Mathematich soluti0n DE c and m ChapterDE c and m chapter 4 sc

Substitution in the first differential equation gives the linear first-order equation 2 2 of these expressions for y and y into the second differential equation yields the second-order

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 back-ground 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  and x x2   , so that x1 x 2

x  x y and 2

23

17. The computation x   yields the single linear second-order equation y x x  x 0

with characteristic equation r2   and general solution 1 0 x t  AcostBsint Then

the original first equation y gives x y t BcostAsint The figure shows a tion field and typical solution curves (obviously circles?) for the given system

x y

Problem 18

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

with characteristic equation r2   and general solution 1 0   t t

x t  Ae Be Then the

original first equation y gives x   t t

y t  Ae Be The figure shows a direction field and some typical solution curves of this system It appears that the typical solution curve

x x x with characteristic equation r2   , characteristic roots r 6 0 r 3

and 2, and general solution   3t 2t

x t  Ae Be Then the original first equation y x

y

Problem 24

24. The computation x   y 10x7y 10x7x yields the single linear second-order

equation x7x10x0 with characteristic equation r27r10 , characteristic 0roots 2r  and 5 , and general solution   2t 5t

x t  Ae Be Then the original first

25. The computation x   y 13x4y 13x4x yields the single linear second-order

equation x4x13x0 with characteristic equation r24r  and characteris-13 0tic roots r 2 3i; hence the general solution is   2  

t

x t e A tB t The tial condition x 0 0 then gives A0, so   2

y

Problem 26

26. The computation x   y 9x 6y  9x 6x yields the single linear second-order

equation x6x9x0 with characteristic equation r2    and repeated char-6r 9 0acteristic root r3, 3, so its general solution is given by     3t

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

or x2 y2 C2, the equation of a circle of radius C  A2B2

(b) Substituting the general solution found in Problem 18, we get

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

or x2 y2 C2, the equation of a circle of radius C  A2B2

(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 et and e we get 2 2t x y 3Aet

30 Looking at Fig 4.1.11 in the text, we see that the first spring is stretched by x , the sec-1

ond spring is stretched by x2 , and the third spring is compressed by x1 x Hence New-2ton's second law gives m x1 1  k x1 1 k2x2x1 and m x2 2  k2x2x1k3 x2

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

x

c  for i1, 2, 3, and each inflow-outflow rate is r10 Hence

,10

1

.10

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

50I 1000Q100, 25I 1000Q 0Then we differentiate each equation and substitute Q   to get the system I1 I2

36. If we write x y ,  for the velocity vector and    2 2

v x  y for the speed, then ,

SECTION 4.2

THE METHOD OF ELIMINATION

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

  2 2

t

y t c e Substitution in the first differential equation gives the linear first-order equation

2 2

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

second-order equation x2x x 0 with general solution

3. From the first differential equation we get 1 

32

32

y xx

Substitu-tion of these expressions for y and y into the second differential equaSubstitu-tion yields the

sec-ond-order equation x x 6x0 with general solution

y t  c e  c e Imposition of the initial conditions x 0 0, y 0 2 now gives the equations

x

y

Problem 4

4. Substitution of 3y x and x y3x —from the first equation—into the second x

equation yields the second-order equation x 4x0 with general solution

  2 2

x t c e c e Substitution of this solution in y3x gives x

  2 2

1 t 5 2 t

y t c e  c e The initial conditions yield the equations

c   Hence the desired particular solution is

5. Substitution of 1 

34

34

y  x x —from the first equation—into the second equation yields the second-order equation x2x5x0 with general solu-tion

x t e c tc t Substitution of this solution in 1 

34

The figure shows a direction field and some typical solution curves

7. Substitution of y x 4x and 2t y x 4x —from the first equation—into the 2

second equation yields the nonhomogeneous second-order equation

x x x  t Substitution of the trial solution x p  A Bt yields 1

18

A , 1

8. Substitution of y x 2x and y x 2x —from the first equation—into the second

equation yields the nonhomogeneous second-order equation x4x3x  Substi-e2t

tution of the trial solution yields A , so 1 2t

2 4 cos 23

y  x x t —from the first equation—into the second equation yields the second-order equation

  1 2  

1

7 cos 2 4 sin 25

x t c e c e  t t

2 2 sin 23

equation yields the second-order equation x4x3x0 with general solution

x t c e c e Substitution in y x 2x now yields

y t  c e c e Imposition of the initial conditions x 0 1, y 0  1 now gives the equations

12. The first equation yields 1 

62

y x x , so 1   4 

62

y x  x Substitution in the ond equation yields

sec-  4

x  x x The characteristic equation is 4 2  2  2 

y x x , so 1   4 

52

y x  x Substitution in the ond equation yields

sec-  4

x  x x The characteristic equation is 4 2  2  2 

14. The first equation x 4xsint has complementary function x cc1cos 2tc2sin 2t,

and substitution of the trial solution x Asint yields the particular solution 1sin

3

x t c tc t t Substitution in the second differential equation gives the equation

with complementary function y c c3cos 2 2 t c4sin 2 2 t Substitution of the trial solution cos 2y p  A tBsin 2tCsint now yields A , c1 B , and c2 4

Similarly, the general solution for y is of the form

y t c tc td td t Now, substitution of these two general solutions in the first equation x3y2x 0and collection of coefficients gives

3a13c2cost3c13a2sint  6b16d2cos 2t3d13b2sin 2t0

Thus we see finally that c1 , a2 c2   , a1 d1  , b2 d2   Hence b1

The associated homogeneous equation has characteristic equation  2  2 

we see (thinking of the operational determinant) that x satisfies a homogeneous

fourth-order equation with characteristic equation

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

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

y t  c c e c e Then the second given equation implies that

LxLyLz Hence

b    If c a and 2 b are chosen arbitrarily, then 2

13

c     Hence the general solution is a b

21 L L1 2  L L2 1 because both sides simplify to the same thing upon multiplying out and

col-lecting terms in the usual fashion of polynomial algebra This “works” because different

powers of D commute—that is, D D i j D D j i because i j i j j i

23. Subtraction of the two equations yields x y e2t e3t We then verify readily that

any two differentiable functions x t  and y t  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 x   But then y t2 t

           2  2

D x D yD xy  xy  t  t  t t   t

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 times the second one) 2

26. Subtraction of the second equation from the first one gives xet Then substitution in

the second equation yields D y2  , so 0 y b t1  Thus there are two arbitrary con- b2stants

27. Subtraction of the second equation from the first one gives x y et Then substitution

in the second equation yields

  2  t

x t D xy e

It follows that y t 0, so there are no arbitrary constants

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

2 t

D D xD y  e ,

which contradicts the first equation Hence the system has no solution

29. Addition of the two given equations yields D x2 et, so   1 2

t

x t e a ta Then the second equation gives D y2 a t1  , so a2

Thus there are four arbitrary constants

30. Substitution of y20x6x and y20x6x —from the first equation—into the

second equation yields the second-order equation 100x45x3x0 with general

r t r t

x t c e c e , where

I  I I —from the first tion—into the second equation yields the second-order equation 3I130I1125I1250with general solution

5 2

32. To solve the system

I Ce Substitution of the trial solution

25 /3 2

1

120 120 cos 60 1778sin 60 ,1321

1

240 240 cos 60 1728sin 60 1321

(Yes, one coefficient of sin 60t is 1778 and the other is 1728.)

33. To solve the system

so x , 1 x , and 2 x all satisfy a third-order homogeneous linear differential equation with 3

imply that b1 100 , a b2    After these substitutions, collection of coefficients b3 a

gives the equations

3 /20 2

3 /20 3

  cos sin

x t  A tB tC Now x 0 0 implies B0, and then x 0 r0 gives A C  Next, r0

2cos

      ,

so y 0  r0 implies that A , and hence that r0 C0 It now follows readily that

the trajectory is the circle

gen-eral form of the solution is

 

 

cos5 sin 5 cos 5 3 sin 5 3

 

 

cos 5 sin 5 cos 5 3 sin 5 3 ,

2 cos 5 2 sin 5 2 cos5 3 2 sin 5 3

(b) In the natural mode with frequency 1  the masses move in the same direction, 5

while in the natural mode with frequency 2 5 3 they move in opposite directions In each case the amplitude of the motion of m is twice that of 2 m 1

38 Looking at Fig 4.2.6 in the text, we see that the first spring is stretched by ,x the second

spring is stretched by y , and the third spring is compressed by x y Hence Newton’s

second law gives m x1   k x1 k2yx and m y2   k2y x k3 y

39 The system has operational determinant 4 2  2  2 

8D 40D 328 D 1 D  Hence 4the general form of the solution is

 

 

In the natural mode with frequency 1 the masses move in the same direction, with 1

the amplitude of motion of the second mass twice that of the first mass (figure a) In the

natural mode with frequency 2  they move in opposite directions with the same am-2

plitude of motion (figure b)

40 The system has operational determinant 4 2  2  2 

2D 250D 50002 D 25 D 100 Hence the general form of the solution is

 

 

2 cos 5 2 sin 5 cos10 sin10

In the natural mode with frequency 1 the masses move in the same direction (figure 1

a), while in the natural mode with frequency 2  they move in opposite directions 3

(figure b) In each case the amplitudes of motion of the two masses are equal

 

 

43. The system has operational determinant 4 2  2  2 

D  D   D  D  Hence the general form of the solution is

     

In the natural mode with frequency 1 the masses move in the same direction (figure 1

a), while in the natural mode with frequency 2  5 they move in opposite directions

(figure b) In each case the amplitudes of motion of the two masses are equal

 

 

In the natural mode with frequency 1 2 the two masses move in the same direction;

in the natural mode with frequency 2  they move in opposite directions In each nat-2ural mode their amplitudes of oscillation are equal

45. The system has operational determinant 4 2  2  2 

2D 20D 322 D 2 D  Hence 8the general form of the solution is

         

In the natural mode with frequency 1 2 the two masses move in the same direction

with equal amplitudes of oscillation (figure a) In the natural mode with frequency

2 8 2 2

   the two masses move in opposite directions with the amplitude of m be-2

ing half that of m (figure b) 1

In the natural mode with frequency 1 the masses move in the same direction with 2equal amplitudes of motion In the natural mode with frequency 2  they move in 4opposite directions with the same amplitude of motion

47 (a) Looking at Fig 4.2.7 in the text, we see that the first spring is stretched by x, the

sec-ond spring is stretched by y  , the third spring is stretched by z y x  , and the fourth

spring is compressed by z Hence Newton's second law gives mx  k x  k yx,

NUMERICAL METHODS FOR SYSTEMS

In Problems 1-8 we first write the given system in the form x  f t x y , ,  y g t x y , ,  Then

we use the template

(with the given values of t , 0 x , and 0 y ) to calculate the Euler approximations 0

Then with both step sizes h0.1 and h0.05 we get the actual value x 1 0.15058

accurate to 5 decimal places

13. With y we want to solve numerically the initial value problem x

When we run Program RK2DIM with step size h0.1 we find that the change of sign in

the velocity v occurs as follows:

t x v

7.6 1050.2 +2.8 7.7 1050.3 -0.4 Thus the bolt attains a maximum height of about 1050 feet in about 7.7 seconds

14. Now we want to solve numerically the initial value problem

Running Program RK2DIM with step size h0.1, we find that the bolt attains a

maxi-mum height of about 1044 ft in about 7.8 sec Note that these values are comparable to those found in Problem 13

15. With y  and with x in miles and in seconds, we want to solve numerically the initial x

Then, using the n-dimensional program rkn with step size 0.1 and initial data

corre-sponding to the indicated initial inclination angles, we got the following results:

Angle Time Range

17. The data in Problem 16 indicate that the range increases when the initial angle is

de-creased below 45 The further data

Angle Range 41.0 352.1 40.5 352.6 40.0 352.9 39.5 352.8 39.0 352.7 35.0 350.8 indicate that a maximum range of about 353 ft is attained with   40

18 We “shoot” for the proper inclination angle by running program rkn (with h0.1) as

follows:

Angle Range

60 287.1

58 298.5 57.5 301.1 Thus we get a range of 300 ft with an initial angle just under 57.5

19 First we run program rkn (with h0.1) with v0 250 ft sec and obtain the following