Chapter 9: DIFFERENTIAL EQUATIONS. Work and Answers

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9.1 Modeling with Differential Equations 1.  = 2 3  + −2 ⇒ 0 = 2 3  − 2−2. To show that  is a solution of the differential equation, we will substitute the e... xpressions for  and 0 in the left-hand side of the equation and show that the left-hand side is equal to the right-hand side. LHS = 0 + 2 = 2 3  − 2−2 + 2 2 3  + −2 = 2 3  − 2−2 + 4 3  + 2−2 = 6 3  = 2 = RHS 2.  = −cos  −  ⇒  = −(−sin) + cos(−1) − 1 =  sin − cos − 1. LHS =    = (sin − cos − 1) = 2 sin −  cos −  = 2 sin +  = RHS, so  is a solution of the differential equation. Also () = − cos  −  = −(−1) −  =  −  = 0, so the initial condition is satisfied. 3. (a)  =  ⇒ 0 =  ⇒ 00 = 2. Substituting these expressions into the differential equation 200 + 0 −  = 0, we get 22 +  −  = 0 ⇒ (22 +  − 1) = 0 ⇒ (2 − 1)( + 1) = 0 [since  is never zero] ⇒  = 1 2 or −1. (b) Let 1 = 1 2 and 2 = −1, so we need to show that every member of the family of functions  = 2 + − is a solution of the differential equation 200 + 0 −  = 0.  = 2 + − ⇒ 0 = 1 2 2 − − ⇒ 00 = 1 4 2 + −. LHS = 200 + 0 −  = 2 1 4 2 + − +  1 2 2 − − − (2 + −) = 1 2 2 + 2− + 1 2 2 − − − 2 − − =  1 2  + 1 2  − 2 + (2 −  − )− = 0 = RHS 4. (a)  = cos  ⇒ 0 = − sin ⇒ 00 = −2 cos. Substituting these expressions into the differential equation 400 = −25, we get 4(−2 cos) = −25(cos) ⇒ (25 − 42)cos = 0 [for all ] ⇒ 25 − 42 = 0 ⇒ 2 = 25 4 ⇒  = ± 5 2 . (b)  = sin +  cos ⇒ 0 =  cos −  sin ⇒ 00 = −2 sin − 2 cos. The given differential equation 400 = −25 is equivalent to 400 + 25 = 0. Thus, LHS = 400 + 25 = 4(−2 sin − 2 cos) + 25(sin +  cos) = −42 sin − 42 cos + 25sin + 25 cos = (25 − 42)sin + (25 − 42) cos = 0 since 2 = 25 4 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. 803 FOR INSTRUCTOR USE ONLYNOT FOR SALE 804 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 5. (a)  = sin ⇒ 0 = cos ⇒ 00 = −sin. LHS = 00 +  = −sin + sin = 0 6= sin, so  = sin is not a solution of the differential equation. (b)  = cos ⇒ 0 = −sin ⇒ 00 = −cos. LHS = 00 +  = −cos + cos = 0 6= sin, so  = cos is not a solution of the differential equation. (c)  = 1 2sin ⇒ 0 = 1 2(cos + sin) ⇒ 00 = 1 2(−sin + cos + cos). LHS = 00 +  = 1 2(−sin + 2cos) + 1 2 sin = cos 6= sin, so  = 1 2 sin is not a solution of the differential equation. (d)  = − 1 2 cos ⇒ 0 = − 1 2(−sin + cos) ⇒ 00 = − 1 2(−cos − sin − sin). LHS = 00 +  = − 1 2(−cos − 2sin) + − 1 2 cos = sin = RHS, so  = − 1 2 cos is a solution of the differential equation. 6. (a)  = ln +   ⇒ 0 =  · (1) − (ln + ) 2 = 1 − ln −  2 . LHS = 20 +  = 2 · 1 − ln −  2 +  · ln +   = 1 − ln −  + ln +  = 1 = RHS, so  is a solution of the differential equation. (b) A few notes about the graph of  = (ln + ): (1) There is a vertical asymptote of  = 0. (2) There is a horizontal asymptote of  = 0. (3)  = 0 ⇒ ln +  = 0 ⇒  = −, so there is an -intercept at −. (4) 0 = 0 ⇒ ln = 1 −  ⇒  = 1−, so there is a local maximum at  = 1−. (c) (1) = 2 ⇒ 2 = ln 1 +  1 ⇒ 2 = , so the solution is  = ln + 2  [shown in part (b)]. (d) (2) = 1 ⇒ 1 = ln 2 +  2 ⇒ 2 + ln 2 +  ⇒  = 2 − ln 2, so the solution is  = ln + 2 − ln 2  [shown in part (b)]. 7. (a) Since the derivative 0 = −2 is always negative (or 0 if  = 0), the function  must be decreasing (or equal to 0) on any interval on which it is defined. (b)  = 1  +  ⇒ 0 = −( +1)2 . LHS = 0 = −( +1)2 = −  +1  2 = −2 = RHS (c)  = 0 is a solution of 0 = −2 that is not a member of the family in part (b). (d) If () = 1  +  , then (0) = 0 +1  = 1 . Since (0) = 05, 1 = 12 ⇒  = 2, so  =  + 2 1 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.1 MODELING WITH DIFFERENTIAL EQUATIONS ¤ 805 8. (a) If  is close to 0, then 3 is close to 0, and hence, 0 is close to 0. Thus, the graph of  must have a tangent line that is nearly horizontal. If  is large, then 3 is large, and the graph of  must have a tangent line that is nearly vertical. (In both cases, we assume reasonable values for .) (b)  = ( − 2)−12 ⇒ 0 = ( − 2)−32. RHS = 3 = [ − 2−12]3 = ( − 2)−32 = 0 = LHS (c) When  is close to 0, 0 is also close to 0. As  gets larger, so does |0|. (d) (0) = ( − 0)−12 = 1√ and (0) = 2 ⇒ √ = 1 2 ⇒  = 1 4, so  =  1 4 − 2−12. 9. (a)   = 121 − 4200  . Now    0 ⇒ 1 − 4200   0 [assuming that   0] ⇒ 4200   1 ⇒   4200 ⇒ the population is increasing for 0    4200. (b)    0 ⇒   4200 (c)   = 0 ⇒  = 4200 or  = 0 10. (a)   = −[2 − (1 + ) + ] = −( − )( − 1), so   = 0 ⇔  = 0, , or 1. (b) With 0    1,  = −( − )( − 1)  0 ⇔   0 or     1, so  is increasing on (−∞0) and (1). (c) With 0    1,  = −( − )( − 1)  0 ⇔ 0     or   1, so  is decreasing on (0 ) and (1 ∞). 11. (a) This function is increasing and also decreasing. But  = ( − 1)2 ≥ 0 for all , implying that the graph of the solution of the differential equation cannot be decreasing on any interval. (b) When  = 1,  = 0, but the graph does not have a horizontal tangent line. 12. The graph for this exercise is shown in the figure at the right. A. 0 = 1 +   1 for points in the first quadrant, but we can see that 0  0 for some points in the first quadrant. B. 0 = −2 = 0 when  = 0, but we can see that 0  0 for  = 0. Thus, equations A and B are incorrect, so the correct equation is C. C. 0 = 1 − 2 seems reasonable since: (1) When  = 0, 0 could be 1. (2) When   0, 0 could be greater than 1. (3) Solving 0 = 1 − 2 for  gives us  = 1 − 0 2 . If 0 takes on small negative values, then as  → ∞,  → 0+, as shown in the figure. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 806 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 13. (a) 0 = 1 + 2 + 2 ≥ 1 and 0 → ∞ as  → ∞. The only curve satisfying these conditions is labeled III. (b) 0 = −2−2  0 if   0 and 0  0 if   0. The only curve with negative tangent slopes when   0 and positive tangent slopes when   0 is labeled I. (c) 0 = 1 1 + 2+2  0 and 0 → 0 as  → ∞. The only curve satisfying these conditions is labeled IV. (d) 0 = sin() cos() = 0 if  = 0, which is the solution graph labeled II. 14. (a) The coffee cools most quickly as soon as it is removed from the heat source. The rate of cooling decreases toward 0 since the coffee approaches room temperature. (b)   = ( − ), where  is a proportionality constant,  is the temperature of the coffee, and  is the room temperature. The initial condition is (0) = 95◦C. The answer and the model support each other because as  approaches ,  approaches 0, so the model seems appropriate. (c) 15. (a)  increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills associated with learning a skill. As  increases, we would expect  to remain positive, but decrease. This is because as time progresses, the only points left to learn are the more difficult ones. (b)   = ( − ) is always positive, so the level of performance  is increasing. As  gets close to ,  gets close to 0; that is, the performance levels off, as explained in part (a). (c) 16. (a)   = (∞ − ). Assuming ∞  , we have   0 and   0 for all . (b) 17. If () = 1 − −1− =  − −1− for   0, where   0,   0, 0    1, and  = (1 − ), then   = 0 − −1− ·   (−1−) = −−1− · (−)(1 − )− = (1− )  −1− =  ( − ). The equation for  indicates that as  increases,  approaches . The differential equation indicates that as  increases, the rate of increase of  decreases steadily and approaches 0 as  approaches . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.2 DIRECTION FIELDS AND EULER’S METHOD ¤ 807 9.2 Direction Fields and Euler's Method 1. (a) (b) It appears that the constant functions  = 05 and  = 15 are equilibrium solutions. Note that these two values of  satisfy the given differential equation 0 = cos. 2. (a) (b) It appears that the constant functions  = 0,  = 2, and  = 4 are equilibrium solutions. Note that these three values of  satisfy the given differential equation 0 = tan 1 2. 3. 0 = 2 − . The slopes at each point are independent of , so the slopes are the same along each line parallel to the -axis. Thus, III is the direction field for this equation. Note that for  = 2, 0 = 0. 4. 0 = (2 − ) = 0 on the lines  = 0 and  = 2. Direction field I satisfies these conditions. 5. 0 =  +  − 1 = 0 on the line  = − + 1. Direction field IV satisfies this condition. Notice also that on the line  = − we have 0 = −1, which is true in IV. 6. 0 = sinsin = 0 on the lines  = 0 and  = 0, and 0  0 for 0    , 0    . Direction field II satisfies these conditions. 7. 8. 9.   0 = 1 2  0 0 0 0 1 05 0 2 1 0 −3 −15 0 −2 −1 Note that for  = 0, 0 = 0. The three solution curves sketched go through (0 0), (0 1), and (0 −1). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 808 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 10.   0 =  −  + 1 −1 0 0 −1 −1 1 0 0 1 0 1 0 0 2 −1 0 −1 2 0 −2 3 1 0 2 1 1 1 Note that 0 = 0 for  =  + 1 and that 0 = 1 for  = . For any constant value of , 0 decreases as  increases and 0 increases as  decreases. The three solution curves sketched go through (0 0), (01), and (0 −1). 11.   0 =  − 2 −2 −2 2 −2 2 6 2 2 −2 2 −2 −6 Note that 0 = 0 for any point on the line  = 2. The slopes are positive to the left of the line and negative to the right of the line. The solution curve in the graph passes through (10). 12.   0 =  − 2 2 3 2 −2 −3 2 ±2 0 −4 0 0 0 2 2 0 0 =  − 2 = ( − ), so 0 = 0 for  = 0 and  = . The slopes are positive only in the regions in quadrants I and III that are bounded by  = 0 and  = . The solution curve in the graph passes through (0 1). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.2 DIRECTION FIELDS AND EULER’S METHOD ¤ 809 13.   0 =  +  0 ±2 ±2 1 ±2 ±4 −3 ±2 ∓4 Note that 0 = ( + 1) = 0 for any point on  = 0 or on  = −1. The slopes are positive when the factors  and  + 1 have the same sign and negative when they have opposite signs. The solution curve in the graph passes through (0 1). 14.   0 =  + 2 −2 ±1 −1 −2 ±2 2 2 ±1 3 0 ±2 4 0 0 0 Note that 0 =  + 2 = 0 only on the parabola  = −2. The slopes are positive “outside”  = −2 and negative “inside”  = −2. The solution curve in the graph passes through (00). 15. 0 = 2 − 1 22 and (0) = 1. In Maple, use the following commands to obtain a similar figure. with(DETools): ODE:=diff(y(x),x)=xˆ2*y(x)-(1/2)*y(x)ˆ2; ivs:=[y(0)=1]; DEplot({ODE},y(x),x=-3..2,y=0..4,ivs,linecolor=black); 16. 0 = cos( + ) and (0) = 1. In Maple, use the following commands to obtain a similar figure. with(DETools): ODE:=diff(y(x),x)=cos(x+y(x)); ivs:=[y(0)=1]; DEplot({ODE},y(x),x=-1.5*Pi..1.5*Pi,y=-1.5*Pi..1.5*Pi, ivs,linecolor=black); °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 810 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 17. The direction field is for the differential equation 0 = 3 − 4.  = lim →∞ () exists for −2 ≤  ≤ 2;  = ±2 for  = ±2 and  = 0 for −2    2. For other values of ,  does not exist. 18. Note that when () = 0 on the graph in the text, we have 0 = () = 0; so we get horizontal segments at  = ±1, ±2. We get segments with negative slopes only for 1  ||  2. All other segments have positive slope. For the limiting behavior of solutions: • If (0)  2, then lim →∞  = ∞ and lim →−∞  = 2. • If 1  (0)  2, then lim →∞  = 1 and lim →−∞  = 2. • If −1  (0)  1, then lim →∞  = 1 and lim →−∞  = −1. • If −2  (0)  −1, then lim →∞  = −2 and lim →−∞  = −1. • If   −2, then lim →∞  = −2 and lim →−∞  = −∞. 19. (a) 0 = ( ) =  and (0) = 1 ⇒ 0 = 0, 0 = 1. (i)  = 04 and 1 = 0 + (0 0) ⇒ 1 = 1 + 04 · 1 = 14. 1 = 0 +  = 0 + 04 = 04, so 1 =  (04) = 14. (ii)  = 02 ⇒ 1 = 02 and 2 = 04, so we need to find 2. 1 = 0 + (0 0) = 1 + 020 = 1 + 02 · 1 = 12, 2 = 1 + (1 1) = 12 + 021 = 12 + 02 · 12 = 144. (iii)  = 01 ⇒ 4 = 04, so we need to find 4. 1 = 0 + (0 0) = 1 + 010 = 1 + 01 · 1 = 11, 2 = 1 + (1 1) = 11 + 011 = 11 + 01 · 11 = 121, 3 = 2 + (2 2) = 121 + 012 = 121 + 01 · 121 = 1331, 4 = 3 + (3 3) = 1331 + 013 = 1331 + 01 · 1331 = 14641. (b) We see that the estimates are underestimates since they are all below the graph of  = . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.2 DIRECTION FIELDS AND EULER’S METHOD ¤ 811 (c) (i) For  = 04: (exact value) − (approximate value) = 04 − 14 ≈ 00918 (ii) For  = 02: (exact value) − (approximate value) = 04 − 144 ≈ 00518 (iii) For  = 01: (exact value) − (approximate value) = 04 − 14641 ≈ 00277 Each time the step size is halved, the error estimate also appears to be halved (approximately). 20. As  increases, the slopes decrease and all of the estimates are above the true values. Thus, all of the estimates are overestimates. 21.  = 05, 0 = 1, 0 = 0, and ( ) =  − 2. Note that 1 = 0 +  = 1 + 05 = 15, 2 = 2, and 3 = 25. 1 = 0 + (0 0) = 0 + 05(1 0) = 05[0 − 2(1)] = −1. 2 = 1 + (1 1) = −1 + 05(15 −1) = −1 + 05[−1 − 2(15)] = −3. 3 = 2 + (2 2) = −3 + 05(2 −3) = −3 + 05[−3 − 2(2)] = −65. 4 = 3 + (3 3) = −65 + 05(25 −65) = −65 + 05[−65 − 2(25)] = −1225. 22.  = 02, 0 = 0, 0 = 1, and ( ) = 2 − 1 22. Note that 1 = 0 +  = 0 + 02 = 02, 2 = 04, 3 = 06, 4 = 08, and 5 = 1. 1 = 0 + (0 0) = 1 + 02(0 1) = 1 + 0202(1) − 1 2(1)2 = 1 + 02− 1 2  = 09. 2 = 1 + (1 1) = 09 + 02(02 09) = 09 + 02(02)2(09) − 1 2(09)2 = 08262. 3 = 2 + (2 2) = 08262 + 02(0408262) = 08262 + 02(04)2(08262) − 1 2(08262)2 = 0784377756. 4 = 3 + (3 3) = 0784377756 + 02(060784377756) ≈ 0779328108. 5 = 4 + (4 4) ≈ 0779328108 + 02(080779328108) ≈ 0818346876. Thus, (1) ≈ 08183. 23.  = 01, 0 = 0, 0 = 1, and ( ) =  + . Note that 1 = 0 +  = 0 + 01 = 01, 2 = 02, 3 = 03, and 4 = 04. 1 = 0 + (0 0) = 1 + 01(0 1) = 1 + 01[1 + (0)(1)] = 11. 2 = 1 + (1 1) = 11 + 01(01 11) = 11 + 01[11 + (01)(11)] = 1221 3 = 2 + (2 2) = 1221 + 01(021221) = 1221 + 01[1221 + (02)(1221)] = 136752. 4 = 3 + (3 3) = 136752 + 01(03136752) = 136752 + 01[136752 + (03)(136752)] = 15452976. 5 = 4 + (4 4) = 15452976 + 01(0415452976) = 15452976 + 01[15452976 + (04)(15452976)] = 1761639264. Thus, (05) ≈ 17616. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 812 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 24. (a)  = 02, 0 = 0, 0 = 0, and ( ) = cos( + ). Note that 1 = 0 +  = 0 + 02 = 02, 2 = 04, and 3 = 06. 1 = 0 + (0 0) = 0 + 02(00) = 02cos(0 + 0) = 02(1) = 02. 2 = 1 + (1 1) = 02 + 02(02 02) = 02 + 02cos(04) ≈ 03842121988. 3 = 2 + (2 2) ≈ 03842 + 02(0403842) ≈ 05258011763. Thus, (06) ≈ 05258. (b) Now use  = 01. For 1 ≤  ≤ 6,  = 0. 1 = 0 + (0 0) = 0 + 01cos(0 + 0) = 01(1) = 01. 2 = 1 + (1 1) = 01 + 01cos(02) ≈ 01980. 3 = 2 + (2 2) ≈ 01980 + 01cos(03980) ≈ 02902. 4 = 3 + (3 3) ≈ 02902 + 01cos(05902) ≈ 03733. 5 = 4 + (4 4) ≈ 03733 + 01cos(07733) ≈ 04448. 6 = 5 + (5 5) ≈ 04448 + 01cos(09448) ≈ 05034. Thus, (06) ≈ 05034. 25. (a)  + 32 = 62 ⇒ 0 = 62 − 32. Store this expression in Y1 and use the following simple program to evaluate (1) for each part, using H =  = 1 and N = 1 for part (i), H = 01 and N = 10 for part (ii), and so forth.  → H: 0 → X: 3 → Y: For(I, 1, N): Y + H × Y1 → Y: X + H → X: End(loop): Display Y. [To see all iterations, include this statement in the loop.] (i) H = 1, N = 1 ⇒ (1) = 3 (ii) H = 01, N = 10 ⇒ (1) ≈ 23928 (iii) H = 001, N = 100 ⇒ (1) ≈ 23701 (iv) H = 0001 N = 1000 ⇒ (1) ≈ 23681 (b)  = 2 + −3 ⇒ 0 = −32−3 LHS = 0 + 32 = −32−3 + 322 + −3 = −32−3 + 62 + 32−3 = 62 = RHS (0) = 2 + −0 = 2 + 1 = 3 (c) The exact value of (1) is 2 + −13 = 2 + −1. (i) For  = 1: (exact value) − (approximate value) = 2 + −1 − 3 ≈ −06321 (ii) For  = 01: (exact value) − (approximate value) = 2 + −1 − 23928 ≈ −00249 (iii) For  = 001: (exact value) − (approximate value) = 2 + −1 − 23701 ≈ −00022 (iv) For  = 0001: (exact value) − (approximate value) = 2 + −1 − 23681 ≈ −00002 In (ii)–(iv), it seems that when the step size is divided by 10, the error estimate is also divided by 10 (approximately). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.2 DIRECTION FIELDS AND EULER’S METHOD ¤ 813 26. (a) We use the program from the solution to Exercise 25 with Y1 = 3 − 3, H = 001, and N = 02−01 0 = 200. With (0 0) = (01), we get (2) ≈ 19000. (b) Notice from the graph that (2) ≈ 19, which serves as a check on our calculation in part (a). 27. (a)    + 1   = () becomes 50 + 1 005  = 60 or 0 + 4 = 12. (b) From the graph, it appears that the limiting value of the charge  is about 3. (c) If 0 = 0, then 4 = 12 ⇒  = 3 is an equilibrium solution. (d) (e) 0 + 4 = 12 ⇒ 0 = 12 − 4. Now (0) = 0, so 0 = 0 and 0 = 0. 1 = 0 + (0 0) = 0 + 01(12 − 4 · 0) = 12 2 = 1 + (1 1) = 12 + 01(12 − 4 · 12) = 192 3 = 2 + (2 2) = 192 + 01(12 − 4 · 192) = 2352 4 = 3 + (3 3) = 2352 + 01(12 − 4 · 2352) = 26112 5 = 4 + (4 4) = 26112 + 01(12 − 4 · 26112) = 276672 Thus, 5 = (05) ≈ 277 C. 28. (a) From Exercise 9.1.14, we have  = ( − ). We are given that  = 20◦C and  = −1◦Cmin when  = 70◦C. Thus, −1 = (70 − 20) ⇒  = − 50 1 and the differential equation becomes  = − 50 1 ( − 20). (b) The limiting value of the temperature is 20◦C; that is, the temperature of the room. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 814 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS (c) From part (a),  = − 50 1 ( − 20). With 0 = 0, 0 = 95, and  = 2 min, we get 1 = 0 + (0 0) = 95 + 2− 50 1 (95 − 20) = 92 2 = 1 + (1 1) = 92 + 2− 50 1 (92 − 20) = 8912 3 = 2 + (2 2) = 8912 + 2− 50 1 (8912 − 20) = 863552 4 = 3 + (3 3) = 863552 + 2− 50 1 (863552 − 20) = 83700992 5 = 4 + (4 4) = 83700992 + 2− 50 1 (83700992 − 20) = 8115295232 Thus, (10) ≈ 8115◦C. 9.3 Separable Equations 1.   = 322 ⇒  2 = 32  [ 6= 0] ⇒  −2  =  32  ⇒ −−1 = 3 +  ⇒ −1  = 3 +  ⇒  = −1 3 +  .  = 0 is also a solution. 2.   = √ ⇒ √ =   [ 6= 0] ⇒  −12  =    ⇒ 212 = 1 22 +  ⇒ √ = 1 42 + 1 2 ⇒  =  1 42 + 2, where  = 1 2.  = 0 is also a solution. 3. 0 = 2 + 1 ⇒    = 2 + 1 ⇒   = 2+ 1  [ 6= 0] ⇒    =   + 1  ⇒ 1 2 2 = 1 22 + ln|| +  ⇒ 2 = 2 + 2 ln|| + 2 ⇒  = ±2 + 2 ln|| + , where  = 2. 4. 0 +  = 0 ⇒   = − ⇒ −  = −  ⇒  −  =  −  ⇒ −− = − 1 22 +  ⇒ − = 1 22 −  ⇒ − = ln 1 22 −  ⇒  = −ln 1 22 −  5. ( − 1)0 = 2 + cos ⇒ ( − 1)   = 2 + cos ⇒ ( − 1) = (2 + cos) ⇒  ( − 1) =  (2 + cos ) ⇒  −  = 2 + sin + . We cannot solve explicitly for . 6.   = 1 + 4 2 + 42 ⇒   = 1 + 4 2( + 4) ⇒ ( + 4)  = 1 +24  ⇒  ( + 4)  =  (−2 + 2) ⇒ 1 2 2 + 1 5 5 = −1  + 1 3 3 + . We cannot solve explicitly for . 7.   = sec 2 ⇒  cos  = −2  ⇒   cos   =  −2  ⇒  sin + cos = − 1 2−2 +  [by parts]. We cannot solve explicitly for . 8.   = 2√1 + 2 ln ⇒ ln 2  = √1 + 2  ⇒  ln 2  =  (1 + 2)12  ⇒ − ln  − 1  = 1 3 (1 + 2)32 +  [by parts]. We cannot solve explicitly for . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.3 SEPARABLE EQUATIONS ¤ 815 9.   = 2 −  + 2 − 1 = (2 − 1) + 1(2 − 1) = ( + 1)(2 − 1) ⇒  + 1 1  = (2 − 1) ⇒   + 1 1  =  (2 − 1) ⇒ ln| + 1| = 1 33 −  +  ⇒ | + 1| = 33−+ ⇒  + 1 = ±33− ⇒  = 33− − 1, where  = ±. Since  = −1 is also a solution,  can equal 0, and hence,  can be any real number. 10.   +  +  = 0 ⇒   = − ⇒  −  = −    ⇒ −− = − +  ⇒ − =  −  ⇒ 1  =  −  ⇒  = 1  −  ⇒  = ln −1   ⇒  = −ln −  11.   =  ⇒ −  =   ⇒  −  =    ⇒ −− = 1 22 + . (0) = 0 ⇒ −−0 = 1 2(0)2 +  ⇒  = −1, so −− = 1 22 − 1 ⇒ − = − 1 22 + 1 ⇒ − = ln1 − 1 22 ⇒  = −ln1 − 1 22. 12.   = sin  ⇒   = sin  ⇒    =  sin  ⇒ 1 22 = −cos + sin +  [by parts]. (0) = −1 ⇒ 1 2(−1)2 = −0cos 0 + sin0 +  ⇒  = 1 2 , so 1 22 = −cos + sin + 1 2 ⇒ 2 = −2cos + 2 sin + 1 ⇒  = −√−2cos + 2 sin + 1 since (0) = −1  0. 13.   = 2 + sec2  2 , (0) = −5.  2  =  2 + sec2   ⇒ 2 = 2 + tan + , where [(0)]2 = 02 + tan 0 +  ⇒  = (−5)2 = 25. Therefore, 2 = 2 + tan + 25, so  = ±√2 + tan + 25. Since (0) = −5  0, we must have  = −√2 + tan + 25. 14.  + 32√2 + 1   = 0 ⇒ 32√2 + 1   = − ⇒ 32  = √−2+ 1  ⇒  32  =  −(2 + 1)−12  ⇒ 3 = −(2 + 1)12 + . (0) = 1 ⇒ 13 = −(02 + 1)12 +  ⇒  = 2, so 3 = −(2 + 1)12 + 2 ⇒  = (2 − √2 + 1)13. 15. ln = 1 + 3 + 2  0, (1) = 1.  ln  =   + 3 + 2   ⇒ 1 22 ln −  1 2  [use parts with  = ln,  = ] = 1 22 + 1 3(3 + 2)32 ⇒ 1 22 ln − 1 42 +  = 1 22 + 1 3(3 + 2)32. Now (1) = 1 ⇒ 0 − 1 4 +  = 1 2 + 1 3(4)32 ⇒  = 1 2 + 8 3 + 1 4 = 41 12 , so 1 2 2 ln − 1 42 + 41 12 = 1 22 + 1 3(3 + 2)32. We do not solve explicitly for . 16.   = √ ⇒ √ = √  ⇒   −12  =  12  ⇒ 2 12 = 2 332 + . (1) = 2 ⇒ 2√2 = 2 3 +  ⇒  = 2√2 − 2 3 , so 2 12 = 2 332 + 2√2 − 2 3 ⇒ √ = 1 332 + √2 − 1 3 ⇒  =  1 332 + √2 − 1 32. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 816 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 17. 0 tan =  + , 0    2 ⇒   =  +  tan ⇒   +  = cot  [ +  6= 0] ⇒   +  =  cos sin  ⇒ ln| + | = ln|sin| +  ⇒ | + | = ln|sin |+ = ln|sin | ·  =  |sin| ⇒  +  =  sin, where  = ±. (In our derivation,  was nonzero, but we can restore the excluded case  = − by allowing  to be zero.) (3) =  ⇒  +  =  sin3  ⇒ 2 =  √23 ⇒  = √43. Thus,  +  = √43 sin and so  = √43 sin − . 18.   = 2 ln ⇒  2 =  ln  ⇒   2 =   ln  ⇒ −1 = ln −    [by parts with  = ln,  =  ] ⇒ − 1  = ln −  +  ⇒  =  − 1ln −  . (1) = −1 ⇒ −1 = 1  −  ln 1 −  ⇒  −  = 1 ⇒  =  + 1. Thus,  =  − ln1 −  − 1. 19.   =   ⇒   =   ⇒    =    ⇒ 1 2 2 = 1 2 2 + . (0) = 2 ⇒ 1 2(2)2 = 1 2(0)2 +  ⇒  = 2, so 1 2 2 = 1 2 2 + 2 ⇒ 2 = 2 + 4 ⇒  = √2 + 4 since (0) = 2  0. 20.  0() =  () −  ⇒   =  −  ⇒   = ( − 1) ⇒  − 1 =   [ 6= 1] ⇒   − 1 =    ⇒ ln| − 1| = 1 2 2 + . (0) = 2 ⇒ ln|2 − 1| = 1 2(0)2 +  ⇒  = 0, so ln| − 1| = 1 2 2 ⇒ | − 1| = 22 ⇒  − 1 = 22 [since (0) = 2] ⇒  = 22 + 1. 21.  =  +  ⇒   () =   ( + ) ⇒   = 1 +   , but   =  +  = , so   = 1 +  ⇒  1 +  =  [ 6= −1] ⇒  1 +  =   ⇒ ln|1 + | =  +  ⇒ |1 + | = + ⇒ 1 +  = ± ⇒  = ± − 1 ⇒  +  = ± − 1 ⇒  =  −  − 1, where  = ± 6= 0. If  = −1, then −1 =  +  ⇒  = − − 1, which is just  =  −  − 1 with  = 0. Thus, the general solution is  =  −  − 1, where  ∈ R. 22. 0 =  +  ⇒ 0 =  +  ⇒   =  + . Also,  =  ⇒  =  ⇒   =    + , so  +  =    +  ⇒   =   [ 6= 0] ⇒    =    ⇒ −− = ln|| +  ⇒ − = −ln|| −  ⇒ − = ln(−ln|| − ) ⇒  = −ln(−ln|| − ) ⇒  = −ln(−ln|| − ). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.3 SEPARABLE EQUATIONS ¤ 817 23. (a) 0 = 2 1 − 2 ⇒   = 2 1 − 2 ⇒ 1 − 2 = 2  ⇒  1 − 2 =  2  ⇒ sin−1  = 2 +  for − 2 ≤ 2 +  ≤ 2 . (b) (0) = 0 ⇒ sin−1 0 = 02 +  ⇒  = 0, so sin−1  = 2 and  = sin(2) for −2 ≤  ≤ 2. (c) For 1 − 2 to be a real number, we must have −1 ≤  ≤ 1; that is, −1 ≤ (0) ≤ 1. Thus, the initial-value problem 0 = 2 1 − 2, (0) = 2 does not have a solution. 24. −0 + cos = 0 ⇔  −  = −  cos  ⇔ −− = −sin + 1 ⇔  = −ln(sin + ). The solution is periodic, with period 2. Note that for   1, the domain of the solution is R, but for −1   ≤ 1 it is only defined on the intervals where sin +   0, and it is meaningless for  ≤ −1, since then sin +  ≤ 0, and the logarithm is undefined. For −1    1, the solution curve consists of concave-up pieces separated by intervals on which the solution is not defined (where sin +  ≤ 0). For  = 1, the solution curve consists of concave-up pieces separated by vertical asymptotes at the points where sin +  = 0 ⇔ sin = −1. For   1, the curve is continuous, and as  increases, the graph moves downward, and the amplitude of the oscillations decreases. 25.   = sin sin , (0) = 2 . So  sin  =  sin  ⇔ −cos = −cos +  ⇔ cos = cos − . From the initial condition, we need cos 2 = cos 0 −  ⇒ 0 = 1 −  ⇒  = 1, so the solution is cos = cos − 1. Note that we cannot take cos−1 of both sides, since that would unnecessarily restrict the solution to the case where −1 ≤ cos − 1 ⇔ 0 ≤ cos, as cos−1 is defined only on [−11]. Instead we plot the graph using Maple’s plots[implicitplot] or Mathematica’s Plot[Evaluate[· · · ]]. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 818 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 26.   =  √2 + 1  ⇔    =   √2 + 1. We use parts on the LHS with  = ,  =  , and on the RHS we use the substitution  = 2 + 1, so  = 2 . The equation becomes  −    = 1 2  √  ⇔ ( − 1) = 1 3(2 + 1)32 + , so we see that the curves are symmetric about the -axis. Every point ( ) in the plane lies on one of the curves, namely the one for which  = ( − 1) − 1 3(2 + 1)32. For example, along the -axis,  = ( − 1) − 1 3 , so the origin lies on the curve with  = − 4 3 . We use Maple’s plots[implicitplot] command or Plot[Evaluate[· · · ]] in Mathematica to plot the solution curves for various values of . It seems that the transitional values of  are − 4 3 and − 1 3 For   − 4 3 , the graph consists of left and right branches. At  = − 4 3 , the two branches become connected at the origin, and as  increases, the graph splits into top and bottom branches. At  = − 1 3 , the bottom half disappears. As  increases further, the graph moves upward, but doesn’t change shape much. 27. (a), (c) (b) 0 = 2 ⇒   = 2 ⇒  −2  =   ⇒ −−1 =  +  ⇒ 1  = − −  ⇒  = 1  − , where  = −.  = 0 is also a solution. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.3 SEPARABLE EQUATIONS ¤ 819 28. (a), (c) (b) 0 =  ⇒   =  ⇒    =    ⇒ ln|| = 1 22 +  ⇒ || = 22 +  = 22 ⇒  = 22, where  = ±. Taking  = 0 gives us the solution  = 0. 29. The curves 2 + 22 = 2 form a family of ellipses with major axis on the -axis. Differentiating gives   (2 + 22) =   (2) ⇒ 2 + 40 = 0 ⇒ 40 = −2 ⇒ 0 = −2. Thus, the slope of the tangent line at any point ( ) on one of the ellipses is 0 = − 2 , so the orthogonal trajectories must satisfy 0 = 2  ⇔   = 2  ⇔   = 2 =   ⇔    = 2   ⇔ ln|| = 2 ln|| + 1 ⇔ ln|| = ln||2 + 1 ⇔ || = ln 2+1 ⇔  = ± 2 · 1 = 2. This is a family of parabolas. 30. The curves 2 = 3 form a family of power functions. Differentiating gives   (2) =   (3) ⇒ 20 = 32 ⇒ 0 = 32 2 = 3(23)2 2 = 3 2 , the slope of the tangent line at ( ) on one of the curves. Thus, the orthogonal trajectories must satisfy 0 = −2 3 ⇔   = − 2 3 ⇔ 3  = −2  ⇔  3  =  −2  ⇔ 3 22 = −2 + 1 ⇔ 32 = −22 + 2 ⇔ 22 + 32 = . This is a family of ellipses. 31. The curves  =  form a family of hyperbolas with asymptotes  = 0 and  = 0. Differentiating gives   () =     ⇒ 0 = −2 ⇒ 0 = − 2 [since  =  ⇒  = ] ⇒ 0 = − . Thus, the slope of the tangent line at any point ( ) on one of the hyperbolas is 0 = −, so the orthogonal trajectories must satisfy 0 =  ⇔   =   ⇔   =   ⇔    =    ⇔ 1 2 2 = 1 2 2 + 1 ⇔ 2 = 2 + 2 ⇔ 2 − 2 = . This is a family of hyperbolas with asymptotes  = ±. 32. The curves  = 1( + ) form a family of hyperbolas with asymptotes  = − and  = 0. Differentiating gives  () =    +1   ⇒ 0 = −( +1)2 ⇒ 0 = −2 [since  = 1( + )]. Thus, the slope of the tangent °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 820 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS line at any point ( ) on one of the hyperbolas is 0 = −2, so the orthogonal trajectories must satisfy 0 = 12 ⇔   = 1 2 ⇔ 2  =  ⇔  2  =   ⇔ 1 33 =  + 1 ⇔ 3 = 3 +  ⇔  = (3 + )13. This is a family of cube root functions with vertical tangents on the -axis [ = 0]. 33. () = 2 + 2[ − ()] ⇒ 0() =  − () [by FTC 1] ⇒   = (1 − ) ⇒  1 −  =    ⇒ −ln|1 − | = 1 22 + . Letting  = 2 in the original integral equation gives us (2) = 2 + 0 = 2. Thus, −ln|1 − 2| = 1 2(2)2 +  ⇒ 0 = 2 +  ⇒  = −2 Thus, −ln|1 − | = 1 22 − 2 ⇒ ln|1 − | = 2 − 1 22 ⇒ |1 − | = 2 − 22 ⇒ 1 −  = ±2 − 22 ⇒  = 1 + 2 − 22 [(2) = 2]. 34. () = 2 + 1  (),   0 ⇒ 0() = 1() ⇒   =  1 ⇒    =  1  ⇒ 1 2 2 = ln +  [  0]. Letting  = 1 in the original integral equation gives us (1) = 2 + 0 = 2. Thus, 1 2(2)2 = ln 1 +  ⇒  = 2. 1 22 = ln + 2 ⇒ 2 = 2 ln + 4 [ 0] ⇒  = √2ln + 4. 35. () = 4 + 0 2() ⇒ 0() = 2() ⇒   = 2√ ⇒  √ =  2  ⇒ 2√ = 2 + . Letting  = 0 in the original integral equation gives us (0) = 4 + 0 = 4. Thus, 2√4 = 02 +  ⇒  = 4. 2√ = 2 + 4 ⇒ √ = 1 22 + 2 ⇒  =  1 22 + 22. 36. (2 + 1) 0() + [()]2 + 1 = 0 ⇒ (2 + 1)   + 2 + 1 = 0 ⇒   = −2 2+ 1 − 1 ⇒  2 + 1 = −  2 + 1 ⇒ arctan = −arctan +  ⇒ arctan + arctan =  ⇒ tan(arctan + arctan) = tan ⇒ tan(arctan) + tan(arctan) 1 − tan(arctan) tan(arctan) = tan ⇒ 1−+  = tan =  ⇒  +  =  −  ⇒  +  =  −  ⇒ (1 + ) =  −  ⇒ () =  =  −  1 + . Since (3) = 2 =  − 3 1 + 3 ⇒ 2 + 6 =  − 3 ⇒ 5 = −5 ⇒  = −1 , we have  = 1 + ( −1 −−1)   =  + 1 − 1. 37. From Exercise 9.2.27,   = 12 − 4 ⇔  12 − 4 =   ⇔ − 1 4 ln|12 − 4| =  +  ⇔ ln|12 − 4| = −4 − 4 ⇔ |12 − 4| = −4−4 ⇔ 12 − 4 = −4 [ = ±−4] ⇔ 4 = 12 − −4 ⇔  = 3 − −4 [ = 4]. (0) = 0 ⇔ 0 = 3 −  ⇔  = 3 ⇔ () = 3 − 3−4. As  → ∞, () → 3 − 0 = 3 (the limiting value). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.3 SEPARABLE EQUATIONS ¤ 821 38. From Exercise 9.2.28,   = − 1 50( − 20) ⇔   −20 =  − 50 1   ⇔ ln| − 20| = − 50 1  +  ⇔  − 20 = −50 ⇔ () = −50 + 20. (0) = 95 ⇔ 95 =  + 20 ⇔  = 75 ⇔ () = 75−50 + 20. 39.   = ( − ) ⇔    −  =  (−) ⇔ ln| − | = − +  ⇔ | − | = −+ ⇔  −  = − [ = ±] ⇔  =  + −. If we assume that performance is at level 0 when  = 0, then (0) = 0 ⇔ 0 =  +  ⇔  = − ⇔ () =  − −. lim →∞ () =  −  · 0 = . 40. (a)   = ( − )( − ),  6= . Using partial fractions, ( − )( 1  − ) = 1(−−) − 1(−−), so  ( −  )( − ) =    ⇒  −1  (−ln| − | + ln| − |) =  +  ⇒ ln    −−     = ( − )( + ). The concentrations [A] =  −  and [B] =  −  cannot be negative, so  −   −  ≥ 0 and   −   −   =  −   −  . We now have ln −−  = ( − )( + ). Since (0) = 0, we get ln  = ( − ). Hence, ln −−  = ( − ) + ln  ⇒  −−  =  (−) ⇒  =  [((−−) ) −−1]1 = [((−−)) −−1] moles L . (b) If  = , then   = ( − )2, so  ( −)2 =    and  −1  =  + . Since (0) = 0, we get  = 1. Thus,  −  = 1  + 1 and  =  − + 1 =  2 + 1 moles L . Suppose  = [C] = 2 when  = 20. Then (20) = 2 ⇒  2 = 202 20 + 1 ⇒ 402 = 202 +  ⇒ 202 =  ⇒  = 201, so  = 2(20) 1 + (20) = 20 1 + 20 =   + 20 moles L . 41. (a) If  = , then   = ( − )( − )12 becomes   = ( − )32 ⇒ ( − )−32  =   ⇒  ( − )−32  =    ⇒ 2( − )−12 =  +  [by substitution] ⇒ 2  +  = √ −  ⇒  +2  2 =  −  ⇒ () =  − ( +4 )2 . The initial concentration of HBr is 0, so (0) = 0 ⇒ 0 =  − 4 2 ⇒ 4 2 =  ⇒ 2 = 4  ⇒  = 2√ [ is positive since  +  = 2( − )−12  0]. Thus, () =  − 4 ( + 2√)2 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 822 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS (b)   = ( − )( − )12 ⇒ ( −  )√ −  =   ⇒  ( −  )√ −  =    (). From the hint,  = √ −  ⇒ 2 =  −  ⇒ 2  = −, so  ( −  )√ −  =  [ −−(2  − 2)] = −2  −  + 2 = −2 √ −  2 + 2 17 = −2√1−  tan−1 √−  So () becomes √−−2  tan−1 √√−− =  + . Now (0) = 0 ⇒  = √−−2  tan−1 √√−  and we have −2 √ −  tan−1 √ −  √ −  =  − √2−  tan−1 √√−  ⇒ √2− tan−1  −  − tan−1 −−  =  ⇒ () = 2  √ −  tan−1  −  − tan−1 −− . 42. If  =   , then   =  22 . The differential equation  22 + 2   = 0 can be written as   + 2  = 0. Thus,   = −2  ⇒   = − 2   ⇒  1  =  −2  ⇒ ln|| = −2ln|| + . Assuming  =   0 and   0, we have  = −2 ln + = ln −2 = −2 [ = ] ⇒  = 1 2  ⇒   = 12  ⇒  = 1 2   ⇒   =  12   ⇒ () = − + . (1) = 15 ⇒ 15 = − +  (1) and (2) = 25 ⇒ 25 = − 1 2 +  (2). Now solve for  and : −2(2) + (1) ⇒ −35 = −, so  = 35 and  = 20, and () = −20 + 35. 43. (a)   =  −  ⇒   = −( − ) ⇒  −  =  − ⇒ (1) ln| − | = − + 1 ⇒ ln| − | = − + 2 ⇒ | − | = −+2 ⇒  −  = 3− ⇒  = 3− +  ⇒ () = 4− + . (0) = 0 ⇒ 0 = 4 +  ⇒ 4 = 0 −  ⇒ () = (0 − )− + . (b) If 0  , then 0 −   0 and the formula for () shows that () increases and lim →∞ () = . As  increases, the formula for () shows how the role of 0 steadily diminishes as that of  increases. 44. (a) Use 1 billion dollars as the -unit and 1 day as the -unit. Initially, there is $10 billion of old currency in circulation, so all of the $50 million returned to the banks is old. At time , the amount of new currency is () billion dollars, so 10 − () billion dollars of currency is old. The fraction of circulating money that is old is [10 − ()]10, and the amount of old currency being returned to the banks each day is 10 − () 10 005 billion dollars. This amount of new currency per day is introduced into circulation, so   = 10 −  10 · 005 = 0005(10 − ) billion dollars per day. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.3 SEPARABLE EQUATIONS ¤ 823 (b)  10 −  = 0005 ⇒ − 10 −  = −0005 ⇒ ln(10 − ) = −0005 +  ⇒ 10 −  = −0005, where  =  ⇒ () = 10 − −0005. From (0) = 0, we get  = 10, so () = 10(1 − −0005). (c) The new bills make up 90% of the circulating currency when () = 09 · 10 = 9 billion dollars. 9 = 10(1 − −0005) ⇒ 09 = 1 − −0005 ⇒ −0005 = 01 ⇒ −0005 = −ln 10 ⇒  = 200 ln 10 ≈ 460517 days ≈ 126 years. 45. (a) Let () be the amount of salt (in kg) after  minutes. Then (0) = 15. The amount of liquid in the tank is 1000 L at all times, so the concentration at time  (in minutes) is ()1000 kgL and   = − 1000 () kg L 10 min L  = −100 () min kg .    = −100 1   ⇒ ln = −100  + , and (0) = 15 ⇒ ln 15 = , so ln = ln 15 − 100  . It follows that ln15   = −100  and 15  = −100, so  = 15−100 kg. (b) After 20 minutes,  = 15−20100 = 15−02 ≈ 123 kg. 46. Let () be the amount of carbon dioxide in the room after  minutes. Then (0) = 00015(180) = 027 m3. The amount of air in the room is 180 m3 at all times, so the percentage at time  (in mimutes) is ()180 × 100, and the change in the amount of carbon dioxide with respect to time is   = (00005)2 min m3  − 180 () 2 min m3  = 0001 − 90  = 9 −9000 100 min m3 Hence,  9 − 100 =  9000  and −100 1 ln|9 − 100| = 9000 1  + . Because (0) = 027, we have − 1 100 ln 18 = , so − 100 1 ln|9 − 100| = 9000 1  − 100 1 ln 18 ⇒ ln|9 − 100| = − 90 1  + ln 18 ⇒ ln|9 − 100| = ln−90 + ln 18 ⇒ ln|9 − 100| = ln(18−90), and |9 − 100| = 18−90. Since  is continuous, (0) = 027, and the right-hand side is never zero, we deduce that 9 − 100 is always negative. Thus, |9 − 100| = 100 − 9 and we have 100 − 9 = 18−90 ⇒ 100 = 9 + 18−90 ⇒  = 009 + 018−90. The percentage of carbon dioxide in the room is () =  180 × 100 = 009 + 018−90 180 × 100 = (00005 + 0001−90) × 100 = 005 + 01−90 In the long run, we have lim →∞ () = 005 + 01(0) = 005; that is, the amount of carbon dioxide approaches 005% as time goes on. 47. Let () be the amount of alcohol in the vat after  minutes. Then (0) = 004(500) = 20 gal. The amount of beer in the vat is 500 gallons at all times, so the percentage at time  (in minutes) is ()500 × 100, and the change in the amount of alcohol with respect to time  is   = rate in − rate out = 0065 min gal  − 500 () 5 min gal  = 03 − 100  = 30100 −  min gal . Hence,  30 −  =  100  and −ln|30 − | = 100 1  + . Because (0) = 20, we have −ln 10 = , so °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 824 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS −ln|30 − | = 100 1  − ln10 ⇒ ln|30 − | = −100 + ln10 ⇒ ln|30 − | = ln−100 + ln 10 ⇒ ln|30 − | = ln(10−100) ⇒ |30 − | = 10−100. Since  is continuous, (0) = 20, and the right-hand side is never zero, we deduce that 30 −  is always positive. Thus, 30 −  = 10−100 ⇒  = 30 − 10−100. The percentage of alcohol is () = ()500 × 100 = ()5 = 6 − 2−100. The percentage of alcohol after one hour is (60) = 6 − 2−60100 ≈ 49. 48. (a) If () is the amount of salt (in kg) after  minutes, then (0) = 0 and the total amount of liquid in the tank remains constant at 1000 L.   = 005 kg L 5 min L  + 004 kg L 10 min L  − 1000 () kg L 15 min L  = 025 + 040 − 0015 = 065 − 0015 = 130 − 3 200 kg min Hence,  130 − 3 =  200  and − 1 3 ln|130 − 3| = 200 1  + . Because (0) = 0, we have − 1 3 ln 130 = , so − 1 3 ln|130 − 3| = 200 1  − 1 3 ln 130 ⇒ ln|130 − 3| = − 200 3  + ln 130 = ln(130−3200), and |130 − 3| = 130−3200. Since  is continuous, (0) = 0, and the right-hand side is never zero, we deduce that 130 − 3 is always positive. Thus, 130 − 3 = 130−3200 and  = 130 3 (1 − −3200) kg. (b) After one hour,  = 130 3 (1 − −3·60200) = 130 3 (1 − −09) ≈ 257 kg. Note: As  → ∞, () → 130 3 = 43 1 3 kg. 49. Assume that the raindrop begins at rest, so that (0) = 0.  =  and ()0 =  ⇒ 0 + 0 =  ⇒ 0 + () =  ⇒ 0 +  =  ⇒   =  −  ⇒   − =   ⇒ −(1)ln| − | =  +  ⇒ ln| − | = − −  ⇒  −  = −. (0) = 0 ⇒  = . So  =  − − ⇒  = ()(1 − −). Since   0, as  → ∞, − → 0 and therefore, lim →∞ () = . 50. (a)    = − ⇒   = −  ⇒ ln|| = −  + . Since (0) = 0, ln|0| = . Therefore, ln   0  = −    ⇒   0  = − ⇒ () = ±0−. The sign is + when  = 0, and we assume  is continuous, so that the sign is + for all . Thus, () = 0−.  = 0− ⇒ () = −0  − + 0. From (0) = 0, we get 0 = −0  + 0, so 0 = 0 +   0 and () = 0 +   0 (1 − −). The distance traveled from time 0 to time  is () − 0, so the total distance traveled is lim →∞ [() − 0] = 0  . Note: In finding the limit, we use the fact that   0 to conclude that lim →∞ − = 0. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.3 SEPARABLE EQUATIONS ¤ 825 (b)    = −2 ⇒  2 = −  ⇒ −1 = −  +  ⇒ 1 =   − . Since (0) = 0,  = − 1 0 and 1  =   + 1 0 . Therefore, () = 1  + 10 = 0 0 + .   = 0 0 +  ⇒ () =     00+ =  ln|0 + | + 0. Since (0) = 0, we get 0 =  ln + 0 ⇒ 0 = 0 −   ln ⇒ () = 0 +  (ln|0 + | − ln) = 0 +  ln  0 +    . We can rewrite the formulas for () and () as () = 0 1 + (0) and () = 0 +  ln  1 + 0    . Remarks: This model of horizontal motion through a resistive medium was designed to handle the case in which 0  0. Then the term −2 representing the resisting force causes the object to decelerate. The absolute value in the expression for () is unnecessary (since , 0, and  are all positive), and lim →∞ () = ∞. In other words, the object travels infinitely far. However, lim →∞ () = 0. When 0  0, the term −2 increases the magnitude of the object’s negative velocity. According to the formula for (), the position of the object approaches −∞ as  approaches (−0): lim →−(0) () = −∞. Again the object travels infinitely far, but this time the feat is accomplished in a finite amount of time. Notice also that lim →−(0) () = −∞ when 0  0, showing that the speed of the object increases without limit. 51. (a) 1 1 1  =  12  2 ⇒   (ln1) =   ( ln2) ⇒    (ln1) =    (ln 2) ⇒ ln1 = ln 2 +  ⇒ 1 = ln  2 +  = ln  2  ⇒ 1 =  2, where  = . (b) From part (a) with 1 = , 2 =  , and  = 00794, we have  =  00794. 52. (a)   = (ln − ln ) ⇒   = − (ln − ln) ⇒  ln( ) = −  ⇒   ln( ) =  −  ⇒  1  =  −    = ln( = (1  ) ) ,  ⇒ ln|| = − +  ⇒ || = − ⇒  = − [where  = ±] ⇒ ln() = − ⇒   = − ⇒  = − with  6= 0. (b)  (0) = 1 ⇒ 1 = −(0) ⇒ 1 =  ⇒  = −, so  = −− = −− = (−−1). 53. (a) The rate of growth of the area is jointly proportional to () and  − (); that is, the rate is proportional to the product of those two quantities. So for some constant ,  =  √( − ). We are interested in the maximum of the function  (when the tissue grows the fastest), so we differentiate, using the Chain Rule and then substituting for °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 826 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS  from the differential equation:    = √(−1)  + ( − ) · 1 2−12    = 1 2−12   [−2 + ( − )] = 1 2 −12 √( − )[ − 3] = 1 22( − )( − 3) This is 0 when  −  = 0 [this situation never actually occurs, since the graph of () is asymptotic to the line  = , as in the logistic model] and when  − 3 = 0 ⇔ () = 3. This represents a maximum by the First Derivative Test, since     goes from positive to negative when () = 3. (b) From the CAS, we get () =  √ √  − + 1 12. To get  in terms of the initial area 0 and the maximum area , we substitute  = 0 and  = 0 = (0): 0 =   − + 112 ⇔ ( + 1)√0 = ( − 1)√ ⇔  √0 + √0 =  √ − √ ⇔ √ + √0 =  √ −  √0 ⇔ √ + √0 = √ − √0  ⇔  = √ √  + − √ √ 0 0 . [Notice that if 0 = 0, then  = 1.] 54. (a) According to the hint we use the Chain Rule:   =    ·   =    = − 2 ( + )2 ⇒    =  (− +2 )2 ⇒ 22 =  + 2 + . When  = 0,  = 0, so 202 = 0 + 2 +  ⇒  = 1 202 −  ⇒ 1 22 − 1 202 = 2  +  − . Now at the top of its flight, the rocket’s velocity will be 0, and its height will be  = . Solving for 0: − 1 202 =  + 2 −  ⇒ 202 = −+2  + (++) =  +  ⇒ 0 = 2 + . (b)  = lim →∞ 0 = lim →∞ 2 +  = lim →∞ ( 2 ) + 1 = √2 (c)  = 2 · 32 fts2 · 3960 mi · 5280 ftmi ≈ 36,581 fts ≈ 693 mis APPLIED PROJECT How Fast Does a Tank Drain? 1. (a)  = 2 ⇒   = 2   [implicit differentiation] ⇒   = 1 2   = 1 2 −√2 = 122 − 12 1 2 √2 · 32√ = − 72 1 √ (b)   = − 1 72 √ ⇒ −12  = − 1 72  ⇒ 2√ = − 72 1  + . (0) = 6 ⇒ 2√6 = 0 +  ⇒  = 2√6 ⇒ () = − 144 1  + √62. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE APPLIED PROJECT HOW FAST DOES A TANK DRAIN? ¤ 827 (c) We want to find  when  = 0, so we set  = 0 = − 144 1  + √62 ⇒  = 144√6 ≈ 5 min 53 s. 2. (a)   =  √ ⇒ −12  =   [ 6= 0] ⇒ 2√ =  +  ⇒ () = 1 4( + )2. Since (0) = 10 cm, the relation 2() =  +  gives us 2√10 = . Also, (68) = 3 cm, so 2√3 = 68 + 2√10 and  = − √10 − √3 34 . Thus, () = 1 42√10 − √1034 − √3 2 ≈ 10 − 0133 + 0000442. Here is a table of values of () correct to one decimal place.  (in s) () (in cm) 10 87 20 75 30 64 40 54 50 45 60 36 (b) The answers to this part are to be obtained experimentally. See the article by Tom Farmer and Fred Gass, Physical Demonstrations in the Calculus Classroom, College Mathematics Journal 1992, pp. 146–148. 3.  () = 2() = 100() ⇒   = 100 and   =     = 100   . Diameter = 25 inches ⇒ radius = 125 inches = 5 4 · 1 12 foot = 48 5 foot. Thus,   = − √2 ⇒ 100   = − 48 5 2 √2 · 32 = −25 288  √ ⇒   = −1152 √ ⇒  −12  =  − 1152 1  ⇒ 2√ = − 1 1152  +  ⇒ √ = − 2304 1  +  ⇒ () = − 2304 1  + 2. The water pressure after  seconds is 625() lbft2, so the condition that the pressure be at least 2160 lbft2 for 10 minutes (600 seconds) is the condition 625 · (600) ≥ 2160; that is,  − 2304 600 2 ≥ 2160 625 ⇒   − 25 96   ≥ √3456 ⇒  ≥ 25 96 + √3456. Now (0) = 2, so the height of the tank should be at least  25 96 + √34562 ≈ 3769 ft. 4. (a) If the radius of the circular cross-section at height  is , then the Pythagorean Theorem gives 2 = 22 − (2 − )2 since the radius of the tank is 2 m. So () = 2 = [4 − (2 − )2] = (4 − 2). Thus, ()   = − √2 ⇒ (4 − 2)   = −(001)2 √2 · 10 ⇒ (4 − 2)   = −00001√20. (b) From part (a) we have (412 − 32) = −00001√20 ⇒ 8 3 32 − 2 5 52 = −00001√20 + . (0) = 2 ⇒ 8 3(2)32 − 2 5(2)52 =  ⇒  =  16 3 − 8 5  √2 = 56 15√2. To find out how long it will take to drain all the water we evaluate  when  = 0: 0 = −00001√20 +  ⇒  =  00001√20 = 56√215 00001√20 = 11,200√10 3 ≈ 11,806 s ≈ 3 h 17 min °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 828 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS APPLIED PROJECT Which Is Faster, Going Up or Coming Down? 1. 0 = − −  ⇒    = −( + ) ⇒   + =  −1  ⇒ 1  ln( + ) = − 1   +  [ +   0]. At  = 0,  = 0, so  = 1  ln(0 + ). Thus, 1  ln( + ) = − 1   + 1  ln(0 + ) ⇒ ln( + ) = −    + ln(0 + ) ⇒  +  = −(0 + ) ⇒  = (0 + )− −  ⇒ () = 0 +   − −   . 2. () =  () =  0 +   − −     = 0 +   −−  −    + . At  = 0,  = 0, so  = 0 +    . Thus, () = 0 +    − 0 +    − −   = 0 +    1 − − −   3. () = 0 ⇒   = 0 +   − ⇒  =  0 + 1 ⇒  = ln 0 + 1 ⇒ 1 =   ln + 0 . With  = 1, 0 = 20,  = 10 1 , and  = 98, we have 1 = 10 ln 11 988 ≈ 186 s. 4. The figure shows the graph of  = 1180(1 − −01) − 98. The zeros are at  = 0 and 2 ≈ 384. Thus, 1 − 0 ≈ 186 and 2 − 1 ≈ 198. So the time it takes to come down is about 012 s longer than the time it takes to go up; hence, going up is faster. 5. (21) = 0 +    (1 − −21) −   · 21 = 0 +   1 − (1)−2 −   · 2 ln0 +   Substituting  = 1 = 0  + 1 = 0 +   (from Problem 3), we get  (21) =  ·    (1 − −2) − 22 · 2ln = 22  − 1 − 2ln. Now   0,   0, 1  0 ⇒  = 1  0 = 1. () =  − 1  − 2ln ⇒  0() = 1 + 1 2 − 2  = 2 − 2 + 1 2 = ( − 1)2 2  0 for   1 ⇒ () is increasing for   1. Since (1) = 0, it follows that ()  0 for every   1. Therefore, (21) = 2 2 () is positive, which means that the ball has not yet reached the ground at time 21. This tells us that the time spent going up is always less than the time spent coming down, so ascent is faster. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.4 MODELS FOR POPULATION GROWTH ¤ 829 9.4 Models for Population Growth 1. (a) Comparing the given equation,   = 0041 −  , to Equation 4,   = 1 −  , we see that the carrying capacity is  = 1200 and the value of  is 004. (b) By Equation 7, the solution of the equation is () =  1 + − , where  =  −00 . Since (0) = 0 = 60, we have  = 1200 − 60 60 = 19, and hence, () = 1200 1 + 19−004 . (c) The population after 10 weeks is (10) = 1200 1 + 19−004(10) ≈ 87. 2. (a)  = 002 − 000004 2 = 002(1 − 0002) = 002(1 − 500). Comparing to Equation 4,  = (1 − ), we see that the carrying capacity is  = 500 and the value of  is 002. (b) By Equation 7, the solution of the equation is () =  1 + − , where  =  −00 . Since (0) = 0 = 40, we have  = 500 − 40 40 = 115, and hence, () = 1 + 11500 5−002 . (c) The population after 10 weeks is (10) = 500 1 + 115−002(10) ≈ 48. 3. (a)  = 005 − 00005 2 = 005(1 − 001) = 005(1 − 100). Comparing to Equation 4,  = (1 − ), we see that the carrying capacity is  = 100 and the value of  is 005. (b) The slopes close to 0 occur where  is near 0 or 100. The largest slopes appear to be on the line  = 50. The solutions are increasing for 0  0  100 and decreasing for 0  100. (c) All of the solutions approach  = 100 as  increases. As in part (b), the solutions differ since for 0  0  100 they are increasing, and for 0  100 they are decreasing. Also, some have an IP and some don’t. It appears that the solutions which have 0 = 20 and 0 = 40 have inflection points at  = 50. (d) The equilibrium solutions are  = 0 (trivial solution) and  = 100. The increasing solutions move away from  = 0 and all nonzero solutions approach  = 100 as  → ∞. 4. (a)  = 6000 and  = 00015 ⇒  = 00015(1 − 6000). (b) All of the solution curves approach 6000 as  → ∞. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 830 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS (c) The curves with 0 = 1000 and 0 = 2000 appear to be concave upward at first and then concave downward. The curve with 0 = 4000 appears to be concave downward everywhere. The curve with 0 = 8000 appears to be concave upward everywhere. The inflection points are where the population grows the fastest. (d) See the solution to Exercise 9.2.25 for a possible program to calculate (50). [In this case, we use X = 0, H = 1, N = 50, Y1 = 00015(1 − 6000), and Y = 1000.] We find that (50) ≈ 1064. (e) Using Equation 7 with  = 6000,  = 00015, and 0 = 1000, we have () =  1 + − = 6000 1 + −00015 , where  =  − 0 0 = 6000 − 1000 1000 = 5. Thus,  (50) = 6000 1 + 5−00015(50) ≈ 10641, which is extremely close to the estimate obtained in part (d). (f ) The curves are very similar. 5. (a)   = 1 −   ⇒ () = 1 +− with  =  −(0) (0). With  = 8 × 107,  = 071, and (0) = 2 × 107, we get the model () = 8 × 107 1 + 3−071 , so (1) = 1 + 3 8 ×10 −0771 ≈ 323 × 107 kg. (b) () = 4 × 107 ⇒ 8 × 107 1 + 3−071 = 4 × 107 ⇒ 2 = 1 + 3−071 ⇒ −071 = 1 3 ⇒ −071 = ln 1 3 ⇒  = ln 3 071 ≈ 155 years 6. (a)   = 04 − 0001 2 = 04(1 − 00025)  0 0001 4 = 00025 = 04 1 − 400   00025−1 = 400 Thus, by Equation 4,  = 04 and the carrying capacity is 400. (b) Using the fact that (0) = 50 and the formula for , we get  0(0) =    =0 = 04(50) − 0001(50)2 = 20 − 25 = 175. (c) From Equation 7,  =  − 0 0 = 400 − 50 50 = 7, so  = 400 1 + 7−04 . The population reaches 50% of the carrying capacity, 200, when 200 = 400 1 + 7−04 ⇒ 1 + 7−04 = 2 ⇒ −04 = 1 7 ⇒ −04 = ln 1 7 ⇒  = ln 1 7(−04) ≈ 486 years. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.4 MODELS FOR POPULATION GROWTH ¤ 831 7. Using (7),  =  − 0 0 = 10,000 − 1000 1000 = 9, so () = 10,000 1 + 9− . (1) = 2500 ⇒ 2500 = 1 + 9 10,000 −(1) ⇒ 1 + 9− = 4 ⇒ 9− = 3 ⇒ − = 1 3 ⇒ − = ln 1 3 ⇒  = ln 3. After another three years,  = 4, and (4) = 10,000 1 + 9−(ln 3)4 = 10,000 1 + 9 (ln 3)−4 = 10,000 1 + 9(3)−4 = 10,000 1 + 1 9 = 10,000 10 9 = 9000. 8. (a) From the graph, we estimate the carrying capacity  for the yeast population to be 680. (b) An estimate of the initial relative growth rate is 1 0   = 1 18 · 39 − 18 2 − 0 = 7 12 = 0583. (c) An exponential model is () = 18712. A logistic model is () = 680 1 + −712 , where  = 68018 − 18 = 331 9 . (d) Time in Hours Observed Values Exponential Model Logistic Model 0 18 18 18 2 39 58 55 4 80 186 149 6 171 596 322 8 336 1914 505 10 509 6147 614 12 597 19,739 658 14 640 63,389 673 16 664 203,558 678 18 672 653,679 679 The exponential model is a poor fit for anything beyond the first two observed values. The logistic model varies more for the middle values than it does for the values at either end, but provides a good general fit, as shown in the figure. (e) (7) = 680 1 + 331 9 −7(712) ≈ 420 yeast cells 9. (a) We will assume that the difference in birth and death rates is 20 million/year. Let  = 0 correspond to the year 2000. Thus,  ≈ 1    = 1 61 billion 20 year million = 305 1 , and   = 1 −   = 305 1 1 − 20   with  in billions. (b)  =  − 0 0 = 20 − 61 61 = 139 61 ≈ 22787. () =  1 + − = 20 1 + 139 61 −305 , so (10) = 20 1 + 139 61 −10305 ≈ 624 billion, which underestimates the actual 2010 population of 69 billion. (c) The years 2100 and 2500 correspond to  = 100 and  = 500, respectively. (100) = 20 1 + 139 61 −100305 ≈ 757 billion and (500) = 20 1 + 139 61 −500305 ≈ 1387 billion. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 832 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 10. (a) Let  = 0 correspond to the year 2000.  =  − 0 0 = 800 − 282 282 = 259 141 ≈ 18369. () =  1 + − = 800 1 + 259 141 − with  in millions. (b) (10) = 309 ⇔ 800 1 + 259 141 −10 = 309 ⇔ 800 309 = 1 + 259 141−10 ⇔ 491 309 = 259 141−10 ⇔ 491 · 141 309 · 259 = −10 ⇔ −10 = ln 491 · 47 103 · 259 ⇔  = − 1 10 ln 23,077 26,677 ≈ 00145. (c) The years 2100 and 2200 correspond to  = 100 and  = 200, respectively. (100) = 800 1 + 259 141 −100 ≈ 559 million and (200) = 800 1 + 259 141 −200 ≈ 727 million. (d) () = 500 ⇔ 800 1 + 259 141 − = 500 ⇔ 800 500 = 1 + 259 141− ⇔ 35 = 259 141− ⇔ 35 ·· 141 259 = − ⇔ − = ln 423 1295 ⇔  = 10 ln(4231295) ln(23,07726,677) ≈ 7718 years. Our logistic model predicts that the US population will exceed 500 million in 7718 years; that is, in the year 2077. 11. (a) Our assumption is that   = (1 − ), where  is the fraction of the population that has heard the rumor. (b) Using the logistic equation (4),   = 1 −  , we substitute  =  ,  = , and   =    , to obtain    = ()(1 − ) ⇔   = (1 − ), our equation in part (a). Now the solution to (4) is () =  1 + − , where  =  −00 . We use the same substitution to obtain  =  1 +  − 0 0 − ⇒  = 0 0 + (1 − 0)− . Alternatively, we could use the same steps as outlined in the solution of Equation 4. (c) Let  be the number of hours since 8 AM. Then 0 = (0) = 1000 80 = 008 and (4) = 1 2, so 1 2 = (4) = 008 008 + 092−4 . Thus, 008 + 092−4 = 016, −4 = 0 0 08 92 = 23 2 , and − =  23 2 14, so  = 008 008 + 092(223)4 = 2 2 + 23(223)4 . Solving this equation for , we get 2 + 2323 2 4 = 2 ⇒ 23 2 4 = 223 −2 ⇒ 23 2 4 = 23 2 · 1 −  ⇒ 23 2 4−1 = 1 −  . It follows that  4 − 1 = ln[(1ln−2)] 23 , so  = 41 + ln((1ln−23 2)). When  = 09, 1 −   = 1 9 , so  = 41 − ln ln 9 23 2  ≈ 76 h or 7 h 36 min. Thus, 90% of the population will have heard the rumor by 3:36 PM. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.4 MODELS FOR POPULATION GROWTH ¤ 833 12. (a) (0) = 0 = 400, (1) = 1200 and  = 10,000. From the solution to the logistic differential equation () = 0 0 + ( − 0)− , we get  = 400 + (9600) 400 (10,000) − = 1 + 24 10,000 − . (1) = 1200 ⇒ 1 + 24− = 100 12 ⇒  = 288 88 ⇒  = ln 36 11. So  = 10,000 1 + 24− ln(3611) = 10,000 1 + 24 · (1136) . (b) 5000 = 10,000 1 + 24(1136) ⇒ 24 11 36  = 1 ⇒ ln 11 36 = ln 24 1 ⇒  ≈ 268 years. 13. (a)   =  1 −   ⇒  22 =  −1    + 1 −      =    − + 1 −   =  1 −  1 − 2  = 2 1 −  1 − 2  (b)  grows fastest when  0 has a maximum, that is, when  00 = 0. From part (a),  00 = 0 ⇔  = 0,  = , or  = 2. Since 0    , we see that  00 = 0 ⇔  = 2. 14. First we keep  constant (at 01, say) and change 0 in the function  = 100 0 + (10 − 0)−01 . (Notice that 0 is the -intercept.) If 0 = 0, the function is 0 everywhere. For 0  0  5, the curve has an inflection point, which moves to the right as 0 decreases. If 5  0  10, the graph is concave down everywhere. (We are considering only  ≥ 0.) If 0 = 10, the function is the constant function  = 10, and if 0  10, the function decreases. For all 0 6= 0, lim →∞  = 10. Now we instead keep 0 constant (at 0 = 1) and change  in the function  = 10 1 + 9− . It seems that as  increases, the graph approaches the line  = 10 more and more quickly. (Note that the only difference in the shape of the curves is in the horizontal scaling; if we choose suitable -scales, the graphs all look the same.) 15. Following the hint, we choose  = 0 to correspond to 1960 and subtract 94,000 from each of the population figures. We then use a calculator to obtain the models and add 94,000 to get the exponential function () = 19097761(10796) + 94,000 and the logistic function () = 33,0864394 1 + 123428−01657 + 94,000.  is a reasonably accurate model, while  is not, since an exponential model would only be used for the first few data points. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 834 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 16. Following the hint, we choose  = 0 to correspond to 1960 and subtract 3500 from each of the population figures. We then use a calculator to obtain the models and add 3500 to get the exponential function () = 1809934(10445) + 3500 and the logistic function () = 13489650 1 + 62784−00721 + 3500.  is a reasonably accurate accurate model, while  is not, since an exponential model would only be used for the first few data points. 17. (a)   =  −  =  −  . Let  =  −  , so   =   and the differential equation becomes   = . The solution is  = 0 ⇒  −   = 0 −   ⇒ () =  + 0 −  . (b) Since   0, there will be an exponential expansion ⇔ 0 −    0 ⇔   0. (c) The population will be constant if 0 −   = 0 ⇔  = 0. It will decline if 0 −   0 ⇔   0. (d) 0 = 8,000,000,  =  −  = 0016,  = 210,000 ⇒   0 (= 128,000), so by part (c), the population was declining. 18. (a)   = 1+ ⇒ −1−  =   ⇒ −− =  + . Since (0) = 0, we have  = −0− . Thus, − − =  + 0− − , or − = 0− − . So  = 1 0− −  = 0 1 − 0 and () = (1 −  00)1 . (b) () → ∞ as 1 − 0 → 0, that is, as  → 1 0 . Define  = 10 . Then →lim − () = ∞. (c) According to the data given, we have  = 001, (0) = 2, and (3) = 16, where the time  is given in months. Thus, 0 = 2 and 16 = (3) = 0 (1 − 0 · 3)1 . Since  = 10 , we will solve for 0. 16 = (1 − 3 2 0)100 ⇒ 1 − 30 =  1 8001 = 8−001 ⇒ 0 = 1 31 − 8−001. Thus, doomsday occurs when  =  = 1 0 = 3 1 − 8−001 ≈ 14577 months or 1215 years. 19. (a) The term −15 represents a harvesting of fish at a constant rate—in this case, 15 fishweek. This is the rate at which fish are caught. (b) (c) From the graph in part (b), it appears that () = 250 and () = 750 are the equilibrium solutions. We confirm this analytically by solving the equation  = 0 as follows: 008(1 − 1000) − 15 = 0 ⇒ 008 − 000008 2 − 15 = 0 ⇒ −000008( 2 − 1000 + 187,500) = 0 ⇒ ( − 250)( − 750) = 0 ⇒  = 250 or 750. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.4 MODELS FOR POPULATION GROWTH ¤ 835 (d) For 0  0  250, () decreases to 0. For 0 = 250, () remains constant. For 250  0  750, () increases and approaches 750. For 0 = 750, () remains constant. For 0  750, () decreases and approaches 750. (e)   = 0081 − 1000   − 15 ⇔ −1008,000 ·   = (008 − 000008 2 − 15) · −1008,000 ⇔ −12,500   =  2 − 1000 + 187,500 ⇔ ( − 250)(  − 750) = −12,1500  ⇔  −1−500 250 + 1−500 750  = −12,1500  ⇔   −1250 −  −1750 = 25 1  ⇔ ln| − 250| − ln| − 750| = 25 1  +  ⇔ ln   − 250  − 750  = 1 25 +  ⇔   − 250  − 750  = 25+ = 25 ⇔  − 250  − 750 = 25 ⇔  − 250 = 25 − 75025 ⇔  − 25 = 250 − 75025 ⇔ () = 250 − 75025 1 − 25 . If  = 0 and  = 200, then 200 = 2501−−750   ⇔ 200 − 200 = 250 − 750 ⇔ 550 = 50 ⇔  = 1 11. Similarly, if  = 0 and  = 300, then  = − 1 9. Simplifying  with these two values of  gives us () = 250(325 − 11) 25 − 11 and () = 750(  25 25 + 9 + 3). 20. (a)  = 0  = 10  = 20  = 21  = 25  = 30 (b) For 0 ≤  ≤ 20, there is at least one equilibrium solution. For   20, the population always dies out. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 836 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS (c)   = 008 − 000008 2 − .   = 0 ⇔  = −008 ± (02( 08) −0200008) − 4(−000008)(−), which has at least one solution when the discriminant is nonnegative ⇒ 00064 − 000032 ≥ 0 ⇔  ≤ 20. For 0 ≤  ≤ 20, there is at least one value of  such that  = 0 and hence, at least one equilibrium solution. For   20,   0 and the population always dies out. (d) The weekly catch should be less than 20 fish per week. 21. (a)   = ()1 −  1 −  . If     , then  = (+)(+)(+) = + ⇒  is increasing. If 0    , then  = (+)(+)(−) = − ⇒  is decreasing. (b)  = 008,  = 1000, and  = 200 ⇒   = 0081 − 1000  1 − 200   For 0  0  200, the population dies out. For 0 = 200, the population is steady. For 200  0  1000, the population increases and approaches 1000. For 0  1000, the population decreases and approaches 1000. The equilibrium solutions are () = 200 and () = 1000. (c)   = 1 −  1 −   = −   −  =  ( − )( − ) ⇔  ( −  )( − ) =   . By partial fractions, ( − )( 1  − ) =  −  +  − , so ( − ) + ( − ) = 1. If  = ,  = 1  − ; if  = ,  =  −1 , so  −1    1−  +  −1   =    ⇒ 1  −  (−ln| − | + ln| − |) =   +  ⇒  −1  ln    −−     =   +  ⇒ ln   −   −   = ( − )    + 1 ⇔  −− = (−)() [ = ±1]. Let  = 0: 0 −   − 0 = . So  −− = 0 −−0 (−)(). Solving for , we get () = ( − 0) + (0 − )(−)()  − 0 + (0 − )(−)() . (d) If 0  , then 0 −   0. Let () be the numerator of the expression for () in part (c). Then (0) = 0( − )  0, and 0 −   0 ⇔ lim →∞ (0 − )(−)() = −∞ ⇒ lim →∞ () = −∞. Since  is continuous, there is a number  such that () = 0 and thus () = 0. So the species will become extinct. 22. (a)   = ln   ⇒   ln( ) =   . Let  = ln  = ln − ln ⇒  = −  ⇒  −  =  +  ⇒ ln|| = − −  ⇒ || = −(+) ⇒ |ln()| = −(+) ⇒ °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.4 MODELS FOR POPULATION GROWTH ¤ 837 ln() = ±−(+). Letting  = 0, we get ln(0) = ±−, so ln() = ±−− = ±−− = ln(0)− ⇒  = ln(0)− ⇒ () = − ln(0)−,  6= 0. (b) lim →∞ () = lim →∞ − ln(0)− = − ln(0)·0 = 0 =  (c) The graphs look very similar. For the Gompertz function, (40) ≈ 732, nearly the same as the logistic function. The Gompertz function reaches  = 900 at  ≈ 617 and its value at  = 80 is about 959, so it doesn’t increase quite as fast as the logistic curve. (d)   = ln  = (ln − ln) ⇒ 2 2 = −1    + (ln − ln)    =    −1 + ln  = [ln()][ln() − 1] = 2 ln()[ln() − 1] Since 0    ,  00 = 0 ⇔ ln() = 1 ⇔  =  ⇔  = .  00  0 for 0     and  00  0 for     , so  0 is a maximum (and  grows fastest) when  = . Note: If   , then ln()  0, so  00()  0. 23. (a)  =  cos( − ) ⇒ () =  cos( − ) ⇒  () =   cos( − ) ⇒ ln = ()sin( − ) + . (Since this is a growth model,   0 and we can write ln instead of ln||.) Since (0) = 0, we obtain ln0 = ()sin(−) +  = −()sin +  ⇒  = ln0 + ()sin. Thus, ln = ()sin( − ) + ln0 + ()sin, which we can rewrite as ln(0) = ()[sin( − ) + sin] or, after exponentiation, () = 0()[sin(−)+sin ]. (b) As  increases, the amplitude increases, but the minimum value stays the same. As  increases, the amplitude and the period decrease. A change in  produces slight adjustments in the phase shift and amplitude. () oscillates between 0()(1+sin ) and 0()(−1+sin ) (the extreme values are attained when  −  is an odd multiple of 2 ), so lim →∞ () does not exist. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 838 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 24. (a)  =  cos2( − ) ⇒ () =  cos2( − ) ⇒  () =   cos2( − ) ⇒ ln =   1 + cos(2( 2  − ))  = 2  + 4 sin(2( − )) + . From (0) = 0, we get ln0 =  4 sin(−2) +  =  − 4 sin 2, so  = ln0 + 4 sin 2 and ln =  2  +  4 sin(2( − )) + ln0 +  4 sin 2. Simplifying, we get ln  0 =  2  +  4 [sin(2( − )) + sin 2] = (), or () = 0(). (b) An increase in  stretches the graph of  vertically while maintaining (0) = 0. An increase in  compresses the graph of  horizontally—similar to changing the period in Exercise 19. As in Exercise 23, a change in  only makes slight adjustments in the growth of , as shown in the figure.  0() = 2 + [(4)][2 cos(2( − ))] = (2)[1 + cos(2( − ))] ≥ 0. Since () = 0(), we have  0() = 0 0()() ≥ 0, with equality only when cos(2( − )) = −1; that is, when  −  is an odd multiple of 2 . Therefore, () is an increasing function on (0 ∞).  can also be written as () = 02(4)[sin(2(−))+sin 2]. The second exponential oscillates between (4)(1+sin 2) and (4)(−1+sin 2), while the first one, 2, grows without bound. So lim →∞ () = ∞. 25. By Equation 7, () =  1 + − . By comparison, if  = (ln) and  = 1 2( − ), then 1 + tanh = 1 +  − −  + − =  + −  + − +  − −  + − = 2  + − · − − = 2 1 + −2 and −2 = −(−) = − = ln − = −, so 1 2 1 + tanh 1 2( − ) = 2 [1 + tanh] = 2 · 1 +2−2 = 1 +−2 = 1 +   − = (). 9.5 Linear Equations 1. 0 + √ = 2 is not linear since it cannot be put into the standard form (1), 0 + () = (). 2. 0 −  =  tan ⇔ 0 + (−tan) =  is linear since it can be put into the standard form (1), 0 + () = (). 3.  =  + √   ⇔ √ 0 −  = − ⇔ 0 − √  = −√ is linear since it can be put into the standard form, 0 + () = (). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.5 LINEAR EQUATIONS ¤ 839 4.   + cos = − ⇔ 0 + cos = − is not linear since it cannot be put into the standard form 0 + () = (). 5. Comparing the given equation, 0 +  = 1, with the general form, 0 + () = (), we see that () = 1 and the integrating factor is () =    ()  =   1  = . Multiplying the differential equation by () gives 0 +  =  ⇒ ()0 =  ⇒  =    ⇒  =  +  ⇒   =   +   ⇒  = 1 + −. 6. 0 −  =  ⇔ 0 + (−1) =  ⇒ () = −1. () =    ()  =   −1  = −. Multiplying the original differential equation by () gives −0 − − = 0 ⇒ (−)0 = 1 ⇒ − =  1 ⇒ − =  +  ⇒  =  +  − ⇒  =  + . 7. 0 =  −  ⇒ 0 +  =  (). () =    ()  =   1  = . Multiplying the differential equation () by () gives 0 +  =  ⇒ ()0 =  ⇒  =    ⇒  =  −  +  [by parts] ⇒  =  − 1 + − [divide by ]. 8. 43 + 40 = sin3  ⇒ (4)0 = sin3  ⇒ 4 =  sin3   ⇒ 4 =  sin(1 − cos2 ) =  (1 − 2)(−)   = cos = − sin ,   =  (2 − 1) = 1 33 −  +  = 1 3(2 − 3) +  = 1 3 cos(cos2  − 3) +  ⇒  = 1 34 cos(cos2  − 3) + 4 9. Since () is the derivative of the coefficient of 0 [() = 1 and the coefficient is ], we can write the differential equation 0 +  = √ in the easily integrable form ()0 = √ ⇒  = 2 332 +  ⇒  = 2 3√ + . 10. 20 +  = 2√ ⇒ 0 + 1 2  = 1 √ [  0] ⇒ () = 1 2 . () =   ()  =  1(2)  = (12) ln|| = (ln )12 = √. Multiplying the differential equation by () gives √ 0 + 1 2√  = 1 ⇒ (√ )0 = 1 ⇒ √  =  1 ⇒ √  =  +  ⇒  = √+ . 11. 0 − 2 = 2 ⇒ 0 − 2   =  ⇒ () = −2  . () =   ()  =  −2  = −2 ln  [  0] = −2 = 1 2 . Multiplying the differential equation by () gives 1 2 0 − 2 3  = 1  ⇒ 12 0 = 1 ⇒ 12  =  1  ⇒ 12  = ln +  ⇒  = 2(ln + ). 12. 0 + 2 = 1 ⇒ () = 2. () =   ()  =  2  = 2. Multiplying the differential equation by () gives 20 + 22 = 2 ⇒ 20 = 2 ⇒ 2 = 0 2  +  [see page 507] ⇒  = −2 0 2  + −2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 840 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 13. 2   + 3 = √1 + 2 ⇒ 0 + 3  = √1 + 2 2 ⇒ () = 3 . () =  () =  3  = 3 ln  [  0] = 3. Multiplying by 3 gives 30 + 32 =  √1 + 2 ⇒ (3)0 = √1 + 2 ⇒ 3 =   √1 + 2  ⇒ 3 = 1 3(1 + 2)32 +  ⇒  = 1 3 −3(1 + 2)32 + −3. 14. ln   +  =  ⇒   +  ln 1   = ln  . () =   ( ln ) = ln(ln ) = ln. Multiplying by ln gives ln   + 1   =  ⇒ [(ln)]0 =  ⇒ (ln) =  +  ⇒  =  +  ln . 15. 20 + 2 = ln ⇒ (2)0 = ln ⇒ 2 =  ln  ⇒ 2 = ln −  +  [by parts]. Since (1) = 2, 12(2) = 1 ln 1 − 1 +  ⇒ 2 = −1 +  ⇒  = 3 so 2 = ln −  + 3, or  = 1  ln − 1  + 3 2 . 16. 3   + 32 = cos ⇒ (3)0 = cos ⇒ 3 =  cos  ⇒ 3 = sin + . Since () = 0, 3(0) = sin +  ⇒  = 0, so 3 = sin, or  = sin 3 . 17.    = 2 + 3 ⇒ 0 − 3  =  (). () =   −3  = −3 ln|| = (ln||)−3 = −3 [  0] = 13 . Multiplying () by () gives 1 3 0 − 34  = 12 ⇒  13 0 = 12 ⇒ 13  =  12  ⇒ 13  = −1 + . Since (2) = 4, 1 32 (4) = −1 2 +  ⇒  = 1, so 1 3  = − 1  + 1, or  = −2 + 3. 18. 0 +  = ln ⇒ ( )0 = ln ⇒  =  ln  ⇒  = 1 2 2 ln − 1 4 2 +  withby parts  = ln   ⇒  = 1 2 ln − 1 4 +   . (1) = 0 ⇒ 0 = 0 − 1 4 +  ⇒  = 14, so  = 12ln − 1 4 + 41 . 19. 0 =  + 2 sin ⇒ 0 − 1   = sin. () =   (−1)  = − ln  = ln −1 = 1  . Multiplying by 1  gives 1  0 − 1 2  = sin ⇒  1 0 = sin ⇒ 1  = −cos +  ⇒  = −cos + . () = 0 ⇒ − · (−1) +  = 0 ⇒  = −1, so  = −cos − . 20. (2 + 1)   + 3( − 1) = 0 ⇒ (2 + 1)0 + 3 = 3 ⇒ 0 + 23+ 1   = 23+ 1  . () =   3(2+1)  = (32) ln|2+1| = ln(2+1)32 = (2 + 1)32. Multiplying by (2 + 1)32 gives (2 + 1)32 0 + 3(2 + 1)12  = 3(2 + 1)12 ⇒ (2 + 1)320 = 3(2 + 1)12 ⇒ (2 + 1)32  =  3(2 + 1)12  = (2 + 1)32 +  ⇒  = 1 + (2 + 1)−32. Since (0) = 2, we have 2 = 1 + (1) ⇒  = 1 and hence,  = 1 + (2 + 1)−32. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.5 LINEAR EQUATIONS ¤ 841 21. 0 + 2 =  ⇒ 0 + 2   =   . () =   (2)  = 2 ln|| = ln||2 = ||2 = 2. Multiplying by () gives 2 0 + 2 =  ⇒ (2)0 =  ⇒ 2 =    = ( − 1) +  [by parts] ⇒  = [( − 1) + ]2. The graphs for  = −5, −3, −1, 1, 3, 5, and 7 are shown.  = 1 is a transitional value. For   1, there is an inflection point and for   1, there is a local minimum. As || gets larger, the “branches” get further from the origin. 22. 0 = 2 + 2 ⇔ 0 − 2 = 2 ⇔ 0 − 2   = . () =   −2  = −2 ln|| = (ln||)−2 = ||−2 = 1 2 . Multiplying by () gives 1 2 0 − 23  = 1 ⇒ 12 0 = 1 ⇒ 12  =  1  ⇒ 1 2  = ln|| +  ⇒  = (ln|| + )2. For all values of , as || → 0,  → 0, and as || → ∞,  → ∞. As || increases from 0, the function decreases and attains an absolute minimum. The inflection points, absolute minimums, and -intercepts all move farther from the origin as  decreases. 23. Setting  = 1−,   = (1 − )−   or   = 1−   =  1 −(1−)  . Then the Bernoulli differential equation becomes (1−) 1 −    + ()1(1−) = ()(1−) or   + (1 − )() = ()(1 − ). 24. Here 0 +  = −2 ⇒ 0 +   = −2, so  = 2, () = 1  and () = −1. Setting  = −1,  satisfies 0 − 1   = 1. Then () =  (−1) = 1  (for   0) and  =  1  +  = (ln|| + ). Thus,  = 1 ( + ln||). 25. Here 0 + 2   = 3 2 , so  = 3, () = 2 and () = 12 . Setting  = −2,  satisfies 0 − 4 = −22 . Then () =   (−4)  = −4 and  = 4 −26  +  = 4525 +  = 4 + 52. Thus,  = ±4 + 52−12. 26. 00 + 20 = 122 and  = 0 ⇒ 0 + 2 = 122 ⇒ 0 + 2   = 12. () =   (2)  = 2 ln|| = ln||2 = ||2 = 2. Multiplying the last differential equation by 2 gives °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 842 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 20 + 2 = 123 ⇒ (2)0 = 123 ⇒ 2 =  123  = 34 +  ⇒  = 32 + 2 ⇒ 0 = 32 + 2 ⇒  = 3 −  + . 27. (a) 2   + 10 = 40 or   + 5 = 20. Then the integrating factor is   5  = 5. Multiplying the differential equation by the integrating factor gives 5   + 55 = 205 ⇒ (5)0 = 205 ⇒ () = −5  205  +  = 4 + −5. But 0 = (0) = 4 + , so () = 4 − 4−5. (b) (01) = 4 − 4−05 ≈ 157 A 28. (a)   + 20 = 40 sin 60, so the integrating factor is 20. Multiplying the differential equation by the integrating factor gives 20   + 2020 = 4020 sin 60 ⇒ (20)0 = 4020 sin 60 ⇒ () = −20  4020 sin 60  +  = −20 4020  4000 1 (20 sin 60 − 60 cos 60) + −20 = sin 60 − 3cos 60 5 + −20 But 1 = (0) = − 3 5 + , so () = sin 60 − 3cos 60 + 8−20 5 . (b) (01) = sin 6 − 3cos 6 + 8−2 5 ≈ −042 A (c) 29. 5   + 20 = 60 with (0) = 0 C. Then the integrating factor is   4  = 4, and multiplying the differential equation by the integrating factor gives 4   + 44 = 124 ⇒ (4)0 = 124 ⇒ () = −4 124  +  = 3 + −4. But 0 = (0) = 3 +  so () = 3(1 − −4) is the charge at time  and  =  = 12−4 is the current at time . 30. 2   + 100 = 10 sin 60 or   + 50 = 5 sin 60. Then the integrating factor is   50  = 50, and multiplying the differential equation by the integrating factor gives 50   + 5050 = 550 sin 60 ⇒ (50)0 = 550 sin 60 ⇒ () = −50 550 sin 60  +  = −50550 6100 1 (50 sin 60 − 60 cos 60) + −50 = 1 122(5 sin 60 − 6 cos 60) + −50 But 0 = (0) = − 122 6 +  so  = 61 3 and () = 5sin 60122 − 6 cos 60 + 361 −50 is the charge at time , while the current is () =   = 150 cos 60 + 180 sin 60 − 150−50 61 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.5 LINEAR EQUATIONS ¤ 843 31.   +  = , so () =     = . Multiplying the differential equation by () gives    +   =  ⇒ ()0 =  ⇒ () = −  +  =  + −,   0. Furthermore, it is reasonable to assume that 0 ≤ (0) ≤ , so − ≤  ≤ 0. 32. Since (0) = 0, we have () = (1 − −). If 1() is Jim’s learning curve, then 1(1) = 25 and 1(2) = 45. Hence, 25 = 1(1 − −) and 45 = 1(1 − −2), so 1 − 251 = − or  = −ln1 − 251  = ln1−125. But 45 = 1(1 − −2) so 45 = 11 − 1−1252 or 45 = 501 −1 625. Thus, 1 = 125 is the maximum number of units per hour Jim is capable of processing. Similarly, if 2() is Mark’s learning curve, then 2(1) = 35 and 2(2) = 50. So  = ln2−2 35 and 50 = 21 − 2−2352 or 2 = 6125. Hence the maximum number of units per hour for Mark is approximately 61. Another approach would be to use the midpoints of the intervals so that 1(05) = 25 and 1(15) = 45. Doing so gives us 1 ≈ 526 and 2 ≈ 518. 33. (0) = 0 kg. Salt is added at a rate of 04 kg L 5 min L  = 2 min kg  Since solution is drained from the tank at a rate of 3 Lmin, but salt solution is added at a rate of 5 Lmin, the tank, which starts out with 100 L of water, contains (100 + 2) L of liquid after  min. Thus, the salt concentration at time  is () 100 + 2 kg L . Salt therefore leaves the tank at a rate of 100 + 2 ()  kg L 3 min L  = 100 + 2 3  min kg . Combining the rates at which salt enters and leaves the tank, we get   = 2 − 100 + 2 3 . Rewriting this equation as   + 100 + 2 3   = 2, we see that it is linear. () = exp  100 + 2 3  = exp 3 2 ln(100 + 2) = (100 + 2)32 Multiplying the differential equation by () gives (100 + 2)32   + 3(100 + 2)12 = 2(100 + 2)32 ⇒ [(100 + 2)32]0 = 2(100 + 2)32 ⇒ (100 + 2)32 = 2 5(100 + 2)52 +  ⇒  = 2 5(100 + 2) + (100 + 2)−32. Now 0 = (0) = 2 5(100) +  · 100−32 = 40 + 1000 1  ⇒  = −40,000, so  =  2 5(100 + 2) − 40,000(100 + 2)−32 kg. From this solution (no pun intended), we calculate the salt concentration at time  to be () = () 100 + 2 = (100 + 2 −40,000 )52 + 25 kg L . In particular, (20) = −140 40,5000 2 + 25 ≈ 02275 kg L and (20) = 2 5(140) − 40,000(140)−32 ≈ 3185 kg. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 844 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 34. Let () denote the amount of chlorine in the tank at time  (in seconds). (0) = (005 gL)(400 L) = 20 g. The amount of liquid in the tank at time  is (400 − 6) L since 4 L of water enters the tank each second and 10 L of liquid leaves the tank each second. Thus, the concentration of chlorine at time  is () 400 − 6 gL . Chlorine doesn’t enter the tank, but it leaves at a rate of 400 (−)6 Lg 10 Ls  = 400 10−(6) gs = 200 5−()3 gs . Therefore,   = −2005− 3 ⇒    =  200 −5−3 ⇒ ln = 5 3 ln(200 − 3) +  ⇒  = exp 5 3 ln(200 − 3) +  = (200 − 3)53. Now 20 = (0) =  · 20053 ⇒  = 20 20053 , so () = 20(200200 −533)53 = 20(1 − 0015)53 g for 0 ≤  ≤ 66 2 3 s, at which time the tank is empty. 35. (a)   +    =  and () =   ()  = (), and multiplying the differential equation by () gives ()   + ()  = () ⇒ ()0 = (). Hence, () = −() ()  +  =  + −(). But the object is dropped from rest, so (0) = 0 and  = −. Thus, the velocity at time  is () = ()[1 − −()]. (b) lim →∞ () =  (c) () =  () = ()[ + ()−()] + 1 where 1 = (0) − 22. (0) is the initial position, so (0) = 0 and () = ()[ + ()−()] − 22. 36.  = ()(1 − −) ⇒   =   0 − − · 2  +  (1 − −) · 1 = −  − +  −  − =   1 − − −  − ⇒     = 1 − 1 +  − = 1 − 1 +  = 1 − 1 + , where  =  ≥ 0. Since   1 +  for all   0, it follows that   0 for   0. In other words, for all   0,  increases as  increases. 37. (a)  = 1  ⇒  = 1 ⇒  0 = −20 . Substituting into  0 = (1 − ) gives us −20 =  1 1 − 1  ⇒ 0 = − 1 − 1  ⇒ 0 = − +  ⇒ 0 +  =  (). (b) The integrating factor is     = . Multiplying () by  gives 0 +  =   ⇒ ()0 =   ⇒  =     ⇒  = 1  +  ⇒  = 1 + −. Since  = 1, we have  = 1 1  + − ⇒  =  1 + − , which agrees with Equation 9.4.7,  = 1 +− , when  = . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.6 PREDATOR-PREY SYSTEMS ¤ 845 38. (a)  = 1  ⇒  = 1 ⇒  0 = −20 . Substituting into   = () 1 − () gives us − 0 2 = ()  1 − (1)  ⇒ 0 = −() 1 − (1)  ⇒ 0 = −() + (()) ⇒   + () = (()) (). (b) The integrating factor is (), where () = 0 (), so that 0() = (). Multiplying () by () gives ()   + ()() = ()()() ⇒ (())0 = 0()() () ⇒ () = 0 0()() ()  + , so  = 1 = 0 0()()(())  +  . Now suppose that  is a constant. Then () = 0 0() ()( ) +  =  () +( ) = 1 +   −() . If 0∞ () = ∞ then lim →∞ () = ∞, so lim →∞ () =  1 +  lim →∞ −() =  1 +  · 0 = . (c) If  is constant, but  varies, then () =  and we get  = 0 ( )  +  ⇒ () = 0 ( )  +   ⇒ () = − 0 ( )  + −. Suppose () has a limit as  → ∞, say lim →∞ () = . Then lim →∞ () = lim →∞ 1 () = lim →∞ 0 (  )  +  = lim →∞  ( )+ 0 l’Hospital’s and FTC 1  = lim →∞ () = . 9.6 Predator-Prey Systems 1. (a)  = −005 + 00001. If  = 0, we have  = −005, which indicates that in the absence of ,  declines at a rate proportional to itself. So  represents the predator population and  represents the prey population. The growth of the prey population, 01 (from  = 01 − 0005), is restricted only by encounters with predators (the term −0005). The predator population increases only through the term 00001; that is, by encounters with the prey and not through additional food sources. (b)  = −0015 + 000008. If  = 0, we have  = −0015, which indicates that in the absence of ,  would decline at a rate proportional to itself. So  represents the predator population and  represents the prey population. The growth of the prey population, 02 (from  = 02 − 000022 − 0006 = 02(1 − 0001) − 0006), is restricted by a carrying capacity of 1000 [from the term 1 − 0001 = 1 − 1000] and by encounters with predators (the term −0006). The predator population increases only through the term 000008; that is, by encounters with the prey and not through additional food sources. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 846 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 2. (a)  = 012 − 000062 + 000001.  = 008 + 000004. The  terms represent encounters between the two species  and . An increase in  makes  (the growth rate of ) larger due to the positive term 000001. An increase in  makes  (the growth rate of ) larger due to the positive term 000004. Hence, the system describes a cooperation model. (b)  = 015 − 000022 − 00006 = 015(1 − 750) − 00006.  = 02 − 0000082 − 00002 = 02(1 − 2500) − 00002. The system shows that  and  have carrying capacities of 750 and 2500. An increase in  reduces the growth rate of  due to the negative term −00002. An increase in  reduces the growth rate of  due to the negative term −00006. Hence, the system describes a competition model. 3. (a)  = 05 − 00042 − 0001 = 05(1 − 125) − 0001.  = 04 − 00012 − 0002 = 04(1 − 400) − 0002. The system shows that  and  have carrying capacities of 125 and 400. An increase in  reduces the growth rate of  due to the negative term −0002. An increase in  reduces the growth rate of  due to the negative term −0001. Hence the system describes a competition model. (b)  = 0 ⇒ (05 − 0004 − 0001) = 0 ⇒ (500 − 4 − ) = 0 (1) and  = 0 ⇒ (04 − 0001 − 0002) = 0 ⇒ (400 −  − 2) = 0 (2). From (1) and (2), we get four equilibrium solutions. (i)  = 0 and  = 0: If the populations are zero, there is no change. (ii)  = 0 and 400 −  − 2 = 0 ⇒  = 0 and  = 400: In the absence of an -population, the -population stabilizes at 400. (iii) 500 − 4 −  = 0 and  = 0 ⇒  = 125 and  = 0: In the absence of -population, the -population stabilizes at 125. (iv) 500 − 4 −  = 0 and 400 −  − 2 = 0 ⇒  = 500 − 4 and  = 400 − 2 ⇒ 500 − 4 = 400 − 2 ⇒ 100 = 2 ⇒  = 50 and  = 300: A -population of 300 is just enough to support a constant -population of 50. 4. Let (), (), and () represent the populations of lynx, hares, and willows at time . Let the ’s and the ’s denote positive constants, so that a plus sign means an increase and a minus sign means a decrease in the corresponding growth rate. “In the absence of hares, the willow population will grow exponentially and the lynx population will decay exponentially” gives us  = +1 and  = −2. “In the absence of lynx and willow, the hare population will decay exponentially” gives us  = −3. “Lynx eat snowshoe hares and snowshoe hares eat woody plants like willows” gives us encounters that lynx win, hares lose and win, and willows lose. In terms of growth rates, this means that  = +1,  = −2 + 3, and  = −4. Putting this information together gives us the following system of differential equations.  = − 2 +1  = − 3 − 2 +3  = +1 − 4 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.6 PREDATOR-PREY SYSTEMS ¤ 847 5. (a) At  = 0, there are about 300 rabbits and 100 foxes. At  = 1, the number of foxes reaches a minimum of about 20 while the number of rabbits is about 1000. At  = 2, the number of rabbits reaches a maximum of about 2400, while the number of foxes rebounds to 100. At  = 3, the number of rabbits decreases to about 1000 and the number of foxes reaches a maximum of about 315. As  increases, the number of foxes decreases greatly to 100, and the number of rabbits decreases to 300 (the initial populations), and the cycle starts again. (b) 6. (a) At  = 0, there are about 600 rabbits and 160 foxes. At  = 1, the number of rabbits reaches a minimum of about 80 and the number of foxes is also 80. At  = 2, the number of foxes reaches a minimum of about 25 while the number of rabbits rebounds to 1000. At  = 3, the number of foxes has increased to 40 and the rabbit population has reached a maximum of about 1750. The curve ends at  = 4, where the number of foxes has increased to 65 and the number of rabbits has decreased to about 950. (b) 7. 8. 9.   = −002 + 000002 008 − 0001 ⇔ (008 − 0001)  = (−002 + 000002)  ⇔ 008 − 0001   = −002 + 0 00002  ⇔  008 − 0001  =  −002 + 000002  ⇔ °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 848 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 008 ln|| − 0001 = −002 ln|| + 000002 +  ⇔ 008 ln + 002 ln = 0001 + 000002 +  ⇔ ln 008002 = 000002 + 0001 +  ⇔  008002 = 000002+0001 + ⇔ 002 008 = 0000020001 ⇔ 002 008 0000020001 = . In general, if   = − −+  , then  =   . 10. (a)  and  are constant ⇒ 0 = 0 and 0 = 0 ⇒ 0 = 2 0 = −0−50+ 0 01 0001  ⇒ 0 = 0 =  ((2 −0−5 + 0 0010001 ) ) So either  =  = 0 or  = 2 001 = 200 and  = 000001 5 = 5000. The trivial solution  =  = 0 just says that if there aren’t any aphids or ladybugs, then the populations will not change. The non-trivial solution,  = 200 and  = 5000, indicates the population sizes needed so that there are no changes in either the number of aphids or the number of ladybugs. (b)   =   = −05 + 00001 2 − 001 (c) The solution curves (phase trajectories) are all closed curves that have the equilibrium point (5000 200) inside them. (d) At 0(1000 200),  = 0 and  = −80  0, so the number of ladybugs is decreasing and hence, we are proceeding in a counterclockwise direction. At 0, there aren’t enough aphids to support the ladybug population, so the number of ladybugs decreases and the number of aphids begins to increase. The ladybug population reaches a minimum at 1(5000 100) while the aphid population increases in a dramatic way, reaching its maximum at 2(14250200). Meanwhile, the ladybug population is increasing from 1 to 3(5000 355), and as we pass through 2, the increasing number of ladybugs starts to deplete the aphid population. At 3 the ladybugs reach a maximum population, and start to decrease due to the reduced aphid population. Both populations then decrease until 0, where the cycle starts over again. (e) Both graphs have the same period and the graph of  peaks about a quarter of a cycle after the graph of . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE SECTION 9.6 PREDATOR-PREY SYSTEMS ¤ 849 11. (a) Letting  = 0 gives us  = 008(1 − 00002).  = 0 ⇔  = 0 or 5000. Since   0 for 0    5000, we would expect the rabbit population to increase to 5000 for these values of . Since   0 for   5000, we would expect the rabbit population to decrease to 5000 for these values of . Hence, in the absence of wolves, we would expect the rabbit population to stabilize at 5000. (b)  and  are constant ⇒  0 = 0 and  0 = 0 ⇒ 0 = 0 = 0 −08 002 (1 −+ 0 00002 00002  ) − 0001  ⇒ 0 = 0 = [0(−08(1 002 + 0 − 00002 00002 )−) 0001] The second equation is true if  = 0 or  = 0000002 02 = 1000. If  = 0 in the first equation, then either  = 0 or  = 1 00002 = 5000 [as in part (a)]. If  = 1000, then 0 = 1000[008(1 − 00002 · 1000) − 0001 ] ⇔ 0 = 80(1 − 02) −  ⇔  = 64. Case (i):  = 0,  = 0: both populations are zero Case (ii):  = 0,  = 5000: see part (a) Case (iii):  = 1000,  = 64: the predator/prey interaction balances and the populations are stable. (c) The populations of wolves and rabbits fluctuate around 64 and 1000, respectively, and eventually stabilize at those values. (d) 12. (a) If  = 0,  = 2(1 − 00001), so  = 0 ⇔  = 0 or  = 00001 1 = 10,000. Since   0 for 0    10,000, we expect the aphid population to increase to 10,000 for these values of . Since   0 for   10,000, we expect the aphid population to decrease to 10,000 for these values of . Hence, in the absence of ladybugs we expect the aphid population to stabilize at 10,000. (b)  and  are constant ⇒ 0 = 0 and 0 = 0 ⇒ 0 = 2 0 = −0(1 5−+ 0 00001 0001  ) − 001  ⇒ 0 = 0 =  ([2(1 −05 + 0 − 00001 0001 ) −) 001] The second equation is true if  = 0 or  = 000001 5 = 5000. If  = 0 in the first equation, then either  = 0 or  = 1 00001 = 10,000. If  = 5000, then 0 = 5000[2(1 − 00001 · 5000) − 001] ⇔ 0 = 10,000(1 − 05) − 50 ⇔ 50 = 5000 ⇔  = 100. The equilibrium solutions are: (i)  = 0  = 0 (ii)  = 0  = 10,000 (iii)  = 5000  = 100 (c)   =   = −05 + 00001 2(1 − 00001) − 001 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 850 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS (d) All of the phase trajectories spiral tightly around the equilibrium solution (5000100). (e) At  = 0, the ladybug population decreases rapidly and the aphid population decreases slightly before beginning to increase. As the aphid population continues to increase, the ladybug population reaches a minimum at about (5000 75). The ladybug population starts to increase and quickly stabilizes at 100, while the aphid population stabilizes at 5000. (f ) The graph of  peaks just after the graph of  has a minimum. 9 Review 1. True. Since 4 ≥ 0, 0 = −1 − 4  0 and the solutions are decreasing functions. 2. True. () =  = ln  ⇒ 0 = 1 − ln 2 . LHS = 20 +  = 2 · 1 − ln 2 +  · ln  = (1 − ln) + ln = 1 = RHS, so  = ln  is a solution of 20 +  = 1. 3. False.  +  cannot be written in the form ()(). 4. True. 0 = 3 − 2 + 6 − 1 = 6 − 2 + 3 − 1 = 2(3 − 1) + 1(3 − 1) = (2 + 1)(3 − 1), so 0 can be written in the form ()(), and hence, is separable. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE CHAPTER 9 REVIEW ¤ 851 5. True. 0 =  ⇒ 0 = − ⇒ 0 + (−−) = 0, which is of the form 0 + () = (), so the equation is linear. 6. False. 0 +  =  cannot be put in the form 0 + () = (), so it is not linear. 7. True. By comparing   = 21 − 5 with the logistic differential equation (9.4.4), we see that the carrying capacity is 5; that is, lim →∞  = 5. 1. (a) (b) lim →∞ () appears to be finite for 0 ≤  ≤ 4. In fact lim →∞ () = 4 for  = 4, lim →∞ () = 2 for 0    4, and lim →∞ () = 0 for  = 0. The equilibrium solutions are () = 0, () = 2, and () = 4. 2. (a) We sketch the direction field and four solution curves, as shown. Note that the slope 0 =  is not defined on the line  = 0. (b) 0 =  ⇔   =   ⇔ 2 = 2 + . For  = 0, this is the pair of lines  = ±. For  6= 0, it is the hyperbola 2 − 2 = −. 3. (a) We estimate that when  = 03,  = 08, so (03) ≈ 08. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 852 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS (b)  = 01, 0 = 0, 0 = 1 and ( ) = 2 − 2. So  = −1 + 012 −1 − 2−1. Thus, 1 = 1 + 0102 − 12 = 09, 2 = 09 + 01012 − 092 = 082, 3 = 082 + 01022 − 0822 = 075676. This is close to our graphical estimate of (03) ≈ 08. (c) The centers of the horizontal line segments of the direction field are located on the lines  =  and  = −. When a solution curve crosses one of these lines, it has a local maximum or minimum. 4. (a)  = 02, 0 = 0, 0 = 1 and ( ) = 22. We need 2. 1 = 1 + 02(2 · 0 · 12) = 1, 2 = 1 + 02(2 · 02 · 12) = 108 ≈ (04). (b)  = 01 now, so 1 = 1 + 01(2 · 0 · 12) = 1, 2 = 1 + 01(2 · 01 · 12) = 102, 3 = 102 + 01(2 · 02 · 1022) ≈ 106162, 4 = 106162 + 01(2 · 03 · 1061622) ≈ 11292 ≈ (04). (c) The equation is separable, so we write  2 = 2  ⇒   2 =  2  ⇔ −1 = 2 + , but (0) = 1, so  = −1 and () = 1 1 − 2 ⇔ (04) = 1 −1016 ≈ 11905. From this we see that the approximation was greatly improved by increasing the number of steps, but the approximations were still far off. 5. 0 = − sin  −  cos ⇒ 0 + (cos) = − sin  (). This is a linear equation and the integrating factor is () =   cos   = sin . Multiplying () by sin  gives sin  0 + sin (cos) =  ⇒ (sin  )0 =  ⇒ sin   = 1 22 +  ⇒  =  1 22 +  − sin . 6.   = 1 −  +  −  = 1(1 − ) + (1 − ) = (1 + )(1 − ) ⇒ 1 +  = (1 − ) ⇒  1 +  =  (1 − ) ⇒ ln|1 + | =  − 1 22 +  ⇒ |1 + | = −22+ ⇒ 1 +  = ±−22 ·  ⇒  = −1 + −22, where  is any nonzero constant. 7. 220 = 2 + 3√ ⇒ 22   = 2 + 3√ ⇒ 22  = 2 + 3√   ⇒  22  =  2 + 3√   ⇒ 2 = 2 + 232 +  ⇒ 2 = ln(2 + 232 + ) ⇒  = ±ln(2 + 232 + ) 8. 20 −  = 23−1 ⇒ 0 − 1 2  = 2−1 (). This is a linear equation and the integrating factor is () =  (−12)  = 1. Multiplying () by 1 gives 1 0 − 1 · 1 2  = 2 ⇒ (1 )0 = 2 ⇒ 1  = 2 +  ⇒  = −1(2 + ). 9.   + 2 =  ⇒   =  − 2 = (1 − 2) ⇒    =  (1 − 2) ⇒ ln|| =  − 2 +  ⇒ || = −2+ = −2. Since (0) = 5, 5 = 0 = . Thus, () = 5−2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE CHAPTER 9 REVIEW ¤ 853 10. (1 + cos )0 = (1 + −)sin ⇒  1 + − = sin  1 + cos ⇒  1 + 1  =  1 + cos sin  ⇒  1 +    =  1 + cos sin  ⇒ ln|1 + | = −ln|1 + cos| +  ⇒ ln(1 + ) = −ln(1 + cos) +  ⇒ 1 +  = − ln(1+cos ) ·  ⇒  = − ln(1+cos ) − 1 ⇒  = ln[− ln(1+cos ) − 1]. Since (0) = 0, 0 = ln[− ln 2 − 1] ⇒ 0 =  1 2 − 1 ⇒  = 4. Thus, () = ln[4− ln(1+cos ) − 1]. An equivalent form is () = ln 3 − cos  1 + cos  . 11. 0 −  = ln ⇒ 0 − 1   = ln. () =   (−1)  = − ln|| = ln||−1 = ||−1 = 1 since the condition (1) = 2 implies that we want a solution with   0. Multiplying the last differential equation by () gives 1  0 − 1 2  = 1  ln ⇒ 1 0 = 1 ln ⇒ 1  =  ln  ⇒ 1  = 1 2(ln)2 +  ⇒  = 1 2(ln)2 + . Now (1) = 2 ⇒ 2 = 0 +  ⇒  = 2, so  = 1 2(ln)2 + 2. 12. 0 = 32 ⇒   = 32 ⇒ −  = 32  ⇒  −  =  32  ⇒ −− = 3 + . Now (0) = 1 ⇒ −−1 = , so −− = 3 − −1 ⇒ − = −3 + −1 ⇒ − = ln(−3 + −1) ⇒  = −ln(−3 + −1). To find the domain, solve −3 + −1  0 ⇒ 3  −1 ⇒   −13, so the domain is (−∞ −13) and  = −13 [≈072] is a vertical asymptote. 13.   () =   () ⇒ 0 =  = , so the orthogonal trajectories must have 0 = −1 ⇒   = −1 ⇒   = − ⇒    = −   ⇒ 1 22 = − +  ⇒  =  − 1 22, which are parabolas with a horizontal axis. 14.   () =   () ⇒ 0 =  =  = ln · , so the orthogonal trajectories must have 0 = − ln   ⇒   = −   ln ⇒  ln  = −  ⇒   ln  = −    ⇒ 1 22 ln − 1 42 [parts with  = ln,  =  ] = − 1 22 + 1 ⇒ 22 ln − 2 =  − 22. 15. (a) Using (4) and (7) in Section 9.4, we see that for   = 011 − 2000   with (0) = 100, we have  = 01,  = 2000, 0 = 100, and  = 2000 − 100 100 = 19. Thus, the solution of the initial-value problem is () = 2000 1 + 19−01 and (20) = 1 + 19 2000−2 ≈ 560. (b)  = 1200 ⇔ 1200 = 2000 1 + 19−01 ⇔ 1 + 19−01 = 2000 1200 ⇔ 19−01 = 53 − 1 ⇔ −01 =  2 319 ⇔ −01 = ln 57 2 ⇔  = −10 ln 57 2 ≈ 335. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 854 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS 16. (a) Let  = 0 correspond to the year 2000. An exponential model is () = . (0) = 61, so () = 61. (10) = 6110 and (10) = 69, so 6110 = 69 ⇔ 69 61 = 10 ⇒ 10 = ln 69 61 ⇒  = 1 10 ln 69 61 ≈ 00123. Thus, () = 61 and (20) = 6120 ≈ 78. Our model predicts that the world population in the year 2020 will be 78 billion. (b) () = 10 ⇔ 61 = 10 ⇔  = 10 61 ⇔  = ln 10 61 ⇔  = 10 ln(1061) ln(6961) ≈ 4011 years. Our exponential model predicts that the world population will exceed 10 billion in 4011 years; that is, in the year 2040. (c)  =  − 0 0 = 20 − 61 61 = 139 61 and from part (a) ,  = 1 10 ln 69 61 , so () =  1 + − = 20 1 + 139 61 − . Thus, (20) = 20 1 + 139 61 −20 ≈ 72 billion, which is less than our prediction of 78 billion from the exponential model in part (a). (d) () = 10 ⇔ 20 1 + 139 61 − = 10 ⇔ 20 10 = 1 + 139 61 − ⇔ 1 = 139 61 − ⇔ 139 61 = − ⇔ ln 61 139 = − ⇔  = −10 ln(61139) ln(6961) ≈ 6683 years. Our logistic model predicts that the world population will exceed 10 billion in 6683 years; that is, in the year 2066, which is considerably later than our prediction of 2040 from the exponential model in part (b). 17. (a)   ∝ ∞ −  ⇒   = (∞ − ) ⇒  ∞−  =    ⇒ −ln|∞ − | =  +  ⇒ ln|∞ − | = − −  ⇒ |∞ − | = −− ⇒ ∞ −  = − ⇒  = ∞ − −. At  = 0,  = (0) = ∞ −  ⇒  = ∞ − (0) ⇒ () = ∞ − [∞ − (0)]−. (b) ∞ = 53 cm, (0) = 10 cm, and  = 02 ⇒ () = 53 − (53 − 10)−02 = 53 − 43−02. 18. Denote the amount of salt in the tank (in kg) by . (0) = 0 since initially there is only water in the tank. The rate at which  increases is equal to the rate at which salt flows into the tank minus the rate at which it flows out. That rate is   = 01kg L × 10min L − 100  kg L × 10min L = 1 − 10  min kg ⇒  10 −  =  10 1  ⇒ −ln|10 − | = 10 1  +  ⇒ 10 −  = −10. (0) = 0 ⇒ 10 =  ⇒  = 10(1 − −10). At  = 6 minutes,  = 10(1 − −610) ≈ 4512 kg. 19. Let  represent the population and  the number of infected people. The rate of spread  is jointly proportional to  and to  − , so for some constant ,   = ( − ) ⇒ () = 0 + ( −00)−  [from the discussion of logistic growth in Section 9.4]. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE CHAPTER 9 REVIEW ¤ 855 Now, measuring  in days, we substitute  = 7,  = 5000, 0 = 160 and (7) = 1200 to find : 1200 = 160 · 5000 160 + (5000 − 160)−5000·7· ⇔ 3 = 160 + 4840 2000−35,000 ⇔ 480 + 14,520−35,000 = 2000 ⇔ −35,000 = 2000 − 480 14,520 ⇔ −35,000 = ln 363 38 ⇔  = 35−,000 1 ln 363 38 ≈ 000006448. Next, let  = 5000 × 80% = 4000, and solve for : 4000 = 160 · 5000 160 + (5000 − 160)−·5000· ⇔ 1 = 160 + 4840 200−5000 ⇔ 160 + 4840−5000 = 200 ⇔ −5000 = 200 − 160 4840 ⇔ −5000 = ln 121 1 ⇔  = −1 5000 ln 121 1 = 1 1 7 ln 363 38 · ln 121 1 = 7 · ln 121 ln 363 38 ≈ 14875. So it takes about 15 days for 80% of the population to be infected. 20. 1    =     ⇒  (ln) =   ( ln) ⇒ ln =  ln +  ⇒  =  ln  +  = ln  ⇒  = , where  =  is a positive constant. 21.   = −     +   ⇒   +   =  −   ⇒  1 +    = −  1 ⇒  +  ln = −   + . This equation gives a relationship between  and , but it is not possible to isolate  and express it in terms of . 22.  = 04 − 0002,  = −02 + 0000008 (a) The  terms represent encounters between the birds and the insects. Since the -population increases from these terms and the -population decreases, we expect  to represent the birds and  the insects. (b)  and  are constant ⇒ 0 = 0 and 0 = 0 ⇒ 0 = 0 = 0 −4 0 2−+ 0 0002 000008    ⇒ 0 = 0 0 = −4 0 2(1 (1 −−00005 00004 ) ) ⇒  = 0 and  = 0 (zero populations) or  = 0005 1 = 200 and  = 000004 1 = 25,000. The non-trivial solution represents the population sizes needed so that there are no changes in either the number of birds or the number of insects. (c)   =   = −02 + 0000008 04 − 0002 (d) At ( ) = (40,000100),  = 8000  0, so as  increases we are proceeding in a counterclockwise direction. The populations increase to approximately (59,646200), at which point the insect population starts to decrease. The birds attain a maximum population of about 380 when the insect population is 25,000. The populations decrease to about (7370200), at which point the insect population starts to increase. The birds attain a minimum population of about 88 when the insect population is 25,000, and then the cycle repeats. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 856 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS (e) Both graphs have the same period and the bird population peaks about a quarter-cycle after the insect population. 23. (a)  = 04(1 − 0000005) − 0002,  = −02 + 0000008. If  = 0, then  = 04(1 − 0000005), so  = 0 ⇔  = 0 or  = 200,000, which shows that the insect population increases logistically with a carrying capacity of 200,000. Since   0 for 0    200,000 and   0 for   200,000, we expect the insect population to stabilize at 200,000. (b)  and  are constant ⇒ 0 = 0 and 0 = 0 ⇒ 0 = 0 = 0 −042(1  + 0 − 0000008 000005 ) − 0002  ⇒ 0 = 0 0 = (4−0[(1 2 + 0 − 0000008 000005)) − 0005] The second equation is true if  = 0 or  = 0000008 02 = 25,000. If  = 0 in the first equation, then either  = 0 or  = 1 0000005 = 200,000. If  = 25,000, then 0 = 04(25,000)[(1 − 0000005 · 25,000) − 0005] ⇒ 0 = 10,000[(1 − 0125) − 0005] ⇒ 0 = 8750 − 50 ⇒  = 175. Case (i):  = 0,  = 0: Zero populations Case (ii):  = 0,  = 200,000: In the absence of birds, the insect population is always 200,000. Case (iii):  = 25,000,  = 175: The predator/prey interaction balances and the populations are stable. (c) The populations of the birds and insects fluctuate around 175 and 25,000, respectively, and eventually stabilize at those values. (d) 24. First note that, in this question, “weighs” is used in the informal sense, so what we really require is Barbara’s mass  in kg as a function of . Barbara’s net intake of calories per day at time  (measured in days) is () = 1600 − 850 − 15() = 750 − 15(), where () is her mass at time . We are given that (0) = 60 kg and   = () 10,000 , so   = 750 − 15 10,000 = 150 − 3 2000 = −3( − 50) 2000 with (0) = 60. From   − 50 =  −2000 3, we get ln| − 50| = − 2000 3  + . Since (0) = 60,  = ln10. Now ln |10 − 50| = −2000 3 , so | − 50| = 10−32000. The quantity  − 50 is continuous, initially positive, and the right-hand side is never zero. Thus,  − 50 is positive for all , and () = 50 + 10−32000 kg. As  → ∞, () → 50 kg. Thus, Barbara’s mass gradually settles down to 50 kg. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE PROBLEMS PLUS 1. We use the Fundamental Theorem of Calculus to differentiate the given equation: [()]2 = 100 + 0 [()]2 + [ 0()]2  ⇒ 2() 0() = [()]2 + [ 0()]2 ⇒ [()]2 + [ 0()]2 − 2() 0() = 0 ⇒ [() −  0()]2 = 0 ⇔ () =  0(). We can solve this as a separable equation, or else use Theorem 9.4.2 with  = 1, which says that the solutions are () = . Now [(0)]2 = 100, so (0) =  = ±10, and hence () = ±10 are the only functions satisfying the given equation. 2. ()0 =  00, where () = 2 ⇒ 20 = 220. Since the student’s mistake did not affect the answer, 20 = 20 + 22 = 220. So (2 − 1)0 = 2, or 0 = 22− 1 = 1 + 21− 1 ⇒ ln|()| =  + 1 2 ln(2 − 1) +  ⇒ () = √2 − 1. 3.  0() = lim →0 ( + ) − ()  = lim →0 ()[() − 1] [since ( + ) = ()()] = () lim →0 () − 1  = () lim →0 () − −  0(0) = () 0(0) = () Therefore,  0() = () for all  and from Theorem 9.4.2 we get () = . Now (0) = 1 ⇒  = 1 ⇒ () = . 4.  ()  () = −1 ⇒   () =  −(1) ⇒ (1) =  (())2 [after differentiating] ⇒  () = ±() [after taking square roots] ⇒ () = ± 0() [after differentiating again] ⇒  =  or  = − by Theorem 9.4.2. Therefore, () =  or () = −, for all nonzero constants , are the functions satisfying the original equation. 5. “The area under the graph of  from 0 to  is proportional to the ( + 1)st power of ()” translates to 0 () = [()]+1 for some constant . By FTC1,   0 () =   [()]+1 ⇒ () = ( + 1)[()] 0() ⇒ 1 = ( + 1)[()]−1 0() ⇒ 1 = ( + 1)−1   ⇒ ( + 1)−1  =  ⇒  ( + 1)−1  =   ⇒ ( + 1) 1   =  + . Now (0) = 0 ⇒ 0 = 0 +  ⇒  = 0 and then (1) = 1 ⇒ ( + 1) 1  = 1 ⇒  =   + 1, so  =  and  = () = 1. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. 857 FOR INSTRUCTOR USE ONLYNOT FOR SALE 858 ¤ CHAPTER 9 PROBLEMS PLUS 6. Let  = () be a curve that passes through the point (1) and whose subtangents all have length . The tangent line at  =  has equation  − () =  0()( − ). Assuming () 6= 0 and  0() 6= 0, it has -intercept  − ()  0() [let  = 0 and solve for ]. Thus, the length of the subtangent is , so   −  − 0(())    =    0(())    =  ⇒ 0(()) = ±1 . Now  0() () = ±1 ⇒  0() = ±1 () ⇒   = ±1  ⇒   = ±1  ⇒  1  = ±1   ⇒ ln|| = ±1   + . Since () = 1, ln 1 = ±1 +  ⇒  = ∓1. Thus,  = ±∓1, or  = ±(−1). One curve is an increasing exponential (as shown in the figure) and the other curve is its reflection about the line  = . 7. Let () denote the temperature of the peach pie  minutes after 5:00 PM and  the temperature of the room. Newton’s Law of Cooling gives us  = ( − ) Solving for  we get   −  =   ⇒ ln| − | =  +  ⇒ | − | = + ⇒  −  = ± ·  ⇒  =  + , where  is a nonzero constant. We are given temperatures at three times. (0) = 100 ⇒ 100 =  +  ⇒  = 100 −  (10) = 80 ⇒ 80 = 10 +  (1) (20) = 65 ⇒ 65 = 20 +  (2) Substituting 100 −  for  in (1) and (2) gives us −20 = 10 −  (3) and −35 = 20 −  (4) Dividing (3) by (4) gives us −20 −35 = 10 − 1 (20 − 1) ⇒ 4 7 = 10 − 1 20 − 1 ⇒ 420 − 4 = 710 − 7 ⇒ 420 − 710 + 3 = 0. This is a quadratic equation in 10. 410 − 310 − 1 = 0 ⇒ 10 = 3 4 or 1 ⇒ 10 = ln 3 4 or ln 1 ⇒  = 10 1 ln 3 4 since  is a nonzero constant of proportionality. Substituting 3 4 for 10 in (3) gives us −20 =  · 3 4 −  ⇒ −20 = − 1 4  ⇒  = 80. Now  = 100 −  so  = 20◦C. 8. Let  be the number of hours before noon that it began to snow,  the time measured in hours after noon, and  = () = distance traveled by the plow at time . Then  = speed of plow. Since the snow falls steadily, the height at time  is () = ( + ), where  is a constant. We are given that the rate of removal is constant, say  (in m3h). If the width of the path is , then  = height × width × speed = () ×  ×   = ( + )   . Thus,   =  + , where  =   is a constant. This is a separable equation.   =     +  ⇒ () =  ln( + ) + . Put  = 0: 0 =  ln +  ⇒  = − ln, so () =  ln( + ) −  ln =  ln(1 + ). Put  = 1: 6000 =  ln(1 + 1) [ = 6 km]. Put  = 2: 9000 =  ln(1 + 2) [ = (6 + 3) km]. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE CHAPTER 9 PROBLEMS PLUS ¤ 859 Solve for : ln(1 + 1) 6000 = ln(1 + 2) 9000 ⇒ 3ln1 + 1  = 2 ln1 + 2  ⇒ 1 + 1 3 = 1 + 2 2 ⇒ 1 + 3  + 3 2 + 1 3 = 1 + 4  + 4 2 ⇒ 1  + 1 2 − 1 3 = 0 ⇒ 2 +  − 1 = 0 ⇒  = −1 ± √5 2 . But   0, so  = −1 +2 √5 ≈ 0618 h ≈ 37 min. The snow began to fall √52− 1 hours before noon; that is, at about 11:23 AM. 9. (a) While running from (0) to ( ), the dog travels a distance  = 1 + ()2  = −1 + ()2 , so   = −1 + ()2. The dog and rabbit run at the same speed, so the rabbit’s position when the dog has traveled a distance  is (0 ). Since the dog runs straight for the rabbit,   =  −  0 −  (see the figure). Thus,  =  −    ⇒   =   −  22 + 1    = − 22 . Equating the two expressions for   gives us   2 2 = 1 +   2, as claimed. (b) Letting  =  , we obtain the differential equation    = √1 + 2, or √1 +  2 =   . Integrating: ln =  √1 +  2 = ln 25  + 1 + 2  + . When  = ,  =  = 0, so ln = ln 1 + . Therefore,  = ln, so ln = ln√1 + 2 +  + ln = ln√1 + 2 +  ⇒  = √1 + 2 +  ⇒ √1 + 2 =   −  ⇒ 1 + 2 =  2 − 2 + 2 ⇒  2 − 2  − 1 = 0 ⇒  = ()2 − 1 2() = 2 − 2 2 =  2 −  2 1  [for   0]. Since  =  ,  = 2 4 −  2 ln + 1. Since  = 0 when  = , 0 =  4 −  2 ln + 1 ⇒ 1 =  2 ln −  4 . Thus,  = 2 4 −  2 ln +  2 ln −  4 = 2 − 2 4 −  2 ln   (c) As  → 0+,  → ∞, so the dog never catches the rabbit. 10. (a) If the dog runs twice as fast as the rabbit, then the rabbit’s position when the dog has traveled a distance  is (0 2). Since the dog runs straight toward the rabbit, the tangent line to the dog’s path has slope   = 2 −  0 −  . Thus,  = 2 − 2   ⇒   = 2   − 2 22 + 2    = −2  22 . From Problem 9(a),   = −1 +   2, so 2 22 = 1 +   2. Letting  =  , we obtain the differential equation 2   = √1 + 2, or √1 + 22 =   . Integrating, we get °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 860 ¤ CHAPTER 9 PROBLEMS PLUS ln =  √1 + 22 = 2ln1 + 2 +  + . [See Problem 9(b).] When  = ,  =  = 0, so ln = 2 ln 1 +  = . Thus, ln = 2 ln√1 + 2 +  + ln = ln√1 + 2 + 2 ⇒  = √1 + 2 + 2 ⇒ √1 + 2 =  −  ⇒ 1 + 2 =  − 2  + 2 ⇒ 2  =  − 1 ⇒   =  = 1 2  − 21 = 2√1 12 − √2 −12 ⇒  = 3√1 32 − √ 12 + 1. When  = ,  = 0, so 0 = 1 3√ 32 − √ 12 + 1 = 3 −  + 1 = 1 − 2 3. Therefore, 1 = 2 3 and  = 32 3√ − √ 12 + 2 3. As  → 0,  → 2 3, so the dog catches the rabbit when the rabbit is at 0 2 3. (At that point, the dog has traveled a distance of 4 3, twice as far as the rabbit has run.) (b) As in the solutions to part (a) and Problem 9, we get  =   = 2 22 − 2 22 and hence  = 632 + 22 − 2 3. We want to minimize the distance  from the dog at ( ) to the rabbit at (0 2). Now  = 1 2 − 1 2   ⇒ 2 =  −  ⇒  − 2 =  = 222 − 222  = 232 − 22 , so  = ( − 0)2 + ( − 2)2 = 2 + 232 − 22 2 = 464 + 22 + 442 = 232 + 22 2 = 3 22 + 2 2 0 = 0 ⇔ 32 22 − 2 22 = 0 ⇔ 2322 = 222 ⇔ 4 = 34 ⇔  = √43,   0,   0. Since 00() = 3 2 + 2 3  0 for all   0, we know that  √43 = ( ·23−214)3 + 2 ·3−2 14) = 3234 is the minimum value of , that is, the closest the dog gets to the rabbit. The positions at this distance are Dog: ( ) =  √43 3754 − 2 3 =  √43 5 √4 39− 6 Rabbit: (02) = 0 8 √493 − 23 = 0 8 √4 39− 6 11. (a) We are given that  = 1 32,  = 60,000 ft3h, and  = 15 = 3 2. So  = 1 3 3 22  = 3 43 ⇒   = 3 4  · 32   = 9 4 2   . Therefore,   = 4(9 2 ) = 240 9 ,000 2  = 803,000 2 () ⇒  32  =  80,000 ⇒ 3 = 80,000 + . When  = 0,  = 60. Thus,  = 603 = 216,000, so °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE CHAPTER 9 PROBLEMS PLUS ¤ 861 3 = 80,000 + 216,000. Let  = 100. Then 1003 = 1,000,000 = 80,000 + 216,000 ⇒ 80,000 = 784,000 ⇒  = 98, so the time required is 98 hours. (b) The floor area of the silo is  =  · 2002 = 40,000 ft2, and the area of the base of the pile is  = 2 =  3 22 = 94 2. So the area of the floor which is not covered when  = 60 is  −  = 40,000 − 8100 = 31,900 ≈ 100,217 ft2. Now  = 94 2 ⇒  = 94 · 2(), and from () in part (a) we know that when  = 60,  = 3(60) 80,0002 = 200 27 fth. Therefore,  = 94(2)(60) 200 27  = 2000 ≈ 6283 ft2h. (c) At  = 90 ft,  = 60,000 − 20,000 = 40,000 ft3h. From () in part (a),   = 4() 92 = 4(40,000) 92 = 160,000 92 ⇒  92  =  160,000 ⇒ 33 = 160,000 + . When  = 0,  = 90; therefore,  = 3 · 729,000 = 2,187,000. So 33 = 160,000 + 2,187,000. At the top,  = 100 ⇒ 3(100)3 = 160,000 + 2,187,000 ⇒  = 813 160, ,000 000 ≈ 51. The pile reaches the top after about 51 h. 12. Let ( ) be any first-quadrant point on the curve  = (). The tangent line at  has equation  −  =  0()( − ), or equivalently,  =  +  − , where  =  0(). If (0 ) is the -intercept, then  =  − . If ( 0) is the -intercept, then  =  −   =  −   . Since the tangent line is bisected at  , we know that || = | |; that is, ( − 0)2 + [ − ( − )]2 = [ − ( − )]2 + ( − 0)2. Squaring and simplifying gives us 2 + 22 = 22 + 2 ⇒ 22 + 24 = 2 + 22 ⇒ 24 + 2 − 22 − 2 = 0 ⇒ 22 − 22 + 1 = 0 ⇒ 2 = 22. Since  is the slope of the line from a positive -intercept to a positive -intercept,  must be negative. Since  and  are positive, we have  = −, so we will solve the equivalent differential equation   = −   ⇒   = −   ⇒    = −    ⇒ ln = −ln +  [   0] ⇒  = − ln + = ln −1 ·  = −1 ·  ⇒  = . Since the point (32) is on the curve, 3 = 2 ⇒  = 6 and the curve is  = 6 with   0. 13. Let ( ) be any point on the curve. If  is the slope of the tangent line at , then  = 0(), and an equation of the normal line at  is  −  = − 1  ( − ), or equivalently,  = − 1   +  +   . The -intercept is always 6, so  +   = 6 ⇒   = 6 −  ⇒  =  6 − . We will solve the equivalent differential equation   = 6 −  ⇒ (6 − ) =   ⇒  (6 − ) =    ⇒ 6 − 1 22 = 1 22 +  ⇒ 12 − 2 = 2 + . Since (32) is on the curve, 12(2) − 22 = 32 +  ⇒  = 11. So the curve is given by 12 − 2 = 2 + 11 ⇒ 2 + 2 − 12 + 36 = −11 + 36 ⇒ 2 + ( − 6)2 = 25, a circle with center (06) and radius 5. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLYNOT FOR SALE 862 ¤ CHAPTER 9 PROBLEMS PLUS 14. Let (0 0) be a point on the curve. Since the midpoint of the line segment determined by the normal line from (0 0) to its intersection with the -axis has -coordinate 0, the -coordinate of the point of intersection with the -axis must be −0. Hence, the normal line has slope 0 − 0 0 − (−0) = 0 20 . So the tangent line has slope −20 0 . This gives the differential equation 0 = −2  ⇒   = −2  ⇒    =  (−2) ⇒ 1 2 2 = −2 +  ⇒ 2 + 1 2 2 =  [  0]. This is a family of ellipses. 15. From the figure, slope  =   . If triangle  is isosceles, then slope  must be −  , the negative of slope . This slope is also equal to  0(), so we have   = −   ⇒    = −    ⇒ ln|| = −ln|| +  ⇒ || = − ln||+ ⇒ || = (ln||)−1 ⇒ || = 1 || ⇒  =  ,  6= 0. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. FOR INSTRUCTOR USE ONLY [Show More]

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