1. = 6x = 6x = 6 x = 6 x x 2 y = 6 2 + C = 3x2 + C General solution: y = 3x 2 + C 3. = 7 x = 7 1 x = 7 1 x General solution: y = 7 ln x + C. = e.2x = e.2x = e.2x (u =.2x, du =.2) y = e u 1.2 du = 1 e u du = 1.2.2 eu + C = e.2x + C General solution: y = e.2x + C 7. = x2 - x; y() = = (x2 - x) y = 1 3 x3-1 2 x2 + C Given y() = : 1 3 ()3-1 2 ()2 + C = Particular solution: y = 1 3 x3 9. = -2xe-x2 ; y() = 3 = -2xe -x2 y = -2xe -x2 C = - 1 2 x2 Let u = -x 2, then du = -2x and -2xe -x2 = e u du = e u + c = e -x2 Thus, y = e -x2 + c Given y() = 3: 3 = e + c 3 = 1 + c c = 2 Particular solution: y = e -x2 + 2 + c EXERCISE 6-3 379
11. = 2 1 + x ; y() = = 2 1 + x = 2 1 1 + x 1 = 2 (u = 1 + x, du = ) 1 + x y = 2 1 du = 2 ln u + C = 2 ln 1 + x + C u Given y() = : = 2 ln 1 + C = C Particular solution: y = 2 ln 1 + x + 13. Figure (B). When x = 1, = 1-1 = for any y. When x =, = - 1 = -1 for any y. When x = 2, = 2-1 = 1 for any y; and so on. These facts are consistent with the slope-field in Figure (B); they are not consistent with the slope-field in Figure (A). 1. = x 1 17. = (x - 1) General solution: y = 1 2 x2 - x + C Given y() = -2: 1 2 ()2 - + C = -2 C = -2 Particular solution: y = 1 2 x2 - x - 2 - y - x 19. = 2y 1 y = 2 1 y = 2 so 1 u du = 2 [u = y, du = = ] ln u = 2t + K [K an arbitrary constant] u = e 2t+K = e K e 2t u = Ce 2t [C = e K, C > ] y = Ce 2t 38 CHAPTER 6 INTEGRATION
21. Now, if we set y(t) = Ce 2t, C ANY constant, then y'(t) = 2Ce 2t = 2y(t), So y = Ce 2t satisfies the differential equation where C is any constant. This is the general solution. Note, the differential equation is the model for exponential growth with growth rate 2. = -.y, y() = 1 1 y = -. 1 y = -. 1 u du = -. [u = y, du = = ] ln u = -.x + K u = e -.x+k = e K e -.x y = Ce -.x, C = e K >. So, general solution: y = Ce -.x, C any constant. Given y() = 1: 1 = Ce = C; particular solution: y = 1e -.x 23. = -x 1 x = - 1 x = - 1 x = - ln x = -t + K x = e -t+k = e K e -t = Ce -t, C = e K >. General solution: x = Ce -t, C any constant. 2. = -t = -t = - t General solution: x = - t2 2 + C EXERCISE 6-3 381
27. Figure (A). When y = 1, = 1-1 = for any x. When y = 2, = 1-2 = -1 for any x; and so on. This is consistent with the slope-field in Figure (A); it is not consistent with the slope-field in Figure (B). 29. y = 1 - Ce -x = d [1 - Ce-x ] = Ce -x From the original equation, Ce -x = 1 - y Thus, we have = 1 - y and y = 1 - Ce -x is a solution of the differential equation for any number c. Given y() = : = 1 - Ce = 1 - c c = 1 Particular solution: y = 1 - e -x 31. y 33. - - x - - 3. y = C " x 2 = (C x 2 ) 1/2 = 1 2 (C x2 ) -1/2 "x (-2x) = = - x C " x 2 y Thus, y = C " x 2 satisfies the differential equation = - x y. Setting x = 3, y = 4, gives 4 = C " 9 16 = C 9 C = 2 The solution that passes through (3, 4) is y = 2 " x 2 382 CHAPTER 6 INTEGRATION
37. y = Cx; C = y x = C = y x Therefore y = Cx satisfies the differential equation = y x. Setting x = -8, y = 24 gives 24 = -8C, C = -3 The solution that passes through (-8, 24) is y = -3x 1 39. y = 1 + ce "t = e t e t (multiply numerator and denominator by e t ) + c = (et + c)e t " e t (e t ) ce t (e t + c) 2 = (e t + c) 2 " e t % # y(1 y) = $ # e t ' + c 1 " e t & % & $ e t ( + c = e t ' e t + c c e t + c = ce t (e t + c) 2 Thus = y(1 y) and y = 1 satisfies the differential "t 1 + ce equation. Setting t =, y = -1 gives -1 = 1, -1 c = 1, c = -2 1 + c The solution that passes through (, -1) is 1 y = 1 " 2e "t 41. y = 1,e.8t t 1, y 3, 3 1 43. p = 1e -.x x 3, p 1 1 3 4. N = 1(1 - e -.t ) t 1, N 1 1 1, 47. N = 1 + 999e!.4t t 4, N 1, 1 1 4 EXERCISE 6-3 383
49. = ky(m - y), k, M positive constants. Set f(y) = ky(m - y) = kmy - ky2. This is a quadratic function which opens downward; it has a maximum value. Now f'(y) = km - 2ky Critical value: km - 2ky = y = M 2 f"(y) = -2k <. Thus, f has a maximum value at y = M 2. 1. In 1967: dq = 3.e.2 3.71 In 1999: dq = 6e.13 6.79 The rate of growth in 1999 is almost twice the rate of growth in 1967. 3. da =.8A and A() = 1, is an unlimited growth model. From 4, the amount in the account after t years is: A(t) = 1e.8t.. da = ra, A() = 8, is an unlimited growth model. From 4, A(t) = 8,e rt. Since A(2) = 9,2, we solve 8,e 2r = 9,2 for r. 8e 2r = 9,2 e 2r = 92 8 2r = ln(92/8) ln(92 8) r =.6 2 Thus, A(t) = 8,e.6t. 7. (A) dp = rp, p() = 1 This is an UNLIMITED GROWTH MODEL. From 4, p(x) = 1e rx. Since p() = 77.88, we have 77.88 = 1e r e r =.7788 r = ln(.7788) r = ln(.7788) -. Thus, p(x) = 1e -.x. (B) p(1) = 1e -.(1) = 1e -. $6.6 per unit (C) 384 CHAPTER 6 INTEGRATION
9. (A) dn = k(l - N); N() = This is a LIMITED GROWTH MODEL. From 4, N(t) = L(1 - e -kt ). Since N(1) =.4L, we have.4l = L(1 - e -1k ) 1 - e -1k =.4 e -1k =.6-1k = ln(.6) k = ln(.6).1!1 Thus, N(t) = L(1 - e -.1t ). (B) N() = L[1 - e -.1() ] = L[1 - e -.2 ].22L Approximately 22.% of the possible viewers will have been exposed after days. (C) Solve L(1 - e -.1t ) =.8L for t: 1 - e -.1t =.8 e -.1t =.2 -.1t = ln(.2) t = ln(.2)!.1 31.6 It will take 32 days for 8% of the possible viewers to be exposed. (D) 61. di = -ki, I() = I This is an exponential decay model. From 4, I(x) = I e -kx with k =.942, we have I(x) = I e -.942x To find the depth at which the light is reduced to half of that at the surface, solve, I e -.942x = 1 2 I for x: e -.942x =. -.942x = ln(.) x = ln(.)!.942 74 feet EXERCISE 6-3 38
63. dq = -.4Q, Q() = Q. (A) This is a model for exponential decay. From 4, Q(t) = Q e -.4t With Q = 3, we have Q(t) = 3e -.4t (B) Q(1) = 3e -.4(1) = 3e -.4 2.1. There are approximately 2.1 milliliters in the bo after 1 hours. (C) 3e -.4t = 1 (D) 3 e -.4t = 1 3 -.4t = ln(1/3) ln(1 3) t =!.4 27.47 It will take approximately 27.47 hours for Q to decrease to 1 milliliter. 9 6. Using the exponential decay model, we have = -ky, y() = 1, k > where y = y(t) is the amount of cesium-137 present at time t. From 4, y(t) = 1e -kt Since y(3) = 93.3, we solve 93.3 = 1e -3k for k to find the continuous compound decay rate: 93.3 = 1e -3k e -3k =.933-3k = ln(.933) k = ln(.933)!3 67. From Example 3: Q = Q e -.1238t.23117 Now, the amount of radioactive carbon-14 present is % of the original amount. Thus,.Q = Q e -.1238t or e -.1238t =.. Therefore, -.1238t = ln(.) -2.997 and t 24,2 years. 69. N(k) = 18e -.11(k-1), 1 k 1 Thus, N(6) = 18e -.11(6-1) = 18e -. 14 times and N(1) = 18e -.11(1-1) = 18e -.99 67 times. 386 CHAPTER 6 INTEGRATION