By Nagle, Saff, Snyder

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4 (page 341) 1 2. x 2 + 2 x 2 1 4. x3 e x 6 6. sec θ – sin θ tan θ + θ sin θ + (cos θ)ln(cos θ) 8. c1 x + c2 x ln x + c3 x3 − x 2 x −1 x4 x 10. ∫ g ( x) dx + ∫ x −5 g ( x)dx − ∫ x −2 g ( x)dx 10 15 6 Chapter 6 Review Problems (page 344) 2. a. Lin. indep. b. Lin. indep. c. Lin. dep. 4. a. c1e−3 x + c2 e− x + c3 e x + c4 xe x ( − 2 + 5 ) x + c e( − 2 − 5 ) x b. c1e x + c2 e c. c1e x + c2 cos x + c3 sin x + c4 x cos x + c5 x sin x d. x x 1 c1e x + c2 e − x + c3 e2 x − e x + + 2 2 4 3 x x x x x/ 2 x/ 2 −x / 2 2 6.

Passes through equilibrium at time 2π 5 20 5 36 CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS 4. 896. 8 b = 16: y (t ) = (1 + 8t )e−8t Figure 13 (b = 16) 4 1 b = 20: y (t ) = e −4t − e −16t 3 3 Figure 14 (b = 20) 37 38 CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS 1 + 2 ( −2 + 2 )t 1 − 2 ( −2 − 2 )t 6. 615 . 2 Figure 17 (k = 6) 8. Never 10. 755 m 12. 9 (page 227) 2. M (γ ) = 1 (3 − 2γ 2 ) 2 + 9γ 2 Figure 18 Response curve for Problem 2 5 4. y (t ) = 1 + t sin t 2 16.

Y = cx 2 / 3 Figure 22 47 48 CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS 16. (0, 0) is a stable node. Figure 23 18. (0, 0) is an unstable node. (0, 5) is a stable node. (7, 0) is a stable node. (3, 2) is a saddle point. 4 20. {vy′′ == −v y ; (0, 0) is a center. Figure 25 y′ = v 22. 3 ; (0, 0) is a center. v ′ = − y Figure 26 49 50 CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS y′ = v 24. 3; v ′ = − y + y (0, 0) is a center. (–1, 0) is a saddle point. (1, 0) is a saddle point.

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