4 374 28 a suppose uk vk converges then so does uk

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Unformatted text preview: L L IeRt/L = I (0) + (a) I (t) = t 1 L dt = eRt/L , d Rt/L V (t) Rt/L (e e I) = , dt L V (u)eRu/L du, I (t) = I (0)e−Rt/L + 0 1 −5t/2 e 4 t 12e5u/2 du = 0 6 −5t/2 5u/2 e e 5 t = 0 1 −Rt/L e L t V (u)eRu/L du. 0 6 1 − e−5t/2 A. 5 6 (b) lim I (t) = A t→+∞ 5 50. From Exercise 49 and Endpaper Table #42, 1 I (t) = 15e−2t + e−2t 3 −2t = 15e t 3e2u sin u du = 15e−2t + e−2t 0 e2u (2 sin u − cos u) 5 t 0 1 1 + (2 sin t − cos t) + e−2t . 5 5 ck dv = − g, v = −c ln(m0 − kt) − gt + C ; v = 0 when t = 0 so 0 = −c ln m0 + C , dt m0 − kt m0 − gt. C = c ln m0 , v = c ln m0 − c ln(m0 − kt) − gt = c ln m0 − kt (b) m0 − kt = 0.2m0 when t = 100 so m0 v = 2500 ln − 9.8(100) = 2500 ln 5 − 980 ≈ 3044 m/s. 0.2m0 51. (a) dv dx dv dv dv dv = = v so m = mv . dt dx dt dx dt dx m mv 2 dv = −dx, ln(kv + mg ) = −x + C ; v = v0 when x = 0 so (b) kv 2 + mg 2k 52. (a) By the chain rule, C= 2 m m m kv0 + mg m 2 2 ln(kv0 + mg ), ln(kv 2 + mg ) = −x + ln(kv0 + mg ), x = ln 2 . 2k 2k 2k 2k kv + mg (c) x = xmax when v = 0 so xmax = 2 3.56 × 10−3 (7.3 × 10−6 )(988)2 + (3.56 × 10−3 )(9.8) m kv0 + mg ln = ln ≈ 1298 m 2k mg 2(7.3 × 10−6 ) (3.56 × 10−3 )(9.8) √ √ dh π = −0.025 h, √ dh = −0.025dt, 2π h = −0.025t + C ; h = 4 when 53. (a) A(h) = π (1)2 = π, π dt h √ √ 0.025 t, h ≈ (2 − 0.003979t)2 . t = 0, so 4π = C, 2π h = −0.025t + 4π, h = 2 − 2π (b) h = 0 when t ≈ 2/0.003979 ≈ 502.6 s ≈ 8.4 min. 54. (a) A(h) = 6 2 4 − (h − 2)2 = 12 4h − h2 , √ √ dh = −0.025 h, 12 4 − h dh = −0.025dt, dt −8(4 − h)3/2 = −0.025t + C ; h = 4 when t = 0 so C = 0, 12 4h − h2 (4 − h)3/2 = (0.025/8)t, 4 − h = (0.025/8)2/3 t2/3 , h ≈ 4 − 0.021375t2/3 ft (b) h = 0 when t = 8 (4 − 0)3/2 = 2560 s ≈ 42.7 min 0.025 2√4 − (h − 2)2 h−2 2 h 349 55. 56. Chapter 10 1 dv 1 1 = −0.04v 2 , 2 dv = −0.04dt, − = −0.04t + C ; v = 50 when t = 0 so − = C, dt v v 50 1 50 dx dx 50 1 cm/s. But v = so = , x = 25 ln(2t + 1) + C1 ; x = 0 − = −0.04t − , v = v 50 2t + 1 dt dt 2t + 1 when t = 0 so C1 = 0, x = 25 ln(2t + 1) cm. √ √ 1 dv = −0.02 v, √ dv = −0.02dt, 2 v = −0.02t + C ; v = 9 when t = 0 so 6 = C , dt v √ dx dx so = (3 − 0.01t)2 , 2 v = −0.02t + 6, v = (3 − 0.01t)2 cm/s. But v = dt dt 100 100 (3 − 0.01t)3 + C1 ; x = 0 when t = 0 so C1 = 900, x = 900 − (3 − 0.01t)3 cm. x=− 3 3 2 dy = − sin x + e−x , y (0) = 1. dx 57. Differentiate to get 58. h(y ) dy = h(y (x))y (x) dx; since h(y ) dy = g (x) it follows that dx h(y ) dy = g (x) dx. EXERCISE SET 10.2 y 1. y 2. 4 3 2 1 y(0) = 2 2 3 2 y 3. 4 1 y(0) = 1 x x 1 4. 2 3 4 1 dy + y = 1, µ = e dx = ex , dx d [yex ] = ex , dx yex = ex + C, y = 1 + Ce−x dy − 2y = −x, µ = e−2 dx 1 y = (2x + 1) + Ce2x 4 dx 2 3 4 y(0) = –1 -1 y 5. 10 y(1) = 1 y(–1) = 0 x -2 (a) −1 = 1 + C, C = −2, y = 1 − 2e−x (b) 1 = 1 + C, C = 0, y = 1 (c) 2 = 1 + C, C = 1, y = 1 + e−x 6. 5 x = e−2x , y(0) = –1 -10 d 1 y e−2x = −xe−2x , ye−2x = (2x + 1)e−2x + C, dx 4 1 1 (2x + 1) + e2x−2 4 4 5 1 (b) −1 = 1/4 + C, C = −5/4, y = (2x + 1) − e2x 4 4 1 1 (c) 0 = −1/4 + Ce−2 , C = e2 /4, y = (2x + 1) + e2x+2 4 4 (a) 1 = 3/4 + Ce2 , C = 1/4e2 , y = 2 Exercise Set 10.2 7. 350 lim y = 1 8. x→+∞ lim y = x→+∞ +∞ if y0 ≥ 1/4 −∞, if y0 < 1/4 9. (a) IV, since the slope is positive for x > 0 and negative for x < 0. (b) VI, since the slope is positive for y > 0 and negative for y < 0. (c) V, since the slope is always positive. (d) II, since the slope changes sign when crossing the lines y = ±1. (e) I, since the slope can be positive or negative in each quadrant but is not periodic. (f ) III, since the slope is periodic in both x and y . 11. (a) y0 = 1, yn+1 = yn + (xn + yn )(0.2) = (xn + 6yn )/5 n xn 1 0.2 1.20 2 3 4 5 0.4 1.48 0.6 1.86 0.8 2.35 1.0 2.98 xn 0 0.2 y ( xn) abs. error perc. error 1 0 0 1.24 0.04 0.4 1.58 0.10 0.6 2.04 0.19 0.8 2.65 0.30 1.0 3.44 0.46 3 7 9 11 13 yn d y e−x = xe−x , dx ye−x = −(x + 1)e−x + C, 1 = −1 + C, C = 2, y = −(x + 1) + 2ex (b) y − y = x, µ = e−x , 0 0 1 y (c) 3 x 0.2 0.4 0.6 0.8 1 12. h = 0.1, yn+1 = (xn + 11yn )/10 n xn yn 0 0 1.00 1 0.1 1.10 2 3 4 5 6 7 8 0.2 1.22 0.3 1.36 0.4 1.53 0.5 1.72 0.6 1.94 0.7 2.20 0.8 2.49 9 0.9 10 1.0 2.82 3.19 In Exercise 11, y (1) ≈ 2.98; in Exercise 12, y (1) ≈ 3.19; the true solution is y (1) ≈ 3.44; so the absolute errors are approximately 0.46 and 0.25 respectively. 13. y0 = 1, yn+1 = yn + √ yn /2 y 9 n xn yn 0 0 1 1 0.5 1.50 2 1 2.11 3 1.5 2.84 4 2 3.68 5 2.5 4.64 6 3 5.72 7 3.5 6.91 8 4 8.23 x 1 2 3 4 351 Chapter 10 y 2 14. y0 = 1, yn+1 = yn + (xn − yn )/4 n xn yn 0 0 1 1 0.25 0.75 2 2 3 4 5 6 7 8 0.50 0.67 0.75 0.68 1.00 0.75 1.25 0.86 1.50 0.99 1.75 1.12 2.00 1.24 1.5 1 0.5 x 0.5 15. y0 = 1, yn+1 = yn + n tn yn 0 0 1 1 0.5 1.42 2 1 1.92 3 1.5 2 y 1 sin yn 2 1.5 2.39 1 3 4 2 2.73 t 3 16. y0 = 0, yn+1 = yn + e...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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