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### nagle_differential_equations_ISM_Part55

Course: MA 221, Spring 2011
School: Stevens Institute of...
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7 2 Chapter 1 K 1 K 2 0 2 4 6 8 10 12 Figure 7K: The graph of y (t) in Problem 32. 1 0 1 2 3 4 5 6 7 Figure 7L: The graph of g (t) in Problem 42. 2 1 0 2.5 5.0 7.5 K 1 Figure 7M: The graph of y (t) in Problem 22. 1.0 0.5 0 2 4 6 8 10 12 K K 0.5 1.0 Figure 7N: The graph of y (t) in Problem 24. 266 CHAPTER 8: Series Solutions of Dierential Equations EXERCISES 8.1: Introduction: The...

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7 2 Chapter 1 K 1 K 2 0 2 4 6 8 10 12 Figure 7K: The graph of y (t) in Problem 32. 1 0 1 2 3 4 5 6 7 Figure 7L: The graph of g (t) in Problem 42. 2 1 0 2.5 5.0 7.5 K 1 Figure 7M: The graph of y (t) in Problem 22. 1.0 0.5 0 2 4 6 8 10 12 K K 0.5 1.0 Figure 7N: The graph of y (t) in Problem 24. 266 CHAPTER 8: Series Solutions of Dierential Equations EXERCISES 8.1: Introduction: The Taylor Polynomial Approximation 2. 2 + 4x + 8x2 + 4. 1 2 x2 + 1 6 1 6 x3 1 120 1 20 x5 + 6. x 8. 1 x3 + x5 + x4 + sin 1 2 x2 + (cos 1)(sin 1) 24 10. (a) p3 (x) = 1x ++ 24 1 24 (b) 3 = 4! (2 )5 (c) x2 x3 + 8 16 4 2 1 11 = 5 0.00823 5 24 2 (3/2) 3 2 1 1 p3 = 0.00260 3 2 384 (d) See Fig. 8A on page 276 12. The dierential equation implies that the functions y (x), y (x), and y (x) exist and continuous. Furthermore, y (x) can be obtained by dierentiating the other terms: y = py p y qy q y + g . Since p, q , and g have derivatives of all orders, subsequent dierentiations display the fact that, in turn, y , y (4) , y (5) , etc. all exist. 14. (a) t + t2 t3 + 2 6 (b) For r = 1, y (t) = 1 (1/2)t2 4t4 + ; For r = 1, y (t) = 1 + (1/2)t2 (49/12)t4 + (c) For small t in part (b), the hard spring recoils but the soft spring extends. 16. 1 t2 t4 + + 2 4 267 Chapter 8 EXERCISES 8.2: 2. (, ) 4. [2, 4] 6. (3, 1) 8. (a) (b) 11 , 22 Power Series and Analytic Functions 11 , 22 (c) (, ) (d) (, ) (e) (, ) (f ) 11 , 22 (n + 2)2 2n+3 + (x 1)n (n + 3)! 2n+1 2 3 x3 + 2 15 x5 + 10. n=0 12. x 14. 1 16. 1 1 2 x+ 1 4 x2 1 24 x3 + 18. k=0 (1)k x 2k = cos x (2k )! n(n 1)an xn2 20. n=2 22. k=0 (1)k x3 x5 2k+1 x =x + (2k + 1)(2k + 1)! 18 600 (k 2)(k 3)ak2 xk 24. k=4 26. k=4 ak 3 k x k 268 Exercises 8.3 30. n=0 (1)n (x 1)n (1)n+1 n x n 1 2 (x 1) 1 8 (x 1)2 + 1 16 (x 1)3 5 128 (x 1)4 + 32. n=1 34. 1 + 36. (a) Always true (b) Sometimes false (c) Always true (d) Always true 38. n=0 x4 x6 (1)n 2n+2 x = x2 + n+1 2 3 Power Series Solutions to Linear Dierential Equations EXERCISES 8.3: 2. 0 4. 1, 0 6. 1 8. No singular points 10. x 1 and x = 2 12. y = a0 x2 x3 1+x+ + + 2! 3! = a0 n=0 xn = a0 ex n! 14. a0 1 16. a0 1 18. a0 1 + x2 x3 + + a1 x + 2 6 x2 x3 x3 + + a1 x + x2 + + 2 3 2 x2 x3 + + a1 x + + 6 27 x 2k x2k+1 + a1 (1)k = a0 cos x + a1 sin x (2k )! (2k + 1)! k=0 269 = a0 ex + (a1 a0 ) xex 20. a0 k=0 (1)k Chapter 8 22. a3k+2 = 0, k = 0, 1, . . . 1 4 (3k 5)(3k 2) 3k x a0 1 + (1)k (3k )! k=1 +a1 x+ k=1 (1)k 2 5 (3k 4)(3k 1) 3k+1 x (3k + 1)! 24. a0 (1)k (2k 3)2 (2k 5)2 32 2k x2 x4 + + x + a1 x 1 2 24 k=3 (2k )! x2 x3 x4 + + + 2 2 3 x2 x3 x4 + 2 3 8 5x2 x3 2 6 26. x + 28. 1 30. 1 + x + 32. (a) If a1 = 0, then y (x) is an even function; (d) a0 = 0, a1 > 0 36. 3 9t2 + t3 + t4 + 2 Equations with Analytic Coecients EXERCISES 8.4: 2. Innite 4. 2 6. 1 8. a0 1 2(x + 1) + 3(x + 1)2 10 (x + 1)3 + 3 1 1 1 1 10. a0 1 (x 2)2 (x 2)3 + + a1 (x 2) + (x 2)2 (x 2)3 + 4 24 4 12 1 2 7 12. a0 1 + (x + 1)2 + (x + 1)3 + + a1 (x + 1) + 2(x + 1)2 + (x + 1)3 + 2 3 3 14. 1 + x + x2 + 16. t + 270 5x3 + 6 t3 t4 t5 + + + 3 12 24
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Stevens Institute of Technology - MA - 221
Exercises 8.618. 1 + x 22. a0 1 24. a0 1 1 + x 2 2 22 1 x 24 24+ x2 x3 x2 + + x + + 2 2 6 3x2 x3 + + a1 x + + 2 622 x2 326. a0 1 x 28. a0 1 + 30. 1 + a1x3 x + + 6x2 + 2 x2 + 2x3 + + a1 (x + ) + 6t2 t3 t4 + + 2 2 4 Cauchy-Euler (Equidimension
Stevens Institute of Technology - MA - 221
Chapter 8FIGURES105K2K1012Figure 8A: The graphs of f (x) = 1/(2 x) and its Taylor polynomial p3 (x)276CHAPTER 9: Matrix Methods for Linear SystemsEXERCISES 9.1: Introduction 2. x y x1 = 01 1 0 1 1 0 0 x y 1 1 0 0 x1 x2 = 1 4. x 3 0 x4 0 1
Stevens Institute of Technology - MA - 221
Exercises 9.46. x1 (t) x2 (t) x1 (t) = 0 1 x1 (t) x2 (t) + 0 t2 0 1 0 010x1 (t) 8. x2 (t) = 0 0 1 x2 (t) + 0 x3 (t) 1 1 0 x3 (t) cos t 10. x1 (t) = 2x1 (t) + x2 (t) + tet ; x2 (t) = x1 (t) + 3x2 (t) + et 12. x1 (t) = x2 (t) + t + 3; x2 (t) = x3 (t) t
Stevens Institute of Technology - MA - 221
Chapter 9 20. x1 (t) = cos t cos 3 t; x2 (t) = cos t + cos 3 t 22. I1 (t) = 16 5 4 5 e2t 1 5 4 5 e8t + 2;I2 (t) = 4e2t e8t + 2; I3 (t) = e2t + e8tEXERCISES 9.7: Nonhomogeneous Linear Systems 2. c1 e3t 2e3t 1 1 + c2 et 2et e4t e4t+t 2 2 sin t4. c1+
Stevens Institute of Technology - MA - 221
CHAPTER 10: Partial Dierential EquationsEXERCISES 10.2: 2. y = Method of Separation of Variables(e10 1) ex + (1 e2 ) e5x e10 e24. y = 2 sin 3x 6. No solution 8. y = ex1 + xex1 10. n = (2n 1)2 and yn = cn cos 4 2n 1 x , where n = 1, 2, 3, . . . and cn s
Stevens Institute of Technology - MA - 221
Chapter 1024. u(x, y ) =n=1An eny sin nx,2 where An = 0f (x) sin nx dx296CHAPTER 11: Eigenvalue Problems and Sturm-Liouville EquationsEXERCISES 11.2: Eigenvalues and Eigenfunctions2. sin x cos x + x + 1 4. ce2x (sin 2x + cos 2x) 6. 2ex + ex x 8.
Stevens Institute of Technology - MA - 221
Exercises 11.6 5s (e es ) (5e6x + e5x ) , 30e6 + 6 5x (e ex ) (5e6s + e5s ) , 30e6 + 6 s(x ) , 0 s x xs 0sx10. G(x, s) =xs112. G(x, s) = x(s ) , x4 3 x y= 12 14. G(x, s) = y= s, x, 0sxxs4 3 x x4 12 sin s sin(x 2) , 0 s x sin 2 16. G(x, s) = sin x s
Stevens Institute of Technology - MA - 221
Chapter 1214. (3, 3) is an unstable saddle point; (2, 2) is an asymptotically stable spiral point. See Fig. 12E on page 310. 16. (0, 0) is an unstable saddle point; (4, 2) is an asymptotically stable spiral point. See Fig. 12F on page 310. EXERCISES 12.4
Stevens Institute of Technology - MA - 221
Figures4y20x4v20xK2 K4Figure 12G: Potential and Phase plane diagrams in Problem 8, Section 12.42yK5K2205xK2xFigure 12H: Potential and Phase plane diagrams in Problem 10, Section 12.4311Chapter 1286y42K 10 3 2 112x3xK1
Stevens Institute of Technology - MA - 221
Chapter 1242K4K2024K2 K4Figure 12S: Phase plane diagram in Problem 6, Review Section4y2K2K1012x32xK2Figure 12T: Potential and Phase plane diagrams in Problem 8, Review Section4 2K4K2024xK2 K4Figure 12U: Phase plane diagram
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kuruvila (lk5992) HW 14 opyrchal (11113) This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A spring stretches 3.3 cm when a 14 g object is hung from i
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kuruvila (lk5992) HW-02 gokce (P111S11) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points The velocity of the transverse waves produced by an earthqu
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kuruvila (lk5992) HW-03 gokce (P111S11) This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points A particle has an initial horizontal velocity
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NJIT - CS - 101
%Livia Kuruvila problem 2.11; x1= input('enter starting x1'); x2= input('enter starting x2'); y1= input('enter starting y1'); y2= input('enter starting y2'); d= sqrt ([x1-x2].^2 + [y1-y2].^2);
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%Livia Kuruvila problem 2.12; x1= input('enter starting x1'); x2= input('enter starting x2'); y1= input('enter starting y1'); y2= input('enter starting y2'); z1= input('enter starting z1'); z2= input('enter starting z2'); d= sqrt ([x1-x2].^2 + [y1-y2].^2
NJIT - CS - 101
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