Download e-book for iPad: Answers, .: to selected exercises for Introductory DE with by Abell M.L., Braselton J.P.

By Abell M.L., Braselton J.P.

ISBN-10: 0123846641

ISBN-13: 9780123846648

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Partsol, {x, 0, 15}] 5 4 3 2 1 2 29. 5 1 numericalsols /. 5 1 2 3 4 5 6 7 Answers, Hints, and Solutions to Selected Exercises 30. 75 Ϫ1 Chapter 2 Review Exercises 1. y = 1 5 √ 3 25 t 6 + 125C 3. y = sinh−1 1 1 2 cosh(x) + 2 C 5. y4 dy = e5t dt ⇒ 15 y5 = 15 e5t + C ⇒ y5 = e5t + C → y = (e5t + C)1/5 √ −2t 7. y(t) = e± −e +2 C 1 1 y 6 y 9. − 20 cos(10 t) + 18 cos(4 t) − 37 e cos (6 y) − 37 e sin (6 y) = C 37 38 CH A P T E R 2: First-Order Equations 11. The equation (y − t) dt + (t + y) dy = 0 is homogeneous of degree 1.

Y = 0 is unstable; y = 4 is stable. 0 1 2 3 4 5 6 5 6 t 21 5. y = 0 is stable; y = 2 is unstable; y = 4 is stable. 5 1 2 3 4 t Ϫ1 7. With y(t) = y0 ekt , y(4) = y0 e4k = 3y0 ⇒ ek = 31/4 so y(t) = y0 · 3t/4 . 86 days. 9. y(t) = y0 ekt ; y(1700) = y0 e1700k = 12 y0 ⇒ ek = (1/2)1/1700 . 98% of y0 ). Answers, Hints, and Solutions to Selected Exercises 11. First we solve y = ky(1000 − y), y(0) = 250: dy = ky(1000 − y) dt 1 dy = k dt y(1000 − y) 1 1000 1 1 + dy = k dt y 1000 − y ln |y| − ln |1000 − y| = 1000kt + C y = Ce1000kt 1000 − y y= 1000Ce1000kt .

7. Using Newton’s Law of Cooling, T (t) = (T0 − Ts )e−kt + Ts , observe that T (t) = (90 − 70)e−kt + 70 = 20e−kt + 70. Because T (3) = 20e−3k + 70 = 80, 20e−3k = 10 ⇒ e−3k = 1/2 ⇒ e−k = (1/2)1/3 so T (t) = 20 (1/2)(t/3) + 70. 3◦ F. 9. T (t) = −150 2 3 t/30 + 300; T (−30) = 75◦ F 11. Using Newton’s Law of Cooling, T (t) = (T0 − Ts )e−kt + Ts = (200 − 68)e−kt + 68. T (2) = 132e−2k + 68 = 170 ⇒ 132e−2k = 102 ⇒ e−2k = 17/22 ⇒ e−k = (17/22)1/2 . Thus, T (t) = 132 (17/22)t/2 + 68. 7 min. 12. This problem is a great class experiment if time allows.

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Answers, .: to selected exercises for Introductory DE with BVP 3ed. by Abell M.L., Braselton J.P.


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