Solutions Manual Introduction Differential

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Solutions Manual toIntroduction to DifferentialEquations with DynamicalSystemsby Stephen L. Campbell and Richard HabermanM. Ziaul HaquePRINCETON UNIVERSITY PRESSPRINCETON AND OXFORD

c 2008 by Princeton University PressCopyright Published by Princeton University Press41 William Street, Princeton, New Jersey 08540In the United Kingdom: Princeton University Press6 Oxford Street, Woodstock, Oxfordshire, 0X20 1TWAll Rights ReservedThis book has been composed in LATEXpress.princeton.edu

ContentsPrefacevChapter 1. First-Order Differential Equations and Their Applications1.11.21.31.41.51.61.7Introduction to Ordinary Differential EquationsDefinite Integral and the Initial Value ProblemFirst-Order Separable Differential EquationsDirection FieldsEuler’s Numerical Method (Optional)First-Order Linear Differential EquationsLinear First-Order Differential Equations with Constant Coeffi cients and Constant Input1.8 Growth and Decay Problems1.9 Mixture Problems1.10 Electronic Circuits1.11 Mechanics II: Including Air Resistance1.12 Orthogonal Trajectories (optional)Chapter 2. Linear Second and Higher-Order Differenial Equations2.12.22.32.4General Solution of Second-Order Linear Differential EquationsInitial Value Problem (For Homogeneous Equation)Reduction of OrderHomogeneous Linear Constant Coefficient Differential Equations(Second Order)2.5 Mechanical Vibrations I: Formulation and Free Response2.6 The Method of Undetermined Coefficients2.7 Mechanical Vibrations II: Forced Response2.8 Linear Electric Circuits2.9 Euler Equation2.10 Variation of Parameters (Second-Order)2.11 Variation of Parameters (nth-Order)Chapter 3. The Laplace Transform3.13.23.33.43.53.63.7Definition and Basic PropertiesInverse Laplace Transforms (Roots, Quadratics, & Partial Fractions)Initial-Value Problems for Differential EquationsDiscontinuous Forcing FunctionsPeriodic FunctionsIntegrals and the Convolution TheoremImpulses and 56870758282869498109114118

ivCONTENTSChapter 4. An Introduction to Linear Systems of Differential Equations andTheir Phase Plane4.14.24.3IntroductionIntroduction to Linear Systems of Differential EquationsPhase Plane for Linear Systems of Differential EquationsChapter 5. Mostly Nonlinear First-Order Differential Equations5.15.25.35.4First-Order Differential EquationsEquilibria and StabilityOne Dimensional Phase LinesApplication to Population Dynamics: The Logistic EquationChapter 6. Nonlinear Systems of Differential Equations in the Plane6.16.26.36.4IntroductionEquilibria of Nonlinear Systems, Linear Stability Analysis of Equi librium, and Phase PlanePopulation ModelsMechanical Systems121121121130142142142143146150150150161178

PrefaceThis Student Solutions Manual contains solutions to the odd-numbered ex ercises in the text Introduction to Differential Equations with DynamicalSystems by Stephen L. Campbell and Richard Haberman.To master the concepts in a mathematics text the students must solve prob lems which sometimes may be challenging. This manual has been writtenfocusing student’s needs and expectations. Instead of providing only theanswer with very few steps, I include a reasonably detailed solution witha fair amount of detail when explaining the solution of the problem. Thesolutions are self-explanatory and consistent with the notations and termi nologies used in the text book. I hope this manual will help students buildproblem-solving skills.I would like to thank many people who have provided invaluable help, inmany ways, in the preparation of this manual. First, I take this opportunityto thank Professor Richard Haberman for his generous expert help, construc tive comments and accuracy checking. I would also like to thank ProfessorStephen L. Campbell for assembling the final manuscript, Professor PeterK. Moore for facilitating support process and Ms. Vickie Kearn of the pub lishing company for her patience and support. Finally, I must appreciatethe patience of my wife, Rukshana, and my daughters, Zareen and Ehramfor their understanding and compromise of summer time that was slightedbecause of my busy schedule.M. Ziaul HaqueSouthern Methodist UniversityDallas, TX, 75275, U.S.A.July, 2007.

Chapter OneFirst-Order Differential Equations and TheirApplications1.1 INTRODUCTION TO ORDINARY DIFFERENTIALEQUATIONSThere are no exercises in this section.1.2 DEFINITE INTEGRAL AND THE INITIAL VALUEPROBLEM1-7. Substitute expression for x into the differential equation3t1. x 2e3t 1. l.h.s. dxdt 6e .3tr.h.s. 3x 3 3(2e 1) 3 6e3t . Hence l.h.s. r.h.s.xt 13. x t 1. l.h.s. dxdt 1. r.h.s t 1 t 1 1. Hence l.h.s. r.h.s.222tt5. x et . l.h.s. dxdt 2te . r.h.s 2tx 2te . Hence l.h.s. r.h.s.dx 2t 2t7. x e . l.h.s. dt 2e .r.h.s. 2e2t x2 2e2t (e 2t )2 2e 2t . Hence l.h.s. r.h.s.tt9. dxdt 3e . Integrating we get, x 3e c.dx11. dt 5 cos 6t. Integrating we get, x 56 sin 6t c.Rt 1 1213. dx). Use of definite integral gives x 8 0 cos t 2 dt c.dt 8 cos(t215. dxdt R ln(4 cos t). Use of definite integral givestx 0 ln(4 cos2 t)dt cdx17. dt t4 ; x(2) 3. Integrating we get x 15 t5 c.171 517t 2 3 325 c c 5 . So x 5 t 5 .ln t19. dxdt 4 cos2 t ; x(2) 5. Use of definite integral givesR t ln tx 5 2 4 cos2 t dt.et1 t ;t21.dxdtex(1) 3. dx 1 tdt. Use of definite integral givesR t etR t etx 3 1 1 t dt x 3 1 1 tdt.23.d2 xdt2 15. Integrating we getdxdxdt 15t c1 ( dt v0 at t 0 c1 v0 .)dxdt 15t v0 . Integrating again we get2x 152 t v0 t c2 (x 0 at t 0 c2 0.)

2CHAPTER 1v0Car stops when dxdt 0 v0 15t 0 t 15 (stopping time).So distance travelledis 2 2v02v0 21 v01 v0x 15() 2 15152 15 75 2 15 v0 15 10 m/sec.2dx25. ddt2x 2500. Integrating we get dxdt 2500t c1 ( dt 60 at t 0dx c1 60). So dt 2500t 60. Integrating again we get2x 25002 t 60t c2 (x 0 at t 0 c2 0.)Car stops when dxdt 0 2500t 60 060 t 2500(stopping time). So distance travelled is60 2602x 25002 ( 2500 ) 2500 0.72 km.2dx27. ddt2x 2500. Integrating we get dxdt 2500t c1 ( dt v0 at t 0 c1 v0 ). So dxdt 2500t v0 . Integrating again we get2t vt c2 (x 0 at t 0 c2 0)x 250002Car stops when dxdt 0 v0 2500t 0v0 t 2500(stopping time). So distance travelled isv2v2v0 200x 25002 ( 2500 ) 2500 5000 km.2dx229. ddt2x 6t. Integrating we get dxdt 3t c1 ( dt 50 at t 2dx2 c1 62). So dt 3t 62. Integrating again we getx t3 62t c2 (x 0 at t 2 c2 116.)2Car stops when dxdt 0 3t 62 0q t 623 (stopping time). So distance travelled isq62x t(62 t2 ) 116 623 (62 3 ) 116q¡ 62 322 116 km. 623 ( 3 )62 116 2 3dy331. (a) V volume, dVdt Q m /h. Let snow depth be y. So dt c y ct c1 (y 0 at t 0 c1 0). Thus y ct. Nowconsider the snowplow has moved Δx over the time Δt and theapproximate change in volume over this time is ΔV. HenceΔxΔV w(Δx)y wctΔx ΔVΔt wct Δt . Now taking limitQdx1dVas Δt 0 we get dt Q wct dt dxdt wct kt withwck Q.RR1dx k1 1t dt (b) dxdt kt . Separating the variables we get,x k1 ln t a. At 11 A.M. t 3 and x(3) 0. So0 k1 ln 3 a a k1 ln 3. Then at noon (t 4),x(4) k1 ln 4 k1 ln 3 k1 ln 34 .33.d2 y2dt2 g 9.8 m/sec . Integrating we getdydydt 9.8t c1 ( dt v0 at t 0 c1 v0 .) Sodydt 9.8t v0 . Integrating again we gety 92.8 t2 v0 t c2 (y 0 at t 0 c2 0.)At maximum height dydt 0 v0 9.8t 0v0 t 9.8(time at maximum height). So maximum height is 22v02v0 21 v01 v01960 m/sec.y 9.82 ( 9.8 ) 9.8 2 9.8 100 2 9.8 v0

3FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS35.2d y2dt2 g 9.8 m/sec . Integrating we getdydydt 9.8t c1 ( dt 0 at t 0 c1 0.)9.8 2So dydt 9.8t. Integrating again we get y 2 t c29.8 2(y 200 at t 0 c2 200). y 2 t 200. Now to2 9.82 tqfall,4009.8sec. 200 t 209.8tR ¡ 37. Since x(t0 ) x0 , the general solution x f t dt c becomesy 0. So 0 200 9.8 22 t0Rt0 ¡ Rt0 ¡ x0 f t dt c c x0 f t dt. Hence the solution is0x Rt00¡ Rt ¡ Rt0 ¡ f t dt x0 f t dt x0 f t dt.t001.3 FIRST-ORDER SEPARABLE DIFFERENTIAL EQUATIONS1.R dt. Integrating we get, t x 1 x 1 ln x 1 ln t c1 ln t c1 t ec1c1 x 1 c x ct 1.t eRRdxt3. dt e . Separating variables gives dx et dt. Integrating we get,x et c.5. dx tx 4x 3t 12 (x 3) (t 4) . Separating variables givesdtRR dx2(t 4) dt. Integrating we get, ln x 3 t2 4t c1x 3 dxdt x 1t .Separating variables givest2Rdxx 1t2 x 3 ec1 e 2 4t x ce 2 R 4t 3 R where c ec1 .dx7. dt 3. Separating variables gives dx 3dt. Integrating we get,x 3t c.R 5R59. dxx dx dt. Integrating we get,dt x . Separating variables gives 4 x 4 t c. Using x(2) 1 we get, 14 2 c c 94 . 1/4Substituting c we get x 4 9 4t x (9 4t).22 2dx cos(t2 )dt. Using11. dxdt x cos(t ). Separating variables gives xRxRt2definite integrals we get, x 2 dx cos(t )dt1 x1 x 1 Rt0102cos(t )dt 1xRt21 cos(t )dt 1 Rt2cos(t )dt0013.dxdt t cos(x 1/2 ). Separating variables givesRxRtdefinite integrals we get, cos(xdx 1/2 ) tdt 15.dudtRx2 2dxcos(x 1/2 )t2 1u2 4 . t22 12 12¡ t2 1 .Separating variables givesdxcos(x 1/2 ) tdt. Using1R¡ R¡2u2 4 du t 1 dt.

4CHAPTER 133Integrating we get, u3 4u t3 t c. Using u(0) 1 we get,1313 4 c c 3 . Substituting c we obtain the solution as33u 12u t 3t 13. ¡ ¡ 17. dx t2 x2 x2 t2 1 x2 1 t2 1 . Separating variables givesdt R dxR¡23 1(x) t3 t cx2 1 ³ t 1 dt. Integrating we get, tan3 x tan t3 t c . Using x(0) 2 we get, c tan 1 (2). Hence³ 3 the solution is x tan t3 t tan 1 (2) .R dxR19. dxdt. Usingdt x (x 1) . Separating variables givesx(x 1) partial fractions to the integral on the left we get,AB1x(x 1) x x 1 1 A (x 1) Bx. Putting x 0 and 1,respectively,have,A 1R weR and B 1. HenceR dxR dxdx dt ln x 1 ln x t cx(x 1)x x 1 x 1x 1 ln x t c x ket where k ec . Solving this1equation for x we obtain the general solution, x 1 ket . Since x 0and x 1 both satisfy the differential equation they are alsosolutions. The solution x 1 corresponds to k 0, however,x 0 is not included in the general solution for any finite k.1Hence the solutions are x 1 ket and x 0.RR2dxdx21. dt (x 1) (x 2) . Separating variables gives (x 1)(x 2)dt.2 Using partial fractions to the integral on the left we get,ABC1 x 1 x 2 (x 2)2 (x 1)(x 2)221 A (x 2) B (x 1) (x 2) C (x 1) . Putting x 1and 2, respectively, we have, A 1 and C 1. Then equating2theof0 BR xdx we haveR 1. HenceR coefficientsR dx A RB dxdx dt. Integrating both2 2x 1x 2(x 2)(x 1)(x 2)1sides we obtain the solution, ln x 1 ln x 2 x 2 t c.Since x 1 and x 2 both satisfy the differential equation theyare also solutions. Hence the solutions are1 t c, x 1 and x 2.ln x 1 ln x 2 x 223. ¡(tx x)dt (tx t)dx 0.Dividing by tx we get, ¡ 1 1t dt 1 x1 dx 0. Now integrating we have,t ln t x ln x c ln tx t x c c t xtxctx¡ 2 e e e ¡ 2 te xe c where c e¡ 2. ¡ 25. t 4 dz z 9 dt 0. Dividing by t 4 z 2 9we get, (z2dz 9) (t2dt 0. Now we use the formula 4) R du u a 1u2 a2 2a ln u a (from integration table) to integrate and 1 t 2 z 3 1/6 t 2 1/4get 61 ln z 3 ln c ln c z 34z 3t 2 ³ t 2³ 1/6 ³ 1/4 z 3 1/6 ³ t 2 1/4 t 2 ec z 3 ec t 2z 3t 2 z 3 ³³³ 1 41 1 3³t 2 4t 2 2z 3 6 ec t 2 ec t 2 z 3z 3t 2z 3 c t 2

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS56where c ( ec ) .27. et x dt e2t 3x dx 0. Dividing by ex e2t we get, 4xe t dt e 4x dx 0. Now integrating we have, e t e 4 c 4x t t e 4e 4c 4x ln ( 4e 4c) x 41 ln ( 4e t k) where k 4c.29. z at bx c. Differentiating with respect to t we get,dzdxdzdt a b dt dt a bf (z) which is a differential equationinand t andR z dzR can be solved by separation of variables asdt. a bf (z)22 (t 4x 1) . Let z t 4x 1. Then dxdt z f (z)dxdz2 1 4 dt 1 4z . Separating the variables we get,andRR dzdtdt. We use the substitution u 2z dz 12 du1 4z 2 RR duto integrate the left hand side. This gives 12 1 udt 2 11 1 1(2z) t c 2 tan (2t 8x 2) t c.2 tan 1t xz 133. dx e(t x) 1. Let z t x. Then dx 1dtdt e zdxdzz 1z 11 ez 1 ez.Separatingtheand dt 1 dt RRvariables we get, ze z dz dt. We use integration by partsto integrate the left Rhand side asR u z du e z dz.R dz and dv z z z Rv e . Then ze dz udv uv vdu ze e z dz ze z e z . This gives e z (z 1) t c.Substituting z t x we get the solution as e (t x) (t x 1) t c e t x (t x 1) t c.31.dxdt1.4 DIRECTION FIELDS1.2x0 20t2

6CHAPTER 13.2x0 20t25.2x0 220t1.4.1 Existence and Uniquenessdxdt1.3.5.7.9.11.13.x1 1 t2 f (t, x) and fx 1 t2 are continuous for all (t, x) . Sounique solution exists for all (t0 , x0 ) .¡ ¡ 4/3dx22 7/3are f (t, x) and fx 73 1 t2 x2dt 1 t xcontinuous for all (t, x) . So unique solution exists for all (t0 , x0 ) .1/5dx1 f (t, x) is continuous for all (t, x) but fx 5(x t)4/5dt (x t)is not continuous for x t 0.So unique solution exists for all(t0 , x0 ) such that x0 t0 6 0.dxcos t cos tdt x 1 f (t, x) and fx (x 1)2 are not continuous at x 1.So unique solution exists for all (t0 , x0 ) such that x0 6 1.¡ ¡ 1/2dx22 3/2 f (t, x) and fx 6x 1 t2 2x2dt 1 t 2xare continuous for 1 t2 2x2 0. So unique solution exists forall (t0 , x0 ) such that t20 2x20 1.dx1dt t1/3 f (t, x) is not continuous at t 0 although fx 0 iscontinuous everywhere. So unique solution exists for all (t0 , x0 )such that t0 6 0.(a) Differentiating t2 x2 c wrt(with respect to) t we get,

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS7dxt222t 2x dxdt 0 dt x . Hence t x c definesa solution.(b) graph(c) At x0 0, as f (t, x) xt and fx xt2 are discontinuous atthis point.15. (a) Differentiating t x2 c wrt t we get, 1 2x dxdt 0.Hence t x2 c defines a solution.(b) graph11(c) Here dxdt 2x f (t, x) and fx 2x2 are discontinuous atx0 0. Hence the theorem fails to hold at this point.17. (a) Differentiating x c sin t wrt t we get,dxxcos tdt c cos t c x cos t cx c sin t x cot t.Hence x c sin t defines a solution.(b) graphcos tcos t(c) Here dxdt x sin t f (t, x) and fx sin t are discontinuouswhen sin t 0 t nπ for n 0, 1, 2, .Hence thetheorem fails to hold at the point t0 nπ for n 0, 1, 2, .11219. (a) Differentiating x t cwrt t we get, dxdt (t c)2 x .1defines a solution.Hence x t c(b) graph2(c) Here dxdt x f (t, x) and fx 2x are continuouseverywhere. Hence there is NO point where the theoremfails to holds.1/521. (a) For x 1, l.h.s. is dx 0.dt 0 and r.h.s. (x 1) 5/4 1/4¡¡44 1, l.h.s. is dx t candFor x 5 t cdt5³¡ 1/5 ¡ 1/45/41/5r.h.s. (x 1) 45 t c 1 1 54 t c.dx(b) By separating the variables we get, (x 1)1/5 dt which, on 5/4¡4 1. Then using theintegration, becomes x 5 t cinitial condition x0 1 for any t0 we get c 45 t0 and thus 5/4¡ 1. Another solution isone solution is x 54 t 45 t0clearly x 1 because it satisfies the initial condition aswell as the differential equation. So there are at least twosolutions through the point (t0 , x0 ) with x0 1.(c) Graph1/5(d) Although f (t, x) (x 1)is continuous everywhere,1fx (x 1)4/5 is not continuous at x0 1. As a result,uniqueness does not hold and two solutions in part (b)is not a surprise.1.5 EULER’S NUMERICAL METHOD (OPTIONAL)1.dxdt x t f (t, x) , t0 0, x0 1, h 0.5. We use recursive formula,

8CHAPTER 1xn 1 xn hf (tn , xn ) xn 0.5 (xn tn ), where tn 1 tn hto approximatex1 x0 h (x0 t0 ) 1 0.5(1 0) 1.5 at t1 0.5x2 x1 h (x1 t1 ) 1.5 0.5(1.5 0.5) 2 at t2 1x3 x2 h (x2 t2 ) 2 0.5(2 1) 2.5 at t3 1.5x4 x3 h (x3 t3 ) 2.5 0.5(2.5 1.5) 3 at t4 2So estimate for x(2) is x4 3.23. dxdt tx f (t, x) , t0 0, x0 1, h 1. We use recursive formula,xn 1 xn hf (tn , xn ) xn tn x2n , where tn 1 tn hto approximatex1 x0 ht0 x20 1 0 1 at t1 1x2 x1 ht1 x21 1 1 0 at t2 2So estimate for x(2) is x2 0.5. dxdt 2x 4t f (t, x) , t0 0, x0 1, h 0.5. We use recursiveformula,xn 1 xn hf (tn , xn ) xn 0.5 (2xn 4tn ) 2 (xn tn ) ,where tn 1 tn h to approximatex1 2 (x0 t0 ) 2(1 0) 2 at t1 0.5x2 2 (x1 t1 ) 2(2 0.5) 3 at t2 1x3 2 (x2 t2 ) 2(3 1) 4 at t3 1.5x4 2 (x3 t3 ) 2(4 1.5) 5 at t4 2So estimate for x(2) is x4 5.7. dxdt sin x f (t, x) , t0 0, x0 0, h 0.5. We use recursiveformula,xn 1 xn hf (tn , xn ) xn 0.5 (sin xn ) , where tn 1 tn hto approximatex1 x0 0.5 sin x0 0 at t1 0.5x2 x1 0.5 sin x1 0 at t2 1Similarly, xi 0 at ti for all i 0, 1, ., 8.So estimate for x(4) is x8 0.9. (a) dxdt 20x f (t, x) , t0 0, x0 1, h 0.2.We use recursive formula,xn 1 xn hf (tn , xn ) xn 0.2 ( 20xn ) 3xn .nHere xn ( 3) at tn 0.2n, n 0, 1, 2, .10So estimate for x(2) is x10 ( 3) 59049.(b) xn oscillates wildly as n . x(2) e 40 4.248 10 18 .So x10 59049 is not a very good approximation.11. (a) dxdt 20x f (t, x) , t0 0, x0 1, h 0.01.We use recursive formula,xn 1 xn hf (tn , xn ) xn 0.01 ( 20xn ) 0.8xn .nHere xn (0.8) at tn 0.01n, n 0, 1, 2, .200So estimate for x(2) is x200 (0.8) 4.1495 10 20 .(b) In this case, xn 0 as n , so the numerical solution behaveslike the actual solution x e 20t and the statementx200 x(2) e 40 4.248 10 18 is not too bad.

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONSdxdt x2 f (t, x) , t0 0, x0 1, h 0.2.We use recursive formula,xn 1 xn hf (tn , xn ) xn 0.2x2n , where tn 1 tn hto approximatex1 x0 0.2x20 1 0.2 1.2 at t1 0.2x2 x1 0.2x21 1.2 0.288 1.488 at t2 0.4x3 x2 0.2x22 1.488 0.443 1.9308 at t3 0.6x4 x3 0.2x23 1.9308 0.7456 2.6764 at t4 0.8x5 x4 0.2x24 2.6764 1.4326 4.109 at t5 1So 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 stepand the estimates for x(1) is x10 6.1289.For h 0.01 the estimates for x(1) is 30.3897 using softwareand for h 0.001 the estimates for x(1) is 193.1368 usingsoftware.215. dxdt 1 2x x f (t, x) , t0 0, x0 5, h 0.2.We use recursive formula,¡ xn 1 xn hf (tn , xn ) xn 0.2 1 2xn x2n2 xn 0.2 (xn 1) , where tn 1 tn h to approximate2x1 x0 0.2 (x0 1) 5 7.2 2.2 at t1 0.22x2 x1 0.2 (x1 1) 2.488 at t2 0.42x3 x2 0.2 (x2 1) 2.931 at t3 0.62x4 x3 0.2 (x3 1) 3.676 at t4 0.82x5 x4 0.2 (x4 1) 5.109 at t5 12x6 x5 0.2 (x5 1) 8.486 at t6 1.22x7 x6 0.2 (x6 1) 19.694 at t7 1.42x8 x7 0.2 (x7 1) 89.591 at t8 1.62x9 x8 0.2 (x8 1) 1659.265 at t9 1.82x10 x9 0.2 (x9 1) 551627.57 at t10 2So estimate for x(2) is x10 551627.57 and it seems to tendto infinity.217. dxdt 1 2x x f (t, x) , t0 0, x0 5, h 0.1.We use recursive formula, ¡xn 1 xn hf (tn , xn ) xn 0.1 1 2xn x2n2 xn 0.1 (xn 1) , where tn 1 tn h to approximate2x1 x0 0.1 (x0 1) 1.4 at t1 0.12x2 x1 0.1 (x1 1)

1.3 First-Order Separable Differential Equations 3 1.4 Direction Fields 5 1.5 Euler’s Numerical Method (Optional) 7 1.6 First-Order Linear Differential Equations 10 1.7 Linear First-Order Differential Equations with Constant Coeffi cients and Constant Input 15 1.8 Growth and Decay Problems 20 1.9 Mixture Problems 23

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