# Uk 1 3 5 2k 1 1 by the ratio test lim 0 converges k

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Unformatted text preview: dt 3 dt dt r = −kt + 4, 3 = −k + 4, k = 1, r = 4 − t m. Chapter 10 Supplementary Exercises 358 10. Assume the tank contains y (t) oz of salt at time t. Then y0 = 0 and for 0 &lt; t &lt; 15, y dy = 5 · 10 − 10 = (50 − y/100) oz/min, with solution y = 5000 + Ce−t/100 . But y (0) = 0 so dt 1000 C = −5000, y = 5000(1 − e−t/100 ) for 0 ≤ t ≤ 15, and y (15) = 5000(1 − e−0.15 ). For 15 &lt; t &lt; 30, y dy = 0− 5, y = C1 e−t/200 , C1 e−0.075 = y (15) = 5000(1 − e−0.15 ), C1 = 5000(e0.075 − e−0.075 ), dt 1000 y = 5000(e0.075 − e−0.075 )e−t/100 , y (30) = 5000(e0.075 − e−0.075 )e−0.3 ≈ 556.13 oz. 11. (a) Assume the air contains y (t) ft3 of carbon monoxide at time t. Then y0 = 0 and for y d 1 t/12000 dy = 0.04(0.1) − (0.1) = 1/250 − y/12000, y et/12000 = e , t &gt; 0, dt 1200 dt 250 yet/12000 = 48et/12000 + C, y (0) = 0, C = −48; y = 48(1 − e−t/12000 ). Thus the percentage y 100 = 4(1 − e−t/12000 ) percent. of carbon monoxide is P = 1200 (b) 0.012 = 4(1 − e−t/12000 ), t = 35.95 min 12. 13. 14. dy = dx, tan−1 y = x + C, π/4 = C ; y = tan(x + π/4) y2 + 1 1 1 + y5 y dy = dx 1 1 , − y −4 + ln |y | = ln |x| + C ; − = C, y −4 + 4 ln(x/y ) = 1 x 4 4 2 dy + y = 4x, µ = e dx x (2/x)dx = x2 , d y x2 = 4x3 , yx2 = x4 + C, y = x2 + Cx−2 , dx 2 = y (1) = 1 + C, C = 1, y = x2 + 1/x2 15. 16. π π dy 1 = 4 sec2 2x dx, − = 2 tan 2x + C, −1 = 2 tan 2 + C = 2 tan + C = 2 + C, C = −3, y2 y 8 4 1 y= 3 − 2 tan 2x y2 dy dy = dx, = dx, − 5y + 6 (y − 3)(y − 2) 1 y−3 1 − dy = dx, ln = x + C1 , y−3 y−2 y−2 3 − 2Cex ln 2 − 3 3 ln 2 − 6 − (2 ln 2 − 6)ex y−3 = Cex ; y = ln 2 if x = 0, so C = ; y= = x y−2 ln 2 − 2 1 − ce ln 2 − 2 − (ln 2 − 3)ex 17. (a) µ = e− ye−x = d y e−x = xe−x sin 3x, dx 3 2 3 1 e−x cos 3x + − x + xe−x sin 3x dx = − x − 10 50 10 25 dx = e−x , 1 = y (0) = − 53 3 + C, C = , y= 50 50 y (c) 4 x -10 -2 -2 − 3 3 x− 10 50 cos 3x + − e−x sin 3x + C ; 2 1 x+ 10 25 sin 3x + 53 x e 50 359 Chapter 10 ln 2 ≈ 0.00012182; T2 = 5730 + 40 = 5770, k2 ≈ 0.00012013. T1 1 y 1 = 684.5, 595.7; t2 = − ln(y/y0 ) = 694.1, 604.1; in With y/y0 = 0.92, 0.93, t1 = − ln k1 y0 k2 1988 the shroud was at most 695 years old, which places its creation in or after the year 1293. 19. (a) Let T1 = 5730 − 40 = 5690, k1 = (b) If the true half-life is T with decay rate k and solution y (t) = y0 e−kt , and if the half-life is taken to be T1 = T (1 + r/100) with decay rate k1 and solution y1 (t) = y0 e−k1 t , then k 100k r ln 2 ln 2 100k = = ; k − k1 = k − =k and the = k1 = T1 T (1 + r/100) 1 + r/100 100 + r 100 + r 100 + r y1 − y = 100 e(k−k1 )t − 1 = 100 ektr/(100+r) − 1 percent. y percentage error is given by 100 20. (a) yn+1 = yn + 0.1(1 + 5tn − yn ), y0 = 5 n tn yn 0 1 5.00 1 1.1 5.10 2 3 4 5 6 7 8 9 1.2 5.24 1.3 5.42 1.4 5.62 1.5 5.86 1.6 6.13 1.7 6.41 1.8 6.72 1.9 7.05 10 2 7.39 (b) The true solution is y (t) = 5t − 4 + 4e1−t , so the percentage errors are given by tn yn y ( tn) abs. error rel. error (%) 1 5.00 5.00 0.00 0.00 1.1 5.10 5.12 0.02 0.38 1.2 5.24 5.27 0.03 0.66 1.3 5.42 5.46 0.05 0.87 1.4 5.62 5.68 0.06 1.00 1.5 5.86 5.93 0.06 1.08 1.6 6.13 6.20 0.07 1.12 1.7 6.41 6.49 0.07 1.13 1.8 6.72 6.80 0.08 1.11 1.9 7.05 7.13 0.08 1.07 2 7.39 7.47 0.08 1.03 21. (b) y = C1 ex + C2 e−x (c) 1 = y (0) = C1 + C2 , 1 = y (0) = C1 − C2 ; C2 = 0, C1 = 1, y = ex 22. (a) 2ydy = dx, y 2 = x + C ; if y (0) = 1 then C = 1, y 2 = x + 1, y = √ C = 1, y 2 = x + 1, y = − x + 1. y 1 x 1 -1 (b) x + 1; if y (0) = −1 then y 1 -1 √ x -1 1 -1 dy 1 = −2x dx, − = −x2 + C, −1 = C, y = 1/(x2 + 1) y2 y y 1 x -1 1 1805 1805 , 25 = y (1) = , −kt 19 + 76e 19 + 76e−k y0 L 5 95 , t ≈ 7.77 yr. k ≈ 0.3567; when 0.8L = y (t) = , 19 + 76e−kt = y0 = 19 + 76e−kt 4 4 23. (a) Use (15) in Section 10.3 with y0 = 19, L = 95: y (t) = Chapter 10 Supplementary Exercises (b) From (13), 360 y dy =k 1− y , y (0) = 19. dt 95 24. (a) y0 = y (0) = c1 , v0 = y (0) = c2 k , c2 = m m v0 , y = y0 cos k (b) l = 0.5, k/m = g/l = 9.8/0.5 = 19.6, √ √ 1 y = − cos( 19.6 t) + 0.25 √ sin( 19.6 t) 19.6 k t + v0 m m sin k k t m 1.1 0 3.1 -1.1 √ √ √ 0.25 19.6 sin 19.6 t + 0.25 cos 19.6 t = 0 when tan 19.6 t = − √ , so 19.6 √ √ √ 0.25 19.6 , cos 19.6 t = √ , sin 19.6 t = ± √ 19.6625 19.6625 √ √ 0.25 0.25 19.6 19.6625 √ +√ =√ |y (t)| = √ 19.6625 19.6 19.6625 19.6 (c) v = y (t) = √ ≈ 1.0016 m is the maximum displacement. 25. y = y0 cos k t, T = 2π m 2πt m , y = y0 cos k T 2π 2πt y0 sin has maximum magnitude 2π |y0 |/T and occurs when T T 2πt/T = nπ + π/2, y = y0 cos(nπ + π/2) = 0. (a) v = y (t) = − 4π 2 2πt has maximum magnitude 4π 2 |y0 |/T 2 and occurs when y cos 20 T T 2πt/T = jπ, y = y0 cos jπ = ±y0 . (b) a = h (t) = − 26. (a) In t years the interest will be compounded nt times at an interest rate of r/n each time. The value at the end of 1 interval is P +...
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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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