Bretscher 33a solution manual

Bretscher_600928X_ISM_TTL.qxd:au_123456_ttl 1/12/09 8:40 AM Page INSTRUCTOR’S SOLUTIONS MANUAL KYLE BURKE Boston University with art by GEORGE WELCH LINEAR ALGEBRA WITH APPLICATIONS FOURTH EDITION Otto Bretscher Colby College Bretscher_600928X_ISM_TTL.qxd:au_123456_ttl 1/12/09 8:40 AM Page The author and publisher of this book have used their best efforts in preparing this book These efforts include the development, research, and testing of the theories and programs to determine their effectiveness The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs Reproduced by Pearson Prentice Hall from electronic files supplied by the author Copyright © 2009 Pearson Education, Inc Publishing as Pearson Prentice Hall, Upper Saddle River, NJ 07458 All rights reserved No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher Printed in the United States of America ISBN-13: 978-0-13-600928-3 ISBN-10: 0-13-600928-X OPM 10 09 08 Table of Contents Chapter 1: Linear Equations ……………………………………………………… Chapter 2: Linear Transformations ………………………………………………… 48 Chapter 3: Subspaces of Rn and Their Dimensions….…………………………… 120 Chapter 4: Linear Spaces………………………………………………… …………171 Chapter 5: Orthogonality and Least Squares …………………………………… 217 Chapter 6: Determinants …………………………………………………………… 266 Chapter 7: Eigenvalues and Eigenvectors…………………………………………… 297 Chapter 8: Symmetric Matrices and Quadratic Forms.…………… …………… 360 Chapter 9: Linear Differential Equations.…………………………………………… 392 Section 1.1 C hap ter Section 1.1 x + 2y = 2x + 3y = 1.1.1 x + 2y = y=1 4x + 3y 7x + 5y 1.1.2 x + 34 y − 14 y −2 × 1st equation −2 × 2nd equation =2 =3 = 12 = − 12 ÷4 → ×(−4) → x + 2y −y → = 7x + 5y =3 x + 43 y y ÷(−1) → x = −1 , so that (x, y) = (−1, 1) y =1 x + 34 y → =1 = −1 −7 × 1st equation = 12 =2 → − 43 × 2nd equation → x = −1 , y = so that (x, y) = (−1, 2) 2x + 4y 3x + 6y 1.1.3 =3 =2 ÷2 → x + 2y 3x + 6y = 32 =2 → x + 2y 3x + 6y =1 =3 −3 × 1st equation → x + 2y = 32 = − 52 → x + 2y =1 =0 So there is no solution 2x + 4y 3x + 6y 1.1.4 =2 =3 ÷2 −3 × 1st equation This system has infinitely many solutions: if we choose y = t, an arbitrary real number, then the equation x + 2y = gives us x = − 2y = − 2t Therefore the general solution is (x, y) = (1 − 2t, t), where t is an arbitrary real number 2x + 3y 4x + 5y 1.1.5 =0 =0 x + 23 y =0 −y =0 ÷2 ÷(−1) → x + 23 y 4x + 5y → x + 32 y y =0 =0 =0 =0 −4 × 1st equation − 23 × 2nd equation → → x =0 , y =0 so that (x, y) = (0, 0)    x + 2y + 3z = x + 2y + 3z 1.1.6  x + 3y + 3z = 10  −I →  y x + 2y + 4z = −I z     x + 3z = −3(III) x=1  y = 2 →  y = , so that z =1 z=1  x + 2y + 3z 1.1.7  x + 3y + 4z x + 4y + 5z   x + 2y + 3z =1 y+z =  −I →  2y + 2z = −I This system has no solution  =8 −2(II) = 2 → =1 (x, y, z) = (1, 2, 1)   x+z = −2(II) → y +z = 2 = −2(II)  = −3 = 2 = −1 Chapter    x + 2y + 3z = x + 2y + 3z =0 =  ÷(−3) → 1.1.8  4x + 5y + 6z =  −4(I) →  −3y − 6z −6y − 11z = 7x + 8y + 10z = −7(I)       x + 2y + 3z = −2(II) x−z =0 +III x =0  y + 2z = 0 →  y + 2z =  −2(III) →  y = , −6y − 11z = +6(II) z =0 z =0  so that (x, y, z) = (0, 0, 0)     x + 2y + 3z = x + 2y + 3z = 1.1.9  3x + 2y + z =  −3(I) →  −4y − 8z = −2  ÷(−4) → 7x + 2y − 3z = −7(I) −12y − 24z = −6     −2(II) x−z =0 x + 2y + 3z =  y + 2z →  y + 2z = 12  = 12  −12y − 24z = −6 +12(II) =0 This system has infinitely many solutions: if we choose z = t, an arbitrary real number, then we get x = z = t and y = 21 − 2z = 12 − 2t Therefore, the general solution is (x, y, z) = t, 21 − 2t, t , where t is an arbitrary real number  x + 2y + 3z 1.1.10  2x + 4y + 7z 3x + 7y + 11z  x + 2y + 3z  y + 2z z   =1 x + 2y + 3z =  −2(I) →  z = −3(I) y + 2z   x−z = −2(II) = 5 →  y + 2z =0 z so that (x, y, z) = (−9, 5, 0) 1.1.11 x − 2y 3x + 5y =2 = 17 −3(I) → x − 2y 11y = −9  =1 Swap : = 0 → II ↔ III =5    +III x = −9 =  −2(III) →  y = , z =0 =0 =2 = 11 ÷11 → x − 2y y =2 =1 so that (x, y) = (4, 1) See Figure 1.1 Figure 1.1: for Problem 1.1.11 1.1.12 x − 2y 2x − 4y =3 =6 −2(I) → x − 2y =3 =0 +2(II) → x =4 , y =1 Section 1.1 This system has infinitely many solutions: If we choose y = t, an arbitrary real number, then the equation x − 2y = gives us x = + 2y = + 2t Therefore the general solution is (x, y) = (3 + 2t, t), where t is an arbitrary real number (See Figure 1.2.) Figure 1.2: for Problem 1.1.12 1.1.13 x − 2y 2x − 4y =3 =8 −2(I) → x − 2y =3 , which has no solutions (See Figure 1.3.) =2 Figure 1.3: for Problem 1.1.13  x + 5z 1.1.14 The system reduces to  y − z planes  =0 = , so that there is no solution; no point in space belongs to all three =1 Compare with Figure 2b   x =0 1.1.15 The system reduces to  y =  so the unique solution is (x, y, z) = (0, 0, 0) The three planes intersect at z =0 the origin  x + 5z 1.1.16 The system reduces to  y − z arbitrary number The three planes 1.1.17 x + 2y 3x + 5y =a =b −3(I) →  =0 =  , so the solutions are of the form (x, y, z) = (−5t, t, t), where t is an =0 intersect in a line; compare with Figure 2a x + 2y −y =a = −3a + b ÷(−1) → x + 2y y =a = 3a − b −2(II) Chapter x = −5a + 2b , so that (x, y) = (−5a + 2b, 3a − b) y = 3a − b     x + 2y + 3z = a x + 2y + 3z =a −2(II) 1.1.18  x + 3y + 8z = b  −I →  y + 5z = −a + b  → x + 2y + 2z = c −I −z = −a + c       x = 10a − 2b − 7c x − 7z = 3a − 2b +7(III) x − 7z = 3a − 2b  y + 5z = −a + b  →  y + 5z = −a + b  −5(III) →  y = −6a + b + 5c , z =a−c z =a−c −z = −a + c ÷(−1) so that (x, y, z) = (10a − 2b − 7c, −6a + b + 5c, a − c) 1.1.19 a Note that the demand D1 for product increases with the increase of price P2 ; likewise the demand D2 for product increases with the increase of price P1 This indicates that the two products are competing; some people will switch if one of the products gets more expensive 70 − 2P1 + P2 105 + P1 − P2 which yields the unique solution P1 = 26 and P2 = 46 = −14 + 3P1 , or = −7 + 2P2 b Setting D1 = S1 and D2 = S2 we obtain the system −5P1 + P2 P1 − 3P2 = −84 , = 112 1.1.20 The total demand for the product of Industry A is 1000 (the consumer demand) plus 0.1b (the demand from Industry B) The output a must meet this demand: a = 1000 + 0.1b Setting up a similar equation for Industry B we obtain the system which yields the unique solution a = 1100 and b = 1000 a = 1000 + 0.1b a − 0.1b or b = 780 + 0.2a −0.2a + b = 1000 , = 780 1.1.21 The total demand for the products of Industry A is 310 (the consumer demand) plus 0.3b (the demand from Industry B) The output a must meet this demand: a = 310 + 0.3b Setting up a similar equation for Industry B we obtain the system a = 310 + 0.3b b = 100 + 0.5a which yields the solution a = 400 and b = 300 1.1.22 Since x(t) = a sin(t) + b cos(t) we can compute dx dt = a cos(t) − b sin(t) and or a − 0.3b −0.5a + b = 310 , = 100 d2 x dt2 = −a sin(t) − b cos(t) Substituting these expressions into the equation ddt2x − dx dt − x = cos(t) and simplifying gives (b − 2a) sin(t) + (−a − 2b) cos(t) = cos(t) Comparing the coefficients of sin(t) and cos(t) on both sides of −2a + b = , so that a = − 51 and b = − 25 See Figure 1.4 the equation then yields the system −a − 2b = Figure 1.4: for Problem 1.1.22 Section 1.1 1.1.23 a Substituting λ = yields the system 7x − y −6x + 8y = 5x = 5y or 2x − y −6x + 3y =0 =0 or 2x − y =0 =0 There are infinitely many solutions, of the form (x, y) = t 2, t , where t is an arbitrary real number b Proceeding as in part (a), we find (x, y) = − 13 t, t c Proceedings as in part (a), we find only the solution (0, 0) 1.1.24 Let v be the speed of the boat relative to the water, and s be the speed of the stream; then the speed of the boat relative to the land is v + s downstream and v − s upstream Using the fact that (distance) = (speed)(time), we obtain the system = (v + s) 31 = (v − s) 23 ← downstream ← upstream The solution is v = 18 and s = x+z 1.1.25 The system reduces to  y − 2z   =1 = −3  =k−7 a The system has solutions if k − = 0, or k = b If k = then the system has infinitely many solutions c If k = then we can choose z = t freely and obtain the solutions (x, y, z) = (1 − t, −3 + 2t, t) x − 3z  y + 2z 1.1.26 The system reduces to (k − 4)z   = =  = k−2 This system has a unique solution if k − = 0, that is, if k = ±2 If k = 2, then the last equation is = 0, and there will be infinitely many solutions If k = −2, then the last equation is = −4, and there will be no solutions 1.1.27 Let x = the number of male children and y = the number of female children Then the statement “Emile has twice as many sisters as brothers” translates into y = 2(x − 1) and “Gertrude has as many brothers as sisters” translates into x = y − Solving the system −2x + y x−y = −2 = −1 gives x = and y = There are seven children in this family Chapter 1.1.28 The thermal equilibrium condition requires  −4T1 + T2 We can rewrite this system as  T1 − 4T2 + T3 T2 − 4T3 The solution is (T1 , T2 , T3 ) = (75, 100, 125) , T2 = that T1 = T2 +200+0+0  = −200 = −200  = −400 T1 +T3 +200+0 , and T3 = T2 +400+0+0 1.1.29 To assure that the graph goes through the point (1, −1), we substitute t = and f (t) = −1 into the equation f (t) = a + bt + ct2 to give −1 = a + b + c   a+b+c = −1 Proceeding likewise for the two other points, we obtain the system  a + 2b + 4c =  a + 3b + 9c = 13 The solution is a = 1, b = −5, and c = 3, and the polynomial is f (t) = − 5t + 3t2 (See Figure 1.5.) Figure 1.5: for Problem 1.1.29  a+b+c =p 1.1.30 Proceeding as in the previous exercise, we obtain the system  a + 2b + 4c = q a + 3b + 9c = r   a = 3p − 3q + r The unique solution is  b = −2.5p + 4q − 1.5r  c = 0.5p − q + 0.5r Only one polynomial of degree goes through the three given points, namely, f (t) = 3p − 3q + r + (−2.5p + 4q − 1.5r)t + (0.5p − q + 0.5r)t2 1.1.31 f (t) is of the form at2 + bt + c So f (1) = a(12 ) + b(1) + c = 3, and f (2) = a(22 ) + b(2) + c = Also, f (t) = 2at + b, meaning that f (1) = 2a + b =   a+b+c=3   So we have a system of equations:  4a + 2b + c =  2a + b =  a=2    which reduces to  b = −3  c=4 Section 1.1 Thus, f (t) = 2t2 − 3t + is the only solution 1.1.32 f (t) is of the form at2 + bt + c So, f (1) = a(12 ) + b(1) + c = and f (2) = 4a + 2b + c = Also, 2 f (t)dt = (at2 + bt + c)dt = a3 t3 + 2b t2 + ct|21 = 83 a + 2b + 2c − ( a3 + b + c) = 73 a + 23 b + c = −1   a+b+c=1 So we have a system of equations:  4a + 2b + c =  3 a + b + c = −1   a=9 which reduces to  b = −28  c = 20 Thus, f (t) = 9t2 − 28t + 20 is the only solution 1.1.33 f (t) is of the form at2 + bt + c f (1) = a + b + c = 1, f (3) = 9a + 3b + c = 3, and f (t) = 2at + b, so f (2) = 4a + b =   a+b+c=1   Now we set up our system to be  9a + 3b + c =  4a + b =   a − 3c =   This reduces to  b + 43 c =  0=0 We write everything in terms of a, revealing c = 3a and b = − 4a So, f (t) = at2 + (1 − 4a)t + 3a for an arbitrary a 1.1.34 f (t) = at2 + bt + c, so f (1) = a + b + c = 1, f (3) = 9a + 3b + c = Also, f (2) = 3, so 2(2)a + b = 4a + b =   a+b+c=1 Thus, our system is  9a + 3b + c =  4a + b = When we reduce this, however, our last equation becomes = 2, meaning that this system is inconsistent 1.1.35 f (t) = ae3t + be2t , so f (0) = a + b = and f (t) = 3ae3t + 2be2t , so f (0) = 3a + 2b = Thus we obtain the system which reveals a+b=1 , 3a + 2b = a=2 b = −1 So f (t) = 2e3t − e2t Section 9.2 −1 k then tr(A) = −2 and det(A) = − k By Theorem 9.2.5, the zero state is stable if k −1 det(A) = − k > 0, that is, if |k| < 9.2.17 If A = 9.2.18 If λ1 , λ2 , λ3 are real and negative, then tr(A) = λ1 + λ2 + λ3 < and det(A) = λ1 λ2 λ3 < If λ1 is real and negative and λ2,3 = p ± iq, where p is negative, then tr(A) = λ1 + 2p < and det(A) = λ1 (p2 + q ) < Either way, both trace and determinant are negative  9.2.19 False, consider A =  0  0  −4 9.2.20 Use Theorem 9.2.6, with p = 0, q = π; a = 1, b = x(t) = [ w v] cos(πt) − sin(πt) sin(πt) cos(πt) = (cos(πt))w + (sin(πt))v See Figure 9.35 ( ) x 12 = v x(0) = x(2) = w x(1) = –w –v Figure 9.35: for Problem 9.2.20 9.2.21 a db dt ds dt = 0.05b+ = s 0.07s b λ1 = 0.07, λ2 = 0.05; v1 = b(0) 1, 000 = s(0) 1, 000 and 50 ; x(0) = 1, 000v1 − 49, 000v2 ; so that , v2 = b(t) = 50, 000e0.07t − 49, 000e0.05t and s(t) = 1, 000e0.07t 9.2.22 λ1 = 3, λ2 = 0.5; E3 = span , E0.5 = span −1 System is discrete so choose VII 9.2.23 λ1,2 = − 12 ± i, r > 1, so that trajectory spirals outwards Choose II 9.2.24 λ1 = 3, λ2 = 0.5, E3 = , E0.5 = −1 System is continuous, so choose I 9.2.25 λ1,2 = − 12 ± i; real part is negative so that trajectories spiral inwards in the counterclockwise direction −1.5 if x = then dx Choose IV dt = 409 Chapter 9.2.26 λ1 = 1, λ2 = −2; E1 = span , E−2 = span −1 System is continuous so choose V +i −1 9.2.27 λ1,2 = ±3i, E3i = span ,v= −1 , so that p = 0, q = 3, w = Now use Theorem 9.2.6: x(t) = e0t −1 cos(3t) sin(3t) 9.2.28 λ1,2 = ±6i, E6i = span x(t) = cos(6t) sin(6t) − sin(3t) cos(3t) a sin(3t) = b − cos(3t) +i , so that − sin(6t) cos(6t) +i 9.2.29 λ1,2 = ± 4i, E2+4i = span x(t) = e2t 1 cos(4t) sin(4t) a sin(6t) = b cos(6t) x(t) = e−2t cos(3t) sin(3t) − sin(3t) cos(3t) v= Now Then x(t) = e−t a b cos(4t) − sin(4t) a b , so that sin(3t) a = e−2t cos(3t) + sin(3t) b +i −1 9.2.31 λ1,2 = −1 ± 2i, E−1+2i = span a b , so that +i 9.2.30 λ1,2 = −2 ± 3i, E−2+3i = span cos(6t) −3 sin(6t) sin(4t) a = e2t cos(4t) b − sin(4t) cos(4t) cos(3t) sin(3t) cos(3t) − sin(3t) + cos(3t) , so that p = −1, q = 2, w = , −1 = x(0) = w + v, so that a = and b = −1 −1 cos(2t) sin(2t) − sin(2t) cos(2t) sin(2t) + cos(2t) = e−t sin(2t) − cos(2t) See Figure 9.36 9.2.32 λ1,2 = ±2i, E2i = span x(t) = cos(2t) sin(2t) +i − sin(2t) cos(2t) 9.2.33 λ1,2 = ±i, Ei = span a = 1, b = 0, so that x(t) = , x(0) = 0 +1 , so that a = and b = 0 cos(2t) = See Figure 9.37 −2 sin(2t) +i 1 1 cos(t) sin(t) − cos(t) cos(t) 410 sin(t) = sin(t) + cos(t) a b Section 9.2 1 Figure 9.36: for Problem 9.2.31 x (0) = x(π) = 1 –1 x π2 = ( ) x π4 = –2 ( ) Figure 9.37: for Problem 9.2.32 = cos(t) + sin(t) See Figure 9.38 1 9.2.34 λ1,2 = ± 2i, E1+2i = span a = 1, b = 0, so that x(t) = et +i −2 −2 cos(2t) sin(2t) − sin(2t) cos(2t) cos(2t) + sin(2t) See Figure 9.39 = et −2 sin(2t) 9.2.35 If z = f + ig and w = p + iq then zw = (f p − gq) + i(f q + gp), so that (zw) = (f p + f p − g q − gq ) + i(f q + f q + g p + gp ) Also z w = (f + ig )(p + iq) = (f p − g q) + i(f q + g p) and zw = (f + ig)(p + iq ) = (f p − gq ) + i(gp + f q ) We can see that (zw) = z w + zw , as claimed 9.2.36 A = −b −c and fA (λ) = λ2 + cλ + b, with eigenvalues λ1,2 = 411 √ −c± c2 −4b Chapter x (0) = x(2π) = 1 x π2 = ( ) x(π) = –1 Figure 9.38: for Problem 9.2.33 Figure 9.39: for Problem 9.2.34 √ 9.2.36 a If c = then λ1,2 = ±i b The trajectories are ellipses See Figure 9.40 v x Figure 9.40: for Problem 9.2.36a The block oscillates harmonically, with period 2π √ b The zero state fails to be asymptotically stable 412 Section 9.2 b λ1,2 = √ −c±i 4b−c2 The trajectories spiral inwards, since Re(λ1 ) = Re(λ2 ) = − 2c < This is the case of a damped oscillation The zero state is asymptotically stable See Figure 9.41 Figure 9.41: for Problem 9.2.36b c This case is discussed in Exercise 9.1.55 The zero state is stable here 9.2.37 a z(t) To find b z w 9.2.38 a is differentiable when z(t) = 0, since both the real and the imaginary parts are differentiable if z = p + iq then z , apply the product rule to the equation z = z w1 z1 z2 z = =z w +z z z2 −z1 z z22 w = = z w − zw w2 λz1 z2 −λz1 z2 z22 = = 1: z z +z z = 0, so that z = − zz2 z w−zw w2 = 0, so that z1 (t) z2 (t) = k, a constant Now z1 (t) = kz2 (t); substituting t = gives = z1 (0) = kz2 (0) = k, so that z1 (t) = z2 (t), as claimed b Let z2 (t) = ept (cos(qt) + i sin(qt)) be the solution constructed in the text (see Page 413) Since z2 (t) = for all t, this is the only solution, by Part a  λ 9.2.39 Let A =  λ 0 c3 (t), dc = dt   We first solve the system λ c3 (t) = k3 , a constant, so that dc2 dt dc dt  = (A − λI3 )c =  0 = k3 and c2 (t) = k3 t + k2 Likewise c1 (t) = 0 k3 2 t   c, or dc1 dt = c2 (t), dc dt = + k2 t + k1 Applying Exercise 9.1.24, with k = −λ, we find that c(t) = e−λt x(t) or x(t) = eλt c(t)   k1 + k2 t + k23 t2  where k1 , k2 , k3 are arbitrary constants The zero state is stable if (and only if) the real k2 + k3 t = eλt  k3 part of λ is negative 9.2.40 a B(t) = 1000(1 + 0.05i)t = 1000(r(cos θ + i sin θ))t = 1000rt (cos(θt) + i sin(θt)), where √ r = + 0.052 > and θ = arctan(0.05) ≈ 0.05 See Figure 9.42 413 z = Chapter b B(t) = 1000e0.05i = 1000(cos(0.05t) + i sin(0.05t)) See Figure 9.42 1000 1000 1000 trajectory to part a slowly spiral outwards trajectory to part b circle 2π ≅ period = 126 (years) 0.05 Figure 9.42: for Problem 9.2.40 c We would choose an account with annual compounding, since the modulus of the balance grows in this case In the case of continuous compounding the modulus of the balance remains unchanged Section 9.3 9.3.1 The characteristic polynomial of this differential equation is λ − 5, so that λ1 = By Theorem 9.3.8 the general solution is f (t) = Ce5t , where C is an arbitrary constant −3t , where C is an arbitrary constant, and the differential 9.3.2 The solutions of dx dt +3x = are of the form x(t) = Ce dx equation dt + 3x = has the particular solution xp (t) = 73 , so that the general solution is x(t) = Ce−3t + 37 (where C is a constant) Alternatively, we could use Theorem 9.3.13 9.3.3 Use Theorem 9.3.13, where a = −2 and g(t) = e3t : f (t) = e−2t e2t e3t dt = e−2t e5t dt = e−2t 5t e +C = 3t e + Ce−2t , where C is a constant 9.3.4 We can look for a sinusoidal solution xp (t) = P cos(3t) + Q sin(3t), as in Example P and Q need to be −2P + 3Q = chosen in such a way that −3P sin(3t) + 3Q cos(3t) − 2P cos(3t) − 2Q sin(3t) = cos(3t) or with −3P − 2Q = dx 2t solution P = − 13 and Q = 13 Since the general solution of dt − 2x = is x(t) = Ce , the general solution of dx 2t dt − 2x = cos(3t) is x(t) = Ce − 13 cos(3t) + 13 sin(3t), where C is an arbitrary constant 9.3.5 Using Theorem 9.3.13, f (t) = et stant 9.3.6 Using Theorem 9.3.13, f (t) = e2t e−t t dt = et (−te−t − e−t + C) = Cet − t − 1, where C is an arbitrary con- e−2t e2t dt = e2t 414 dt = e2t (t + C), where C is an arbitrary constant Section 9.3 9.3.7 By Definition 9.3.6, pT (λ) = λ2 + λ − 12 = (λ + 4)(λ − 3) Since pT (λ) has distinct roots λ1 = −4 and λ2 = 3, the solutions of the differential equation are of the form f (t) = c1 e−4t + c2 e3t , where c1 and c2 are arbitrary constants (by Theorem 9.3.8) 9.3.8 pT (λ) = λ2 + 3λ − 10 = (λ + 5)(λ − 2) = x(t) = c1 e−5t + c2 e2t , where c1 , c2 are arbitrary constants 9.3.9 pT (λ) = λ2 − = (λ − 3)(λ + 3) = f (t) = c1 e3t + c2 e−3t , where c1 , c2 are arbitrary constants 9.3.10 pT (λ) = λ2 + = has roots λ1,2 = ±i By Theorem 9.3.9, f (t) = c1 cos(t) + c2 sin(t), where c1 , c2 are arbitrary constants 9.3.11 pT (λ) = λ2 − 2λ + = has roots λ1,2 = ± i By Theorem 9.3.9, x(t) = et (c1 cos(t) + c2 sin(t)), where c1 , c2 are arbitrary constants 9.3.12 pT (λ) = λ2 − 4λ + 13 = has roots λ1,2 = ± 3i By Theorem 9.3.9, f (t) = e2t (c1 cos(3t) + c2 sin(3t)), where c1 , c2 are arbitrary constants 9.3.13 pT (λ) = λ2 + 2λ + = (λ + 1)2 = has the double root λ = −1 Following Summary 9.3.15 in the text, we find f (t) = e−t (c1 t + c2 ), where c1 , c2 are arbitrary constants 9.3.14 pT (λ) = λ2 + 3λ = λ(λ + 3) = has roots λ1 = 0, λ2 = −3 9.3.15 By integrating twice we find f (t) = c1 + c2 t, where c1 , c2 are arbitrary constants 9.3.16 By Theorem 9.3.10, the differential equation has a particular solution of the form fp (t) = P cos(t) + Q sin(t) Plugging fp into the equation we find (−P cos(t) − Q sin(t)) + 4(−P sin(t) + Q cos(t)) + 13(P cos(t) + Q sin(t)) = cos(t) or 12P + 4Q = , so −4P + 12Q = P = 40 Q= 40 Therefore, fp (t) = 40 cos(t) + 40 sin(t) Next we find a basis of the solution space of f (t) + 4f (t) + 13f (t) = pT (λ) = λ2 + 4λ + 13 = has roots −2 ± 3i By Theorem 9.3.9, f1 (t) = e−2t cos(3t) and f2 (t) = e−2t sin(3t) is a basis of the solution space By Theorem 9.3.4, the solutions of the original differential equation are of the form f (t) = c1 f1 (t) + c2 f2 (t) + cos(t) + 40 sin(t), where c1 , c2 are arbitrary constants fp (t) = c1 e−2t cos(3t) + c2 e−2t sin(3t) + 40 415 Chapter 9.3.17 By Theorem 9.3.10, the differential equation has a particular solution of the form fp (t) = P cos(t) + Q sin(t) Plugging fp into the equation we find (−P cos(t) − Q sin(t)) + 2(−P sin(t) + Q cos(t)) + P cos(t) + Q sin(t) = sin(t) P = − 12 2Q = or , so −2P = Q=0 Therefore, fp (t) = − 21 cos(t) Next we find a basis of the solution space of f (t) + 2f (t) + f (t) = In Exercise 13 we see that f1 (t) = e−t , f2 (t) = te−t is such a basis By Theorem 9.3.4, the solutions of the original differential equation are of the form f (t) = c1 f1 (t) + c2 f2 (t) + fp (t) = c1 e−t + c2 te−t − 12 cos(t), where c1 , c2 are arbitrary constants 9.3.18 We follow the approach outlined in Exercises 16 and 17 • Particular solution fp = 10 cos(t) + 10 sin(t) • Solutions of f (t) + 3f (t) + 2f (t) = are f1 (t) = e−t and f2 (t) = e−2t • The solutions of the original differential equation are of the form f (t) = c1 e−t + c2 e−2t + where c1 and c2 are arbitrary constants 10 cos(t) + 10 sin(t), 9.3.19 We follow the approach outlined in Exercise 17 • Particular solution xp (t) = cos(t) • Solutions of d2 x dt2 √ √ + 2x = are x1 (t) = cos( 2t) and x2 (t) = sin( 2t) √ √ • The solutions of the original differential equation are of the form x(t) = c1 cos( 2t) + c2 sin( 2t) + cos(t), where c1 and c2 are arbitrary constants 9.3.20 pT (λ) = λ3 − 3λ2 + 2λ = λ(λ − 1)(λ − 2) = has roots λ1 = 0, λ2 = 1, λ3 = By Theorem 9.3.8, the general solution is f (t) = c1 + c2 et + c3 e2t , where c1 , c2 , c3 are arbitrary constants 9.3.21 pT (λ) = λ3 + 2λ2 − λ − = (λ − 1)(λ + 1)(λ + 2) = has roots λ1 = 1, λ2 = −1, λ3 = −2 By Theorem 9.3.8, the general solution is f (t) = c1 et + c2 e−t + c3 e−2t , where c1 , c2 , c3 are arbitrary constants 9.3.22 pT (λ) = λ3 − λ2 − 4λ + = (λ − 1)(λ − 2)(λ + 2) = has roots λ1 = 1, λ2 = 2, λ3 = −2 By Theorem 9.3.8, the general solution is f (t) = c1 et + c2 e2t + c3 e−2t , where c1 , c2 , c3 are arbitrary constants 9.3.23 General solution f (t) = Ce5t Plug in: = f (0) = Ce0 = C, so that f (t) = 3e5t 9.3.24 General solution x(t) = Ce−3t + (see Exercise 2) Plug in: = x(0) = C + 37 , so that C = − 37 and x(t) = − 73 e−3t + 37 9.3.25 General solution f (t) = Ce−2t Plug in: = f (1) = Ce−2 , so that C = e2 and f (t) = e2 e−2t = e2−2t 416 Section 9.3 9.3.26 General solution f (t) = c1 e3t + c2 e−3t (see Exercise 9), with f (t) = 3c1 e3t − 3c2 e−3t Plug in: = f (0) = c1 + c2 and = f (0) = 3c1 − 3c2 , so that c1 = 61 , c2 = − 16 , and f (t) = 16 e3t − 61 e−3t 9.3.27 General solution f (t) = c1 cos(3t) + c2 sin(3t) (Theorem 9.3.9) Plug in: = f (0) = c1 and = f π = −c2 , so that c1 = 0, c2 = −1, and f (t) = − sin(3t) 9.3.28 General solution f (t) = c1 e−4t + c2 e3t , with f (t) = −4c1 e−4t + 3c2 e3t Plug in: = f (0) = c1 + c2 and = f (0) = −4c1 + 3c2 , so that c1 = c2 = and f (t) = 9.3.29 General solution f (t) = c1 cos(2t) + c2 sin(2t) + (use the approach outlined in Exercise 17) sin(t), so that f (t) = −2c1 sin(2t) + 2c2 cos(2t) + Plug in: = f (0) = c1 and = f (0) = 2c2 + 13 , so that c1 = 0, c2 = − 61 , and f (t) = − 61 sin(2t) + 3 cos(t) sin(t) 9.3.30 a k is a positive constant that depends on the rate of cooling of the coffee (it varies with the material of the cup, for example) A is the room temperature b T (t) + kT (t) = kA Constant particular solution: Tp (t) = A General solution of T (t) + kT (t) = is T (t) = Ce−kt General solution of the original differential equation: T (t) = Ce−kt + A Plug in: T0 = T (0) = C + A, so that C = T0 − A and T (t) = (T0 − A)e−kt + A 9.3.31 dv dt + k mv =g constant particular solution: vp = General solution of dv dt + k mv mg k k = is v(t) = Ce− m t k General solution of the original differential equation: v(t) = Ce− m t + Plug in: = v(0) = C + lim v(t) = t→∞ mg k mg k , so that C = − mg k and v(t) = mg k k − e− m t (the “terminal velocity”) See Figure 9.43 Figure 9.43: for Problem 9.3.31 417 mg k Chapter dB dt 9.3.32 = kB − r or dB dt − kB = −r ↑ ↑ interest withdrawals constant particular solution Bp = General solution of dB dt r k − kB = is B(t) = Cekt General solution of the original differential equation: B(t) = Cekt + Plug in: B0 = B(0) = C + kr , so that C = B0 − r k and B(t) = B0 − r k r k ekt + r k if B0 > r k then interest will exceed withdrawals and balance will grow if B0 < r k then withdrawals will exceed interest and account will eventually be depleted if B0 = r k then the balance will remain the same The graph in Figure 9.44 shows the three possible scenarios Figure 9.44: for Problem 9.3.32 √ 9.3.33 By Theorem 9.3.9, x(t) = c1 cos g Lt g Lt + c2 sin √ = P = 2π √Lg or L = g π2 2π √L It is required that , with period P = √ g = 2π g L ≈ 0.994 (meters) 9.3.34 a We will take downward forces as positive Let g = acceleration due to gravity, ρ = density of block a = length of edge of block Then (weight of block) = (mass of block) · g = (density of block)(volume of block) g buoyancy = (weight of displaced water) = (mass of displaced water) · g = (density of water) (volume of displaced water) g = 1a2 x(t)g = a2 gx(t) b Newton’s Second Law of Motion tells us that m ddt2x = F = weight − buoyancy = ρa3 g − a2 gx(t), where m = ρa3 is the mass of the block ρa3 ddt2x = ρa3 g − a2 gx(t) 418 = ρa3 g Section 9.3 d2 x dt2 =g− d2 x dt2 + g ρa x g ρa x(t) =g constant solution xp = ρa g ρa t general solution (use Theorem 9.3.9): x(t) = c1 cos + c2 sin g ρa t + ρa Now c2 = since block is at rest at t = Plug in: a = x(0) = c1 + ρa, so that c1 = a − ρa and x(t) = (a − ρa) cos g ρa t c The period is P = √2πg = ρa + ρa ≈ cos(11t) + (measured in centimeters) √ 2π ρa √ g Thus the period increases as ρ or a increases (denser wood or larger block), or as g decreases (on the moon) The period is independent of the initial state 9.3.35 a pT (λ) = λ2 + 3λ + = (λ + 1)(λ + 2) = with roots λ1 = −1 and λ2 = −2, so x(t) = c1 e−t + c2 e−2t b x (t) = −c1 e−t − 2c2 e−2t Plug in: = x(0) = c1 + c2 and = x (0) = −c1 − 2c2 , so that c1 = 2, c2 = −1 and x(t) = 2e−t − e−2t See Figure 9.45 Figure 9.45: for Problem 9.3.35b c Plug in: = x(0) = c1 + c2 and −3 = x (0) = −c1 − 2c2 , so that c1 = −1, c2 = 2, and x(t) = −e−t + 2e−2t See Figure 9.46 Figure 9.46: for Problem 9.3.35c d The oscillator in part (b) never reaches the equilibrium, while the oscillator in part (c) goes through the equilibrium once, at t = ln(2) Take another look at Figures 9.45 and 9.46 419 Chapter 9.3.36 fT (λ) = λ2 + 2λ + 101 = has roots λ1,2 = −1 ± 20i By Theorem 9.3.9, x(t) = e−t (c1 cos(20t) + c2 sin(20t)) Any nonzero solution goes through the equilibrium infinitely many times See Figure 9.47 Figure 9.47: for Problem 9.3.36 9.3.37 fT (λ) = λ2 + 6λ + = (λ + 3)2 has roots λ1,2 = −3 Following the method of Example 10, we find the general solution x(t) = e−3t (c1 + c2 t) with x (t) = e−3t (c2 − 3c1 − 3c2 t) Plug in: = x(0) = c1 , and = x (0) = c2 − 3c1 , so that c1 = 0, c2 = 1, and x(t) = te−3t See Figure 9.48 Figure 9.48: for Problem 9.3.37 The oscillator does not go through the equilibrium at t > 9.3.38 a (D − λ)(p(t)eλt ) = [p(t)eλt ] − λp(t)eλt = p (t)eλt + λp(t)eλt − λp(t)eλt = p (t)eλt , as claimed b Applying the result from part (a) m times we find (D − λ)m (p(t)eλt ) = p(m) (t)eλt = 0, since p(m) (t) = for a polynomial of degree less than m c By Theorem 9.3.3, we are looking for m linearly independent functions By part (b), the functions eλt , teλt , t2 eλt , , tm−1 eλt the job d Note that the kernel of (D − λi )mi is contained in the kernel of (D − λ1 )m1 · · · (D − λr )mr , for any ≤ i ≤ r Therefore, we have the following basis: eλ1 t , teλ1 t , , tm1 −1 eλ1 t , eλ2 t , teλ2 t , , tm2 −1 eλ2 t , eλr t , teλr t , , tmr −1 eλr t 420 Section 9.3 9.3.39 fT (λ) = λ3 + 3λ2 + 3λ + = (λ + 1)3 = has roots λ1,2,3 = −1 In other words, we can write the differential equation as (D + 1)3 f = By Exercise 38, part (c), the general solution is f (t) = e−t (c1 + c2 t + c3 t2 ) 9.3.40 fT (λ) = λ3 + λ2 − λ − = (λ + 1)2 (λ − 1) = has roots λ1,2 = −1, λ3 = In other words, we can write the differential equation as (D + 1)2 (D − 1) = By Exercise 38, part (d), the general solution is x(t) = e−t (c1 + c2 t) + c3 et 9.3.41 We are looking for functions x such that T (x) = λx, or T (x) − λx = Now T (x) − λx is an nth-order linear differential operator, so that its kernel is n-dimensional, by Theorem 9.3.3 Thus λ is indeed an eigenvalue of T , with an n-dimensional eigenspace 9.3.42 a We need to solve the second-order differential equation T x = D2 x = ddt2x = λx This differential equation has a two-dimensional solution space Eλ for any λ, so that all λ are eigenvalues of T √ if λ > then Eλ = span e √ λt λt , e− if λ = then Eλ = span(1, t) if λ < then Eλ = span sin √ √ −λt , cos −λt b Among the eigenfunctions of T we found in part (a), we seek those of period In the case λ < the shortest period is P = √2π Now is a period if P = √2π = k1 for some positive integer k, or, λ = −4π k Then −λ −λ Eλ = span(cos(2πkt), (sin(2πkt)) In the case λ > there are no periodic solutions In the case λ = we have the constant solutions, so that λ = is an eigenvalue with E0 = span(1) Summary: λ = −4π k is an eigenvalue, for k = 1, 2, 3, , with Eλ = span(cos(2πkt), (sin(2πkt)) λ = is an eigenvalue, with E0 = span(1) 9.3.43 a Using the approach of Exercise 17, we find x(t) = c1 e−2t + c2 e−3t + b For large t, x(t) ≈ 10 cos t + 10 10 cos t + 10 sin t sin t 9.3.44 a Using the approach of Exercises 16 and 17 we find x(t) = e−2t (c1 cos t + c2 sin t) − b For large t, x(t) ≈ − 40 cos(3t) + 9.3.45 We can write the system as 40 sin(3t) dx1 dt = x1 + 2x2 = x2 dx2 dt with x1 (0) = 1, x2 (0) = −1 The solution of the second equation, with the given initial value, is x2 (t) = −et Now the first equation takes the form dx1 dt − x1 = −2et Using Example (with a = and c = −2) we find x1 (t) = et (−2t + C) 421 40 cos(3t) + 40 sin(3t) Chapter plug in: = x1 (0) = C, so that x1 (t) = et (1 − 2t) and x(t) = et  dx1 9.3.46 We can write the system as  dt dx2 dt dx3 dt = = = We solve for x2 and x3 as in Exercise 45: − 2t −1  2x1 + 3x2 + x3 x2 + 2x3  with x1 (0) x3 = 2, x2 (0) = 1, x3 (0) = −1 x2 (t) = et (1 − 2t) x3 (t) = −et Now the first equation takes the form dx1 dt − 2x1 = 3et (1 − 2t) − et = et (2 − 6t), x1 (0) = We use Theorem 9.3.13 to solve this differential equation: x1 (t) = e2t e−2t et (2 − 6t) dt = e2t (2e−t − 6te−t ) dt = e2t [−2e−t + 6te−t + 6e−t + C] plug in: = x1 (0) = (−2 + + c), so that c = −2 and x1 (t) = e2t (4e−t + 6te−t − 2) = 4et + 6tet − 2e2t   t 4e + 6tet − 2e2t  x(t) =  et − 2tet t −e 9.3.47 a We start with a preliminary remark that will be useful below: If f (t) = p(t)eλt , where p(t) is a polynomial, then f (t) has an antiderivative of the form q(t)eλt , where q(t) is another polynomial We leave this remark as a calculus exercise The function xn (t) satisfies the differential equation sired form dxn dt = ann xn , so that xn = Ceann t , which is of the de- Now we will show that xk is of the desired form, assuming that xk+1 , , xn have this form xk satisfies the dxk k differential equation dx dt = akk xk + ak,k+1 xk+1 + · · · + akn xn or dt − akk xk = ak,k+1 xk+1 + · · · + akn xn Note that, by assumption, the function on the right-hand side has the form p1 (t)eλ1 t + · · · + pm (t)eλm t If we set λ1 t k akk = a for simplicity, we can write dx + · · · + pm (t)eλm t dt − axk = p1 (t)e By Theorem 9.3.13, the solution is xk (t) = eat = eat e−at (p1 (t)eλ1 t + · · · + pm (t)eλm t ) dt (p1 (t)e(λ1 −a)t + · · · + pm (t)e(λm −a)t ) dt = eat (q1 (t)e(λ1 −a)t + · · · + qm (t)e(λm −a)t + C) = q1 (t)eλ1 t + · · · + qm (t)eλm t + Ceat as claimed (note that a is one of the λi ) The constant C is determined by xk (0) Note that we used the preliminary remark in the second to last step b It is shown in introductory calculus classes that lim (tm eλt ) = if and only if λ is negative (here m is a fixed t→∞ positive integer) In light of part (a), this proves the claim 422 Section 9.3 9.3.48 a By Exercise 47, the system solution x(t) = Su(t) du dt = Bu has a unique solution u(t) Then the system b It suffices to note that lim x(t) = if and only if lim u(t) = 0, where u = S −1 x t→∞ t→∞ 423 dx dt = Ax has the unique ... upstream The solution is v = 18 and s = x+z 1.1.25 The system reduces to  y − 2z   =1 = −3  =k−7 a The system has solutions if k − = 0, or k = b If k = then the system has infinitely many solutions... shown that any solution of the “old” system is also a solution of the “new.” To see that, conversely, any solution of the new system is also a solution of the old system, note that elementary row... This system has infinitely many solutions: If we choose y = t, an arbitrary real number, then the equation x − 2y = gives us x = + 2y = + 2t Therefore the general solution is (x, y) = (3 + 2t,