36 p0 p1 a sin p1 p2 a sin cos p2 p3 a sin cos2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e−x 2 2 , 3 = −1 + C , C = 4, y = −1 + 4ex = et , et y = 29. (y + cos y ) dy = 4x2 dx, C= 2 /2 /2 + C, /2 2et dt = 2et + C , y = 2 + Ce−t , 1 = 2 + C , C = −1, y = 2 − e−t 4 4 4 y2 π2 π2 + sin y = x3 + C, + sin π = (1)3 + C, = + C, 2 3 2 3 2 3 π2 4 − , 3y 2 + 6 sin y = 8x3 + 3π 2 − 8 2 3 1 dy = (x + 2)ey , e−y dy = (x + 2)dx, −e−y = x2 + 2x + C , −1 = C , dx 2 1 1 1 −e−y = x2 + 2x − 1, e−y = − x2 − 2x + 1, y = − ln 1 − 2x − x2 2 2 2 31. 2(y − 1) dy = (2t + 1) dt, y 2 − 2y = t2 + t + C, 1 + 2 = C, C = 3, y 2 − 2y = t2 + t + 3 32. y + sinh x y = cosh x, µ = e cosh x (sinh x/ cosh x)dx = eln cosh x = cosh x, 1 1 1 (cosh 2x + 1)dx = sinh 2x + x + C = 2 4 2 1 1 1 1 1 y = sinh x + x sech x + C sech x, = C , y = sinh x + x sech x + 2 2 4 2 2 (cosh x)y = 33. (a) cosh2 x dx = dx 1 dy = , ln |y | = ln |x| + C1 , |y | = C |x|1/2 ; y 2x 2 1 1 sinh x cosh x + x + C , 2 2 1 sech x 4 y y = 2.5x 2 y = 2x 2 2 y = x2 by inspection y = 0 is also a solution. (b) 2 = C (1)2 , C = 2, x = 2y 2 y=0 -2 x 2 y = – 0.5x 2 -2 y = –3x 2 x2 y2 = − + C1 , y = ± C 2 − x2 34. (a) y dy = −x dx, 2 2 √ (b) y = 25 − x2 3 y y = –1.5x 2 y = √ 9 – x2 y = √ 2.25 – x 2 y = √ 0.25 – x 2 x -3 3 y = – √ 1 – x2 y = – √ 4 – x2 -3 y = – √ 6.25 – x 2 Exercise Set 10.1 35. 346 x dx dy =− 2 , y x +4 1 ln |y | = − ln(x2 + 4) + C1 , 2 C y=√ x2 + 4 36. y + 2y = 3et , µ = e2 = e2t , d y e2t = 3e3t , ye2t = e3t + C, dt y = et + Ce−2t 100 C=2 C=1 C=0 1.5 -2 C=2 C=1 C=0 -2 dt 2 C = –1 C = –2 2 C = –1 C = –2 -100 -1 37. (1 − y 2 ) dy = x2 dx, y− 38. x3 y3 = + C1 , x3 + y 3 − 3y = C 3 3 1 +y y yey 2 /2 dy = dx, ln |y | + y2 = x + C1 , 2 = ±eC1 ex = Cex y y 3 2 x -5 5 x -2 -3 2 -2 41. dy x2 = xey , e−y dy = x dx, −e−y = + C, x = 2 when y = 0 so −1 = 2 + C, C = −3, x2 + 2e−y = 6 dx 2 42. 3x2 dy = , 2y dy = 3x2 dx, y 2 = x3 + C, 1 = 1 + C, C = 0, dx 2y 2 y 2 = x3 , y = x3/2 passes through (1, 1). 0 1.6 0 43. dy = rate in − rate out, where y is the amount of salt at time t, dt 1 dy 1 dy y = (4)(2) − (2) = 8 − y, so + y = 8 and y (0) = 25. dt 50 25 dt 25 µ=e (1/25)dt = et/25 , et/25 y = 8et/25 dt = 200et/25 + C , y = 200 + Ce−t/25 , 25 = 200 + C , C = −175, (a) y = 200 − 175e−t/25 oz (b) when t = 25, y = 200 − 175e−1 ≈ 136 oz 347 44. Chapter 10 y 1 dy 1 dy = (5)(10) − (10) = 50 − y, so + y = 50 and y (0) = 0. dt 200 20 dt 20 µ=e 1 20 dt = et/20 , et/20 y = 50et/20 dt = 1000et/20 + C , y = 1000 + Ce−t/20 , 0 = 1000 + C , C = −1000; (a) y = 1000 − 1000e−t/20 lb (b) when t = 30, y = 1000 − 1000e−1.5 ≈ 777 lb 45. The volume V of the (polluted) water is V (t) = 500 + (20 − 10)t = 500 + 10t; if y (t) is the number of pounds of particulate matter in the water, dt y 1 dy 1 dy = 0 − 10 = − y, + y = 0; µ = e 50+t = 50 + t; then y (0) = 50, and dt V 50 + t dt 50 + t d [(50 + t)y ] = 0, (50 + t)y = C, 2500 = 50y (0) = C, y (t) = 2500/(50 + t). dt The tank reaches the point of overflowing when V = 500 + 10t = 1000, t = 50 min, so y = 2500/(50 + 50) = 25 lb. 46. The volume of the lake (in gallons) is V = 246πr2 h = 246π (15)2 3 = 166, 050π . Let y (t) denote y dy y = 0 − 103 =− lb/h and the number of pounds of mercury salts at time t, then dt V 166.05π dt t dy =− , ln y = − + C1 , y = Ce−t/(166.05π) , and y0 = 10−5 V = 1.6605π lb; y 166.05π 166.05π C = y (0) = y0 = 1.6605π , y = 1.6605πe−t/(166.05π) . t 1 2 3 4 5 6 7 8 9 10 11 12 y(t) 5.2066 5.1967 5.1867 5.1768 5.1669 5.1570 5.1471 5.1372 5.1274 5.1176 5.1078 5.0980 c dv d gm ct/m + v = −g, µ = e(c/m) dt = ect/m , v ect/m = −gect/m , vect/m = − e + C, dt m dt c gm gm gm gm −ct/m gm + Ce−ct/m , but v0 = v (0) = − + C, C = v0 + ,v = − + v0 + e v=− c c c c c mg with vτ and −ct/m with −gt/vτ in (23). (b) Replace c vτ (c) From part (b), s(t) = C − vτ t − (v0 + vτ ) e−gt/vτ ; g vτ vτ vτ s0 = s(0) = C − (v0 + vτ ) , C = s0 +(v0 + vτ ) , s(t) = s0 − vτ t + (v0 + vτ ) 1 − e−gt/vτ g g g 47. (a) 48. Given m = 240, g = 32, vτ = mg/c: with a closed parachute vτ = 120 so c = 64, and with an open parachute vτ = 24, c = 320. (a) Let t denote time elapsed in seconds after the moment of the drop. From Exercise 47(b), while the parachute is closed v (t) = e−gt/vτ (v0 + vτ ) − vτ = e−32t/120 (0 + 120) − 120 = 120 e−4t/15 − 1 and thus v (25) = 120 e−20/3 − 1 ≈ −119.85, so the parachutist is falling at a speed of 119.85 ft/s 120 120 1 − e−4t/15 , when the parachute opens. From Exercise 47(c), s(t) = s0 − 120t + 32 s(25) = 10000 − 120 · 25 + 450 1 − e−20/3 ≈ 7449.43 ft. (b) If t denotes time elapsed after the parachute opens, then, by Exercise 47(c), 24 (−119.85 + 24) 1 − e−32t/24 = 0, with the solution (Newton’s s(t) = 7449.43 − 24t + 32 Method) t = 310.42 s, so the sky diver is in the air for about 25 + 310 = 335 s. Exercise Set 10.1 49. 348 R V (t) dI + I= , µ = e(R/L) dt...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online