Solutions Manual to Introduction to Dierential Equations with Dynamical Systems by Stephen L. Campbell and Richard Haberman M. Ziaul Haque PRINCETON UNIVERSITY PRESS PRINCETON AND OXFORD Copyright c 2008 by Princeton University Press Published by Princeton University Press 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press 6 Oxford Street, Woodstock, Oxfordshire, 0X20 1TW All Rights Reserved This book has been composed in LATEX press.princeton.edu Contents Preface vChapter 1. First-Order Dierential Equations and Their Applications 11.1 Introduction to Ordinary Dierential Equations 11.2 Denite Integral and the Initial Value Problem 11.3 First-Order Separable Dierential Equations 31.4 Direction Fields 51.5 Eulers Numerical Method (Optional) 71.6 First-Order Linear Dierential Equations 101.7 Linear First-Order Dierential Equations with Constant Coecients and Constant Input 151.8 Growth and Decay Problems 201.9 Mixture Problems 231.10 Electronic Circuits 251.11 Mechanics II: Including Air Resistance 261.12 Orthogonal Trajectories (optional) 27Chapter 2. Linear Second and Higher-Order Dierenial Equations 292.1 General Solution of Second-Order Linear Dierential Equations 292.2 Initial Value Problem (For Homogeneous Equation) 302.3 Reduction of Order 322.4 Homogeneous Linear Constant Coecient Dierential Equations(Second Order) 352.5 Mechanical Vibrations I: Formulation and Free Response 392.6 The Method of Undetermined Coecients 452.7 Mechanical Vibrations II: Forced Response 582.8 Linear Electric Circuits 652.9 Euler Equation 682.10 Variation of Parameters (Second-Order) 702.11 Variation of Parameters (nth-Order) 75Chapter 3. The Laplace Transform 823.1 Denition and Basic Properties 823.2 Inverse Laplace Transforms (Roots, Quadratics, & Partial Fractions) 863.3 Initial-Value Problems for Dierential Equations 943.4 Discontinuous Forcing Functions 983.5 Periodic Functions 1093.6 Integrals and the Convolution Theorem 1143.7 Impulses and Distributions 118iv CONTENTS Chapter 4. An Introduction to Linear Systems of Dierential Equations andTheir Phase Plane 1214.1 Introduction 1214.2 Introduction to Linear Systems of Dierential Equations 1214.3 Phase Plane for Linear Systems of Dierential Equations 130Chapter 5. Mostly Nonlinear First-Order Dierential Equations 1425.1 First-Order Dierential Equations 1425.2 Equilibria and Stability 1425.3 One Dimensional Phase Lines 1435.4 Application to Population Dynamics: The Logistic Equation 146Chapter 6. Nonlinear Systems of Dierential Equations in the Plane 1506.1 Introduction 1506.2 Equilibria of Nonlinear Systems, Linear Stability Analysis of Equilibrium, and Phase Plane 1506.3 Population Models 1616.4 Mechanical Systems 178Preface This Student Solutions Manual contains solutions to the odd-numbered exercises in the text Introduction to Dierential Equations with Dynamical Systems by Stephen L. Campbell and Richard Haberman. To master the concepts in a mathematics text the students must solve problems which sometimes may be challenging. This manual has been written focusing students needs and expectations. Instead of providing only the answer with very few steps, I include a reasonably detailed solution with a fair amount of detail when explaining the solution of the problem. The solutions are self-explanatory and consistent with the notations and terminologies used in the text book. I hope this manual will help students build problem-solving skills. I would like to thank many people who have provided invaluable help, in many ways, in the preparation of this manual. First, I take this opportunity to thank Professor Richard Haberman for his generous expert help, constructive comments and accuracy checking. I would also like to thank Professor Stephen L. Campbell for assembling the nal manuscript, Professor Peter K. Moore for facilitating support process and Ms. Vickie Kearn of the publishing company for her patience and support. Finally, I must appreciate the patience of my wife, Rukshana, and my daughters, Zareen and Ehram for their understanding and compromise of summer time that was slighted because of my busy schedule. M. Ziaul HaqueSouthern Methodist UniversityDallas, TX, 75275, U.S.A.July, 2007.Chapter OneFirst-Order Dierential Equations and Their Applications 1.1 INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS There are no exercises in this section. 1.2 DEFINITE INTEGRAL AND THE INITIAL VALUE PROBLEM 1-7. Substitute expression for x into the dierential equation 1. x = 2e3t + 1. l.h.s. = dx = 6e3t .dt r.h.s. = 3x 3 = 3(2e3t + 1) 3 = 6e3t . Hence l.h.s. = r.h.s. 3. x = t 1. l.h.s. = dx = 1. r.h.s = x = t1 = 1. Hence l.h.s. = r.h.s. dt t1 t1 5. x = et2 . l.h.s. = dx = 2tet2 . r.h.s = 2tx = 2tet2 . Hence l.h.s. = r.h.s. dt dx 7. x = e2t . l.h.s. = dt = 2e2t . r.h.s. = 2e2tx2 = 2e2t(e2t)2 = 2e2t . Hence l.h.s. = r.h.s. dx t t9. dt = 3e . Integrating we get, x = 3e + c. 11. dx = 5 cos 6t. Integrating we get, x = 5 sin 6t + c.dt 6 dx 1 1 13.dt = 8 cos(t 2 ). Use of denite integral gives x = 8 _ 0 t cos t 2 dt + c. 15. dx = ln(4 + cos2 t). Use of denite integral gives x dt = _ 0 t ln(4 + cos2 t)dt + cdx 117.dt = t4; x(2) = 3. Integrating we get x = 5 t5 + c.17 1 17t = 2 = 3 = 32 + c = 5 . So x = t5 5 c = 5 5 . 19. dx = ln t ; x(2) = 5. Use of denite integral gives dt 4+cos2 tln tx = 5 + _ t dt. 2 4+cos2 t 21. dx = e t ; x(1) = 3. dx = e t dt. Use of denite integral gives dt 1+t 1+t t t e e x 3 = _ t dt = x = 3 + _ tdt. 1 1+t 1 1+td2 x23.dt2 = 15. Integrating we getdx = 15t + c1 ( dxdx dt dt = v0 at t = 0 = c1 = v0.)= 15t + v0. Integrating again we getdt 15 t2x = 2 + v0t + c2 (x = 0 at t = 0 = c2 = 0.) _2 CHAPTER 1 v0Car stops when dx = 0 = v0 15t = 0 = t = (stopping time). dt 15 So distance travelled is v0 = 1510 m/sec. 20 20 20 15 1 v 15 2 15 75 = 1v v( v0 )2 +15 x = = = = 2 15 2 25. d2 x = 2500. Integrating we get dx = 2500t + c1( dx = 60 at t = 0 dt2 dt dt = c1 = 60). So dx = 2500t + 60. Integrating again we get 2500 dt x = t2 + 60t + c2(x = 0 at t = 0 = c2 = 0.)2 Car stops when dx = 0 =2500t + 60 = 0 dt = t = 60 (stopping time). So distance travelled is 2500 2500 60 602 x = 2 (2500 )2 + 2500 = 0.72 km. x27. d2 = 2500. Integrating we get dx = 2500t + c1( dx = v0 at t = 0 dt2 dt dt = c1 = v0). So dx = 2500t + v0. Integrating again we get 2500 t2 dt x = 2 + v0t + c2(x = 0 at t = 0 = c2 = 0) Car stops when dx = 0 = v0 2500t = 0 v0 dt = t = 2500 (stopping time). So distance travelled is 20 20 2500 v0 v v)2( km. + x = = 2 2500 2500 5000 29. d2 x = 6t. Integrating we get dx dt2 dt = 3t2 + c1( dx dt = 3t2 + 62. Integrating again we get = 50 at t = 2 c1 = 62). So dx dt = x = t3 + 62t + c2(x = 0 at t = 2 =Car stops when dx = 0 =3t2 + 62 = 0 dt c2 = 116.) _ 62 = t = (stopping time). So distance travelled is x = t(62 3 t2) 116 = _ 62 (62 62 ) 116 3 3 _ 62 3 2( 2 3 116 km. = 3 )62 116 = 2 _ 62 3 dV So dy 31. (a) V =volume, dt = Q m3/h. Let snow depth be y. dt = c y = ct + c1(y = 0 at t = 0 c1 = 0). Thus y = ct. Now consider the snowplow has moved x over the time t and the approximate change in volume over this time is V. Hence V xV = w(x)y = wctx t = wct t . Now taking limit as t 0 we get dV = Q = wct dx dx = Q = 1 with wc dt dt dt wct kt k = . (b) dx = Q 11 . Separating the variables we get, _ dx = 1 _ 1 dt dt kt k t x = k ln t + a. At 11 A.M. t = 3 and x(3) = 0. So 0 = k 1 ln 3 + a 1 a = k 1 ln 3. Then at noon (t = 4), x(4) = 1 ln 4 ln 3 = 1 ln 4 .k k k 3 33. d2 y = g = 9.8 m/sec2 . Integrating we get dt2 dy = 9.8t + c1( dy dy = v0 at t = 0 = c1 = v0.) So dt dt = 9.8t + v0. Integrating again we get dt y = 92 .8 t2 + v0t + c2(y = 0 at t = 0 = c2 = 0.) At maximum height dy = 0 = v0 9.8t = 0 v0 dt = t = 9.8 (time at maximum height). So maximum height is = 1960 m/sec. 20 20 20y = 9.8 2 ( v0 )2 +9.8 = 1 v 100 = 1 2 v v = = v09.8 2 9.8 9.8 FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 3 35.d2 y = g = 9.8 m/sec2 . Integrating we get dt2 dy = 9.8t + c1( dy = 0 at t = 0 = c1 = 0.)dt dt So dy = 9.8t. Integrating again we get y = 9.8 t2 + c2dt 2 (y = 200 at t = 0 = c2 = 200). y = 9.8 t2 + 200. Now to fall, 2 9.8 9.8_ 400 20y = 0. So 0 = 2 t2 + 200 2 t2 = 200 = t = 9.8 = 9.8 sec. 37. Since x(t0) = x0, the general solution x = _tf _t_ dt + c becomes 0 x0 = _t0 f _t_ dt + c = c = x0 _t0 f _t_ dt. Hence the solution is 0 0 x = _tf _t_ dt + x0 _t0 f _t_ dt = x0 + _tf _t_ dt. 0 0 t0 1.3 FIRST-ORDER SEPARABLE DIFFERENTIAL EQUATIONS dx x+1 dx dt 1.dt = t . Separating variables gives _ x+1 = _ t . Integrating we get, ln x + 1 = ln t + c1 ln |x+1| = c1 x+1 c1 t|x+1 | c1| | |t| = edx tt = e = c x = ct 1. 3.dt = e . Separating variables gives _ dx = _ etdt. Integrating we get, x = et + c. 5.dx = tx + 4x + 3t + 12 = (x + 3) (t + 4) . Separating variables gives dtdx t2_ x+3 = _ (t + 4) dt. Integrating we get, ln x + 3 = 2 + 4t + c1 2 | |2 2. dx |x + 3| = ec1 e t2 +4t x = ce t+4t 3 where c = ec1 7.dt = 3. Separating variables gives _ dx = _ 3dt. Integrating we get, x = 3t + c. dx 59.dt = x . Separating variables gives _ x5dx = _ dt. Integrating we get, x4 = t + c. Using x(2) = 1 we get, 1 = 2 + c c = 9 .4 4 4 Substituting c we get x4 = 9 4t x = (9 4t)1/4 . dx11.dt = x2 cos(t2). Separating variables gives x2dx = cos(t2)dt. Using 2 denite integrals we get, _xx2dx = _tcos(t )dt 1 0 2 2 1 + 1 = _ 0 tcos(t )dt 1 = 1 _ 0 tcos(t )dt x x 1 x = 21_tcos(t )dt 0 13. dx = t cos(x1/2). Separating variables gives dx = tdt. Using dt cos(x1/2) dxdenite integrals we get, _x= _ttdt cos(x1/2)2 1 2 _ 2 xdx = t 1 = 1 _t2 1_ . cos(x1/2) 2 2 2 du t2+1 15. dt = u2+4 . Separating variables gives _ _u2 + 4_ du = _ _t2 + 1_ dt. 4 CHAPTER 1 3 3 Integrating we get, u 3 + 4u = t3 + t + c. Using u(0) = 1 we get, 1 13 + 4 = c c = . Substituting c we obtain the solution as u33 + 12u = t3 + 3t 3 + 13.dx 2 217.dt = t2x + x + t2 +1 = _x2 + 1_ _t2 + 1_ . Separating variables gives dx3_ x2+1 = _ _t2 + 1_ dt. Integrating we get, tan1 (x) = t3 + t + c 3 x = tan _ t3 + t + c_ . Using x(0) = 2 we get, c = tan1(2). Hence _ t_the solution is x = tan 3 3 + t + tan1(2) . dx dx 19.dt = x (x 1) . Separating variables gives _ x(x1) = _ dt. Using partial fractions to the integral on the left we get, 1 A B x(x1) = x + x1 1 = A (x 1) + Bx. Putting x = 0 and 1, respectively, we have, A = 1 and B = 1. Hence dx dx dx _ x(x1) = _ x1 _ x = _ dt ln |x 1| c ln |x| = t + c x1ln x1 = t + c = ket where k = e . Solving this x x 1equation for x we obtain the general solution, x = 1ket . Since x = 0 and x = 1 both satisfy the dierential equation they are also solutions. The solution x = 1 corresponds to k = 0, however, x = 0 is not included in the general solution for any nite k. Hence the solutions are x = 1 and x = 0.1ket dx 2 dx 21.dt = (x 1) (x 2) . Separating variables gives _ (x1)(x2)2 = _ dt. Using partial fractions to the integral on the left we get, 1 A B C(x1)(x2)2 = x1 + x2 + (x2)2 21 = A (x 2) + B (x 1) (x 2) + C (x 1) . Putting x = 1 and 2, respectively, we have, A = 1 and C = 1. Then equating the coecients of x2 we have A + B = 0 B = 1. Hence dx dx dx dx _ (x1)(x2)2 = _ x1 _ x2 + _ (x2)2 = _ dt. Integrating both sides we obtain the solution, ln ln 1 = t + c. |x 1| |x 2| x2 Since x = 1 and x = 2 both satisfy the dierential equation they are also solutions. Hence the solutions are ln ln 1 = t + c, x = 1 and x = 2. |x 1| |x 2| x2 23. (tx + x) dt + (tx + t) dx = 0. Dividing by tx we get,_1 + 1 t _ dt + _1 + x 1 _ dx = 0. Now integrating we have,t + ln t + x + ln x = c ln tx | | | | x | | = t x + c c tx = ecetex tetxe = c where c = e . 25._t2 4_ dz + _z2 9_ dt = 0. Dividing by _t2 4_ _z2 9_ dz dt we get, (z29) + (t24) = 0. Now we use the formula du 1 ua_ u2a2 = 2a ln u+a (from integration table) to integrate and 1/4 get 1 ln + 1 ln = c ln 1/6 = cz3 t2 z3 t2 6 z+3 4 t+2 z+3 t+2 _1/4_z3 _1/6 _ t2 _1/4 c _ z3 _1/6 _ t2 c = e = e z+3 t+2 z+3 t+2 6 4 4 2_ z3 _ 1 c _ t2 _ 1 c _ t+2 _ 1 z3 _ t+2 _ 3 z+3 = et+2 = et2 z+3 = c t2 FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 5 where c = (ec)6 . 27. et+xdt + e2t3xdx = 0. Dividing by exe2t we get,etdt + e4xdx = 0. Now integrating we have, et e4x = c4 e4x = 4et 4c 4x = ln (4et 4c) 1 x = 4 ln (4et + k) where k = 4c. 29. z = at + bx + c. Dierentiating with respect to t we get,dz = a + b dx dz = a + bf (z) which is a dierential equationdt dt dt in z and t and can be solved by separation of variables as dz _ = _ dt. a+bf (z) 31.dx = (t + 4x 1)2 . Let z = t + 4x 1. Then dx = z2 = f(z)dt dt and dz = 1 + 4 dx = 1 + 4z2 . Separating the variables we get, dz 1_ 1+4dt z2 = _ dt. dt We use the substitution u = 2z dz = 2 du du to integrate the left hand side. This gives 12 _ 1+u2 = _ dt 1 tan1 (2z) = t + c 1 tan1 (2t + 8x 2) = t + c. dx z33.2 = et+x (t + x)1 1. 2 Let z = t + x. Then dx = e z1 dt dt 1 z zand dz = 1 + dx = 1 + e z1 1 = e z1 . Separating the dt dt variables we get, _ zez dz = _ dt. We use integration by parts to integrate the left hand side as u = z du = dz and dv = ezdz. v = ez . Then _ zez dz = _ udv = uv _ vdu = zez + _ ez dz = zez ez . This gives ez (z + 1) = t + c. Substituting z = t + x we get the solution as e(t+x) (t + x + 1) = t + c etx (t + x + 1) = t + c. 1.4 DIRECTION FIELDS 1. 2 0 2 0 2 x t

6 CHAPTER 1 3. 2 0 2 0 2 t x 5. 2 0 2 0 2 x t 1.4.1 Existence and Uniqueness 1.dx = x = f (t, x) and fx = 1 are continuous for all (t, x) . Sodt 1+t21+t2 unique solution exists for all (t0, x0) .dx 2_7/3 7 2_4/33.dt = _1 t2 x = f (t, x) and fx = 3 _1 t2 x are continuous for all (t, x) . So unique solution exists for all (t0, x0) . dx 1/515.dt = (x + t) = f (t, x) is continuous for all (t, x) but fx = 5(x+t)4/5 is not continuous for x + t = 0.So unique solution exists for all (t0, x0) such that x0 + t0 = 0. dx cos t

cos t7.dt = x1 = f (t, x) and fx = (x1)2 are not continuous at x = 1. So unique solution exists for all (t0, x0) such that x0 = 1. dx 9. = _1 t2 2x2_3/2 = f (t, x) and fx = 6x _1 t

2 2x2_1/2 dt are continuous for 1 t2 2x2 > 0. So unique solution exists for all (t0, x0) such that t20 + 2x02 < 1. dx 111.dt = t1/3 = f (t, x) is not continuous at t = 0 although fx = 0 is continuous everywhere. So unique solution exists for all (t0, x0) such that t0 = 0. 13. (a) Dierentiating t2 + x2 = c wrt(with respect to) t we get, FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 7 2t + 2x dx = 0 dx t . Hence t2 + x2 = c denesdt dt = xa solution.(b) graph (c) At x0 = 0, as f (t, x) = t and fx = t are discontinuous at x x2 this point. 2 dx 15. (a) Dierentiating t + x = c wrt t we get, 1 + 2x dt = 0.Hence t + x2 = c denes a solution.(b) graph (c) Heredx 1 = f (t, x) and fx = 1 are discontinuous at 2dt = 2x 2xx0 = 0. Hence the theorem fails to hold at this point. 17. (a) Dierentiating x = c sin t wrt t we get, dx x cos t = c cos t = c cos t = cx = x cot t. dt x c sin t Hence x = c sin t denes a solution. (b) graph (c) Heredx = x cos t = f (t, x) and fx = cos t are discontinuous dt sin t sin t when sin t = 0 t = n for n = 0, 1, 2, ...Hence the theorem fails to hold at the point t0 = n for n = 0, 1, 2, ... 19. (a) Dierentiating x = 1 wrt t we get, dx 1 2 . t+c dt = (t+c)2 = xHence x = 1 denes a solution. t+c (b) graph (c) Heredx = x2 = f (t, x) and fx = 2x are continuous dt everywhere. Hence there is NO point where the theorem fails to holds. 21. (a) For x = 1, l.h.s. is dx = 0 and r.h.s. = (x 1)1/5 = 0.dt For x = _ 4 t + c_5/4 + 1, l.h.s. is dx = _ 4 t + c_1/4 and 5 dt 5 r.h.s. = (x 1)1/5 = __ 4 t + c_5/4 + 1 1_1/5 = _ 4 t + c_1/4 .5 5 (b) By separating the variables we get, dx = dt which, on (x1)1/5 integration, becomes x = _ 45 t + c_5/4 + 1. Then using the initial condition x0 = 1 for any t0 we get c = 4 t0 and thus 5 one solution is x = _ 45 t 54 t0_5/4 + 1. Another solution is clearly x = 1 because it satises the initial condition as well as the dierential equation. So there are at least two solutions through the point (t0, x0) with x0 = 1. (c) Graph (d) Although f (t, x) = (x 1)1/5 is continuous everywhere, fx = 1 is not continuous at x0 = 1. As a result, (x1)4/5 uniqueness does not hold and two solutions in part (b) is not a surprise. 1.5 EULERS NUMERICAL METHOD (OPTIONAL) 1. dx = x t = f (t, x) , t0 = 0, x0 = 1, h = 0.5. We use recursive formula, dt 8 CHAPTER 1 xn+1 = xn + hf (tn, xn) = xn + 0.5 (xn tn), where tn+1 = tn + h to approximate x1 = x0 + h (x0 t0) = 1 + 0.5(1 0) = 1.5 at t1 = 0.5 x2 = x1 + h (x1 t1) = 1.5 + 0.5(1.5 0.5) = 2 at t2 = 1 x3 = x2 + h (x2 t2) = 2 + 0.5(2 1) = 2.5 at t3 = 1.5 x4 = x3 + h (x3 t3) = 2.5 + 0.5(2.5 1.5) = 3 at t4 = 2 So estimate for x(2) is x4 = 3. 3. dx = tx2 = f (t, x) , t0 = 0, x0 = 1, h = 1. We use recursive formula, dt xn+1 = xn + hf (tn, xn) = xn tnx2 , where tn+1 = tn + hnto approximatex1 = x0 ht0x02 = 1 0 = 1 at t1 = 1x2 = x1 ht1x21 = 1 1 = 0 at t2 = 2So estimate for x(2) is x2 = 0.5. dx = 2x 4t = f (t, x) , t0 = 0, x0 = 1, h = 0.5. We use recursive dt formula, xn+1 = xn + hf (tn, xn) = xn + 0.5 (2xn 4tn) = 2 (xn tn) , where tn+1 = tn + h to approximate x1 = 2 (x0 t0) = 2(1 0) = 2 at t1 = 0.5 x2 = 2 (x1 t1) = 2(2 0.5) = 3 at t2 = 1 x3 = 2 (x2 t2) = 2(3 1) = 4 at t3 = 1.5 x4 = 2 (x3 t3) = 2(4 1.5) = 5 at t4 = 2 So estimate for x(2) is x4 = 5. 7. dx = sin x = f (t, x) , t0 = 0, x0 = 0, h = 0.5. We use recursive dt formula, xn+1 = xn + hf (tn, xn) = xn + 0.5 (sin xn) , where tn+1 = tn + h to approximate x1 = x0 + 0.5 sin x0 = 0 at t1 = 0.5 x2 = x1 + 0.5 sin x1 = 0 at t2 = 1 Similarly, xi = 0 at ti for all i = 0, 1, ..., 8. So estimate for x(4) is x8 = 0. 9. (a) dx = 20x = f (t, x) , t0 = 0, x0 = 1, h = 0.2.dt We use recursive formula, xn+1 = xn + hf (tn, xn) = xn + 0.2 (20xn) = 3xn. Here xn = (3)n at tn = 0.2n, n = 0, 1, 2, ... 10 So estimate for x(2) is x10 = (3) = 59049. (b) xn oscillates wildly as n . x(2) = e40 = 4.248 1018 . So x10 = 59049 is not a very good approximation. 11. (a) dx = 20x = f (t, x) , t0 = 0, x0 = 1, h = 0.01.dt We use recursive formula, xn+1 = xn + hf (tn, xn) = xn + 0.01 (20xn) = 0.8xn. Here xn = (0.8)n at tn = 0.01n, n = 0, 1, 2, ... 200 So estimate for x(2) is x200 = (0.8) = 4.1495 1020 . (b) In this case, xn 0 as n , so the numerical solution behaves like the actual solution x = e20t and the statement x200 x(2) = e40 = 4.248 1018 is not too bad. 9 FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 13. dx = x2 = f (t, x) , t0 = 0, x0 = 1, h = 0.2.dt We use recursive formula, xn+1 = xn + hf (tn, xn) = xn + 0.2xn2 , where tn+1 = tn + h to approximate x1 = x0 + 0.2x02 = 1 + 0.2 = 1.2 at t1 = 0.2 x2 = x1 + 0.2x12 = 1.2 + 0.288 = 1.488 at t2 = 0.4 x3 = x2 + 0.2x22 = 1.488 + 0.443 = 1.9308 at t3 = 0.6 x4 = x3 + 0.2x23 = 1.9308 + 0.7456 = 2.6764 at t4 = 0.8 x5 = x4 + 0.2x42 = 2.6764 + 1.4326 = 4.109 at t5 = 1 So estimate for x(1) is x5 = 4.109. For h = 0.1 use the same formula and procedure as above. There are 11 points this time, that is 10 steps after initial step and the estimates for x(1) is x10 = 6.1289. For h = 0.01 the estimates for x(1) is 30.3897 using software and for h = 0.001 the estimates for x(1) is 193.1368 using software. 15. dx = 1 2x + x2 = f (t, x) , t0 = 0, x0 = 5, h = 0.2.dt We use recursive formula,xn+1 = xn + hf (tn, xn) = xn + 0.2 _1 2xn + x2 _n 2 = xn + 0.2 (xn 1) , where tn+1 = tn + h to approximate2 x1 = x0 + 0.2 (x0 1) = 5 + 7.2 = 2.2 at t1 = 0.22 x2 = x1 + 0.2 (x1 1) = 2.488 at t2 = 0.42 x3 = x2 + 0.2 (x2 1) = 2.931 at t3 = 0.6x4 = x3 + 0. 22 (x3 1) = 3.676 at t4 = 0.8x5 = x4 + 0. 22 (x4 1) = 5.109 at t5 = 12 x6 = x5 + 0.2 (x5 1) = 8.486 at t6 = 1.2x7 = x6 + 0. 22 (x6 1) = 19.694 at t7 = 1.4x8 = x7 + 0. 22 (x7 1) = 89.591 at t8 = 1.62 x9 = x8 + 0.2 (x8 1) = 1659.265 at t9 = 1.82 x10 = x9 + 0.2 (x9 1) = 551627.57 at t10 = 2 So estimate for x(2) is x10 = 551627.57 and it seems to tend to innity. 17. dx = 1 2x + x2 = f (t, x) , t0 = 0, x0 = 5, h = 0.1.dt We use recursive formula, xn+1 = xn + hf (tn, xn) = xn + 0.1 _1 22xn + xn _ = xn + 0.1 (xn 1)2, where tn+1 = tn + h to approximate x1 = x0 + 0.1 (x0 1)2= 1.4 at t1 = 0.1 2x2 = x1 + 0.1 (x1 1) = 0.824 at t2 = 0.2 x3 = x2 + 0.1 (x 22 1) = 0.4913 at t3 = 0.3 2x4 = x3 + 0.1 (x3 1) = 0.2689 at t4 = 0.4 x5 = x4 + 0.1 (x4 1)2= 0.1079 at t5 = 0.5 x6 = x5 + 0.1 (x5 1)2= 0.0148 at t6 = 0.6 x7 = x6 + 0.1 (x6 1)2= 0.1119 at t7 = 0.7 x8 = x7 + 0.1 (x7 1)2= 0.1908 at t8 = 0.8 _ _ _ 10 CHAPTER 12x9 = x8 + 0.1 (x8 1) = 0.2563 at t9 = 0.9x10 = x9 + 0.1 (x9 1)2= 0.3116 at t10 = 1x11 = x10 + 0.1 (x10 1)2= 0.359 at t11 = 1.12x12 = x11 + 0.1 (x11 1) = 0.4 at t12 = 1.2x13 = x12 + 0.1 (x122 1) = 0.436 at t13 = 1.3x14 = x13 + 0.1 (x13 1)2= 0.4678 at t14 = 1.4x15 = x14 + 0.1 (x14 1)2= 0.4961 at t15 = 1.52x16 = x15 + 0.1 (x15 1) = 0.5215 at t16 = 1.6x17 = x16 + 0.1 (x16 1)2= 0.5444 at t17 = 1.72x18 = x17 + 0.1 (x17 1) = 0.5652 at t18 = 1.8x19 = x18 + 0.1 (x18 1)2= 0.5841 at t19 = 1.92x20 = x19 + 0.1 (x19 1) = 0.6014 at t20 = 2So estimate for x(2) is x20 = 0.6014.1.6 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS 1.6.1 Form of the General Solution There are no exercises in this subsection. 1.6.2 Solutions of Homogeneous First-Order Linear Dierential Equations dx dx 1.dt = 3x. By separation we obtain _ x = 3 _ dt. Integration yields ln x = 3t + c1 x = ec1 e3t x = ce3t , where c = ec1 . dx | | | | dx 3. dt= 2tx. By separation we obtain _ x = _ 2tdt. Integration yields ln |x| = t2 + c1 |x| = ec1 et2 x = cet2 , where c = ec1 . dx dx 1 _ dt 5. 2t dt + x = 0. By separation we obtain _ x = 2 . Integration 1yields ln |x| = ln |t| + c1 ln |x| + ln t1/2 = c1 t ln xt1/2 = c1 c1 c1 c1xt1/2 = e2 xt1/2 = e x = ct1/2 where c = e . dx 7. + (cos t) x = 0 . By separation we obtain _ dx = _ cos tdt. dt xIntegration yields ln |x| = sin t + c1 |x| = ec1 e sin t x = ce sin t , where c = ec1 . dx 9. + _cos(t1/2)_ x = 0. By separation we obtain dx _dt = _ cos(t 12 )dt. Denite integration yieldst tln x |x| = _ 0 cos(s 1 12 )ds + c1 |x| = ec1 exp _ _ 0 cos(s 2 )ds _ t1 x = c exp cos(s 2 )ds , where c = ec1 . 0 dx dx 11. = 5x. By separation we obtain _ = 5 _ dt. Integration dt xyields ln |x| = 5t + c1 x = ec1 e5t = ce5t . Using the initial condition x(0) = 9 we get, c = 9 and the solution is x = 9e5t . dx dx 13. dt= 9x. By separation we obtain _ x = 9 _ dt. Integration 11 FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS yields ln x = 9t + c1 c1 e9t = ce9t . Using the initial | | x = e27 condition x(3) = 7 we get, 7 = ce c = 7e27 and the solution is x = 7e9t27 = 7e9(t3). dx sin t dx sin t15. dt + 4+et x = 0. By separation we obtain _ x = _ 4+et dt. sin sDenite integration yields ln |x| = _ 5 t4+es ds + c1 _ sin s _ x = c exp _ 5 t4+es ds . Using the initial condition x(5) = 10 _ sin s _ we get, c = 10 and the solution is x = 10 exp _ 5 t4+es ds . dx + t2 dx 117.dt x = 0. By separation we obtain _ x = _ t2 dt. Integration yields ln |x| = 1 t + c1 x = ec1 e1/t = ce1/t. Using the initial condition x(1) = 3 we get, 3 = ce c = 3e1 and the solution is x = 3e( 1 t 1). 1.6.3 Integrating Factors for First-Order Linear Dierential Equations dx dx e t 19. t dt + x = et dt + xt = t , p(t) = 1 t . The integrating factor is e _ p(t)dt = eln|t| = t. Multiplying by integrating factor we get dx t d t tt dt + x = edt (xt) = e . Integration yields xt = e + c. Using the initial condition x(1) = 1 we get, 1 = e + c c = 1 e. Hence the solution is xt = et + 1 e x = e t+1e . dx t t21. = 3e . By integration we obtain _ dx = 3 _ etdt x = 3et + c. dx 2t 1 2t23. _dt t2 + 1_ dx + 2tx = 1 + x = , p(t) = .dt dt (t2+1) _ (t2+1) (t2+1) 2t dt _ p(t)dt (t2+1)The integrating factor e = e . Using the substitution 2tu = t2 + 1 we have _ (t2+1) dt = ln t2 + 1 and then the integrating ln t2+1factor is e | | = t2 + 1 = t2 + 1 (since positive for real t). Multiplying by integrating factor we getd_t2 + 1_ dx + 2tx = 1 _x _t2 + 1__ = 1. Integration yields x _t2 + 1_dt = t + c x = dt t + c . dx t2+1 t2+1 _ p(t)dt 4t25.dt + 4x = t, p(t) = 4. The integrating factor is e = e .Multiplying by integrating factor we get4t dx 4t d edt + 4e x = te4t dt _xe4t_ = te4t . Integration yields 4t xe = _ te4tdt. We use integration by parts on the r.h.s. as u = t du = dt and dv = e4tdt v = 1 e4t . Then r.h.s.= _ udv = uv _ vdu = 1 te4t 1 _4 e4tdt = 1 te4t 1 e4t .4 4 4 16 4t 1 1 4t 1 1 So xe = 4 te4t 16 e + c x = 4 t 16 + ce4t 1 . Using the initial condition x(0) = 0 we have, c = 16 .1 1 1Hence the solution is x = 4 t 16 + 16 e4t . 27. t dx = 2x dx 2 x = 0, p(t) = 2 . The integrating factor is e dt = edt 2 _ t= t2 t . _ p(t)dt 1 dt = e2 ln t t Multiplying by integrating factor we get 12 CHAPTER 1 t2 dx dt x 2 t3 x = 0 d dt _ x t2 _ = 0. Integration yields = c. Using the initial condition x(1) = 4 we have, c = 4. t2 Hence the solution is x = 4t2 . This can also be done by separation. 29. t2 dx + tx = 1 dx + 1 x = t2, p(t) = 1 . The integrating dt dt t t_ p(t)dt _ 1 dt ln tfactor is e = t. = e t = eMultiplying by integrating factor we get t dx + x = 1 d (xt) = 1 . Integration yields xt dt = ln t + t c dt x = t1 ln t t + ct1 . 31. t dx + 3x = t dx + 3 x = 1, p(t) = 3 . The integrating dt dt 3 t_ t_ p(t)dt 1 dt 3 ln t = t3factor is e = e t = e . Multiplying by integrating factor we get t3 dx + 3t2x = t3 d _t3x_ = t3 . Integration yields dt dt 4 t3x = t4 + c x = 14 t + ct3 .One solution, 14 t, is continuous at (0, 0) . All other solutionsapproaches as t 0.dx dx 1 133. t x = t_ 2 p(t)dt = e x _ = t, p(t) = . The integrating 1 t dt dt t t dt = e ln t = t1factor is e . Multiplying by integrating factor we get t1 dx + t2x = 1 d _t1x_ = 1. Integration yields dt dt t1x = t + c x = t2 + ct. All solutions are continuous and pass through (0, 0) . x t dx 35. dt + 2tx = 1, p(t) = 2t. The integrating factor is 2_ p(t)dt _ 2tdt te = e = e . Multiplying by integrating factor 2 2 2 2t2 dx t d _ t_ twe get edt + 2tetx = edt e x = e . Denite 2 2 2 2 2t s sintegration yields e x = _te ds + c x = et _te ds + cet. 0 0 37. dx + t2x = t, p(t) = t2 . The integrating factor is dt 1 3_ p(t)dt t3e = e . Multiplying by integrating factor we get 13 FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 3 3 3 3e 1 t3 _ dx + t2x_ = te 1 t3 d _e 1 t3 x_ = te 1 t3 .dt dt 3 3Denite integration yields e 1 t3 x = _tse s 3 ds + c 0 s 3 3 3x = e t3 _tse 3 ds + ce t3 . 0 39. dx + etx = 3, p(t) = et . The integrating factor is dt t_ p(t)dt ee = e . Multiplying by integrating factor we get t t t t ee _ dx + etx_ = 3ee d _ee x_ = 3ee .dt dt t s e eDenite integration yields e x = 3 _te ds + c 0 t s t ex = 3ee _te ds + cee . 0 41. dx + x = 1 , x(2) = 1, p(t) = 1. The integrating factor is dt t+1 _ p(t)dt te = e . Multiplying by integrating factor we get t t et _ dx + x_ = e d (etx) = e . Denite integration yields dt t+1 dt t+1 st e e x = _ts+1 ds + c. Using initial condition we get,2e s 2tc = e2 and the solution is x = et _ts+1 ds + e .243. 3t dx x = t sin t dx 1 x = 1 sin t, x(5) = 0, p(t) = 1 .dt dt 3t 3 3t 1 The integrating factor is e _ p(t)dt = e 3 ln t = t1/3 . Multiplying 1 1by integrating factor we get t1/3 _ dx x_ = t1/3 sin tdt 3t 3 d _t1/3x_ = 1 t1/3 sin t. Denite integration yields dt t3 1t1/3x = _ 3 s1/3 sin sds + c. Using initial condition we get, 5 1 t1/3c = 0 and the solution is x = 3 _ts1/3 sin sds.5dx t dx 1 1 e t 145. 7t dt + x = edt + 7t x = 7 t , p(t) = 7t . The integrating 1 factor is e _ p(t)dt = e 7 ln t = t1/7 . Multiplying by integrating factor 1 e t d e t we get t1/7 _ dx + x_ = _t1/7x_ = 7t6/7 . Denite dt 7t 7t6/7 dt 1integration yields t1/7x = _t7 s6/7esds + c0 x = 71 t1/7 _ts6/7esds + ct1/7 . 0dx 1 dt dt47. = = x + t dx t = x, p(x) = 1. The integrating dt x+t dx factor e _ p(x)dx = ex . Multiplying by integrating factor we get ex _ dt d dx t_ = xex dx (tex) = xex . Integration yields tex = _ xexdx + c. We use integration by parts on the r.h.s. as u = x du = dx and dv = exdx v = ex . Then r.h.s.= _ udv = uv _ vdu = xex + _ exdx = xex ex . So tex = xex ex + c t = x 1 + cex . 14 CHAPTER 1 49. Let u = e _ p(t)dt be the integrating factor of dx + p(t)x = f(t). If we dt introduce an arbitrary constant c1 while we compute _ p(t)dt then new integrating factor is u1 = e _ p(t)dt+c1 = ec1 e _ p(t)dt = c2e _ p(t)dt where c2 = ec1 > 0. Multiplying the dierential equation by this new integrating factor we get, u1 _ dx + p(t)x_ = u1f(t) ddt (u1x) = u1f(t) which, on integration, becomes udt 1x = _ u1f(t)dt + c3 c2e _ p(t)dtx = c2 _ e _ p(t)dtf(t)dt + c3. _ p(t)dt Dividing by c2e we get x = e _ p(t)dt _ e _ p(t)dtf(t)dt + cc23 e _ p(t)dt . Now writing cc32 = c we have the same general solution x = e _ p(t)dt _ f(t)e _ p(t)dtdt + ce _ p(t)dt as (23) . 51. Let u = e _ p(t)dt be the integrating factor of dx + p(t)x = f(t). So d (ux) = u dx + x du = u dx + xp(t)u = u _ dx dt + xp(t)_ . Now dt dt dt dt dtdsince k is a constant (kux) = k d (ux) = ku _ dx + xp(t)_ dt dt dt which implies that ku is also an integrating factor of (19) as the product d (kux) of l.h.s. of (19) and ku is easily integrable with dt respect to t. 53. dx + _ sin t x = q(t) dx + p(t)x = q(t) where p(t) = sin t .dt t dt t_ p(t)dt Multiplying this equation by the integrating factor e we get e _ p(t)dt _ dx + p(t)x_ = q(t)e _ p(t)dt d _xe _ dt p(t)dt_ = q(t)e _ p(t)dt . Denite integral yields dt e _ p(t)dtx(t) = c + _tq _t_ e _ p(t)dtdt. Particular solution with 0 xp(0) = 0 gives c = 0. Then xp = e _ p(t)dt _tq _t_ e _ p(t)dtdt 0 xp = _te _ p(t)dte _ p(t)dtq _t_ dt = _tG _t, t_ q _t_ dt where 0 0G _t, t_ = e _ p(t)dte _ p(t)dt = e _ p(t)dt_ p(t)dt .Using denite integrals yields _ t t_ _ t_ G _t, t_ = exp _ p(s)ds _ p(s)ds = exp _ p(s)ds 0 0 t _ t_ _ t_ sin s sin s = exp_ s ds = exp _ s ds . t t 55. dx + p(t)x = f(t). From the change of variable x = u(t)x1(t)dtwe have dx = du x1 + u dx1 . But since x1 = e _ p(t)dt ,dt dt dt dx1 = x1p(t). Thus dx = du x1 ux1p. Plugging in the dt dt dt original equation we get, du x1 ux1p + ux1p = f(t)dt du e _ p(t)dt = f(t) du = e _ p(t)dtf(t). Integrating we get, dt dt u = _ f(t)e _ p(t)dtdt. Then x = e _ p(t)dt _ f(t)e _ p(t)dtdt. Comparison: From the original equation, the integrating factor is d _ _ p(t)dt_ _ p(t)dt e _ p(t)dt and so dt e x = f(t)e . Integrating we get, FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 15 e _ p(t)dtx = _ f(t)e _ p(t)dtdt x = e _ p(t)dt _ f(t)e _ p(t)dtdt. 1.7 LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS AND CONSTANT INPUT dx rt rt 1.dt 8x = 0. Substituting x = e we get, re 8ert = 0. Dividing by ert = 0, we have, r = 8. Thus the general solution is x = ce8t . dx

rt rt rt 3. = 2x. Substituting x = e we get, re = 2e . Dividing dt by ert = 0, we have, r = 2. Thus the general solution is x = ce2t . dx

rt 5. dt + 7x = 0. Substituting x = e we get, rert + 7ert = 0. Dividing by ert = 0, we have, r = 7. Thus the general solution is x = ce7t . dx

rt rt 7. dt + x = 0. Substituting x = e we get, re + ert = 0. Dividing by ert = 0, we have, r = 1. Thus the general solution is x = cet . dx

rt rt 9.dt = 5x. Substituting x = e we get, rert = 5e . Dividing by ert = 0, we have, r = 5. Thus the general solution is x = ce5t . dx

811.dt + 3x = 8. Substituting xp = A we get, 3A = 8 A = 3 . Thus 8 rt xp = 3 . Let the associated homogeneous solution be x1 = e . Then rert + 3ert = 0. Dividing by 8 ert = 0 , we have, r = 3. Thus the general solution is x = 3 + ce3t . 13. dx 4x = 9. Substituting xp = A we get, 4A = 9 A = 9 .dt 4 9 rt Thus xp = 4 . Let the associated homogeneous solution be x1 = e . Then rert 4ert = 0. Dividing by ert = 0, we have, r = 4. 9

4tThus the general solution is x = 4 + ce . 15. dx 4 x = 3. Substituting xp = A we get, 4 A = 3 15 .dt 5 5 A = 4 Thus xp 15 . Let the associated homogeneous solution be x1 = ert . = 4 Then rert 4 ert = 0. Dividing by e15 rt = 0 , t we have, r = 4 .5 5 4 5Thus the general solution is x = dx 2 4 + ce .4 2 417. dt + 3 x = 3 . Substituting xp = A we get, Thus xp A = A = 2.3 3 rt 2 = 2. Let the associated homogeneous solution be x1 Then rert + 32 ert = 0. Dividing by ert = 0 , we have, r = 2 t = e . .3 Thus the general solution is x = 2 + cedx 3 . 19. = 2x + 18. Substituting xp = A we get, 2A + 18 = 0 dt A = 9. rt = 9. Let the associated homogeneous solution be x1 Then rert = 2ert . Dividing by ert = 0, we have, r = 2. 2tThus the general solution is x = 9 + ce . dx 17 17 17 21.dt = x 3 . Substituting xp = A we get, A 3 = 0 A = 3 . 17 rt Thus xp = 3 . Let the associated homogeneous solution be x1 = e .Then rert = ert . Dividing by ert = 0 , we have, r = 1.Thus the general solution is x = 17 + cet .3 dx 23.dt + 7x = 8e4t . Substituting xp = Ae4t we get, 4Ae4t + 7Ae4t = 8e4t . Dividing by e4t = 0, we have, 4A + 7A = 8 A = 38 . Thus 8 e4t rt xp = 3 . Let the associated homogeneous solution be x1 = e . Thus xp = e . 16 25. 27. 29. 31. 33. 35. 37. 39. 41. CHAPTER 1 Then rert + 7ert = 0. Dividing by ert = 0, we have, r = 7.Thus the general solution is x = 83 e4t

+ ce7t .dx 2x = 3e5t . Substituting xp = Ae5t we get,dt 5Ae5t 2Ae5t = 3e5t . Dividing by e5t = 0, we have, 5A 2A = 3 A = 73 . Thus 3 rt xp = 7 e5t . Let the associated homogeneous solution be x1 = e .Then rert 2ert = 0. Dividing by ert = 0 , we have, r = 2.Thus the general solution is x = 37 e5t + ce2t .dx +4x = 3e4t . Substituting xp = Ae4t we get, 4Ae4t + 4Ae4t = 3e4t .dt Dividing by e4t = 0, we have, 4A + 4A = 3 A = 83 . Thus 3 4t rt xp = 8 e . Let the associated homogeneous solution be x1 = e .Then rert + 4ert = 0. Dividing by 3 ert = 0 , we have, r = 4.Thus the general solution is x = 8 e4t + ce4t .dx + 3x = 3e2t . Substituting xp = Ae2t we get,dt 2Ae2t + 3Ae2t = e2t . Dividing by e2t = 0, we have, 2A + 3A = 1 A = 1. Thus = e2t rt xp . Let the associated homogeneous solution be x1 = e . Then rert + 3ert = 0. Dividing by ert = 0 , we have, r = 3. Thus the general solution is x = e2t + ce3t . dx +7x = 3+5t. Substituting xp = At+B we get, A+7At+7B = 3+5t. dt Equating the coecients of t and 1 we have 7A = 5 A = 57 and 5 16 16 A + 7B = 3 7B = 3 = B = 5 16 7 7 49 Thus xp = 7 t + 49 . dx +2x = 14t. Substituting xp = At + B we get, A +2At +2B = 14t. dt Equating the coecients of t and 1 we have 2A = 14 A = 7 and 7 A + 2B = 0 2B = A = 7 B = 2 . 7 Thus xp = 7t 2 .dx + x = 2t2 + 5t 8. Substituting xp = At2 + Bt + C we get,dt 2At + B + At2 + Bt + C = 2t2 + 5t 8.Equating the coecients of t2, t and 1 we have A = 2,2A+B = 5 B = 52A = 1, and B +C = 8 C = 8B = 9 Thus xp = 2t2 + t 9.dx + 3x = t3 . Substituting xp = At3 + Bt2 + Ct + D we get,dt 3At2 + 2Bt + C + 3At3 + 3Bt2 + 3Ct + 3D = t3 .Equating the coecients of t3, t2, t and 1 we have 3A = 1 A = 1 ,1 2 2 3 3A + 3B = 0 B = A = 3 , 2B + 3C = 0 C = 3 B = 9 , 1 2 C + 3D = 0 D = 3 C = 27 1 1 2 2Thus xp = t3 t2 + dx 3 3 9 t 27 . + 5x = t. Substituting xp = At + B we get, A + 5B + 5At = t. dt Equating the coecients of t and 1 we have 5A = 1 A = 1 and 1 1 5 A + 5B = 0 B = 5 A = 25 .1 1Thus xp = 5 t 25 . dx + 2x = 3 sin 6t. Substituting xp = A sin 6t + B cos 6t we get, dt 6A cos 6t 6B sin 6t + 2A sin 6t + 2B cos 6t = 3 sin 6t. Equating the coecients of cos 6t and sin 6t we have FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 17 cos 6t : 6A + 2B = 0 sin 6t : 2A 6B = 3 Solving these equations for A and B (multiplying rst equation by 3 and adding with the second) we get A = 3 and B 9 20 = 20 . Thus xp = 3 sin 6t 9 cos 6t. 20 20 43.dx 5x = 2 cos t. Substituting xp = A cos t + B sin t we get, dt A sin t + B cos t 5A cos t 5B sin t = 2 cos t. Equating the coecients of sin t and cos t we have sin t : A 5B = 0 cos t : 5A + B = 2 Solving these equations for A and B (multiplying second equation by 5 and adding with the rst) we get A = 5 and B = 1 .13 13 Thus xp = 1 sin t 5 cos t. 13 13 45. dx + 4x = 3 cos 2t + 5 sin 2t. Substituting xp = A cos 2t + B sin 2tdt we get, 2A sin 2t + 2B cos 2t + 4A cos 2t + 4B sin 2t = 3 cos 2t + 5 sin 2t. Equating the coecients of sin 2t and cos 2t we have sin 2t : 2A + 4B = 5 cos 2t : 4A + 2B = 3 Solving these equations for A and B (multiplying rst equation by 2 13 1and adding with the second) we get B = 10 and A = 10 . Thus xp = 1 cos 2t + 13 sin 2t. 10 10 dx 47.dt + 6x = cos t + sin 5t. Substituting xp = A cos t + B sin t + C sin 5t +D cos 5t we get, A sin t + B cos t + 5C cos 5t 5D sin 5t + 6A cos t +6B sin t + 6C sin 5t + 6D cos 5t = cos t + sin 5t. Equating the coecients we have cos t : 6A + B = 1 sin t : A + 6B = 0 cos 5t : 5C + 6D = 0 sin 5t : 6C 5D = 1 Solving the rst pair of equations for A and B (multiplying second 1 6equation by 6 and adding with the rst) we get B = 37 and A = 37 . Similarly, to solve the second pair of equations for C and D, 6multiply the rst by 5 and the second by 6 and add to get C = 61 and D = 5 .61 Thus xp = 6 cos t + 1 sin t + 6 sin 5t 5 cos 5t. 37 37 61 61 49. dx 9x = 5 + 2 sin 3t. Substituting xp = A + B sin 3t + C cos 3tdt we get, 3B cos 3t 3C sin 3t 9A 9B sin 3t 9C cos 3t = 5+2 sin 3t. Equating the coecients we have Non-t : 9A = 5 A = 95 cos 3t : 3B 9C = 0 sin 3t : 9B 3C = 2 Solving last pair of equations for B and C (multiplying the rst equation by 3 and adding with the second) we get 1 1 5 1 1C = 15 and B = 5 . Thus xp = 9 5 sin 3t 15 cos 3t. 18 51. 53. 55. 57. 59. 61. CHAPTER 1 dx + x = 2e3t + sin t. Substituting xp = Ae3t + B sin t + C cos tdt we get, 3Ae3t + B cos t C sin t +Ae3t +B sin t + C cos t = 2e3t +sin t. Equating the coecients we have e3t : 3A + A = 2 A = 1 2 cos t : B + C = 0 sin t : B C = 1 Solving last pair of equations for B and C (adding ) we get 1 1 1B = 2 and C = B = 2 .Thus xp = 2 _e3t + sin t cos t_ . dx + 3x = 8e3t . Substituting xp = Ae3t we get, dt 3Ae3t + 3Ae3t = 8e3t 8e3t = 0 which is impossible becasue e3t = 0 , t. So a simple exponential, Ae3t , does not work as a particular solution becasue e3t is a solution of the associated homogeneous equation. Solve by the integrating factor method: The integrating factor _ 3dt 3tis e = e . Multiplying the equation by the integrating factor 3t _ dx dwe get edt + 3x_ = 8 dt _xe3t_ = 8. Integrating we have xe3t = 8t + c x = 8te3t + ce3t . Conjecture: Notice that the particular solution part of this general solution is a constant multiple of t times the exponential. This indicates that we may nd a particular solution by substituting Ate3t , when simple exponential forcing e3t is a solution of the associated homogeneous equation. dx 2x = 7e2t . Substituting xp = Ae2t we get, dt 2Ae2t 2Ae2t = 7e2t 7e2t = 0 which is impossible 2t becasue e = 0 , t. So a simple exponential, Ae2t , does not work as a particular solution becasue e2t is a solution of the associated homogeneous equation. Substituting x = ve2t we get, dv e2t + 2ve2t 2ve2t = 7e2t dt Dividing by e2t = 0 we have dv = 7 v = 7t + c. Thus dt the general solution is x = 7te2t + ce2t . We can make the same conjecture as in #53. dx t rt x = 4e . Let x1 = e be the solution of the associated dt homogeneous equation. On substitution we get, rert ert = 0 r = 1. So x1 = c1et (similar to the forcing function). So let xp = Atet . Substituting this in the equation we get, Aet + Atet Atet = 4ett A = 4. Thus the general solution is x = 4tet + ce . rt dx + x = 5et . Let x1 = e be the solution of the associated dt homogeneous equation. On substitution we get, rert + ert = 0 r = 1. So x1 = c1et (similar to the forcing function). So let xp = Atet . Substituting this in the equation we get, Aet Atet + Atet = 5et A = 5. Thus the general solution is x = 5tet + cet . dx 7t rt 7x = 8e . Let x1 = e be the solution of the associated dt 19 FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS homogeneous equation. On substitution we get, rert 7ert = 0 r = 7. So x1 = c1e7t (similar to the forcing function). So let xp = Ate7t . Substituting this in the equation we get, Ae7t + 7Ate7t 7Ate7t = 8e7t A = 8. Thus the general solution is x = 8te7t + ce7t . 63. dx + p(t)x = f(t), 0 t 2, f(t) = 0, x(0) = 2.dtFor 0 t < 1, p(t) = 2 and so dx + 2x = 0dt x(t) = c1e2t . Using x(0) = 2 we get c1 = 2. So x(t) = 2e2t for 0 t < 1. For 1 t 2, p(t) = 1 and so dx + x = 0 x(t) = c2et . In order for x to be continuous dt at t = 1 we must have lim x(t) = x(1) 2e2 = c2e1 t 1 c2 = 2e1 . Thus the solution is _2e2t , 0 t < 1 x(t) = 2e1et , 1 t 2 2 x 1 0 1 2 t 65. dx + p(t)x = f(t), 0 t 2, p(t) = 0, x(0) = 0.dtFor 0 t < 1, f(t) = 1 and so dx = 1 x(t) = t + c1.dt Using x(0) = 0 we get c1 = 0. So x(t) = t for 0 t < 1.For 1 t 2, f(t) = 1 and so dx x(t) = t + c2.dt = 1 In order for x to be continuous at t = 1 we must have lim x(t) = x(1) 1 = 1 + c2 c2 = 2. Thus the solution is t 1 _t, 0 t < 1 x(t) = 2 t, 1 t 2 x t 0 1 1 67. dx + p(t)x = f(t), 0 t 2, x(0) = 2. For 0 t < 1,dtp(t) = 1, f(t) = 0 and so dx + x = 0 x(t) = c1et .dt Using x(0) = 2 we get c1 = 2. So x(t) = 2et for 0 t < 1. For 1 t 2, p(t) = 0, f(t) = 1 and so dx = 1 dt 20 CHAPTER 1 x(t) = t + c2. In order for x to be continuous at t = 1 we must have lim x(t) = x(1) 2e1 = 1 + c2 t 1 c2 = 2e1 1. Thus the solution is _2et , 0 t < 1 x(t) = t + 2e1 1, 1 t 2 x 0 1 2 1 2 t 1.8 GROWTH AND DECAY PROBLEMS The growth rate k of a population P (t) is given by 1 dP = k dP = kP P dt dt whose solution with initial population P (0) is P (t) = P (0)ekt . The population will be doubled when P (t) = 2P (0). Then 2P (0) = P (0)ekt ekt = 2 kt = ln 2 t = ln 2 which is known as kdoubling time, denoted by td as = ln 2 , and we will use this tdk formula throughout this section. 1. In this problem, the growth rate is k = 1.5% = 0.015 and sodoubling time, td = ln 2 = ln 2 46.2 years.k 0.015 3. Using td = 8 hours = 31 day in the doubling time formula we get, 1 ln 2 = k = 3 ln 2 2.08 = 208% per day. 3 k 5. Here, P (0) = 1500, t = 1 hour, P (1) = 2000. So P (t) = P (0)ekt 2000 = 1500ek ek = 43 k = ln _ 43 _ . After 4 hour (i.e. t = 4), 3P (t) = 1500e4k = 1500e 4 ln(4 ) = 1500 _ 4 _4 .3 A B10 0.02 Q c 5 0.01 0 100 200 0 100 200 tt 7. td = ln 2 = 3 k = ln 2 . Here, P (t) = 10P (0) and so k 3 kt 10P (0) = P (0)e ekt = 10 kt = ln 10 ln 10 3 ln 10 t = = 9.97 years. k ln 2 FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 21 9. Let Q be the number of organisms.Birth: dQ = k1Q, Death: dQ = k2Q, Addition: dQ = k.dt dt dt dQ dt = k1Q k2Q + k. 11. With the interest rate of 3% per year:Doubling time, td = ln 2 = ln 2 23.10 years.k 0.03 With the 3% yield:Yield = ek 1 = 0.03 ek = 1.03 k = ln (1.03) . So, in this caseln 2 ln 2 Doubling time, td = = ln(1.03) 23.45 years. k 13. Here, the growth after one year isek 1 = 0.064 ek = 1.064 k = ln (1.064) .ln 2 Then td = ln(1.064) 11.17 years. 15. This is the case of exponential growth. So x(t) = x(0)ekt, k > 0. The cost of living rose from x(0) = 10, 000 to x(1) = 11, 000 in one year, that is, t = 1. So 11000 = 10000ek ek = 1.1 k = ln (1.1) 0.0953 = 9.53% per year. 17. Using the half-life formula T = ln 2 = 16 days, we get k = ln 2 .k 16 At the end of t = 30 days, x(t) = 30 g. Then x(t) = x(0)ekt ln 2 ln 2 16 16 30 = x(0)e (30) x(0) = 30e (30) 110.04 g. 19. dx = kx x(t) = x(0)ekt . When t = 10 years, x(t) = 0.80x(0).dt So 0.801 x(0) = x(0)e10k e10k = 0.8 10k = ln (0.8) k = 10 ln (0.8) . ln 2 ln 2 (a) Half-life: T = = 10 ln(0.8) 31.063 years. k (b) x(t) = 0.15x(0). So 0.15x(0) = x(0)ekt ekt = 0.15 kt = ln (0.15) t = 1 ln (0.15) = 10 ln(0.15) 85.018 years. k ln(0.8) Thus it will take 85.018 10 = 75.018 additional years. 21. Newtons law of cooling dT = k (T Q0) has particular solution Q0dt so the general solution is T (t) = Q0 + cekt . Here Q0 = 30. Using the initial temperature T (0) = 200, we get 200 = 30 + c c = 170. Thus T (t) = 30 + 170ekt . Again, using T (10) = 180, 15 we get, 180 = 30 + 170e10k 170e10k = 150 e10k = 17 10k = ln _ 15 _ 1 ln _ 15 ln 15ln 17 = ln 17ln 15 . 17 k = 10 17 _ = 10 10 ekt 1(a) T will be 400C if 40 = 30+170ekt 170ekt = 10 = 17 kt = ln (17) t = 1 ln (17) = 10 ln(17) 226.4 minutes. k ln 17ln 15 (b) T (t) = 30 + cekt , where we previously had k = ln 17ln 15 .10 At t = 0, T = 200, so c = 170 and T (t) = 30 + 170ekt .Now we solve this equation for t :T (t) 30 = 170ekt ln (T 30) = ln (170) kt1 10 t = [ln (T 30) ln (170)] = ln _ T 30 _ . k ln 15ln 17 170 23. Newtons law of cooling dT = k (T Q0) has particular solution dt Q0 so the general solution is T (t) = Q0 + cekt . Here Q0 = 40. Using the initial temperature T (0) = 100, we get 100 = 40 + c c = 60. Thus T (t) = 40 + 60ekt . Again, using T (10) = 60, e10k 1we get, 60 = 40 + 60e10k 60e10k = 20 = 3 1 10k = ln _ 1 _ 10k = ln 3 k = ln 3 0.1099.3 10 22 CHAPTER 1 25. (a) Newtons law of cooling dT = k (T Q0) = k [T (20 + 10t)] .dt (b) Substituting k = 1 in the equation we get, dT = T + 20 + 10tdt dT rt dt + T = 20 + 10t. Let T1 = rt e be the solution of the homogeneous equation. Then re + ert = 0 r = 1. Thus T1 = cet . Substituting Tp = A + Bt we get, B+ A + Bt = 20 + 10t. Equating the coecients of t and 1 we have B = 10 and A + B = 20 A = 10. Thus the general solution is T (t) = 10 + 10t + cet . Using the initial temperature T (0) = 40 we get, 40 = 10 + c c = 30. Hence the solution is T (t) = 10 + 10t + 30et . 27. (a) Let the amount invested be $A. The 6% interest rate will eect the amount in excess of $500. So the amount of interest per year will be $0.06 (A 500) . Thus the rate of change of money can be written as the dierential equation dA = 0.06 (A 500) with dt initial condition A(0) = 2000. (b) dA 0.06A = 30. Let Ap= 30 B = 500 dt= B 0.06B Ap = 500 and Ah = ce0.06t . So the general solution is 0.06tA(t) = 500 + ce . Using initial condition A(0) = 2000 we get,c = 1500. Thus A(t) = 500 + 1500e0.06t .After 10 years the amount will be A(10) = $3233.18.(c) The full amount earns interest when dA = 0.06A dA dt 0.06A = 0, a homogeneous equation with general solution dt A(t) = ce0.06t which becomes A(t) = 2000e0.06t for the initial condition. After 10 years the amount will be A(10) = $3644.24 which is $411.06 more than that in part (b) . 29. Constant deposit rate is $B/day means $365B/year and 0.08P is the amount of interest per year with 8% interest rate. So dP dP = 0.08P + 365B 0.08P = 365B. Let Pp = Adt dt 365 365 0.08A = 365B A = 0.08 B Pp = 0.08 B and 0.08t 365 0.08tPh = ce . So the general solution is P (t) = B + ce .0.08 Using initial condition P (0) = 1000 we get, c = 1000 + 365 B. 365 0.08tThus P (t) = B + _1000 + 365 B_ e . 0.08 0.08 0.08 (a) After t = 5 years P = $10, 000. 365 0.4So 10000 = B + _1000 + 365 B_ e80(10 e 0.4) 0.08 0.08 0.4 0.4800 = 365B _e 1_ + 80e B = 1) $3.79. 365(e0.4365 0.08t(b) Solve 10000 = B(t) + _1000 + 365 B(t)_ e for B(t) : 0.08t 0.08t800 = B(t) _365e0.08 365_ + 80e0.080.08t 0.08tB(t) _365e 365_ = 800 80e 0.08tB(t) = _800 80e0.08t_ _365e 365_1 . dQ dQ 31. = 0.2Q + 400 cos (2t) 0.2Q = 400 cos (2t) .dt dt Substituting Qp = A cos (2t) + B sin (2t) we get2A sin (2t) + 2B cos (2t) 0.2A cos (2t) 0.2B sin (2t)= 400 cos (2t) .Equating the coecients of cos (2t) and sin (2t) we have23 FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS cos (2t) : 2B 0.2A = 400 sin (2t) : 0.2B 2A = 0 Solving these equations for A and B (Multiplying the rst equation by 0.2 and the second equation by 2 and adding and simplifying) 20 200we get, A = 2+0.01 and B = 2+0.01 . Thus the particular solution is = 1 (20 cos (2t) + 200 sin (2t)) . Qp2+0.01Q1 = ce0.02t . So the general solution isQ(t) = 1 (20 cos (2t) + 200 sin (2t)) + ce0.02t .2+0.01 20 Using the initial condition Q(0) = 100 we get c = 100 + 2+0.01 . Hence Q(t) = 1 (20 cos (2t) + 200 sin (2t))+_100 + 20 _ e0.02t .2+0.01 2+0.01 1.9 MIXTURE PROBLEMS 1. (a) Let Q be the amount of salt in the tank of volume V = 300. This volume is constant as water is owing in and out at the same rate (2 gal/min). So the concentration of salt is C = Q . The 300 inow and outow rate of salt are 2(0.4) and 2 Q 300 , respectively. Thus the rate of change in salt can be written as dQ dQ Q= 2(0.4) 2 Q + = 0.8. Substituting Qp = Adt 300 dt 150 rt 1 t we get, A = 120 and Q1= e Q1 = et r = 150 150 .Thus Q(t) = 120 + ke 150 . Using initial conditionQ(0) = 300(0.2) = 60 we have, k = 60. Hencet Q(t) 2 1 t 150 . Q(t) = 120 60e 150 and C(t) = 300 = 5 5 e (b) When does C(t) = 0.3? t t 150 0.3 = 0.4 0.2e 150 = 1 t = ln 2 e 2 150 t = 150 ln 2 104 minutes. 3. (a) Let Q be the amount of salt in the tank of volume V = 100. Thisvolume is constant as water is owing in and out at the sameQrate (5 L/h). So the concentration of salt is C = 100 . Theinow and outow rate of salt are 5(0.2) and 4 Q (evaporated100 water contains no salt so that Q = 0), respectively.Thus the rate of change in salt can be written asdQ dQ Q= 5(0.2) 4 Q + = 1. Substituting Qp = Adt 100 dt 25 t we get, A = 25 and Q1 = ert r = 1 Q1 = e 25 .t 25 Thus Q(t) = 25 + ke 25 . Using initial conditionQ(0) = 100(0.1) = 10 we have, k = 15. HenceQ(t) = 25 15e t and C(t) = Q(t) = 0.25 0.15e t25 25 .100 (b) No. lim C(t) = 0.25 > 0.2. t 5. (a) Let Q be the amount of iodine in the tank. Water is owing inand out at a dierent rate and so the volume is changing asdV = 6 1 = 5 V (t) = 5t + 500 since the initial half-volume is dt Q500 gal. Concentration of iodine is C = 5t+500 . The inow and 24 CHAPTER 1 outow rate of iodine are 0 (Pure water has no concentration) and Q1 5t+500 , respectively. Thus the rate of change in iodine can be written as dQ Q dQ dt dt = 5t+500 Q = 5t+500 1ln Q = ln (5t + 500) + k1 Q(t) = k2 (5t + 500)1/5 5 = k (t + 100)1/5 . Using initial condition Q(0) = 10 we have, 10 = k (100)1/5 k = 10 (100)1/5 . _ 100 _1/5 Hence Q(t) = 10 t+100 . Since volume increases as t increases, tank will overow when V = 1000. That is, tank overows when 5t + 500 = 1000 means at t = 100. So for 0 t 100, _ 100 _1/5 Q(t)Q(t) = 10 t+100 and C(t) = 5t+500 . (b) Tank overows when t = 100. After overow, V = 1000 and C = Q . During overow, mixture is leaving at the rate of 6 gal/min 1000 dQ 6Q(as pure water is entering at this rate) and so for t 100, dt = 1000 . 3 3 500 500 Since it has constant coecients, Q(t) = c2e t = c2e (t100). In order for Q(t) to be continuous at t = 100 we must have lim Q(t) = lim Q(t) 10 _ 100 _1/5 = c2 t 100 t 100+ 200 Q(t)So for t 100, Q(t) = 10 (2)1/5 e3(t100)/500 and C(t) = .1000 7. Let Q be the amount of pollutant in the lake. Water is owing in and out at a dierent rate and so the volume is changing as dV = 5 2 = 3 V (t) = 3t + 1000 since the initial volume is 1000 dt QkL. Concentration of pollutant is C = 3t+1000 . The inow and outow rate of pollutant are 5 7 and 2 Q , respectively. Thus 3t+1000 the rate of change in pollutant can be written as dQ = 35 2Q 2 dt 3t+1000 dQ + 2Q = 35. The integrating factor is e _ 3t+1000 dt = eln(3t+1000)2/3 Q 3t+1000 = (3t + 1000)2/3 . Multiplying by integrating factor and rearranging d 2/3we have dt _Q (3t + 1000)2/3_ = 35 (3t + 1000) 2/3 2/3Q (3t + 1000) = 35 _ (3t + 1000) dt + k Q(t) = (3t + 1000)2/3 _7 (3t + 1000)5/3 + k_ . Using initial condition Q(0) = 2000 we have, 2000 = (10)2 _7 (10)5 + k_ 57 (10) + k = 200000 k = 500000. Thus 5/3Q(t) = (3t + 1000)2/3 _7 (3t + 1000) 500000_ = 7 (3t + 1000) 500000 (3t + 1000)2/3 andC(t) = Q(t) = 7 500000 (3t + 1000)5/3 .3t+1000 9. Let V1 and V2 be the volume of tank 1 and tank 2, respectively. V1 = (13 7) t + 150 = 6t + 150; V2 = (7 28) t + 250 = 21t + 250 Let S1 and S2 be the amount of salt in tank 1 and tank 2, respectively. Then dS1 = 13 3 7 S1 dS1 = 39 7S1 dt V1 dt 6t+150and dS2 = 7S1 dS2 = 7S1 28S2 .dt 6t+150 28 SV22 dt 6t+150 21t+250FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 25 11. Let V1 and V2 be the volume of tank 1 and tank 2, respectively.V1 = (21 18) t + 230 = 3t + 230; Tank 2 receives brine at therate 1 (18) = 9 gal/s. So V2 = (9 4) t + 275 = 5t + 275.2 Let S1 and S2 be the amount of salt in tank 1 and tank 2, respectively. Then dS1 = 21 5 18 S1 dS1 = 105 18S1 dt V1 dt 3t+230 and dS2 = 9S1 dS2 = 9S1 4S2 .dt 3t+230 4 SV22 dt 3t+230 5t+275 13. Let V1, V2 and V3 be the volume of tank 1, tank 2 and tank 3, respectively. V1 = (11 18) t + 100 = 7t + 100; V2 = 200 (same rate of inow and outow). V3 = 300 (brine gets in and overow so that volume remains the same). Let S1, S2 and S3 be the amount of salt in tank 1, tank 2 and tank 3, respectively. Then dS1 dS1 18S1 dt = 11 5 18 SV11 dt = 55 7t+100 dS2 18S1 dS2 18S1 18S2= = and dt 7t+100 18SV22 dt 7t+100 200dS3 18S2= .dt 200 1.10 ELECTRONIC CIRCUITS 1. The dierential equation is L di + Ri = e. In this problem, dt Ri = v = 2i R = 2, L = 1, e = 1. Substituting all these we get, di 1+ 2i = 1. Let ip = A. So 2A = 1 A = .dt 2 1 rt So ip = 2 . Taking i1 = e we have r = 2 i1 = ce2t . Thus i (t) = 1 + ce2t . At t = 0, i(0) = 1 + c c = i(0) 1 .2 2 2 Hence i (t) = 1 + _i(0) 1 e2t 1 . Alternatively, 2 2 . As t , i(t) 21for an equilibrium solution di = 0 so that 2i = 1 i = .dt 2 3. The dierential equation is L di + Ri = e. In this problem, dt Ri = v = i, L = 1, e = sin t. Substituting all these we get, di + i = sin t. Substituting ip = A sin t + B cos t in the dierential dt equation we get A cos t B sin t + A sin t + B cos t = sin t.Equating the coecients of cos t and sin t we get:cos t : A + B = 0sin t : A B = 1Solving these equations for A and B by adding we get,A = 1 , B = 1 . Thus ip = 1 sin t 1 cos t. Taking2 2 2 2 i1 = ert in the dierential equation we have, r = 1 i1 = cet . Thus i (t) = 1 sin t 1 cos t + cet . At t = 0, i(0) = 1 + c 1 c = i(0) + 21 . Hence 2 i (t) = 1 sin t cos t + _i(0) + 21 et . 2 2 2 2 5. The dierential equation is L di + Ri = e. For 0 t < 10, dt Ri = v = i R = 1, L = 1, e = 9. Substituting all these we get, di rt + i = 9. Let ip = A. So A = 9 ip = 9. Taking i1 = edt we have r = 1 i1 = c1et . Soi (t) = 9 + c1et . Using the initial condition i(0) = 0 we get 0 = 9 + c1 c1 = 9.Thus for 0 t < 10, i (t) = 9 9et . For 10 t 20, Ri = v = i, L = 1, e = 0._ 26 CHAPTER 1 Substituting all these we get, di + i = 0. This is a homogeneous dt equation and, in fact, the homogeneous part of the above equation. So i(t) = c2et . In order for i(t) to be continuous at t = 10 we must have lim i(t) = i(10) 9 9e10 = c2e10 t 10 10 c2 = 9e 9. Thus the solution is _9(1 et), 0 t < 10 i(t) = 9(e10 1)et , 10 t 20. 7. Capacitance, C = 0.5, Ri = v = 2i R = 2, voltage source qe = 6 sin t. In order to nd charge q we can use: C + Ri = e q + R dq = e (since i = dq ) q + 2 dq = 6 sin t dq C dt dt 0.5 dt + q = 3 sin t. Substituting qp = A sin t + B cos t in this dt equation we get A cos t B sin t + A sin t + B cos t = 3 sin t. Equating the coecients of cos t and sin t we get: cos t : A + B = 0 sin t : A B = 3 Solving these equations for A and B by adding we get, A = 32 , B = 23 . Thus qp = 32 sin t 32 cos t. Taking q1 = ert in the dierential equation we have r = 1 q1 = cet . Thus q (t) = 3 (sin t cos t) + cet . At t = 0 , q(0) = 1 = 3 + c2 2 c = 1 + 3 = 25 . Hence q (t) = 3 (sin t cos t) + 5 et . 2 2 2 1.11 MECHANICS II: INCLUDING AIR RESISTANCE 1. (a) The dierential equation is m dv = mg kv dv dt 1 20 dv = 20 (980) 10v + v = 980. Substituting dt dt 2 1 vp = A we get, A = 980 A = 1960. So vp = 1960.2 1 = cet . Thus v (t) = 1960 + ceUsing initial condition v(0) = 0 we have, c = 1960. Hence trt r = 2 2v1 = e v1 .2 _1 et t v (t) = 1960 + 1960e(b) v (10) = 1960 _e5 = 1960 1_ = 1946.79. 2 2 . (c) lim v(t) = 1960. tdv 3. The dierential equation is m = mg kv dt 70 dv = 70 (980) 1 7v dv + 1 v = 980. Substituting dt dt 10 vp = A we get, A = 980 A = 9800. So vp = 9800.10 1 = ce t r = 10 . Thus v (t) = 9800 + ceUsing the condition v(5) = 12600 we have, t 10 . rt v1 = e v110 1 212600 = 9800 + ce c = 36931.36. So v (0) = 9800 + 36931.36 = 27131.36. 5. (a) The dierential equation is m dv = mg + v2 . Here, weight is dt mg = 32 so that m = 32 = 1. Then dv = 32 + v2 g dt FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 27 dv = _32 v2_ _ dv = _ dt. Integrating using tables gives dt 32v2 1 v+32 v+32 232c 232 ln v32 = t + c v32 = ke232t , where k = e . 1000+32 Using the initial condition v(0) = 1000 we have k = . v+32 = ke232t _1 ke232t_ 100032 So v32 v = 32 _1 + ke232t_ _ 1+ke232t _ 1000+32 v(t) = 32 1ke232t , where k = 100032 . (b) lim v(t) = 32. t 1.12 ORTHOGONAL TRAJECTORIES (OPTIONAL) 1. x = t + c x t = c. Dierentiating with respect to (wrt) t: dx dx 1 = 0 = 1. The slope of the orthogonal family is thus dt dt dx = 1. Integrating we have, x(t) = t + c2.dt t x dx dx x3. xt = c. Dierentiating wrt t: x + t dt = 0 dt = t . The slope of the orthogonal family is thus dx = t 2 2 dt x x t 2_ xdx = _ tdt. Integrating we have, 2 = 2 + c x t2 = c2, where c2 = 2c. t x x = tc ln x5. ln x = c ln t ln t = c. Dierentiating wrt t: 1 dx 1 x dt ln tln x t dx x ln x (ln t)2 = 0 dt = t ln t . The slope of the orthogonal t ln tfamily is thus dx _ x ln xdx = _ t ln tdt. Integration by dt = x ln x 1 2 1 1parts on both sides yields, 12 x2 ln |x| 4 x = 2 t2 ln |t| + 4 t2 + c 28 CHAPTER 1 2x2 ln |x| x2 = t2 2t2 ln |t| + c2, where c2 = 4c. 7. x = t2 + c x t2 = c. Dierentiating wrt t: dx dx 2t = 0 = 2t. The slope of the orthogonal family is thus dt dt dx 1 _ dx = 1 _ 1 dt. Integration yields, x = 1 ln tdt = 2t 2 2 t 2 | | + c2. 9. t = (x c) x c = t x t = c. Dierentiating wrt t:dx 1 dx 1 dt 2t = 0 dt = 2t . The slope of the orthogonal family is thus dx = 2 t _ dx = 2 _ tdt. Integrating we have, dt 4 t3/2 x = 3 + c2.11. x = tan (t + c) tan1 x = t + c. Dierentiating wrt t: 1 dx = 1 1+x2 dt dx = 1 + x2 . The slope of the orthogonal family is thus dt 1 dx dt = 1+x_ _1 + x2_ dx = _ dt. Integrating we have, 32 x + x = t + c2.3 13. x = c cos t. Dierentiating wrt t: dx = c sin t. Since c = x ,dt cos t dx x dt = cos t sin t = x tan t. The slope of the orthogonal family is 1thus dx = cot t _ xdx = _ cot tdt. Integrating we have, 2 dt x x 2 = ln sin t + c x2 = 2 ln sin t + c2, where c2 = 2c. | | |dx | x dx 2x15. x = ct2 . Dierentiating wrt t: dt = 2ct. Since c = t2 , dt = t . The tslope of the orthogonal family is thus dx _ 2xdx = _ tdt. 2 dt = 2x Integrating we have, x2 = t2 + c 2x2 = t2 + c2, where c2 = 2c. 17. x3 + t2 = c. Dierentiating wrt t: 3x2 dx + 2t = 0 dx 2t 2 .dt 2 dt = 3xThe slope of the orthogonal family is thus dx = 3x 3 dt 1 dt 32t _ x2dx = _ . Integrating we have, = ln t + c2 t x 2 1 3 3 _1 | | x = _ 2 ln |t| c_ x = _c2 2 ln |t| , where c2 = c. Chapter TwoLinear Second and Higher-Order Dierenial Equations 2.1 GENERAL SOLUTION OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 1. x = sin t x = cos t x = sin t. So x + x = sin t + sin t = 0. x = cos t x = sin t x = cos t. So x + x = 0. Thus {sin t, cos t} is a set of solutions for the associated homogeneous equation. Now xp = 1 xp = xp = 0 and so x p + xp = 1 =r.h.s. Thus the general solution is x(t) = 1 + c1 sin t + c2 cos t. Using the initial condition x(0) = 0 we have, 0 = 1 + c2 c2 = 1. Dierentiating the general solution we get, x(t) = c1 cos t c2 sin t. The initial condition x(0) = 0 gives c1 = 0. Thus the solution of the initial value problem is x(t) = 1 cos t. 3. x = et x = et x = et . So x 3x + 2x = et t + 2et 2t 2t 2t 3e= 0. For x = e , x = 2e and x = 4e . So x 3x + 2x 2t t= 4e 6e2t + 2e2t = 0. Thus _e , e2t_ is a set of solutions for the associated homogeneous equation. Now xp = t + 23 x p = 1, x = 0 and so x p 3x p + 2xp = 3 + 2t + 3 = 2t =r.h.s. p Thus the general solution is x(t) = t + 32 + c1et + c2e2t .Dierentiating this we get, x(t) = 1 + c1et + 2c2e2t . Using theinitial conditions x(0) = 1 and x(0) = 0 we have, 1 = 32 + c1 + c2and 0 = 1 + c1 + 2c2. Solving these equations for c1 and c2 bysubtracting them we have c2 = 21 and then c1 = 0.Thus the solution of the initial value problem is x(t) = t + 32 1 e2t .2 5. x = et cos t x = et cos t et sin t x = et cos t + et sin t + et sin t et cos t. So x + 2x + 2x = et cos t + et sin t + et sin t et cos t 2et cos t 2et sin t + 2et cos t = 0. Similarly it can be shown that for x = et sin t, x + 2x + 2x = 0. Thus {et cos t, et sin t} is a set of solutions for the associated homogeneous equation. Now xp = 3 xp = x p = 0 and so x p + 2xp + 2xp = 6 =r.h.s. Thus the general solution is x(t) = 3+c1et cos t+c2et sin t. Using the initial condition x(0) = 1 30 CHAPTER 2 we have, 1 = 3 + c1 c1 = 2. Dierentiating the general solution we get, x(t) = c1et cos tc1et sin tc2et sin t+c2et cos t. Using x(0) = 1 and c1 = 2 we have, 1 = 2 + c2 c2 = 1. Thus the solution of the initial value problem isx(t) = 3 2et cos t et sin t.7. x = sin t x = cos t x = sin t. So x + x = sin t + sin t = 0. x = cos t x = sin t x = cos t. So x + x = 0. Thus {sin t, cos t} is a set of solutions for the associated homogeneous equation. Now xp = t sin t, x p = sin t + t cos t, x p = cos t + cos t t sin t and so x p + xp = 2 cos t t sin t + t sin t = 2 cos t =r.h.s. Thus the general solution is x(t) = t sin t + c1 cos t + c2 sin t. Using the initial condition x(0) = 1 we have, 1 = c1. Dierentiating the general solution we get, x(t) = sin t + t cos t c1 sin t + c2 cos t. Using x(0) = 1 we have, 1 = c2. Thus the solution of the initial value problem is x(t) = t sin t sin t + cos t. 9. (a) x = et x = et x = et . So x 3x + 2x = et 3et + 2et 2t 2t 2t= 0. For x = e , x = 2e and x = 4e . So x 3x + 2x 2t t= 4e 6e2t + 2e2t = 0. Thus _e , e2t_ is a set of solutions for the associated homogeneous equation. xp = 2 cosh t x = 2 sinh t, p x p = 2 cosh t. So x p 3x + 2xp = 2 cosh t 6 sinh t + 4 cosh t = p 6 (cosh t sinh t) = 6et =r.h.s. The general solution is thus x = c1et + c2e2t + 2 cosh t. For xp = et , x p = et , x = et . So x p 3x + 2xp = p p et + 3et + 2et = 6et =r.h.s. The general solution is thus x = c1et + c2e2t + et . (b) Using the initial conditions x(0) = 4, x(0) = 3 for the rst solution we get, 4 = c1 + c2 + 2 and 3 = c1 + 2c2. Solving these equations for for c1 and c2 we have c1 = c2 = 1. Thus the solution of the initial value problem is x = et + e2t + 2 cosh t = et + e2t + et + et = 2et + e2t + et . Using the same initial conditions for the second solution we get, 4 = c1 + c2 + 1 and 3 = c1 + 2c2 1. Solving these equations for for c1 and c2 we have c1 = 2, c2 = 1.Thus the solution of the initial value problem is x = 2et + e2t + et , which is the same as before. 2.2 INITIAL VALUE PROBLEM (FOR HOMOGENEOUS EQUATION) 1. x1 = sin t x1 = cos t x 1 = sin t = x1 x 1 + x1 = 0. x2 = cos t x2 = sin t x 2 = cos t = x2 x2 + x2 = 0. Thus x1 and x2 both are solutions of x + x = 0._ x1 x2 _ _ sin t cos t _ The Wronskian is W [x1, x2] = det = det x 1x2 cos t sin t 31 LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS = _sin2 t + cos2 t_ = 1 = 0 . Hence {sin t, cos t} is a fundamental set of solutions. 3. Let x = tr . Then x = rtr1, x = r (r 1) tr2 . So t2x tx + x = 0 tr [r (r 1) r + 1] = 0 (r 1) (r 1) = 0 (since tr = 0) r = 1(repeated). So we have two solutions: x = t which are the _ t t _ same. The Wronskian is W [x1, x2] = det = 0. They dont 1 1 form a fundamental set of solutions. _ x1(1) x2(1) _ _ 1 0 _ 5. (a) The Wronskian is W (1) = det = det x 1(1) x2 (1) 1 1 = 0. Hence {x1, x2} is a fundamental set of solutions. = 1 (b) x3 = c1x1 + c2x2 x3 = c1x1 + c2x 2. Using all three sets of initial conditions we have, 2 = x3(1) = c1x1(1) + c2x2(1) = c1 and 0 = x 3(1) = c1x1 (1) + c2x 2(1) = c1 c2 = 2 c2 c2 = 2. Thus x3 = 2x1 + 2x2. _ x1(t0) x2(t0) _ _ 1 0 _ 7. The Wronskian is W (t0) = det = det x 1(t0) x2 (t0) 0 1 = 1 = 0 . Hence {x1, x2} is a fundamental set of solutions. Now x3 = c1x1 + c2x2 x 3 = c1x 1 + c2x 2. Using all three sets of initial conditions we have, = x3(t0) = c1x1(t0) + c2x2(t0) = c1 1 + c2 0 = c1 = x3 (t0) = c1x 1(t0) + c2x2 (t0) = c1 0 + c2 1 = c2Thus x3 = x1 + x2.9. Let x1 and x2 be two solutions of Airys equation x + tx = 0,so that x 1 + tx1 = 0 and x 2 + tx2 = 0.Then the Wronskian is W (t) = det _ xx1 1 xx2 2 _ = x1x 2 x1 x2. dW dt = x 1x 2 + x1x2 x 1x 2 x1 x2 = x1x2 x 1 x2= x1 (tx2) x2 (tx1)= 0So W (t) is constant.11. x1 = sin t x 1 = cos t x 1 = sin t = x1 x1 + x1 = 0. x2 = cos t x2 = sin t x 2 = cos t = x2 x2 + x2 = 0. Thus x1 and x2 both are solutions of x + x = 0._ x1 x2 _ _ sin t cos t _ The Wronskian is W [x1, x2] = det = det = _sin2 t + cos2 t_ = 1 = 0 .x 1 x2 cos t sin t t_ 0 tp(s)ds According to (13), W (t) = W (t0)e for the equation x + p(t)x + q(t)x = 0. In this problem p(t) = 0 so that W (t) = W (t0) =constant. 13. x1 t = et x 1 = x 1 = et = x1. Then l.h.s. = x 1 3x1 + 2x1e 3et + 2et = 0 =r.h.s.x2 = e2t x 2 = 2e2t x 2 = 4e2t . Thenl.h.s.= x 2 3x2 + 2x2 = 4e2t 6e2t + 2e2t = 0 =r.h.s.Thus et and e2t both are solutions of x 3x + 2x = 0.32 CHAPTER 2 t 2t The Wronskian is W (t) = det _ et e2t _ = 2e3t e3t = e3t . e 2e_tp(s)ds According to (13), W (t) = W (t0)e t0 for the equation x + p(t)x + q(t)x = 0. In this problem p(t) = 3 so that W (t) = W (t0)e3t3t0 = e3t0 e3t3t0 = e3t . 15. x1 = t1 x 1 = t2, x 1 = 2t3 . Then l.h.s. = t2x 1 + 4tx 1 + 2x1 = 2t1 4t1 + 2t1 = 0 =r.h.s. x2 = t2 x 2 = 2t3, x 2 = 6t4 . Then l.h.s. = t2 x2 + 4tx 2 + 2x2 = 6t2 8t2 + 2t2 = 0 =r.h.s. Thus t1 and t2 both are solutions of t2x + 4tx + 2x = 0._ t1 t2 _ The Wronskian is W (t) = det t2 2t3 = 2t4 + t4 = t4 . _tp(s)ds According to (13), W (t) = W (t0)e t0 for the equation x + p(t)x + q(t)x = 0. In this problem p(t) = 4 t so that _ t _4 W (t) = W (t0)e4(ln tln t0) = t04 lnt0 = t4e . 2.3 REDUCTION OF ORDER 1. x1 = t1 x1 = t2 and x1 = 2t3 so that 2t1 3t1 + t1 = 0. Let x = vx1 = vt1 and substitute this in the original equation to get t2 _vt1_ + 3t _vt1_ + vt1 = 0 t2 _vt1 2vt2 + 2vt3_ + 3t _vt1 vt2_ + vt1 = 0 vt + v = 0, since the v terms cancel. Writing w = v and then dw 1 dw dt dividing by t we get dt + t w = 0. By separation, _ w = _ t , which, on integration, yields ln |w| = ln t + c w = c2t1 . But w = v , so that v = c2t1 . Now integrating again we get v = c2 ln t + c1. Thus the general solution is x = vx1 = vt1 = c2t1 ln t + c1t1 . A fundamental set of solutions would be _t1, t1 ln t_ . 3. x1 = e5t x1 = 5e5t and x 1 = 25e5t so that 25e5t 50e5t + 25e5t = 0. Substitute x = vx1 = ve5t in the original equation to get _ve5t_ +10 _ve5t_ +25ve5t = 0 _v e5t 10v e5t + 25ve5t_+10 _v e5t 5ve5t_+25ve5t = 0 ve5t = 0, since the v (and v) terms cancel v = 0 since e5t = 0 . Now integrating twice we get v = c2t + c1. Thus the general solution is x = vx1 = ve5t = c2te5t + c1e5t . A fundamental set of solutions would be _e5t, te5t_ . 5. x1 = et x1 = et and x 1 = et, so tet (t 1) et et = 0. Let x = vx1 = vet and substitute this in the original equation to get t (vet) + (t 1) (vet) vet = 0 t (vet 2vet + vet) + (t 1) (vet vet) vet = 0 _ _ 33 LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS tvet (t + 1) vet = 0,since the v terms cancel. Dividing by tet = 0 we get, v _1 + 1 _ v = 0. Now w = v yields t dw _1 + 1 _ w = 0. By separation, _ dw = _ _1 + 1 _ dt, which, dt t w t on integration, yields ln |w| = t + ln |t| + c w = c2tet . But w = v ,so that v = c2tet . Now using integration by parts we getv = c2 (t 1) et + c1.Thus the general solution is x = vx1 = vet = (c2 (t 1) et + c1) et= c2 (t 1)+c1et . A fundamental set of solutions would be {et, t 1} .7. x1 = t x1 = 1 and x 1 = 0 so that x1 + tx 1 x1 = t t = 0. Let x = vx1 = vt and substitute this in the original equation to get (vt) + t (vt) vt = 0 (vt + 2v) + t (vt + v) vt = 0 v t + _2 + t2_ v = 0, since the v terms cancel. Writing w = v and dw + _ 2then dividing by t we get dt t + t_ w = 0.dwBy separation, _ = _ _ 2 + t_ dt, which, on integration, yields w t 2 2 1 11 w = c2e2 ln|t|t2 + c2 = c2t2e t tln |w| = 2 ln |tBut w = v , so that v| 2 2 . = c2t2e1 2t2 . Now integrating by _t2 1 2s2e susing denite integral we get, v = c2 ds + c1. 1 _ts2e 2 221sThus the general solution is x = vx1 ds + c1t. = vt = c2t 1 _ts2e1 2sA fundamental set of solutions would be ds . t, t 1 9. Let x = tr . Then x = rtr1 and x = r (r 1) tr2 . Substituting these into t2x 3tx + 4x = 0 we get, r (r 1) tr 3rtr + 4tr = 0 r2 4r + 4 = 0 (since tr = 0). Thus (r 2)2 = 0 r = 2. So x1 = t2 is a solution. Let x = vx1 = vt2 and substitute this in the original equation to get t2 _vt2_ 3t _vt2_ + 4vt2 = 0 t2 _t2v + 4tv + 2v_ 3t _t2v + 2tv_ + 4vt2 = 0 t4v + t3v = 0, since the v terms cancel. v + 1 t v = 0. Writing w = v we get dw + 1 w = 0. By separation, _ dw 1 dt ln w = ln t + cdt t w = _ t | | | | w = c2e ln|t| = c2t1 . But w = v , so that v = c21 t . Now integrating again we get, v = c2 ln t + c1. Thus the general solution is x = vt2 = (c2 ln t + c1) t2 = c2t2 ln t + c1t2 (i.e., the second solution is t2 ln t). 11. Let x = tr . Then x = rtr1 and x = r (r 1) tr2 . Substituting these into t2x + 7tx + 9x = 0 we get, r (r 1) tr + 7rtr + 9tr = 0 r2 + 6r + 9 = 0 (since tr = 0). Thus (r + 3)2 = 0 r = 3. So x1 = t3 is a solution. Let x = vx1 = vt3 and substitute this in the original equation to get t2 _vt3_ + 7t _vt3_ + 9vt3 = 0 t2 _t3v 6t4v + 12t5v_ + 7t _t3v 3t4v_ + 9vt3 = 0, since the v terms cancel, t1v + t2v = 0 v + 1 v = 0. dw 1 t dw 1Writing w = v we get dt + t w = 0. By separation, _ w = _ t dt ln |w| = ln |t| + c w = c2e ln|t| = c2t1 . But w = v , so that 34 CHAPTER 2 v = c21 t . Now integrating again we get, v = c2 ln t + c1. Thus the general solution is x = vt3 = (c2 ln t + c1) t3 = c2t3 ln t + c1t3 (i.e., the second solution is t3 ln t). 13. Let x = sin (rt) . Then x = r cos (rt) and x = r2 sin (rt) . Substituting these into x + 4x = 0 we get, _4 r2_ sin (rt) = 0 2 = 0 (since sin (rt) = 0 as nontrivial solution). Thus 4 r r = 2. So x1 = sin (2t) is a solution. Let x = vx1 = v sin (2t) and substitute this in the original equation to get (v sin (2t)) + 4v sin (2t) = 0 (v sin (2t) + 4v cos (2t) 4v sin (2t)) + 4v sin (2t) = 0, since the v terms cancel, v sin (2t) + 4v cos (2t) = 0 v + 4 cos(2t) v = 0. sin(2t) Writing w = v we get dw + 4 cos(2t) w = 0. By integrating factors, dt sin(2t) cos(2t) sin(2t) 2 dt ln(sin(2t))) 1we have w = ce4 _ = ce4( 1 = c sin2(2t) = c csc2 (2t) . But w = v , so that v = c csc2 (2t) . Now using integration table we get, v = c cot (2t) + c1 = c2 cos(2t) + c1. Thus the general solution 2 sin(2t) is x = v sin (2t) = _ cos(2t) + c1 _ sin (2t) = c1 sin (2t) + c2 cos (2t) . c2 sin(2t) 15. Let x = ert x = rert and x = r2ert . Substituting these into x 4x + 4 x = 0 we get, _r2 4r + 4_ ert = 0 r2 4r + 4 = 0 (since ert = 0). Thus (r 2)2 = 0 r = 2. So x1 = e2t is a solution. 2t Let x = vx1 = ve and substitute this in the original equation to get _ve2t_ 4 _ve2t_ + 4ve2t = 0 ve2t + 4ve2t + 4ve2t 4ve2t 8ve2t + 4ve2t = 0, since the v (and v) terms cancel, ve2t = 0 v = 0(since e2t = 0). Integrating twice we have, v = c1 + c2t. Thus the general solution is x = ve2t = (c1 + c2t) e2t = c1e2t + c2te2t . 17. Let x = ert x = rert and x = r2ert . Substituting these into x + 2x + x = 0 we get, _r2 + 2r + 1_ ert = 0 r2 + 2r + 1 = 0 (since ert = 0). Thus (r + 1)2 = 0 r = 1. So x1 = et is a solution. Let x = vx1 = vet and substitute this in the original equation to get (vet) + 2 (vet) + vet = 0 vet 2vet + vet + 2vet 2vet + vet = 0, since the v (and v ) terms cancel, v et = 0 v = 0(since et = 0). Integrating twice we have, v = c1 + c2t. Thus the general solution is x = vet = (c1 + c2t) et = c1et + c2tet . 19. Let x = ert x = rert and x = r2ert . Substituting these into 2 2 rt x 4x = 0 we get, _r 4_ ert = 0 r 4 = 0 (since e = 0). 2t 2tThus r2 = 4 r = 2. So x1 = e is a solution. Let x = vx1 = veand substitute this in the original equation to get2t_ve2t_ 4ve2t = 0 ve2t + 4ve2t + 4ve 4ve2t = 0, since the v 2tterms cancel, v e2t + 4v e2t = 0 v + 4v = 0(since e = 0). Writing w = v we get dw + 4w = 0. So w = ce4t . But w = v , so dt that v = ce4t . Now integrating again we get, v = 1 ce4t + c1 = 4c2e4t + c1. Thus the general solution is

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS 35 2t 2t 2tx = ve = _c2e4t + c1_ e = c1e + c2e2t . 21. x2 = vx1 where v must be a non-constant function of t (otherwise x2 is a scalar multiple of x1 and so they are linearly dependent). _ x1 x2 _ _ x1 vx1 _ The Wronskian is W [x1, x2] = det = det x 1x 2 x 1 v x1 + vx 1 = v (x1)2 + vx1x 1 vx1x = v (x1)2 = 0 (since v = 0 as v is a 1 non-constant function of t and x1 = 0 as a nontrivial solution). Hence {x1, x2} forms a fundamental set of solutions. 2.4 HOMOGENEOUS LINEAR CONSTANT COEFFICIENT DIFFERENTIAL EQUATIONS (SECOND ORDER) 1. Substituting x = ert into the equation x + x 6x = 0 we get r2 + r 6 = 0 (r + 3) (r 2t 2) = 0 r = 2, 3.Thus the general solution is x = c1e + c2e3t . 3. Substituting x = ert into the equation x + x = 0 we get r2 +1 = 0 r = i. Thus the general solution is x = c1 cos t+c2 sin t. rt 5. Substituting x = e into the equation x + 4x + 5x = 0 we get r2 +4r +5 = 0 r = 41620 = 2 i.Thus the general solution 2 is x = e2t (c1 cos t + c2 sin t) . 7. Substituting x = ert into the equation x 3x + 2x = 0 we get r2 3r + 2 = 0 (r 1) (r 2) r = 1, 2.Thus the general t 2t solution is x = c1e + c2e . 9. Substituting x = ert into the equation x x = 0 we get r2 r = 0 r (r 1) = 0 rt r = 0, 1.Thus the general solution is x = c1 +c2et . 11. Substituting x = e into the equation 3x = 0 we get 3r2 = 0r = 0, a repeated root with multiplicity 2. Thus the general solution is x = c1 + c2t. 13. Substituting x = ert into the equation 3x + 2x x = 0 we get3r2 + 2r 1 = 0 (r + 1) (3r 1) = 0 1 r = 1, 1 . Thus3the general solution is x = c1et + c2e t3 . 15. Substituting x = ert into the equation x + x 2x = 0 we getr2 + r 2 = 0 (r + 2) (r 1) = 0 r = 1, 2. Thus t + c2e2t tthe general solution is x = c1e . So x = c1e 2c2e2t .Using the initial conditions x(0) = 0 and x (0) = 1 we getc1 + c2 = 0 and c1 2c2 = 1. Solving these equations forc1 and c2, we have c1 = 13 and c2 = 13 .Hence x = 31 et 31 e2t . rt 17. Substituting x = e into the equation 2x + 4x = 0 we get 22r2 + 4 = 0 r = 2 r = i2. Thus the general solution is x = c1 cos 2t + c2 sin 2t. 19. Substituting x = ert into the equation 2x + 8x + 6x = 0 we get2r2 + 8r + 6 = 0 r2 + 4r + 3 = 0 (r + 3) (r + 1) = 0 36 CHAPTER 2 r = 1, 3.Thus the general solution is x = c1et + c2e3t . So x = c1et 3c2e3t . Using the initial conditions x(0) = 2 and x(0) = 0 we get c1 + c2 = 2 and c1 3c2 = 0. Solving these equations for c1 and c2, we have c1 = 3 and c2 = 1. Hence x = 3et e3t . 21. Substituting x = ert into the equation x + 10x + 25x = 0 weget, r2 + 10r + 25 = 0 (r + 5)2 = 0 r = 5, repeated twice. Thus the general solution is x = c1e5t + c2te5t . 23. Substituting x = ert into the equation x 14x + 49x = 0 weget, r2 14r + 49 = 0 (r 7)2 = 0 r = 7, repeated 7ttwice. Thus the general solution is x = c1e + c2te7t . 25. Substituting x = ert into the equation x 6x + 25x = 0 we get r2 6r + 25 = 0 r = 636100 = 3 4i.Thus the general 2 solution is x = e3t (c1 cos 4t + c2 sin 4t) . 27. Substituting x = ert into the equation x 12x = 0 we get 2r 12 = 0 + cr 2e= 12t 12. Thus the general solution is12t x = c1e . 29. Substituting x = ert into the equation x + 4x + 8x = 0 we getr2 + 4r + 8 = 0 r = 421632 = 2 2i.Thus the generalsolution is x = e2t (c1 cos 2t + c2 sin 2t) .31. Substituting x = ert into the equation x + 8x = 0 we getr2 + 8 = 0 r = 8 = 8i.Thus the generalsolution is x = c1 cos 8t + c2 sin 8t.33. Substituting x = ert into the equation x + 6x + 7x = 0 we getr2 + 6r + 7 = 0 r = 63628 = 3 2.Thus the general (3+2)t 2(32)t 2tsolution is x = c1e + c2e = e3t _ c1e + c2e2t_ . 35. The Wronskian W [et cos t, et sin t]_ et cos t et sin t _= det et ( cos t sin t) et ( sin t + cos t) = e2t _ sin t cos t + cos2 t sin t cos t + sin2 t_ = e2t _cos2 t + sin2 t_ = e2t = 0 (since = 0). t So {e cos t, et sin t} forms a fundamental set of solutions. 37. Repeated roots have b2 4ac = 0 c = b2 , so that b2 ar2 + br + c = ar2 + br + b2 = a _ r2 + b 4r a + 2 _ = a _r + b _2 .4a a 4a 2a So repeated root is r = b .2a 39. Choose #21. This problem has a repeated root, r = 5. So one solution is x1 = e5t . Let x = vx1 = ve5t and substitute this in the original equation to get _ve5t_ + 10 _ve5t_ + 25ve5t = 0 _ve5t 10ve5t + 25ve5t_ +10 _ve5t 5ve5t_ + 25ve5t = 0 v e5t = 0 v = 0 since e5t = 0. Now integrating twice we get v = c2t + c1. Thus the general solution isx = vx1 = ve5t = c2te5t + c1e5t .37 LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS 41. c1et + c2e2t will be the general solution if r = 1 and r = 2 are the roots of the characteristic equation. One such characteristic equation would be (r + 1) (r + 2) = r2 +3r +2 = 0. A corresponding dierential equation is x + 3x + 2x = 0. 43. c1e3t + c2te3t will be the general solution if r = 3 is a repeated (twice) root of the characteristic equation. One such characteristic equation would be (r 3)2 = r2 6r + 9 = 0. A corresponding dierential equation is x 6x + 9x = 0. 45. c1 sin 4t + c2 cos 4t will be the general solution if r = 4i is a complex root of the characteristic equation. One such characteristic equation would be (r + 4i) (r 4i) = r2 + 16 = 0. A corresponding dierential equation is x + 16x = 0. 47. c1 sin 3t + c2 cos 3t will be the general solution if r = 3i is a complex root of the characteristic equation. One such characteristic equation would be (r + 3i) (r 3i) = r2 + 9 = 0. A corresponding dierential equation is x + 9x = 0. 49. c1 + c2t will be the general solution if r = 0 is a repeated rootof the characteristic equation. One such characteristicequation would be r2 = 0. A correspondingdierential equation is x = 0.51. c1et sin t + c2et cos t will be the general solution if r = 1 i is a complex root of the characteristic equation. One such characteristic equation would be (r 1 i) (r 1 + i) = r2 2r + 2 = 0. A corresponding dierential equation is x 2x + 2x = 0. 2.4.1 Homogeneous Linear Constant Coecient Dierential Equations (nth-Order) 1. Substituting x = ert into the equation x 6x + 12x 8x = 0 we get, r3 6r2 + 12r 8 = 0 (r 2)3 = 0 r = 2 is a repeated root of multiplicity 3.Thus the general solution is x = c1e2t + c2te2t + c3t2e2t . 3. Substituting x = ert into the equation x 5x + 4x = 0 we get, 4 2 2r 5r2 + 4 = 0 _r 1_ _r 4_ = 0 r = 1, 2. t 2t + c4e2tThus the general solution is x = c1e + c2et + c3e . 5. Substituting x = ert into the equation x + x 2x = 0 we get, r3 + r2 2r = 0 r (r 1) (r + 2) = 0 r = 0, 1, 2. + c3e2tThus the general solution is x = c1 + c2et . 7. Substituting x = ert into the equation x +4x +6x +4x + x = 0 we get, r4 + 4r3 + 6r2 + 4r + 1 = 0 (r + 1)4 = 0 r = 1 is a repeated root of multiplicity 4. Thus the general solution is x = c1et + c2tet + c3t2et + c4t3et . 9. Substituting x = ert into the equation x 3x = 0 we get, r 3 = 0 r = 3. Thus the general solution is x = c1e3t . 38 CHAPTER 2 11. Substituting x = ert into the equation x + x 2x = 0 we get, 3 2r + r 2 = 0 (r 1) _r2 + 2r + 2_ = 0 r = 1, 1 i. t Thus the general solution is x = c1e + et (c2 cos t + c3 sin t) . 13. Substituting x = ert into the equation x +4x +8x +8x +4x = 0 we get, r4 + 4r3 + 8r2 + 8r + 4 = 0_r4 + 4r3 + 4r2_ + 4r2 + 8r + 4 = 0 _r2 + 2r_2 + 4 _r2 + 2r_ + 4 = 0 _r2 + 2r + 2_2 = 0 r = 1 i is a repeated root of multiplicity 2. Thus the general solution is x = et (c1 cos t + c2 sin t) + tet (c3 cos t + c4 sin t) . 15. Substituting x = ert into the equation x(4)4x(3)+6x(2)4x(1) +x = 0 we get, r4 4r3 + 6r2 4r + 1 = 0 (r 1)4 = 0 r = 1 is a repeated root of multiplicity 4. Thus the general solution is x = c1et + c2tet + c3t2et + c4t3et . 17. Substituting x = ert into the equation x(6) 3x(4) + 3x(2) x = 0 we get, r6 3r4 + 3r2 1 = 0_r2_3 3 _r2_2 .1 + 3 _r2_ .12 13 = 0 2_r 1_3 = 0 r = 1, each repeated with multiplicity 3. Thus the general solution isx = c1et + c2tet + c3t2et + c4et + c5tet + c6t2et .19. Substituting x = ert into the equation x(4) x = 0 we get, r4 1 = 0 2 2_r 1_ _r2 + 1_ = 0 r2 = 1 and r = 1 r = 1, i. Thus the general solution is x = c1et + c2et + c3 cos t + c4 sin t. 21. Substituting x = ert into the equation x + 50x + 625x = 0we get, r4 + 50r2 + 625 = 0 _r2 + 25_2 = 0 r = 5i repeated twice. Thus the general solution isx = c1 cos 5t + c2 sin 5t + c3t cos 5t + c4t sin 5t.23. Substituting x = ert into the equation x + 3x + x 5x = 0 we 2get, r3 +3r + r 5 = 0 (r 1) _r2 + 4r + 5_ = 0 r = 1, 2 i. t Thus the general solution is x = c1e + e2t (c2 cos t + c3 sin t) . 25. c1e2t + c2te2t + c3 will be the general solution if r = 2 is a root with multiplicity 2 and r = 0 is a simple root of the characteristic 2equation. One such characteristic equation would be r (r 2)= r3 4r2 + 4r = 0. A corresponding dierential equation is x 4x + 4x = 0. 27. c1 sin 5t + c2 cos 5t + c3t sin 5t + c4t cos 5t will be the general solution if r = 5i is a complex root with multiplicity 2 of the characteristic 2equation. One such characteristic equation would be ((r 5i) (r + 5i))= _r2 + 25_2 = r4 + 50r2 + 625 = 0. A corresponding dierential equation is x + 50x + 625x = 0. 29. c1e2t + c2te2t + c3t2e2t will be the general solution if r = 2 is a root with multiplicity 3 of the characteristic equation. One such characteristic equation would be (r + 2)3 = r3 + 6r2 + 12r + 8 = 0. A corresponding dierential equation is x + 6x + 12x + 8x = 0. LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS 39 31. c1et sin t + c2et cos t + c3tet sin t + c4tet cos t will be the general solution if r = 1 i is a complex root with multiplicity 2 of the characteristic equation. One such characteristic equation would be 2 2((r 1 i) (r 1 + i)) = _r 2r + 2_2 =r4 4r3 + 8r2 8r + 4 = 0. A corresponding dier