CHAPTER 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 SECTION 4.1 FIRST-ORDER SYSTEMS AND APPLICATIONS Let x1 x and x2 x1 x , so that x2 x 7 x 3x t Equivalent system: x1 x2 , x2 7 x1 3x2 t Let x1 x and x2 x1 x , so that x2 x 4 x x Equivalent system: x1 x2 , x2 4 x1 x13 Let x1 x and x2 x1 x , so that x2 x 26 x x 82 cos 4t Equivalent system: x1 x2 , x2 26 x1 x2 82 cos 4t Let x1 x , x2 x1 x , and x3 x2 x , so that x3 x x x tet Equivalent system: x1 x2 , x2 x3 , x3 x3 x2 tet Let x1 x , x2 x1 x , x3 x2 x , and x4 x3 x , so that x4 x 3x x et sin 2t Equivalent system: x1 x2 , x2 x3 , x3 x4 , x4 3x2 x1 et sin 2t Let x1 x, x2 x1 x , x3 x2 x , and x4 x3 x , so that x4 x (4) x 3x x cos 3t Equivalent system: 241 Copyright © 2015 Pearson Education, Inc 242 FIRST-ORDER SYSTEMS AND APPLICATIONS x1 x2 , x2 x3 , x3 x4 , x4 x1 3x2 x3 cos 3t 1 t x tx x x , so that x x Let x1 x and x2 Equivalent system: t2 x1 x2 , t x2 1 t x1 tx2 Let x1 x , x2 x1 x , and x3 x2 x , so that x3 x 5 x 3tx 2t x ln t t Equivalent system: x1 x2 , x2 x3 , t x3 x1 3tx2 2t x3 ln t Let x1 x , x2 x1 x , and x3 x2 x , so that x3 x x cos x Equivalent system: x1 x2 , x2 x3 , x3 x2 cos x1 10 Let x1 x , x2 x1 x , y1 y , and y2 y1 y , so that x2 x x y and y2 y 4 x y Equivalent system: x1 x2 , x2 x1 y1 , 11 y2 4 x1 y1 Let x1 x , x2 x1 x , y1 y , and y2 y1 y , so that x2 x y2 y x ky y2 32 kx x2 y2 32 and Equivalent system: x1 x2 , x2 12 y1 y2 , kx1 x12 y 3/2 , y1 y2 , y2 ky1 x12 y12 Let x1 x , x2 x1 x , y1 y , and y2 y1 y , so that x2 x 3/2 y and y2 y x Equivalent system: x1 x2 , x2 13 y2 , y1 y2 , y2 x2 Let x1 x , x2 x1 x , y1 y , and y2 y1 y , so that x2 x 75 x 25 y and y2 y 50 x 50 y 50cos5t Equivalent system: Copyright © 2015 Pearson Education, Inc Section 4.1 243 x1 x2 , x2 75 x1 25 y1 , y1 y2 , y2 50 x1 50 y1 50cos5t Let x1 x , x2 x1 x , y1 y , and y2 y1 y , so that x2 x 4 x y 3x and y2 y 3x y y cos t Equivalent system: 14 x1 x2 , x2 4 x1 y1 3x2 , y1 y2 , y2 3x1 y1 y2 cos t Let x1 x , x2 x1 x , y1 y , y2 y1 y , z1 z , and z2 z1 z , so that x2 x 3x y z , y2 y x y z , and z2 z x y z Equivalent system: 15 x1 x2 , x2 3x1 y1 z1 y1 y2 , y2 x1 y1 z1 z1 z2 , z2 x1 y1 z1 Let x1 x , x2 x1 x , y1 y , and y2 y1 y , so that x2 x x 1 y and 16 y2 y y 1 x Equivalent system: x1 x2 , x2 x1 1 y1 , y1 y2 , y2 y1 1 x1 The computation x y x yields the single linear second-order equation x x with characteristic equation r and general solution x t A cos t B sin t Then 17 the original first equation y x gives y t B cos t A sin t The figure shows a direction field and typical solution curves (obviously circles?) for the given system Problem 17 Problem 18 5 4 3 2 y y 0 −1 −1 −2 −2 −3 −3 −4 −4 −5 −5 −5 −4 −3 −2 −1 −5 −4 −3 −2 −1 x x Copyright © 2015 Pearson Education, Inc 244 FIRST-ORDER SYSTEMS AND APPLICATIONS The computation x y x yields the single linear second-order equation x x with characteristic equation r and general solution x t Aet Be t Then the 18 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 The computation x 2 y 4 x yields the single linear second-order equation x x with characteristic equation r and general solution x t A cos 2t B sin 2t Then the original first equation y x gives y t B cos 2t A sin 2t Finally, the condition x implies that A , and then 19 the condition y gives B Hence the desired particular solution is given by x t cos 2t , y t sin 2t The figure shows a direction field and some typical circular solution curves for the given system Problem 19 Problem 20 5 4 3 2 y y 0 −1 −1 −2 −2 −3 −3 −4 −4 −5 −5 −5 −4 −3 −2 −1 −5 −4 −3 −2 −1 x 20 x The computation x 10 y 100 x yields the single linear second-order equation x 100 x with characteristic equation r 100 and general solution x t A cos10t B sin10t Then the original first equation y x gives 10 y t B cos10t A sin10t Finally, the condition x implies that A , and then the condition y gives B Hence the desired particular solution is given by x t 3cos10t 4sin10t , y t cos10t 3sin10t The typical solution curve is a circle, as the figure suggests Copyright © 2015 Pearson Education, Inc Section 4.1 245 y 4 x yields the single linear second-order equation x x with characteristic equation r and general solution x t A cos 2t B sin 2t Then the original first equation y x gives The computation x 21 y t B cos 2t A sin 2t The figure shows a direction field and some typical ellipti- cal solution curves Problem 21 Problem 22 5 4 3 2 y y 0 −1 −1 −2 −2 −3 −3 −4 −4 −5 −5 −5 −4 −3 −2 −1 −5 −4 −3 −2 −1 x x 22 The computation x y 16 x yields the single linear second-order equation x 16 x with characteristic equation r 16 and general solution x t A cos 4t B sin 4t Then the original first equation y x gives B A y t cos 4t sin 4t The typical solution curve is an ellipse The figure shows a 2 direction field and some typical solution curves 23 The computation x y x y x x yields the single linear second-order equation x x x with characteristic equation r r , characteristic roots r 3 and 2, and general solution x t Ae 3t Be t Then the original first equation y x gives y t 3 Ae 3t Be t Finally, the initial conditions x A B 1, y 3 A B imply that A and B , so the desired particular solution is given by x t e2t , y t 2e t The figure shows a direction field and some typical solution curves Copyright © 2015 Pearson Education, Inc 246 FIRST-ORDER SYSTEMS AND APPLICATIONS Problem 23 Problem 24 5 4 3 2 y y 0 −1 −1 −2 −2 −3 −3 −4 −4 −5 −5 −5 −4 −3 −2 −1 −5 −4 −3 −2 −1 x 24 x The computation x y 10 x y 10 x x yields the single linear second-order equation x x 10 x with characteristic equation r 7r 10 , characteristic roots r 2 and 5 , and general solution x t Ae 2 t Be 5t Then the original first equation y x gives y t Ae 2 t 5Be 5t Finally, the initial conditions x A B 2, imply that A y A 5B 7 17 11 and B , so the desired particular solution is given by 3 1 17e 2 t 11e 5t , y t 34e 2 t 55e 5t 3 It appears that the typical solution curve is tangent to the straight line y x The figure shows a direction field and some typical solution curves x t 25 The computation x y 13x y 13x x yields the single linear second-order equation x x 13x with characteristic equation r 4r 13 and characteristic roots r 3i ; hence the general solution is x t e t A cos 3t B sin 3t The initial condition x then gives A , so x t Be t sin 3t Then the original first equation y x gives y t e t 3B cos 3t B sin 3t Finally, the initial condition y gives B 1 , so the desired particular solution is given by x t e t sin 3t , y t e t 3cos 3t 2sin 3t The figure shows a direction field and some typical solution curves Copyright © 2015 Pearson Education, Inc Section 4.1 247 Problem 25 Problem 26 5 4 3 2 y y 0 −1 −1 −2 −2 −3 −3 −4 −4 −5 −5 −5 −4 −3 −2 −1 −5 −4 −3 −2 −1 x 26 x The computation x y 9 x y 9 x x yields the single linear second-order equation x x x with characteristic equation r 6r 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 A B 3Bt e3t It appears that the typical solution curve is tangent to the straight line y 3x The figure shows a direction field and some typical solution curves 27 (a) Substituting the general solution found in Problem 17 we get x y A cos t B sin t B cos t A sin t 2 A2 B cos2 t sin t A2 B , or x y C , the equation of a circle of radius C A2 B (b) Substituting the general solution found in Problem 18, we get x y Aet Be t Aet Be t AB , 2 the equation of a hyperbola 28 (a) Substituting the general solution found in Problem 19 we get x y A cos 2t B sin 2t B cos 2t A sin 2t A2 B cos2 2t sin 2t A2 B , Copyright © 2015 Pearson Education, Inc 248 FIRST-ORDER SYSTEMS AND APPLICATIONS or x y C , the equation of a circle of radius C A2 B (b) Substituting the general solution found in Problem 21 we get 16 x y 16 A cos 2t B sin 2t B cos 2t A sin 2t 2 16 A2 B cos2 2t sin 2t 16 A2 B , or 16x y C , the equation of an ellipse with semi-axes and 29 When we solve Equations (20) and (21) in the text for e t and e 2t we get x y Ae t and x y 3Be t Hence x y x y Ae t 2 3Be t 27 A2 B C Clearly y x or y x if C , and expansion gives the equation x 3xy y C 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 Newton's second law gives m1 x1 k1 x1 k2 x2 x1 and m2 x2 k2 x2 x1 k3 x2 31 Looking at Fig 4.1.12 in the text, we see that my1 T sin 1 T sin T tan 1 T tan Ty1 T y2 y1 L L and T ( y2 y1 ) Ty2 L L L We get the desired equations when we multiply each of these equations by and set T mL k T my2 T sin T sin 3 T tan T tan 32 The concentration of salt in tank i is ci rate is r 10 Hence xi for i 1, 2,3, and each inflow-outflow 100 Copyright © 2015 Pearson Education, Inc Section 4.1 249 x1 x3 , 10 x2 rc1 rc2 x1 x2 , 10 x3 rc2 rc3 x2 x3 10 x1 rc1 rc3 33 We apply Kirchhoff's law to each loop in Figure 4.1.14 in the text, and immediately get the equations I1 I 2 50 I1 100sin 60t , I 2 I1 25I 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 50 I1 1000Q 100, 25I 1000Q Then we differentiate each equation and substitute Q I1 I to get the system I1 20 I1 I , I 2 40 I1 I 35 If is the polar angular coordinate of the point x, y and we write F k k 2, x y r then Newton’s second law gives mx F cos 36 k x kx k y ky , my F sin r r r r r r If we write x, y for the velocity vector and v x y for the speed, then x y , is a unit vector pointing in the direction of the velocity vector, and so the comv v ponents of the air resistance force Fr are given by x y Fr kv , kvx, kvy v v 37 If r x, y , z is the particle's position vector, then Newton's law mr F gives i mr qv × B q x j k y z qBy i qBxj qB y, x,0 B Copyright © 2015 Pearson Education, Inc 250 THE METHOD OF ELIMINATION SECTION 4.2 THE METHOD OF ELIMINATION The second differential equation y y has the exponential solution y t c2 e t Substitution in the first differential equation gives the linear first-order equation x x 3c2e t with integrating factor et Solution of this equation in the usual way gives x t e t c1 c2e3t c1e t c2 e2 t The figure shows a direction field and some typical solution curves Problem Problem 5 4 3 2 y y 0 −1 −1 −2 −2 −3 −3 −4 −4 −5 −5 −5 −4 −3 −2 −1 −5 −4 −3 −2 −1 x 2 x 1 x x , so y x x Substitution 2 of these expressions for y and y into the second differential equation yields the secondorder equation x x x with general solution From the first differential equation we get y x t c1 c2t e t Substitution in y x x now yields y t c1 c2 c2t e t The figure shows a direction field and some typical solution curves Copyright © 2015 Pearson Education, Inc Section 4.2 265 x t a1 cos5t a2 sin 5t b1 cos5 3t b2 sin 3t , y t 2a1 cos5t 2a2 sin 5t 2b1 cos5 3t 2b2 sin 3t (b) In the natural mode with frequency 1 the masses move in the same direction, while in the natural mode with frequency 2 they move in opposite directions In each case the amplitude of the motion of m2 is twice that of m1 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 x , and the third spring is compressed by y Hence Newton’s second law gives m1 x k1 x k2 y x and m2 y k2 y x k3 y 38 The system has operational determinant 8D 40 D 32 D 1 D Hence 39 the general form of the solution is x t a1 cos t a2 sin t b1 cos 2t b2 sin 2t , y t c1 cos t c2 sin t d1 cos 2t d sin 2t Upon substitution in either differential equation we see that c1 2a1 and c2 2a2 , and d1 b1 and d b2 This gives x t a1 cos t a2 sin t b1 cos 2t b2 sin 2t , y t 2a1 cos t 2a2 sin t b1 cos 2t b2 sin 2t In the natural mode with frequency 1 the masses move in the same direction, with 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 amplitude of motion (figure b) Problem 39a Problem 39b x1 = cos(t) x, y x, y x2 = cos(2t) y1 = cos(t) y2 = − cos(2t) t t Copyright © 2015 Pearson Education, Inc 266 40 THE METHOD OF ELIMINATION The system has operational determinant D 250 D 5000 D 25 D 100 Hence the general form of the solution is x t a1 cos5t a2 sin 5t b1 cos10t b2 sin10t , y t c1 cos5t c2 sin 5t d1 cos10t d sin10t Upon substitution in either differential equation we see that c1 2a1 and c2 2a2 , and d1 b1 and d b2 This gives x t a1 cos5t a2 sin 5t b1 cos10t b2 sin10t , y t 2a1 cos5t 2a2 sin 5t b1 cos10t b2 sin10t In the natural mode with frequency 1 the masses move in the same direction, with the amplitude of motion of the second mass twice that of the first mass In the natural mode with frequency 2 10 they move in opposite directions with the same amplitude of motion 41 The system has operational determinant D 10 D D 1 D Hence the general form of the solution is x t a1 cos t a2 sin t b1 cos 3t b2 sin 3t , y t c1 cos t c2 sin t d1 cos 3t d sin 3t Upon substitution in either differential equation we see that c1 a1 and c2 a2 , and d1 b1 and d b2 This gives x t a1 cos t a2 sin t b1 cos 3t b2 sin 3t , y t a1 cos t a2 sin t b1 cos3t b2 sin 3t In the natural mode with frequency 1 the masses move in the same direction (figure a), while in the natural mode with frequency 2 they move in opposite directions (figure b) In each case the amplitudes of motion of the two masses are equal Copyright © 2015 Pearson Education, Inc Section 4.2 267 Problem 41a Problem 41b x, y x2 = cos(3t) x1 = y1 = cos(t) x, y y2 = − cos(3t) t 42 t The system has operational determinant D 10 D D 1 D Hence the general form of the solution is x t a1 cos t a2 sin t b1 cos 2t b2 sin 2t , y t c1 cos t c2 sin t d1 cos 2t d sin 2t Upon substitution in either differential equation we see that c1 a1 and c2 a2 , and 1 d1 b1 and d b2 This gives 2 x t a1 cos t a2 sin t b1 cos 2t b2 sin 2t , 1 y t a1 cos t a2 sin t b1 cos 2t b2 sin 2t 2 In the natural mode with frequency 1 the two masses move in the same direction with equal amplitudes of oscillation In the natural mode with frequency 2 the two masses move in opposite directions with the amplitude of m2 being half that of m1 43 The system has operational determinant D D D 1 D 5 Hence the general form of the solution is 5t b sin 5t , y t c cos t c sin t d cos 5t d sin 5t x t a1 cos t a2 sin t b1 cos 2 Upon substitution in either differential equation we see that c1 a1 and c2 a2 , and d1 b1 and d b2 This gives Copyright © 2015 Pearson Education, Inc THE METHOD OF ELIMINATION 268 5t b sin 5t , y t a cos t a sin t b cos 5t b sin 5t x t a1 cos t a2 sin t b1 cos 2 In the natural mode with frequency 1 the masses move in the same direction (figure a), while in the natural mode with frequency 2 they move in opposite directions (figure b) In each case the amplitudes of motion of the two masses are equal Problem 43a Problem 43b x1 = y1 = cos(t) x, y √ x2 = cos( 5t) x, y √ y2 = − cos( 5t) t 44 t The system has operational determinant D D D D Hence the general form of the solution is x t a1 cos t a2 sin t b1 cos 2t b2 sin 2t , y t c1 cos t c2 sin t d1 cos 2t d sin 2t Upon substitution in either differential equation we see that c1 a1 and c2 a2 , and d1 b1 and d b2 This gives 2t a sin 2t b cos 2t b sin 2t, y t a cos 2t a sin 2t b cos 2t b sin 2t x t a1 cos 2 2 In the natural mode with frequency 1 the two masses move in the same direction; in the natural mode with frequency 2 they move in opposite directions In each natural mode their amplitudes of oscillation are equal 45 The system has operational determinant D 20 D 32 D D Hence the general form of the solution is Copyright © 2015 Pearson Education, Inc Section 4.2 269 2t a sin 2t b cos 8t b sin 8t , y t c cos 2t c sin 2t d cos 8t d sin 8t x t a1 cos 2 2 Upon substitution in either differential equation we see that c1 a1 and c2 a2 , and 1 d1 b1 and d b2 This gives 2 2t a sin 2t b cos 8t b sin 8t , 1 y t a cos 2t a sin 2t b cos 8t b sin 8t 2 x t a1 cos 2 2 In the natural mode with frequency 1 the two masses move in the same direction with equal amplitudes of oscillation (figure a) In the natural mode with frequency 2 2 the two masses move in opposite directions with the amplitude of m2 being half that of m1 (figure b) Problem 45a Problem 45b √ x1 = y1 = cos( 2t) x, y x, y √ x1 = y1 = cos( 8t) t 46 t The system has operational determinant D 20 D 64 D D 16 Hence the general form of the solution is x t a1 cos 2t a2 sin 2t b1 cos 4t b2 sin 4t , y t c1 cos 2t c2 sin 2t d1 cos 4t d sin 4t Upon substitution in either differential equation we see that c1 a1 and c2 a2 , and d1 b1 and d b2 This gives x t a1 cos 2t a2 sin 2t b1 cos 4t b2 sin 4t , y t a1 cos 2t a2 sin 2t b1 cos 4t b2 sin 4t Copyright © 2015 Pearson Education, Inc 270 THE METHOD OF ELIMINATION In the natural mode with frequency 1 the masses move in the same direction with equal amplitudes of motion In the natural mode with frequency 2 they move in opposite 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 second spring is stretched by y x , the third spring is stretched by z y , and the fourth spring is compressed by z Hence Newton's second law gives mx k x k y x , my k y x k z y , and mz k z y k z (b) The operational determinant is D 2 D 1 D D D , and the characteristic equation r r has roots i and i 48 The given system has operational determinant D 10 D D 1 D Hence the general form of the solution is x t a1 cos t a2 sin t b1 cos 3t b2 sin 3t , y t c1 cos t c2 sin t d1 cos 3t d sin 3t Upon substitution in either differential equation we see that c1 a2 and c2 a1 , and d1 b2 and d b1 This gives x t a1 cos t a2 sin t b1 cos 3t b2 sin 3t , y t a2 cos t a1 sin t b2 cos 3t b1 sin 3t When we impose the initial conditions x and y x y we find that a1 , b1 , and a2 b2 SECTION 4.3 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 h 0.1; x1 x0 hf t0 , x0 , y0 ; x2 x1 hf t1 , x1 , y1 ; t1 t0 h y1 y0 hg t0 , x0 , y0 y2 y1 hg t1 , x1 , y1 Copyright © 2015 Pearson Education, Inc Section 4.3 271 (with the given values of t0 , x0 , and y0 ) to calculate the Euler approximations x1 x 0.1 and y1 y 0.1 , and x2 x 0.2 and y2 y 0.2 , in part (a) We give these approximations and the actual values xact x 0.2 , yact y 0.2 in tabular form We use the template h 0.2; u1 x0 hf t0 , x0 , y0 ; x1 x0 h f t0 , x0 , y0 f t1 , u1 , v1 t1 t0 h v1 y0 hg t0 , x0 , y0 y1 y0 h g t0 , x0 , y0 g t1 , u1 , v1 to calculate the improved Euler approximations u1 x 0.2 and u1 y 0.2 , and x1 x 0.2 and y1 y 0.2 , in part (b) We give these approximations and the actual values xact x 0.2 , yact y 0.2 in tabular form We use the template h 0.2; F1 f t0 , x0 , y0 ; G1 g t0 , x0 , y0 ; h h h h h F2 f t0 , x0 F1 , y0 hG1 ; G2 g t0 , x0 F1 , y0 G1 ; 2 2 2 h h h h h h F3 f t0 , x0 F2 , y0 G2 ; G3 g t0 , x0 F2 , y0 G2 ; 2 2 F4 f t0 h, x0 hF3 , y0 hG3 ; G4 g t0 h, x0 hF3 , y0 hG3 ; h h x1 x0 F1 F2 F3 F4 ; y1 y0 G1 2G2 2G3 G4 6 to calculate the intermediate slopes and Runge-Kutta approximations x1 x 0.2 and y1 y 0.2 for part (c) Again, we give the results in tabular form (a) x1 y1 x2 y2 xact yact 0.4 2.2 0.88 2.5 1.0034 2.6408 u1 v1 x1 y1 xact yact 0.8 2.4 0.96 2.6 1.0034 2.6408 (b) Copyright © 2015 Pearson Education, Inc 272 NUMERICAL METHODS FOR SYSTEMS (c) F1 G1 F2 G2 F3 G3 F4 G4 4.8 5.08 3.26 6.32 4.684 x1 y1 xact yact 1.0027 2.6401 1.0034 2.6408 (a) x1 y1 x2 y2 xact yact 0.9 –0.9 0.81 –0.81 0.8187 –0.8187 u1 v1 x1 y1 xact yact 0.8 –0.8 0.82 –0.82 0.8187 –0.8187 (b) (c) F1 G1 F2 G2 F3 G3 F4 G4 –1 –0.9 0.9 –0.91 0.91 –0.818 0.818 x1 y1 xact yact 0.8187 –0.8187 0.8187 –0.8187 (a) x1 y1 x2 y2 xact yact 1.7 1.5 2.81 2.31 3.6775 2.9628 u1 v1 x1 y1 xact yact 2.4 3.22 2.62 3.6775 2.9628 (b) Copyright © 2015 Pearson Education, Inc Section 4.3 273 (c) F1 G1 F2 G2 F3 G3 F4 G4 11.1 8.1 13.57 9.95 23.102 17.122 x1 y1 xact yact 3.6481 2.9407 3.6775 2.9628 (a) x1 y1 x2 y2 xact yact 1.9 –0.6 3.31 –1.62 4.2427 -2.4205 u1 x1 y1 xact yact 2.8 –1.2 3.82 –2.04 4.2427 -2.4205 (b) (c) F1 G1 F2 G2 F3 G3 F4 G4 –6 14.1 –10.2 16.59 –12.42 26.442 –20.94 x1 y1 xact yact 4.2274 –2.4060 4.2427 -2.4205 (a) x1 y1 x2 y2 xact yact 0.9 3.2 –0.52 2.92 -0.5793 2.4488 u1 v1 x1 y1 xact yact –0.2 3.4 –0.84 2.44 -0.5793 2.4488 (b) Copyright © 2015 Pearson Education, Inc 274 NUMERICAL METHODS FOR SYSTEMS (c) F1 G1 F2 G2 F3 G3 F4 G4 –11 –14.2 –2.8 –12.44 –3.12 –12.856 –6.704 x1 y1 xact yact –0.5712 2.4485 -0.5793 2.4488 (a) x1 y1 x2 y2 xact yact –0.8 4.4 –1.76 4.68 -1.9025 4.4999 u1 v1 x1 y1 xact yact –1.6 4.8 –1.92 4.56 -1.9025 4.4999 (b) (c) F1 G1 F2 G2 F3 G3 F4 G4 –8 –9.6 2.8 –9.52 2.36 –10.848 0.664 x1 y1 xact yact –1.9029 4.4995 -1.9025 4.4999 (a) x1 y1 x2 y2 xact yact 2.5 1.3 3.12 1.68 3.2820 1.7902 u1 v1 x1 y1 xact yact 1.6 3.24 1.76 3.2820 1.7902 (b) Copyright © 2015 Pearson Education, Inc Section 4.3 275 (c) F1 G1 F2 G2 F3 G3 F4 G4 6.2 3.8 6.48 8.088 5.096 x1 y1 xact yact 3.2816 1.7899 3.2820 1.7902 (a) x1 y1 x2 y2 xact yact 0.9 –0.9 2.16 –0.63 2.5270 -0.3889 u1 v1 x1 y1 xact yact 1.8 –0.8 2.52 –0.46 2.5270 -0.3889 (b) (c) F1 G1 F2 G2 F3 G3 F4 G4 12.6 2.7 12.87 3.25 16.02 5.498 x1 y1 xact yact 2.5320 –0.3867 2.5270 -0.3889 In Problems 9-11 we use the same Runge-Kutta template as in part (c) of Problems 1–8 above, and give both the Runge-Kutta approximate values with step sizes h 0.1 and h 0.05 , and also the actual values With h 0.1 : x 1 3.99261 y 1 6.21770 With h 0.05 : x 1 3.99234 y 1 6.21768 Actual values: x 1 3.99232 y 1 6.21768 Copyright © 2015 Pearson Education, Inc 276 NUMERICAL METHODS FOR SYSTEMS 10 With h 0.1 : x 1 1.31498 y 1 1.02537 With h 0.05 : x 1 1.31501 y 1 1.02538 Actual values: x 1 1.31501 y 1 1.02538 With h 0.1 : x 1 0.05832 y 1 0.56664 With h 0.05 : x 1 0.05832 y 1 0.56665 Actual values: x 1 0.05832 y 1 0.56665 11 12 We first convert the given initial value problem to the two-dimensional problem x 0, x y , y x sin t , y Then with both step sizes h 0.1 and h 0.05 we get the actual value x 1 0.15058 accurate to decimal places 13 With y x we want to solve numerically the initial value problem x y , y 32 0.04 y , x 0, y 288 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 x y , y 32 0.0002 y , x 0, y 288 Copyright © 2015 Pearson Education, Inc Section 4.3 277 Running Program RK2DIM with step size h 0.1 , we find that the bolt attains a maximum 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 x and with x in miles and in seconds, we want to solve numerically the initial value problem x 0, x y , 95485.5 , y x 7920 x 15681600 We find (running RK2DIM with h ) that the projectile reaches a maximum height of about 83.83 miles in about 168 sec 48 sec y 16 We first defined the MATLAB function function xp = fnball(t,x) % Defines the baseball system % x1 = x = x3, x3 = -cvx % x2 = y = x4, x4 = -cvy- g % with air resistance coefficient c g = 32; c = 0.0025; xp = x; v = sqrt(x(3).^2) + x(4).^2); xp(1) = x(3); xp(2) = x(4); xp(3) = -c*v*x(3); xp(4) = -c*v*x(4) - g; Then, using the n-dimensional program rkn with step size 0.1 and initial data corresponding to the indicated initial inclination angles, we got the following results: Angle Time Range 40 5.0 352.9 45 5.4 347.2 50 5.8 334.2 We have listed the time to the nearest tenth of a second, but have interpolated to find the range in feet 17 The data in Problem 16 indicate that the range increases when the initial angle is decreased below 45 The further data Copyright © 2015 Pearson Education, Inc 278 NUMERICAL METHODS FOR SYSTEMS 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 results: t x y 5.0 457.43 103.90 6.0 503.73 36.36 Interpolation gives x 494.4 when y 50 Then a run with v0 255 ft sec gives the following results: t x y 5.5 486.75 77.46 6.0 508.86 41.62 Finally, a run with v0 253 ft sec gives these results: Copyright © 2015 Pearson Education, Inc Section 4.3 279 t x y 5.5 484.77 75.44 6.0 506.82 39.53 Now x 500ft when y 50ft Thus Babe Ruth’s home run ball had an initial velocity of 253 ft sec 20 A run of program rkn with h 0.1 and with the given data yields the following results: t 5.5 5.6 x 989 1005 y 539 539 v 162 161 0.95 0.18 11.5 11.6 1868 1881 16 214 216 52 53 1 The first two lines of data above indicate that the crossbow bolt attains a maximum height of about 1005 ft in about 5.6 sec About sec later (total time 11.6 sec) it hits the ground, having traveled about 1880 ft horizontally 21 A run with h 0.1 indicates that the projectile has a range of about 21, 400ft 4.05mi and a flight time of about 46 sec It attains a maximum height of about 8970 ft in about 17.5 sec At time t 23sec it has its minimum velocity of about 368 ft/sec It hits the ground ( t 23sec ) at an angle of about 77 with a velocity of about 518 ft/sec Copyright © 2015 Pearson Education, Inc ... y1 xact yact 2.8 –1.2 3.82 –2. 04 4. 242 7 -2 .42 05 (b) (c) F1 G1 F2 G2 F3 G3 F4 G4 –6 14. 1 –10.2 16.59 –12 .42 26 .44 2 –20. 94 x1 y1 xact yact 4. 22 74 –2 .40 60 4. 242 7 -2 .42 05 (a) x1 y1 x2 y2 xact yact... 2 .44 88 u1 v1 x1 y1 xact yact –0.2 3 .4 –0. 84 2 .44 -0.5793 2 .44 88 (b) Copyright © 2015 Pearson Education, Inc 2 74 NUMERICAL METHODS FOR SYSTEMS (c) F1 G1 F2 G2 F3 G3 F4 G4 –11 – 14. 2 –2.8 –12 .44 ... –12.856 –6.7 04 x1 y1 xact yact –0.5712 2 .44 85 -0.5793 2 .44 88 (a) x1 y1 x2 y2 xact yact –0.8 4. 4 –1.76 4. 68 -1.9025 4. 4999 u1 v1 x1 y1 xact yact –1.6 4. 8 –1.92 4. 56 -1.9025 4. 4999 (b) (c) F1 G1 F2