 1 ( x

) now yields

THE

OF ELIMINATION

250

METHOD

y (t) = c e2t .

x  + x

gives

 = et . Solution

x (t) = e t (c + c e3t ) = c e t + c e2t . 1 2 1 2

= 3

e2t

Problem 1 5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x Problem 2 5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x



 x ) Substitution

x (t) = (c + c t) e t .

1 2

x 

y (t) =  c 1 c + c t  e t

) ,

y = 1 ( x

Substitutionin

The figure shows a direction field and some typical solution curves.

3. From the first differential equation we get y =

(3x +

) .

of these expressions for y and y  into the second differential equation yields the second-orderequation

= 0 with general solution

4. Substitutionof y = 3x x

from the first equation into the second equation yields the second-order equation

and y

with general solution

Pearson Education, Inc. 2 2 1 2 2 Section 4.2 251

2015



y  = 1 (3x 



Substitution

x  x  6x

x (t) = c e 2t + c e3t . Substitutionin y = 1 (3x + x ) now yields y (t) = 1 c e 2t + 3c e3t . 2 1 2 Imposition of

initial conditions x (0) = 0 , y (0) = 2 now

the

c + c = 0, 1 c + 3c = 2 1 2 2 1 2 with solution c = 4 , c = 4 These coefficients give the desired particular

1 5 2 5 x (t) = 4 (e3t e 2t ), y (t) = 2 (6e3t e 2t ) . 5 5 The

some typical solution curves. Problem 3 5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x Problem 4 5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x

) , so

+ x

the

gives

equations

solution

figure shows a direction field and



 = 3x  x 

x  4x = 0

Copyright © 2015 Pearson Education, Inc. 1 2 4 4 1 2 4 252 THE METHOD OF ELIMINATION y (t) = c e2t + 5c e 2t . The initial conditions yield the equations c1 + c2 = 1, c1 + 5c2 = 1 with solution c = 3 , c = 1 . Hence the desired particular solution is 1 2 2 2 x (t) = 1 (3e2t e 2t ), y (t) = 1 (3e2t 5e 2t ) 2 2

figure

a direction field and some typical solution curves.

Substitutionof y = 1 ( x  + 3x) and y  = 1 ( x  + 3x ) from the first equation into the second

equation tion x  + 2x  + 5x = 0 with general solux (t) = e t (c cos2t + c sin2t) . Substitution of this solution in y = 1 ( x  + 3x) gives y (t) = 1 e t  (c + c )cos2t + (c c )sin2t . 2  1 2 1 2  The figure

a direction field and some typical solution curves. Problem 5 5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x Problem 6 5 4 3 2 1 y 0 −1 −2 −3 −4 −5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x 6. Substitutionof y = 1 (x  x) and y  = 1 ( x  x ) from the first equation

The

shows

equation yields the second-order

shows

into the

Copyright © 2015 Pearson Education, Inc. 1 2 9 p 1 2 1 2 Section 4.2 253 x (t) = e 2t (c cos3t + c sin3t) . Substitutionin y = 1 (x  x) now yields y (t) = 1 e 2t ( c + c )cos3t + ( c c )sin3t 3  1 2 1 2  Imposition of the initial conditions x (0) = 3, y (0) = 2 now gives the equations c1 = 3, c1 + c2 = 2 3 3 with solution c1 = 3, c2 = 9 These coefficients give the desired particular solution x (t) = e 2t (3cos3t + 9sin3t), y (t) = e 2t (2cos3t 4sin3t) . The figure shows a direction field and some typical solution curves. 7. Substitutionof y = x  4x 2t and y  = x  4x  2 from the first equation into the second equation yields the nonhomogeneous second-order equation x  5x  + 6x = 2 2t Substitution of the trial solution xp = A + Bt yields A = 1 , 18 B = 1 , so x = 1 t . Hence the general solution for x is 3 p 18 3 x (t) = c e2t + c e3t t + 1 . Substitutionin y = x  4x 2t 1 2 nowyields 3 18 y (t) = 2c e2t c e3t 2 t 5 . 1 2 3 9 8. Substitutionof y = x  2x and y  = x  2x  from the first equation into the second equation yields the nonhomogeneous second-order equation x  4x  + 3x = e2t Substitution of the trial solution yields A = 1, so x = e2t Hence the general solution for x is Substitutionin y = x  2x x (t) = c et + c e3t + e2t now yields y (t) = c et + c e3t

Copyright © 2015 Pearson Education, Inc. 3 3 9. Substitutionof y = 1 ( x  + 2x + 2sin2t) and y  = 1 ( x  + 2x  + 4cos2t) from the first equation into the second equation yields the second-order equation x  x = 7cos2t + 4sin2t with general solution

10. First we solve the given equations for the normal-form first-order equations

the first equation into the second equation yields the second-order

11. First we solve the given equations for the normal-form first-order equations

first equation into the second equation yields the nonhomogeneous

Copyright © 2015 Pearson Education, Inc. 3 1 2 1 2 9 9 2 p 254 THE METHOD OF ELIMINATION x (t) = c e t + c et 1 (7cos2t + 4sin2t) . Substitutionin 1 2 y = 1 ( x  + 2x + 2sin2t) 5 now yields y (t) = c e t + 1 c et 1 (2cos2t + 4sin2t) . 1 3 2 5

x  = 2x + y, y  = x + 2 y . Substitutionof y = x  2x and y  = x  2x  from

equation x  4x  + 3x = 0 with general solution x (t) = c et + c e3t . Substitutionin y = x  2x now yields y (t) = c et + c e3t . Imposition of the initial conditions x (0) = 1, y (0) = 1 now gives the equations c1 + c2 = 1, c1 + c2 = 1 with

c1 = 1, c2 = 0 . These coefficients give the desired particular solution x (t) = et , y (t) = et

solution

x  = 3x 9 y + e t + 2et , y  = 2x 3y + 1 e t + 3 et . 2 2 Substitutionof y = 1 ( x  + 3x + e t + 2et ) and y  = 1 ( x  + 3x  e t + 2et ) from the

tion x  + 9x = 1 (5e t + 11et ). Substitution of the trial solution x = Ae t + Bet yields A = 1 , 4 B = 11 , so 20 x = 1 e t 11 et Hence the general solution for x is p 4 20 x(t) = c1 cos3t + c2 sin3t 1 e t 11 et . 4 20

second-order equa-

Copyright © 2015 Pearson Education, Inc. 2 2 1 2 3 4 2 2 2 2 12. The first equation yields ond equation yields y = 1 (x  6x) , so Section 4.2 255 y  = 1 (x(4) 6x ) Substitution in the secThe characteristic equation is tion for x is x(4) 13x  + 36x = 0 . r4 13r2 + 36 = (r2 4)(r2 9) = 0, so the general solux (t) = c e2t + c e 2t + c e3t + c e 3t . Substitutionin y = 1 ( x  6x) now gives y (t) = c e2t c e 2t + 3 c e3t + 3 c e 3t . 1 2 2 3 2 4 13. The first equation yields ond equation yields y = 1 (x  + 5x) , so y  = 1 (x(4) + 5x ). Substitution in the secx(4) + 13x  + 36x = 0. The characteristic equation is r4 + 13r2 + 36 = (r2 + 4)(r2 + 9) = 0, so the general solution for x is Substitution in x (t) = a1 cos2t + a2 sin2t + b1 cos3t + b2 sin3t y = 1 ( x  + 5x) now gives y (t) = 1 a cos2t + 1 a cos2t 2b cos3t 2b sin3t 2 1 2 2 1 2 14. The first equation x  + 4x = sin t has complementary function xc = c1 cos2t + c2 sin2t , and substitution of the trial solution Hence the general solution for x is x = Asin t yields the particular solution x = 1 sin t p 3 x(t) = c1 cos2t + c2 sin2t + 1 sin t . 3

Substitution in the second differential equation gives the equation

15. If we write the given differential equations in operator notation as

then we see that the system has operational determinant

the general solution for y is of the form

Now, substitution of these two general solutions in the first equation and collection of coefficients gives

see finally that

16. In operational form our system is

When we operate on the first equation with tract, the result is

Copyright © 2015 Pearson Education, Inc. 256 THE METHOD OF ELIMINATION with complementary function yc = c3 cos(2 2t) + c4 sin(2 2t) Substitution of the trial solution yp = Acos2t + Bsin2t + C sin t nowyields A = c1 , B = c2 , and C = 4 , so the 21 general solution for y is y (t) = c cos2t + c sin2t + c cos(2 2t) + c sin(2 2t) + 4 sin t . 1 2 3 4 21

(D2 2) x 3Dy = 0 3Dx + (D2 2) y = 0,

(D2 2)2 + 9D2 = D4 + 5D2 + 4 = (D2 + 1)(D2 + 4). Therefore (as in

we see

x satisfies the fourth-order differential equation (D2 + 1)(D2 + 4) x = 0 with characteristic equation (r2 + 1)(r2 + 4) = 0 and general solution x (t) = a1 cost + a2 sint + b1 cos2t + b2 sin2t Similarly,

y (t) = c1 cost + c2 sin t + d1 cos2t + d2 sin2t

x  3y  2x = 0 ( 3a1 3c2 )cost + (3c1 3a2 )sint + ( 6b1 6d2 )cos2t + (3d1 3b2 )sin2t = 0 Thus we

c1 = a2 , c2 = a1 , d1 = b2 , d2 = b1 Hence y (t) = a2 cost a1 sin t + b2 cos2t b1 sin2t .

Example 3)

that

(D2 4) x + 2Dx + 13Dy (D2 9) y = 6sin t, = 0.

D2 9 , on the second with

D ,

sub(D4 + 13D2 + 36) x = 60sin t .

and

The associated homogeneous equation has characteristic equation (r2 + 4)(r2 + 9) = 0 and complementary function x

t . Upon substi-

tuting the trial solution xp = Asin t we find that A = 5 . Thus the general solution for x is

We find similarly that

When we substitute these expressions into either of the original differential equations we find that

17. If we write the given equations in the operational form

we see (thinking of the operational determinant) that x satisfies a homogeneous fourthorder equation with characteristic equation

Hence the general solution for x is

similarly, the general solution for y is

To determine the relations between the arbitrary constants in these two general solutions, we substitute them in the first of the original differential equations and get

t + a2 sin2t + b1

= a

cos2

cos3

sin3

2 x (t) = a cos2t + a sin2t + b cos3t + b sin3t + 5 sin t 1 2 1 2 2

y (t) = c cos2t + c sin2t + d cos3t + d sin3t + 1 cost . 1 2 1 2 2

c =

a , c = 4 a , d = 1 b , and d = 1 b . Therefore 1 13 2 2 13 1 1 3 2 2 3 1 y (t) = 4 ( a cos2t + a sin2t) + 1 ( b cos3t + b sin3t) + 1 cost 13 2 1 3 2 1 2

(D2 3D 2) x (2D2 9D 4) x + (D2 D + 2) y + (3D2 2D + 6) y = 0, = 0,

(r2 3r 2)(3r2 2r + 6) (r2 r + 2)(2r2 9r 4) = r4 3r2 4 = (r2 + 1)(r2 4) = (r2 + 1)(r + 2)(r 2) = 0.

x (t) = a cost + a sin t + b e 2t + b e2t ; 1 2 1 2

y (t) = c cost + c sin t + d e 2t + d e2t . 1 2 1 2

( a cost a sin t + 4b e 2t + 4b e2t ) + ( c cost c sin t + 4d e 2t + 4d e2t ) +

Copyright © 2015 Pearson Education, Inc. 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 (3a sin t 3a cost + 6b e 2t 6b e2t ) + (c sin t c cost + 2d e 2t 2d e2t ) + ( 2a cost 2a sin t 2b e 2t 2b e2t ) + (+2c cost + 2c sin t + 2d e 2t + 2d e2t ) = 0.

If we collect coefficients of the trigonometric and exponential terms we get the equations

18. Adding the first and third equations gives equation gives

4, so that

Then the second given equation implies that

Finally, the first given equation completes the solution:

19. The operational determinant of the given system is

y, and

all satisfy a third-order homogeneous linear differential equation with characteristic equation r3 12r

0 The corresponding general solu-

If we substitute x (t) and y (t) in the first differential equation x

2y and collect

OF ELIMINATION

METHOD

3a1 3a2 + c1 c2 = 0, 3a1 3a2 + c1 + c2 = 0 and 8b1 + 8d1 = 0, 4b2 + 4d2 = 0.

c1 = 3a2 and c2 = 3a1 , while the

give d1 = b1 and d2 = b2 We

see finally that y (t) = 3a cost + 3a sin t b e 2t + b e2t .

The first two of these equations imply that

latter two

therefore

x  + z  = 0. Then differentiation of

x  = + 2y  + = 2 y  Then

twice and substituting 2y  for x  gives y  = 6x  y  = 12y  y  , or y  + y  12 y  = 0 , with characteristic equation r3 + r2 12r = 0 and characteristic roots r = 0,3,

y (t) = c + c e3t + c e 4t . 0 1 2

the first

differentiating the second equation

x (t) = 1 ( y + y ) = 1 (c + 4c e3t 3c e 4t ) . 6 6 0 1 2

z (t) = x  x 2 y = 1 ( 13c 4c e3t + 3c e 4t ) . 6 0 1 2

L = 4 D 4 2 = D3 12D2 + 32D , 0 4 D 4

r 4)(r 8)

x (t) = a + a e4t + a e8t , y (t) = b + b e4t + b e8t , z (t) = c + c e4t + c e8t .

4 2 0

2 + 32r =

(

tions are



x  z 

coefficients of like terms, we find quickly that b1 = 2a1 , b2 = 0 , and b3 = 2a3 Similarly, we find by substitution in the other two equations that c1 = 2a1, c2 = 2a2 , and c3 = 2a3 . Thus y and z are given by

Education, Inc.

20. The operational determinant of the given system is

and we find that

1 because both sides simplify to the same thing upon multiplying out and collecting terms in the usual fashion of polynomial algebra. This “works” because different powers of D commute that is,

22. The following calculations show that L

in general:

L = D3 3D 2 = (D + 1)2 (D 2) ,

Hence Lx = Ly = Lz = 0. x = a e2t + a e t + a te t , y = b e2t + b e t + b te t , z = c e2t + c e t + c te t . When

three

efficients of e2t

a = b = c . When

coefficients of

1 1 1 that a + b + c = 0. Comparison of coefficients of e t yields 3 3 3 a2 + b2 + c2 = a3 1 = b3 = c3 . It follows that a = 2 and b = c = 1 . If a and b are chosen arbitrarily, then 3 3 3 3 3 2 2 c = a b 1 . Hence the general solution is 2 2 2 3 x = a e2t + a e t + 2 te t , 1 2 3 y = a e2t + b e t 1 te t , 1 2 z = a e2t  a 3 + b + 1  e t 1 te t 1  2 2 3  3 21. L1L2   =

i D j = D j Di because Di (D j x) = Di+ j x = D j (Di x).

we substitute these expressions in the

differential equations and compare co-

, we find that

we compare

te t we find

1L2x  L2L1x

whereas L1 (L2x) = (tD + 1)(Dx + tx) = tD (Dx + tx) + (Dx + tx) = t (D2 x + tDx + x) + (Dx + tx) = tD2x + t2Dx + Dx + 2tx,

23. Subtraction of the two equations yields x +

. We then verify readily that any two differentiable functions x

satisfying this condition will constitute a solution of the given system, which thus has infinitely many solutions.

24. Subtraction of one equation from the other yields

But then

Thus the given system has no solution.

25. Infinitely many solutions, because any solution of the second equation also satisfies the first equation (because it is D + 2 times the second one).

26. Subtraction of the second equation from the first one gives x = e t . Then substitution in the second equation yields stants.

Thus there are two arbitrary con-

27. Subtraction of the second equation from the first one gives in the second equation yields x + y = e t Then substitution It follows that

there are no arbitrary constants.

28. Differentiation of the difference of the two given equations yields

which contradicts the first equation. Hence the system has no solution.

29. Addition of the two given equations yields D2x =

(

= e t + a t + a Then the second equation gives D2 y = a t + a , so

= e 2t e 3t

y (t)

(

and

x + y = t2 t

(D + 2) x + (D + 2) y = D ( x + y) + 2( x + y) = (2t 1) + 2(t2 t) = 2t2 1  t

D2 y = 0 , so y = b1t + b2 .

x (t) = D2 ( x + y) = e t . y (t)  0, so

(D2 + D) x + D2 y = 2e t ,

y (t) = 1 a t3 + 1 a t2 + a t + a .

Thus there are four arbitrary constants. 30.

Substitution

y = 20x  + 6x and y  = 20x  + 6x  from

100x  + 45x  + 3x = 0

solution x(t) = c er1t + c er2t , where

the first equation into the second equation yields the second-order equation

with general

tion into the second equation yields the second-order equation

Copyright © 2015 Pearson Education, Inc. 1 2 rt r t Section 4.2 261 r = 9 + 33 , r = 9 33 Substitutionin 1 40 2 40 y = 20x  + 6x now yields y (t) = 1 (3 + 33)c ert + 1 (3 33)c er t . 2 1 2 2 Imposition of the initial conditions x (0) = 50, y (0) = 100 now gives the equations c1 + c2 = 50, (3 + 33) c1 + (3 33) c2 = 200 withsolution c = 25 (1 + 33), c = 25 ( 1 + 33) . These coefficients give the de1 33 2 sired particular solution x (t) = (1 + 33)e 1 + ( 1 + 33)e 2  , rt r t   y (t) = (11 + 3 33)e 1 + (11 3 33)e 2  . 11   31. Substitutionof I2 = (I1  + 25I1 50) and I2  = (I1+ 25I1 ) from the first equa-

3I1+ 30I1  + 125I1 = 250

I (t) = 2 + e 5t  c cos  5 6 t  + c sin  5 6  1  1  3  2  3         Substitutionin I2 = (I1  + 25I1 50) now yields I (t) = 1 e 5t (12c + 6c )cos  5 6 t  + (12c 6c )sin  5 6 t  2 15  1 2  3  2 1  3         Imposition of the initial conditions I1 (0) = 0, I2 (0) = 0 now gives the equations c + 2 = 0, 4 c + 6 c = 0 1 5 1 15 2 t 33 25 33 50 1 25 1 25 1 25

with general solution

. These coefficients give the desired particular solution

c2 = 4

I (t) = 2 + e 5t  2cos  5 6 t  + 4 6sin  5 6 t  , 1   3   3         I (t) = e 5t sin  5 6  2  3    t 6

with solution c

32. To solve the system

= 2I1  + 240cos60t . Substitution of these into the second equation gives the firstorder equation

I1 (0) = 0 gives C = A Consequently we find that

(Yes, one coefficient of sin60t is 1778 and the other is 1728.)

33. To solve the system

Substitution of this into the first equation gives the simple

34. The operational determinant of the system

2(I1  I2  ) + 50I1 = 100sin60t, 2(I2  I1 ) + 25I2 = 0 we

2 ( I1  I2  ) = 25I2 , so the

50I1 = 100sin60t , hence I2 = 2I1 + 4sin60t, and



I = Ce

first note from the second equation that

first equation then tells us that 25

2 

6I1

+ 50I1 = 480cos60t + 100sin60t with complementary function

/3 . Substitution of the trial solution I1 p = Acos60t + Bsin60t yields A = 120 , 1321

= 1778 . Finally, the initial condition 1321

I t

120e 25t/3 120cos60t + 1778sin60t , 1 1321

( 240e 25t/3 + 240cos60t + 1728sin60t). 2 1321

= 1

I (

) = 1

I1  = 20(I1 I2 ), I2  = 40(I1 I2 ) we first note that I2  = 2I1  , so I2 = 2I1 + K . Then K = 2I1 (0) + I2 (0) = 2  2 + 0 = 4, so I2  = 2I1  , so I2

first-order

I1  + 60I1 = 80

I (t) = 4 + ce 60t

initial

0) = 2

c = 2 , so I (t) = 2 (2 + e 60t ), I (t) = 4 (1 e 60t ) .

= 2I

+ 4

linear equation

with general solution

. The

condition

(

gives

10x1  = x1 + x3, 10x2  = x1 x2, 10x3  = x2 x3 is

so x1, x2 , and x3 all satisfy a third-order homogeneous linear differential equation with characteristic equation

Substituting in the original differential equations, we see that the constant terms are all equal:

1000r3 + 300r2 + 30r = 10r (100r2 + 30r + 3) = 0 and characteristic roots r = 0, 1 3  i 20 3) The corresponding general solutions are x (t) = a + e 3t/20  b cos  3 t  + c sin  3  1 1  1  20  1  20         x (t) = a + e 3t/20  b cos  3 t  + c sin  3  2 2  2  20  2  20         x (t) = a + e 3t/20  b cos  3 t  + c sin  3  3 3  3  20  3  20        

a1 = a2 = a3 = a

conditions x1 (0) = 100, x2 (0) = x3 (0) = 0 imply that b1 = 100 a , b2 = b3 = a . After

collection of coefficients

the equations 3 a + 3 c 50 = 0, 3 a 1 c c 50 = 0, 2 2 1 2 2 1 3 3 a + 3 c 100 = 0, 3 a c 1 c = 0 2 2 2 2 1 2 2 that we solve for a = 100 , c = 0, c = 100 , and c = 100 . The resulting solution of 3 1 2 3 the original system is given by x1 (t)  1+ 2e 3t/20 cos  3  t  , 3   20   x2 (t) = 100   3t /20  cos t  + 3sin  3   3     20   20   x3 (t) = 100   3t /20  cos t  3sin  3   t , t , t 3 3

(say). Then the initial

these substitutions,

gives

35. The two given equations yield

The general solution

mx(3) = 2 2 qBy = x  , m so x(3) + 2x  = 0.

36. With  = qB

our differential equations are

in Eqs. (3) of Section 4.1, we get the system

If we set

with operational determinant

form of the solution is

that y  = x  = A2 cost , A = r0 , and hence that C = 0. It now follows readily that the trajectory is the circle x (t) = r0 cost, y (t) = r0 sint

implies

x  = y  + qE , m y  = x  Eliminationgives so x(4) + 2x  = y(4) + 2 y  = 0, x = a1 + b1t + c1 cost + d1 sint, y = a2 + b2t + c2 cost + d2 sin t. The initial conditions x (0) = x (0) = 0 = y (0) = y (0) yield c1 = a1, b1 = d1 , c2 = a2 , and b2 = d2 .

in y  = x  yields d1 = b1 = a2 = c2 = 0 and d2 = a1

in x  = y  + qE m yields a = a1 = E , so the solution is B x (t) = a (1 cost), y (t) = a (t sin t) .

Then substitution

Finally, substitution

1 2 2 1 2 2x  = 100x + 25y, 1 y  = 25x 25y 2

eral

D4 + 100D2 + 1875 = (D2 + 25)(D2 + 75) . Hence the gen-

37. (a)

m = 2, m = 1 , k = 75, and k