Differential Equations

6 Differential Equations In this chapter, ou will stu one of the most important applications of calculus differential equations. You will learn several methods for solving different tpes of differential equations, such as homogeneous, first-order linear, and Bernoulli. Then ou will appl these methods to solve differential equations in applied problems. In this chapter, ou should learn the following. How to sketch a slope field of a differential equation, and find a particular solution. (6.) How to use an eponential function to model growth and deca. (6.) How to use separation of variables to solve a differential equation. (6.3) How to solve a first-order linear differential equation and a Bernoulli differential equation. (6.4) Dr. Dennis Kunkel/Gett Images Depending on the tpe of bacteria, the time it takes for a culture s weight to double can var greatl from several minutes to several das. How could ou use a differential equation to model the growth rate of a bacteria culture s weight? (See Section 6.3, Eercise 84.) A function f is a solution of a differential equation if the equation is satisfied when and its derivatives are replaced b f and its derivatives. One wa to solve a differential equation is to use slope fields, which show the general shape of all solutions of a differential equation. (See Section 6..) 405

406 Chapter 6 Differential Equations 6. Slope Fields and Euler s Method Use initial conditions to find particular solutions of differential equations. Use slope fields to approimate solutions of differential equations. Use Euler s Method to approimate solutions of differential equations. General and Particular Solutions In this tet, ou will learn that phsical phenomena can be described b differential equations. Recall that a differential equation in and is an equation that involves,, and derivatives of. In Section 6., ou will see that problems involving radioactive deca, population growth, and Newton s Law of Cooling can be formulated in terms of differential equations. A function f is called a solution of a differential equation if the equation is satisfied when and its derivatives are replaced b f and its derivatives. For eample, differentiation and substitution would show that e is a solution of the differential equation It can be shown that ever solution of this differential equation is of the form Ce 0. General solution of where C is an real number. This solution is called the general solution. Some differential equations have singular solutions that cannot be written as special cases of the general solution. However, such solutions are not considered in this tet. The order of a differential equation is determined b the highest-order derivative in the equation. For instance, is a first-order differential equation. First-order linear differential equations are discussed in Section 6.4. In Section 4., Eample 8, ou saw that the second-order differential equation s t 3 has the general solution 4 0 s t 6t C t C General solution of which contains two arbitrar constants. It can be shown that a differential equation of order n has a general solution with n arbitrar constants. s t 3 EXAMPLE Verifing Solutions Determine whether the function is a solution of the differential equation a. sin b. 4e c. Ce Solution a. Because sin, and it follows that cos, sin sin sin 0. So, sin is not a solution. b. Because 4e, and it follows that 4e, 4e 4e 0. So, 4e is a solution. c. Because Ce, and it follows that Ce, Ce Ce 0. sin, 4e, Ce, 0. So, Ce is a solution for an value of C.

6. Slope Fields and Euler s Method 407 C = General solution: = C Geometricall, the general solution of a first-order differential equation represents a famil of curves known as solution curves, one for each value assigned to the arbitrar constant. For instance, ou can verif that ever function of the form C = C = C = Solution curves for Figure 6. C = C = C = 0 C = C General solution of is a solution of the differential equation Figure 6. shows four of the solution curves corresponding to different values of C. As discussed in Section 4., particular solutions of a differential equation are obtained from initial conditions that give the values of the dependent variable or one of its derivatives for particular values of the independent variable. The term initial condition stems from the fact that, often in problems involving time, the value of the dependent variable or one of its derivatives is known at the initial time t 0. For instance, the second-order differential equation s t 3 having the general solution s t 6t C t C General solution of s t 3 might have the following initial conditions. s 0 80, s 0 64 Initial conditions In this case, the initial conditions ield the particular solution s t 6t 64t 80. 0. Particular solution 0 EXAMPLE Finding a Particular Solution For the differential equation verif that C 3 is a solution, and find the particular solution determined b the initial condition when 3. Solution You know that C 3 is a solution because and 3 3C 3 C 3 3 0, 3C 0. Furthermore, the initial condition when 3ields C 3 General solution C 3 3 Substitute initial condition. 7 C Solve for C. and ou can conclude that the particular solution is 3 7. Particular solution Tr checking this solution b substituting for and in the original differential equation. NOTE To determine a particular solution, the number of initial conditions must match the number of constants in the general solution. The icon indicates that ou will find a CAS Investigation on the book s website. The CAS Investigation is a collaborative eploration of this eample using the computer algebra sstems Maple and Mathematica.

408 Chapter 6 Differential Equations Slope Fields Solving a differential equation analticall can be difficult or even impossible. However, there is a graphical approach ou can use to learn a lot about the solution of a differential equation. Consider a differential equation of the form F, Differential equation where F, is some epression in and. At each point, in the -plane where F is defined, the differential equation determines the slope of the solution at that point. If ou draw short line segments with slope F, at selected points, in the domain of F, then these line segments form a slope field, or a direction field, for the differential equation Each line segment has the same slope as the solution curve through that point. A slope field shows the general shape of all the solutions and can be helpful in getting a visual perspective of the directions of the solutions of a differential equation. F,. F, Figure 6. EXAMPLE 3 Sketching a Slope Field Sketch a slope field for the differential equation 0,, and,. for the points,, Solution The slope of the solution curve at an point, is F,. So, the slope at, is the slope at 0, is and the slope at, is Draw short line segments at the three points with their respective slopes, as shown in Figure 6.., 0. 0, EXAMPLE 4 Identifing Slope Fields for Differential Equations Match each slope field with its differential equation. a. b. c. Figure 6.3 i. ii. iii. Solution a. In Figure 6.3(a), ou can see that the slope at an point along the -ais is 0. The onl equation that satisfies this condition is So, the graph matches equation (ii). b. In Figure 6.3(b), ou can see that the slope at the point, is 0. The onl equation that satisfies this condition is So, the graph matches equation (i)... c. In Figure 6.3(c), ou can see that the slope at an point along the -ais is 0. The onl equation that satisfies this condition is So, the graph matches equation (iii)..

6. Slope Fields and Euler s Method 409 F, A solution curve of a differential equation is simpl a curve in the -plane whose tangent line at each point, has slope equal to F,. This is illustrated in Eample 5. EXAMPLE 5 Sketching a Solution Using a Slope Field Sketch a slope field for the differential equation. Use the slope field to sketch the solution that passes through the point,. Solution Make a table showing the slopes at several points. The table shown is a small sample. The slopes at man other points should be calculated to get a representative slope field. 0 0 5 3 3 3 3 5 Net draw line segments at the points with their respective slopes, as shown in Figure 6.4. Slope field for Figure 6.4 Particular solution for passing through, Figure 6.5 After the slope field is drawn, start at the initial point, and move to the right in the direction of the line segment. Continue to draw the solution curve so that it moves parallel to the nearb line segments. Do the same to the left of,. The resulting solution is shown in Figure 6.5. In Eample 5, note that the slope field shows that increases. increases to infinit as NOTE Drawing a slope field b hand is tedious. In practice, slope fields are usuall drawn using a graphing utilit.

40 Chapter 6 Differential Equations 0 0 Figure 6.6 Eact solution curve Euler approimation h Slope F( 0, 0 ) 0 + h (, ) hf( 0, 0 ) (, ) Euler s Method Euler s Method is a numerical approach to approimating the particular solution of the differential equation F, that passes through the point 0, 0. From the given information, ou know that the graph of the solution passes through the point 0, 0 and has a slope of F 0, 0 at this point. This gives ou a starting point for approimating the solution. From this starting point, ou can proceed in the direction indicated b the slope. Using a small step h, move along the tangent line until ou arrive at the point,, where 0 h and 0 hf 0, 0 as shown in Figure 6.6. If ou think of, as a new starting point, ou can repeat the process to obtain a second point,. The values of and are as follows. 0 h h n n h 0 hf 0, 0 hf, n n hf n, n i i NOTE You can obtain better approimations of the eact solution b choosing smaller and smaller step sizes..0 0.8 Eact solution EXAMPLE 6 Approimating a Solution Using Euler s Method Use Euler s Method to approimate the particular solution of the differential equation passing through the point 0,. Use a step of h 0.. 0.6 0.4 0. 0. Figure 6.7 0.4 0.6 Approimate solution 0.8.0 Solution Using h 0., 0 0, 0, and F,, ou have 0 0, 0., 0., 3 0.3,..., and 0 hf 0, 0 0. 0 0.9 hf, 0.9 0. 0. 0.9 0.8 3 hf, 0.8 0. 0. 0.8 0.758. The first ten approimations are shown in the table. You can plot these values to see a graph of the approimate solution, as shown in Figure 6.7. n 0 3 4 5 6 7 8 9 0 n 0 0. 0. 0.3 0.4 0.5 0.6 0.7 0.8 0.9.0 n 0.900 0.80 0.758 0.7 0.68 0.663 0.657 0.66 0.675 0.697 NOTE For the differential equation in Eample 6, ou can verif the eact solution to be e. Figure 6.7 compares this eact solution with the approimate solution obtained in Eample 6.

6. Slope Fields and Euler s Method 4 6. Eercises See www.calcchat.com for worked-out solutions to odd-numbered eercises. In Eercises 8, verif the solution of the differential equation.. Ce 4 4. e 3. C 4. 5. C sin C cos 6. C e cos C e sin 7. cos ln sec tan 8. 5 e 4 e In Eercises 9, verif the particular solution of the differential equation. 9. 0... Solution ln Solution sin cos cos cos 3 4e 6 4 0 0 5 sin e cos 0 4 In Eercises 3 0, determine whether the function is a solution of the differential equation 4 6 0. 3. 3 cos 4. sin 5. 3 cos 6. 3 sin 7. e 8. 5 ln 9. C e C e C 3 sin C 4 cos 0. 3e 4 sin Differential Equation 3 5 e d 0 0 tan 4 e Differential Equation and Initial Condition sin sin In Eercises 8, determine whether the function is a solution of the differential equation 3e... 3 3. e 4. e 5. sin 6. cos 7. ln 8. e 5 In Eercises 9 3, some of the curves corresponding to different values of C in the general solution of the differential equation are given. Find the particular solution that passes through the point shown on the graph. 9. 30. 3. 3. Ce C C 3 C 0 Figure for 9 Figure for 30 4 3 3 4 Solution Figure for 3 Figure for 3 In Eercises 33 and 34, the general solution of the differential equation is given. Use a graphing utilit to graph the particular solutions for the given values of C. 33. 34. 4 0 4 C C 0, C ±, C ±4 In Eercises 35 40, verif that the general solution satisfies the differential equation. Then find the particular solution that satisfies the initial condition. 35. Ce 36. 3 C 0 3 when 0 3 when 37. C sin 3 C cos 3 38. C C ln 9 0 when 6 0 when when 6 when (0, 3) 3 (4, 4) 3 4 5 6 7 Differential Equation 0 3 0 0 4 4 3 4 0 C C 0, C, C 4 3 0 0 4 3 3 4 (0, ) 4 (3, 4) 3 4

4 Chapter 6 Differential Equations 39. C 40. e 3 C 3 C C 3 3 0 0 when when 4 4 when 0 0 when 3 In Eercises 4 5, use integration to find a general solution of the differential equation. 4. 6 d 4. 43. d 44. 45. d 46. 47. sin d 48. 49. 6 d 50. 5. e d 5. Slope Fields In Eercises 53 56, a differential equation and its slope field are given. Complete the table b determining the slopes (if possible) in the slope field at the given points. 4 0 4 8 0 4 4 6 8 /d 53. 54. d 4 9 4 0 d 3 3 d e 4 e cos d d tan 3 d 5e d d 0 In Eercises 57 60, match the differential equation with its slope field. [The slope fields are labeled (a), (b), (c), and (d).] (a) (c) (b) (d) 57. sin d 58. 59. e d 60. Slope Fields In Eercises 6 64, (a) sketch the slope field for the differential equation, (b) use the slope field to sketch the solution that passes through the given point, and (c) discuss the graph of the solution as and. Use a graphing utilit to verif our results. 6. 6. 63. 64. 3 3 3 3, 4, 3, 4,,,, 3 0, 4 3 3 d cos d 3 3 3 3 0 6 55. cos 56. d 8 4 0 8 6 d tan 6 8 8 65. Slope Field Use the slope field for the differential equation where > 0, to sketch the graph of the solution that satisfies each given initial condition. Then make a conjecture about the behavior of a particular solution of as. To print an enlarged cop of the graph, go to the website www.mathgraphs.com., 3 0 0 8 8 3 6 6 8 (a), 0 (b),

6. Slope Fields and Euler s Method 43 66. Slope Field Use the slope field for the differential equation where > 0, to sketch the graph of the solution that satisfies each initial condition. Then make a conjecture about the behavior of a particular solution of as. To print an enlarged cop of the graph, go to the website www.mathgraphs.com., 6 0 0. 0.4 0.6 0.8 (eact) h 0. h 0. Table for 79 8 (a) CAS Slope Fields In Eercises 67 7, use a computer algebra sstem to (a) graph the slope field for the differential equation and (b) graph the solution satisfing the specified initial condition. 67. 68. 69. 70. 7. 7. Euler s Method In Eercises 73 78, use Euler s Method to make a table of values for the approimate solution of the differential equation with the specified initial value. Use n steps of size h. 73. 74. 75. 76. 77. 78. 3 0, 0.5, d 4, d 0., d 0.4 3, d,, 3, 0.5 3, e, 0 4 0, cos sin, 0 6 0.0 0, d d e 8 sin 4, 0, 0, 3 0 3, 0, n 0, 0 5, In Eercises 79 8, complete the table using the eact solution of the differential equation and two approimations obtained using Euler s Method to approimate the particular solution of the differential equation. Use h 0. and h 0. and compute each approimation to four decimal places. (b) 0 0 9 0 0 n 0, n 0, n 0,, n 5, h 0. h 0.05 h 0. n 0, h 0.05 h 0.4 h 0. 79. 80. 8. Differential Equation d d cos d 8. Compare the values of the approimations in Eercises 79 8 with the values given b the eact solution. How does the error change as h increases? 83. Temperature At time t 0 minutes, the temperature of an object is 40 F. The temperature of the object is changing at the rate given b the differential equation dt 7. (a) Use a graphing utilit and Euler s Method to approimate the particular solutions of this differential equation at t,, and 3. Use a step size of h 0.. (A graphing utilit program for Euler s Method is available at the website college.hmco.com.) (b) Compare our results with the eact solution 7 68e t. Initial Condition 0, 3 0, 0, 0 Eact Solution 3e 4 sin cos e (c) Repeat parts (a) and (b) using a step size of Compare the results. CAPSTONE h 0.05. 84. The graph shows a solution of one of the following differential equations. Determine the correct equation. Eplain our reasoning. (a) 4 (b) (c) (d) 4 4

44 Chapter 6 Differential Equations WRITING ABOUT CONCEPTS 85. In our own words, describe the difference between a general solution of a differential equation and a particular solution. 86. Eplain how to interpret a slope field. 87. Describe how to use Euler s Method to approimate a particular solution of a differential equation. 88. It is known that Ce k is a solution of the differential equation Is it possible to determine C or k from the information given? If so, find its value. 0.07. 94. Errors and Euler s Method Repeat Eercise 93 for which the eact solution of the differential equation d where 0, is e. 95. Electric Circuits The diagram shows a simple electric circuit consisting of a power source, a resistor, and an inductor. R True or False? In Eercises 89 9, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 89. If f is a solution of a first-order differential equation, then f C is also a solution. 90. The general solution of a differential equation is 4.9 C C. To find a particular solution, ou must be given two initial conditions. 9. Slope fields represent the general solutions of differential equations. 9. A slope field shows that the slope at the point, is 6. This slope field represents the famil of solutions for the differential equation 93. Errors and Euler s Method The eact solution of the differential equation d 4. where 0 4, is 4e. (a) Use a graphing utilit to complete the table, where is the eact value of the solution, is the approimate solution using Euler s Method with h 0., is the approimate solution using Euler s Method with h 0., e is the absolute error, e is the absolute error, and r is the ratio e e. 0 0. 0.4 0.6 0.8 e e r (b) What can ou conclude about the ratio r as h changes? (c) Predict the absolute error when h 0.05. E A model of the current I, in amperes A, at time t is given b the first-order differential equation L di RI E t dt where E t is the voltage V produced b the power source, R is the resistance, in ohms, and L is the inductance, in henrs H. Suppose the electric circuit consists of a 4-V power source, a - resistor, and a 4-H inductor. (a) Sketch a slope field for the differential equation. (b) What is the limiting value of the current? Eplain. 96. Think About It It is known that e kt is a solution of the differential equation Find the values of k. 97. Think About It It is known that A sin t is a solution of the differential equation Find the values of. 6 0. 6 0. PUTNAM EXAM CHALLENGE 98. Let f be a twice-differentiable real-valued function satisfing f f g f where g 0 for all real. Prove that is bounded. 99. Prove that if the famil of integral curves of the differential equation p q, d p q 0 f is cut b the line k, the tangents at the points of intersection are concurrent. These problems were composed b the Committee on the Putnam Prize Competition. The Mathematical Association of America. All rights reserved. L

6. Differential Equations: Growth and Deca 45 6. Differential Equations: Growth and Deca Use separation of variables to solve a simple differential equation. Use eponential functions to model growth and deca in applied problems. Differential Equations In the preceding section, ou learned to analze visuall the solutions of differential equations using slope fields and to approimate solutions numericall using Euler s Method. Analticall, ou have learned to solve onl two tpes of differential equations those of the forms and In this section, ou will learn how to solve a more general tpe of differential equation. The strateg is to rewrite the equation so that each variable occurs on onl one side of the equation. This strateg is called separation of variables. (You will stu this strateg in detail in Section 6.3.) f f. EXAMPLE Solving a Differential Equation d d Original equation Multipl both sides b. Integrate with respect to. d d STUDY TIP You can use implicit differentiation to check the solution in Eample. Appl Power Rule. C C Rewrite, letting C C. So, the general solution is given b C. EXPLORATION In Eample, the general solution of the differential equation is C. Use a graphing utilit to sketch the particular solutions for C ±, C ±, and C 0. Describe the solutions graphicall. Is the following statement true of each solution? The slope of the graph at the point, is equal to twice the ratio of and. Eplain our reasoning. Are all curves for which this statement is true represented b the general solution? Notice that when ou integrate both sides of the equation in Eample, ou don t need to add a constant of integration to both sides. If ou did, ou would obtain the same result. d C C 3 C 3 C C Some people prefer to use Leibniz notation and differentials when appling separation of variables. The solution of Eample is shown below using this notation. d d d C C

46 Chapter 6 Differential Equations Growth and Deca Models In man applications, the rate of change of a variable is proportional to the value of. If is a function of time t, the proportion can be written as follows. Rate of change of is proportional to. k dt The general solution of this differential equation is given in the following theorem. THEOREM 6. EXPONENTIAL GROWTH AND DECAY MODEL k If is a differentiable function of t such that > 0 and for some constant k, then Ce kt. C is the initial value of, and k is the proportionalit constant. Eponential growth occurs when k > 0, and eponential deca occurs when k < 0. PROOF 7 6 (3, 5.657) 5 = e 0.3466t 4 (, 4) 3 (0, ) t 3 4 If the rate of change of is proportional to, then follows an eponential model. Figure 6.8 STUDY TIP Using logarithmic properties, note that the value of k in Eample can also be written as ln. So, the model becomes e ln t, which can then be rewritten as t. k dt k dt k dt EXAMPLE Using an Eponential Growth Model The rate of change of is proportional to. When t 0,, and when t, 4. What is the value of when t 3? Solution Because ou know that and t are related b the equation Ce kt. You can find the values of the constants C and k b appling the initial conditions. Ce 0 k k, C Write original equation. Separate variables. Integrate with respect to t. ln kt C e kt e C Ce kt Find antiderivative of each side. Solve for. Let C e C. differentiate the function with respect to t to verif that So, all solutions of are of the form Ce kt. Remember that ou can k dt When t 0,. k. 4 k ek ln 0.3466 When t, 4. So, the model is e 0.3466t. When t 3, the value of is e 0.3466 3 5.657 (see Figure 6.8).

6. Differential Equations: Growth and Deca 47 TECHNOLOGY Most graphing utilities have curve-fitting capabilities that can be used to find models that represent data. Use the eponential regression feature of a graphing utilit and the information in Eample to find a model for the data. How does our model compare with the given model? Radioactive deca is measured in terms of half-life the number of ears required for half of the atoms in a sample of radioactive material to deca. The rate of deca is proportional to the amount present. The half-lives of some common radioactive isotopes are shown below. Uranium 38 U Plutonium 39 Pu Carbon 4 C Radium 6 Ra Einsteinium 54 Es Nobelium 57 No 4,470,000,000 ears 4,00 ears 575 ears 599 ears 76 das 5 seconds EXAMPLE 3 Radioactive Deca LAZARENKO NIKOLAI/ITAR-TASS/Landov NOTE The eponential deca model in Eample 3 could also be written as 0 t 4,00. This model is much easier to derive, but for some applications it is not as convenient to use. Suppose that 0 grams of the plutonium isotope 39 Pu was released in the Chernobl nuclear accident. How long will it take for the 0 grams to deca to gram? Solution Let represent the mass (in grams) of the plutonium. Because the rate of deca is proportional to, ou know that Ce kt where t is the time in ears. To find the values of the constants C and k, appl the initial conditions. Using the fact that 0 when t 0, ou can write 0 Ce k 0 Ce 0 which implies that Net, using the fact that the half-life of 39 C 0. Pu is 4,00 ears, ou have 0 5 when t 4,00, so ou can write 5 0e k 4,00 4,00 ln k 0.0000876 k. So, the model is e4,00k 0e 0.0000876t. Half-life model To find the time it would take for 0 grams to deca to gram, ou can solve for t in the equation 0e 0.0000876t. The solution is approimatel 80,059 ears. From Eample 3, notice that in an eponential growth or deca problem, it is eas to solve for C when ou are given the value of at t 0. The net eample demonstrates a procedure for solving for C and k when ou do not know the value of at t 0.

48 Chapter 6 Differential Equations EXAMPLE 4 Population Growth Number of fruit flies 300 75 50 5 00 75 50 5 00 75 50 5 Figure 6.9 (0, 33) = 33e 0.5493t (, 00) 3 4 Time (in das) (4, 300) t Suppose an eperimental population of fruit flies increases according to the law of eponential growth. There were 00 flies after the second da of the eperiment and 300 flies after the fourth da. Approimatel how man flies were in the original population? Solution Let Ce kt be the number of flies at time t, where t is measured in das. Note that is continuous whereas the number of flies is discrete. Because 00 when t and 300 when t 4, ou can write 00 Ce k and 300 Ce 4k. From the first equation, ou know that C 00e k. Substituting this value into the second equation produces the following. 300 00e k e 4k 300 00e k ln 3 k ln 3 k 0.5493 k So, the eponential growth model is Ce 0.5493t. To solve for C, reappl the condition 00 when t and obtain 00 Ce 0.5493 C 00e.0986 33. So, the original population (when t 0) consisted of approimatel C 33 flies, as shown in Figure 6.9. Units sold (in thousands) 00 90 80 70 60 50 40 30 0 0 Figure 6.0 (0, 00,000) (4, 80,000) (6, 7,500) = 00,000e 0.0558t 3 4 5 6 7 8 Time (in months) t EXAMPLE 5 Declining Sales Four months after it stops advertising, a manufacturing compan notices that its sales have dropped from 00,000 units per month to 80,000 units per month. If the sales follow an eponential pattern of decline, what will the be after another months? Solution Use the eponential deca model Ce kt, where t is measured in months. From the initial condition t 0, ou know that C 00,000. Moreover, because 80,000 when t 4, ou have 80,000 00,000e 4k 0.8 e 4k ln 0.8 4k 0.0558 k. So, after more months t 6, ou can epect the monthl sales rate to be 00,000e 0.0558 6 7,500 units. See Figure 6.0.

6. Differential Equations: Growth and Deca 49 In Eamples through 5, ou did not actuall have to solve the differential equation k. (This was done once in the proof of Theorem 6..) The net eample demonstrates a problem whose solution involves the separation of variables technique. The eample concerns Newton s Law of Cooling, which states that the rate of change in the temperature of an object is proportional to the difference between the object s temperature and the temperature of the surrounding medium. Temperature (in F) 40 0 00 80 60 40 0 Figure 6. (0, 00) (0, 90) (4.09, 80) = 60 + 40e 0.0877t 5 0 5 0 5 Time (in minutes) t EXAMPLE 6 Newton s Law of Cooling Let represent the temperature in F of an object in a room whose temperature is kept at a constant 60. If the object cools from 00 to 90 in 0 minutes, how much longer will it take for its temperature to decrease to 80? Solution From Newton s Law of Cooling, ou know that the rate of change in is proportional to the difference between and 60. This can be written as k 60, To solve this differential equation, use separation of variables, as follows. k 60 Differential equation dt k dt Separate variables. 60 Integrate each side. 60 k dt ln 60 kt C Find antiderivative of each side. Because > 60, 60 60, and ou can omit the absolute value signs. Using eponential notation, ou have 60 e kt C Using 00 when t 0, ou obtain 00 60 Ce k 0 60 C, which implies that C 40. Because 90 when t 0, 90 60 40e k 0 30 40e 0k k 0 ln 3 4 0.0877. So, the model is 60 40e 0.0877t and finall, when 80, ou obtain 80 60 40e 0.0877t 0 40e 0.0877t e 0.0877t ln 0.0877t t 4.09 minutes. 80 00. 60 Ce kt. C e C Cooling model So, it will require about 4.09 more minutes for the object to cool to a temperature of 80 (see Figure 6.).

40 Chapter 6 Differential Equations 6. Eercises See www.calcchat.com for worked-out solutions to odd-numbered eercises. In Eercises 0, solve the differential equation.. 3. 6 d d 3. 3 4. 6 d d 5. 6. 7 7. 8. 9. 0 0. 00 In Eercises 4, write and solve the differential equation that models the verbal statement.. The rate of change of Q with respect to t is inversel proportional to the square of t.. The rate of change of P with respect to t is proportional to 5 t. 3. The rate of change of N with respect to s is proportional to 500 s. 4. The rate of change of with respect to varies jointl as and L. Slope Fields In Eercises 5 and 6, a differential equation, a point, and a slope field are given. (a) Sketch two approimate solutions of the differential equation on the slope field, one of which passes through the given point. (b) Use integration to find the particular solution of the differential equation and use a graphing utilit to graph the solution. Compare the result with the sketch in part (a). To print an enlarged cop of the graph, go to the website www.mathgraphs.com. 5. 6, 0, 0 6., 0, d d 5 5 9 5 In Eercises 7 0, find the function f t passing through the point 0, 0 with the given first derivative. Use a graphing utilit to graph the solution. 7. 8. dt t dt 3 4 t 9. 0. dt 3 dt 4 4 4 4 4 In Eercises 4, write and solve the differential equation that models the verbal statement. Evaluate the solution at the specified value of the independent variable.. The rate of change of is proportional to. When 0, 6, and when 4, 5. What is the value of when 8?. The rate of change of N is proportional to N. When t 0, N 50, and when t, N 400. What is the value of N when t 4? 3. The rate of change of V is proportional to V. When t 0, V 0,000, and when t 4, V,500. What is the value of V when t 6? 4. The rate of change of P is proportional to P. When t 0, P 5000, and when t, P 4750. What is the value of P when t 5? In Eercises 5 8, find the eponential function Ce kt that passes through the two given points. 5. 6. 5 4 3 7. 8. 6 5 4 3 ) 0, ) 3 4 5 (, 5) (5, 5) (5, ) 3 4 5 6 WRITING ABOUT CONCEPTS t t 4 3 5 4 3 (0, 4) 3 4 5 ) 3, ) (4, 5) 3 4 5 5, ) 9. Describe what the values of C and k represent in the eponential growth and deca model, Ce kt. 30. Give the differential equation that models eponential growth and deca. In Eercises 3 and 3, determine the quadrants in which the solution of the differential equation is an increasing function. Eplain. (Do not solve the differential equation.) 3. 3. d d ) t t

6. Differential Equations: Growth and Deca 4 Radioactive Deca In Eercises 33 40, complete the table for the radioactive isotope. Isotope Half-Life in ears 33. 599 0 g 34. 6 Ra 599.5 g 35. 6 Ra 599 0. g 36. 4 C 575 3 g 37. 4 C 575 5 g 38. 4 C 575.6 g 39. 39 Pu 4,00. g 40. 39 Pu 4,00 0.4 g 6 Ra 4. Radioactive Deca Radioactive radium has a half-life of approimatel 599 ears. What percent of a given amount remains after 00 ears? 4. Carbon Dating Carbon-4 dating assumes that the carbon dioide on Earth toda has the same radioactive content as it 4 did centuries ago. If this is true, the amount of C absorbed b a tree that grew several centuries ago should be the same as the amount of 4 C absorbed b a tree growing toda. A piece of ancient charcoal contains onl 5% as much of the radioactive carbon as a piece of modern charcoal. How long ago was the tree burned to make the ancient charcoal? (The half-life of 4 C is 575 ears.) Compound Interest In Eercises 43 48, complete the table for a savings account in which interest is compounded continuousl. Initial Investment Annual Rate Initial Quantit Time to Double Amount After 000 Years 43. $4000 6% 44. $8,000 5 % 45. $750 46. $,500 7 3 4 r 5 r 47. $500 $9.85 48. $000 $5436.56 Compound Interest In Eercises 49 5, find the principal P that must be invested at rate r, compounded monthl, so that $,000,000 will be available for retirement in t ears. 49. r 7 %, t 0 50. r 6%, 5. r 8%, t 35 5. r 9%, Compound Interest In Eercises 53 56, find the time necessar for $000 to double if it is invested at a rate of r compounded (a) annuall, (b) monthl, (c) dail, and (d) continuousl. 53. r 7% 54. r 6% 55. r 8.5% 56. r 5.5% Amount After 0 Years Amount After 0,000 Years t 40 t 5 Population In Eercises 57 6, the population (in millions) of a countr in 007 and the epected continuous annual rate of change k of the population are given. (Source: U.S. Census Bureau, International Data Base) (a) Find the eponential growth model P Ce kt for the population b letting t 0 correspond to 000. (b) Use the model to predict the population of the countr in 05. (c) Discuss the relationship between the sign of k and the change in population for the countr. Countr 007 Population 57. Latvia.3 0.006 58. Egpt 80.3 0.07 59. Paragua 6.7 0.04 60. Hungar 0.0 0.003 6. Uganda 30.3 0.036 CAPSTONE 6. (a) Suppose an insect population increases b a constant number each month. Eplain wh the number of insects can be represented b a linear function. (b) Suppose an insect population increases b a constant percentage each month. Eplain wh the number of insects can be represented b an eponential function. 63. Modeling Data One hundred bacteria are started in a culture and the number N of bacteria is counted each hour for 5 hours. The results are shown in the table, where t is the time in hours. t 0 3 4 5 N 00 6 5 98 43 97 (a) Use the regression capabilities of a graphing utilit to find an eponential model for the data. (b) Use the model to estimate the time required for the population to quadruple in size. 64. Bacteria Growth The number of bacteria in a culture is increasing according to the law of eponential growth. There are 5 bacteria in the culture after hours and 350 bacteria after 4 hours. (a) Find the initial population. (b) Write an eponential growth model for the bacteria population. Let t represent time in hours. (c) Use the model to determine the number of bacteria after 8 hours. (d) After how man hours will the bacteria count be 5,000? 65. Learning Curve The management at a certain factor has found that a worker can produce at most 30 units in a da. The learning curve for the number of units N produced per da after a new emploee has worked t das is N 30 e kt. After 0 das on the job, a particular worker produces 9 units. k

4 Chapter 6 Differential Equations (a) Find the learning curve for this worker. (b) How man das should pass before this worker is producing 5 units per da? 66. Learning Curve If the management in Eercise 65 requires a new emploee to produce at least 0 units per da after 30 das on the job, find (a) the learning curve that describes this minimum requirement and (b) the number of das before a minimal achiever is producing 5 units per da. 67. Modeling Data The table shows the populations P (in millions) of the United States from 960 to 000. (Source: U.S. Census Bureau) Year 960 970 980 990 000 Population, P 8 05 8 50 8 (a) Use the 960 and 970 data to find an eponential model P for the data. Let t 0 represent 960. (b) Use a graphing utilit to find an eponential model P for all the data. Let t 0 represent 960. (c) Use a graphing utilit to plot the data and graph models P and P in the same viewing window. Compare the actual data with the predictions. Which model better fits the data? (d) Estimate when the population will be 30 million. 68. Modeling Data The table shows the net receipts and the amounts required to service the national debt (interest on Treasur debt securities) of the United States from 00 through 00. The ears 007 through 00 are estimated, and the monetar amounts are given in billions of dollars. (Source: U.S. Office of Management and Budget) Year 00 00 003 004 005 Receipts 99.4 853.4 78.5 880.3 53.9 Interest 359.5 33.5 38. 3.7 35.3 Year 006 007 008 009 00 Receipts 407.3 540. 66.5 798.3 954.7 Interest 405.9 433.0 469.9 498.0 53. (a) Use the regression capabilities of a graphing utilit to find an eponential model R for the receipts and a quartic model I for the amount required to service the debt. Let t represent the time in ears, with t corresponding to 00. (b) Use a graphing utilit to plot the points corresponding to the receipts, and graph the eponential model. Based on the model, what is the continuous rate of growth of the receipts? (c) Use a graphing utilit to plot the points corresponding to the amounts required to service the debt, and graph the quartic model. (d) Find a function P t that approimates the percent of the receipts that is required to service the national debt. Use a graphing utilit to graph this function. 69. Sound Intensit The level of sound (in decibels) with an intensit of I is I 0 log 0 I I 0, where I 0 is an intensit of 0 6 watt per square centimeter, corresponding roughl to the faintest sound that can be heard. Determine I for the following. (a) I 0 4 watt per square centimeter (whisper) (b) I 0 9 watt per square centimeter (bus street corner) (c) I 0 6.5 watt per square centimeter (air hammer) (d) I 0 4 watt per square centimeter (threshold of pain) 70. Noise Level With the installation of noise suppression materials, the noise level in an auditorium was reduced from 93 to 80 decibels. Use the function in Eercise 69 to find the percent decrease in the intensit level of the noise as a result of the installation of these materials. 7. Forestr The value of a tract of timber is V t 00,000e 0.8 t, where t is the time in ears, with t 0 corresponding to 008. If mone earns interest continuousl at 0%, the present value of the timber at an time t is A t V t e 0.0t. Find the ear in which the timber should be harvested to maimize the present value function. 7. Earthquake Intensit On the Richter scale, the magnitude R of an earthquake of intensit I is R ln I ln I 0 ln 0 where I 0 is the minimum intensit used for comparison. Assume that I 0. (a) Find the intensit of the 906 San Francisco earthquake R 8.3. (b) Find the factor b which the intensit is increased if the Richter scale measurement is doubled. (c) Find dr di. 73. Newton s Law of Cooling When an object is removed from a furnace and placed in an environment with a constant temperature of 80 F, its core temperature is 500 F. One hour after it is removed, the core temperature is 0 F. Find the core temperature 5 hours after the object is removed from the furnace. 74. Newton s Law of Cooling A container of hot liquid is placed in a freezer that is kept at a constant temperature of 0 F. The initial temperature of the liquid is 60 F. After 5 minutes, the liquid s temperature is 60 F. How much longer will it take for its temperature to decrease to 30 F? True or False? In Eercises 75 78, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 75. In eponential growth, the rate of growth is constant. 76. In linear growth, the rate of growth is constant. 77. If prices are rising at a rate of 0.5% per month, then the are rising at a rate of 6% per ear. 78. The differential equation modeling eponential growth is d k, where k is a constant.

6.3 Separation of Variables and the Logistic Equation 43 6.3 Separation of Variables and the Logistic Equation Recognize and solve differential equations that can be solved b separation of variables. Recognize and solve homogeneous differential equations. Use differential equations to model and solve applied problems. Solve and analze logistic differential equations. Separation of Variables Consider a differential equation that can be written in the form M N d 0 where M is a continuous function of alone and N is a continuous function of alone. As ou saw in the preceding section, for this tpe of equation, all terms can be collected with d and all terms with, and a solution can be obtained b integration. Such equations are said to be separable, and the solution procedure is called separation of variables. Below are some eamples of differential equations that are separable. Original Differential Equation 3 d 0 sin cos e EXAMPLE Separation of Variables Rewritten with Variables Separated 3 d cot d e d NOTE Be sure to check our solutions throughout this chapter. In Eample, ou can check the solution C 4 b differentiating and substituting into the original equation. Find the general solution of 4. d Solution To begin, note that 0 is a solution. To find other solutions, assume that 0 and separate variables as shown. 4 d Differential form Separate variables. 4 d Now, integrate to obtain 4 d ln ln 4 C Integrate. 4 d C 4 4? C 4 C 4 C 4 So, the solution checks. ln ln 4 C ec 4 ±e C 4. Because 0 is also a solution, ou can write the general solution as C 4. General solution C ± e C

44 Chapter 6 Differential Equations In some cases it is not feasible to write the general solution in the eplicit form f. The net eample illustrates such a solution. Implicit differentiation can be used to verif this solution. FOR FURTHER INFORMATION For an eample (from engineering) of a differential equation that is separable, see the article Designing a Rose Cutter b J. S. Hartzler in The College Mathematics Journal. To view this article, go to the website www.matharticles.com. EXAMPLE Finding a Particular Solution Given the initial condition 0, find the particular solution of the equation d e 0. Solution Note that 0 is a solution of the differential equation but this solution does not satisf the initial condition. So, ou can assume that 0. To separate variables, ou must rid the first term of and the second term of e. So, ou should multipl b e and obtain the following. d e 0 e d e d From the initial condition 0, ou have 0 C, which implies that C. So, the particular solution has the implicit form ln e ln e. ln e C You can check this b differentiating and rewriting to get the original equation. 0 = 3e EXAMPLE 3 Finding a Particular Solution Curve Find the equation of the curve that passes through the point, 3 and has a slope of at an point,. Solution Because the slope of the curve is given b, ou have d with the initial condition 3. Separating variables and integrating produces d, ln C 0 6 4 (, 3) = 3e ( )/ 4 6 8 0 Figure 6. e C Ce. Because 3 when, it follows that 3 Ce and C 3e. So, the equation of the specified curve is 3e e 3e, > 0. Because the solution is not defined at 0 and the initial condition is given at, is restricted to positive values. See Figure 6..

6.3 Separation of Variables and the Logistic Equation 45 NOTE The notation f, is used to denote a function of two variables in much the same wa as f denotes a function of one variable. You will stu functions of two variables in detail in Chapter 3. Homogeneous Differential Equations Some differential equations that are not separable in and can be made separable b a change of variables. This is true for differential equations of the form where f is a homogeneous function. The function given b f, is homogeneous of degree n if f t, t t n f, Homogeneous function of degree n where n is an integer. f,, EXAMPLE 4 Verifing Homogeneous Functions a. f, 4 3 3 is a homogeneous function of degree 3 because f t, t t t 4 t 3 3 t t t 3 t 3 4 3 t 3 3 t 3 4 3 3 t 3 f,. b. f, e sin is a homogeneous function of degree because f t, t te t t t sin t t t e sin tf,. c. f, is not a homogeneous function because f t, t t t t t t n. d. f, is a homogeneous function of degree 0 because f t, t t t t 0. DEFINITION OF HOMOGENEOUS DIFFERENTIAL EQUATION A homogeneous differential equation is an equation of the form M, d N, 0 where M and N are homogeneous functions of the same degree. EXAMPLE 5 Testing for Homogeneous Differential Equations a. d 0 is homogeneous of degree. b. 3 d 3 is homogeneous of degree 3. c. d 0 is not a homogeneous differential equation.

46 Chapter 6 Differential Equations To solve a homogeneous differential equation b the method of separation of variables, use the following change of variables theorem. THEOREM 6. CHANGE OF VARIABLES FOR HOMOGENEOUS EQUATIONS If M, d N, 0 is homogeneous, then it can be transformed into a differential equation whose variables are separable b the substitution v where v is a differentiable function of. EXAMPLE 6 Solving a Homogeneous Differential Equation STUDY TIP The substitution v will ield a differential equation that is separable with respect to the variables and v. You must write our final solution, however, in terms of and. General solution of d 3 0 Figure 6.3 C = C = C = 3 C = 4 ( + ) 3 = C Find the general solution of d 3 0. Solution Because and 3 are both homogeneous of degree, let v to obtain dv v d. Then, b substitution, ou have v d 3 v dv v d 0 v d 3v dv 0. Dividing b and separating variables produces v d 3v dv d 3v v dv ln 3 4 ln v C 4 ln 3 ln 4 ln ln v ln C C v 3 4 C v 3. Substituting for v produces the following general solution. 3 4 C v d 3 3 v dv 0 3 C 4 C 3 General solution You can check this b differentiating and rewriting to get the original equation. TECHNOLOGY If ou have access to a graphing utilit, tr using it to graph several solutions of the equation in Eample 6. For instance, Figure 6.3 shows the graphs of 3 C for C,, 3, and 4.

6.3 Separation of Variables and the Logistic Equation 47 Applications EXAMPLE 7 Wildlife Population franzfoto.com/alam The rate of change of the number of cootes N t in a population is directl proportional to 650 N t, where t is the time in ears. When t 0, the population is 300, and when t, the population has increased to 500. Find the population when t 3. Solution Because the rate of change of the population is proportional to 650 N t, ou can write the following differential equation. dn dt k 650 N You can solve this equation using separation of variables. Differential form dn k dt Separate variables. 650 N ln Integrate. ln 650 N kt C 650 N kt C 650 N e kt C Assume N < 650. N 650 Ce kt General solution Using N 300 when t 0, ou can conclude that C 350, which produces N 650 350e kt. dn k 650 N dt Then, using N 500 when t, it follows that 500 650 350e k e k 3 7 So, the model for the coote population is N 650 350e 0.436t. Model for population When t 3, ou can approimate the population to be N 650 350e 0.436 3 55 cootes. k 0.436. The model for the population is shown in Figure 6.4. Note that N 650 is the horizontal asmptote of the graph and is the carring capacit of the model. You will learn more about carring capacit later in this section. N 700 Number of cootes 600 500 400 300 00 (0, 300) (, 500) (3, 55) N = 650 350e 0.436t 00 3 4 5 6 Time (in ears) t Figure 6.4

48 Chapter 6 Differential Equations Each line K is an orthogonal trajector of the famil of circles. Figure 6.5 A common problem in electrostatics, thermonamics, and hdronamics involves finding a famil of curves, each of which is orthogonal to all members of a given famil of curves. For eample, Figure 6.5 shows a famil of circles C Famil of circles each of which intersects the lines in the famil K Famil of lines at right angles. Two such families of curves are said to be mutuall orthogonal, and each curve in one of the families is called an orthogonal trajector of the other famil. In electrostatics, lines of force are orthogonal to the equipotential curves. In thermonamics, the flow of heat across a plane surface is orthogonal to the isothermal curves. In hdronamics, the flow (stream) lines are orthogonal trajectories of the velocit potential curves. EXAMPLE 8 Finding Orthogonal Trajectories Describe the orthogonal trajectories for the famil of curves given b C for C 0. Sketch several members of each famil. Solution First, solve the given equation for C and write C. Then, b differentiating implicitl with respect to, ou obtain the differential equation 0 d d. Differential equation Slope of given famil Given famil: = C Orthogonal famil: = K Because represents the slope of the given famil of curves at,, it follows that the orthogonal famil has the negative reciprocal slope. So, d. Slope of orthogonal famil Now ou can find the orthogonal famil b separating variables and integrating. d Orthogonal trajectories Figure 6.6 K C The centers are at the origin, and the transverse aes are vertical for K > 0 and horizontal for K < 0. If K 0, the orthogonal trajectories are the lines ±. If K 0, the orthogonal trajectories are hperbolas. Several trajectories are shown in Figure 6.6.

6.3 Separation of Variables and the Logistic Equation 49 L = L Logistic curve Note that as t, L. Figure 6.7 t Logistic Differential Equation In Section 6., the eponential growth model was derived from the fact that the rate of change of a variable is proportional to the value of. You observed that the differential equation dt k has the general solution Ce kt. Eponential growth is unlimited, but when describing a population, there often eists some upper limit L past which growth cannot occur. This upper limit L is called the carring capacit, which is the maimum population t that can be sustained or supported as time t increases. A model that is often used to describe this tpe of growth is the logistic differential equation dt k L Logistic differential equation where k and L are positive constants. A population that satisfies this equation does not grow without bound, but approaches the carring capacit L as t increases. From the equation, ou can see that if is between 0 and the carring capacit L, then dt > 0, and the population increases. If is greater than L, then dt < 0, and the population decreases. The graph of the function is called the logistic curve, as shown in Figure 6.7. EXPLORATION Use a graphing utilit to investigate the effects of the values of L, b, and k on the graph of L be kt. Include some eamples to support our results. EXAMPLE 9 Deriving the General Solution Solve the logistic differential equation Solution Begin b separating variables. L kdt ln ln L Solving this equation for L produces be kt. Write differential equation. Separate variables. Integrate each side. Rewrite left side using partial fractions. Find antiderivative of each side. Multipl each side b and simplif. Eponentiate each side. Let ±e C b. From Eample 9, ou can conclude that all solutions of the logistic differential equation are of the general form kdt L L kdt kt C ln L kt C L L L be kt. dt k L e kt C e C e kt be kt dt k L.

430 Chapter 6 Differential Equations EXAMPLE 0 Solving a Logistic Differential Equation 5000 0 80 0 Slope field for EXPLORATION Eplain what happens if p 0 L. dp dt 0.94p 4000 p and the solution passing through 0, 40 Figure 6.8 A state game commission releases 40 elk into a game refuge. After 5 ears, the elk population is 04. The commission believes that the environment can support no more than 4000 elk. The growth rate of the elk population p is dp dt kp 4000 p, where t is the number of ears. a. Write a model for the elk population in terms of t. b. Graph the slope field for the differential equation and the solution that passes through the point 0, 40. c. Use the model to estimate the elk population after 5 ears. d. Find the limit of the model as t. Solution a. You know that L 4000. So, the solution of the equation is of the form p 4000 be kt. Because p 0 40, ou can solve for b as follows. 40 4000 be k 0 40 4000 b Then, because p 04 when t 5, ou can solve for k. 04 4000 So, a model for the elk population is given b p 99e 0.94t. b. Using a graphing utilit, ou can graph the slope field for dp dt 0.94p 4000 p and the solution that passes through 0, 40, as shown in Figure 6.8. c. To estimate the elk population after 5 ears, substitute 5 for t in the model. p 4000 99e k 5 4000 99e 0.94 5 b 99 4000 99e.9 66 40 p 4000 k 0.94 Substitute 5 for t. Simplif. 4000 d. As t increases without bound, the denominator of gets closer and 99e 0.94t closer to. So, lim t 4000 4000. 0.94t 99e

6.3 Separation of Variables and the Logistic Equation 43 6.3 Eercises See www.calcchat.com for worked-out solutions to odd-numbered eercises. In Eercises 4, find the general solution of the differential equation.. 3. d d 3. 4. d 3 5 d 0 6 5. dr dr 0.75r 6. 0.75s ds ds 7. 3 8. 9. 0.. 4. 6 3. ln 0 4. 4 sin In Eercises 5 4, find the particular solution that satisfies the initial condition. Differential Equation 5. 6. 0 7. 0 8. 9. 0 0. 0 du. uv sin v dv dr. er s ds 3. dp kp dt 0 4. dt k T 70 dt 0 e 0 ln 0 8 cos 7e 0 Initial Condition 0 3 9 0 3 0 u 0 r 0 0 P 0 P 0 T 0 40 35. f, ln 36. f, tan 37. f, ln 38. f, tan In Eercises 39 44, solve the homogeneous differential equation. 39. 40. 4. 4. 43. 44. In Eercises 45 48, find the particular solution that satisfies the initial condition. 45. e d 0 46. d 0 47. Differential Equation sec d 0 48. d 0 Slope Fields In Eercises 49 5, sketch a few solutions of the differential equation on the slope field and then find the general solution analticall. To print an enlarged cop of the graph, go to the website www.mathgraphs.com. 49. 50. d d 3 3 3 Initial Condition 0 0 0 4 In Eercises 5 8, find an equation of the graph that passes through the point and has the given slope. 5. 0,, 6.,, 4 7. 9,, 8. 8,, 9 In Eercises 9 and 30, find all functions f having the indicated propert. 9. The tangent to the graph of f at the point, intersects the -ais at, 0. 30. All tangents to the graph of f pass through the origin. 3 6 5. 4 5. d 8 4 4 0.5 4 d 8 4 In Eercises 3 38, determine whether the function is homogeneous, and if it is, determine its degree. 3. f, 3 4 3 3. f, 3 3 33. f, 34. f, 3 3 4 4 3 3 4

43 Chapter 6 Differential Equations Euler s Method In Eercises 53 56, (a) use Euler s Method with a step size of h 0. to approimate the particular solution of the initial value problem at the given -value, (b) find the eact solution of the differential equation analticall, and (c) compare the solutions at the given -value. 53. 54. 55. 56. 57. Radioactive Deca The rate of decomposition of radioactive radium is proportional to the amount present at an time. The half-life of radioactive radium is 599 ears. What percent of a present amount will remain after 50 ears? 58. Chemical Reaction In a chemical reaction, a certain compound changes into another compound at a rate proportional to the unchanged amount. If initiall there is 40 grams of the original compound, and there is 35 grams after hour, when will 75 percent of the compound be changed? Slope Fields In Eercises 59 6, (a) write a differential equation for the statement, (b) match the differential equation with a possible slope field, and (c) verif our result b using a graphing utilit to graph a slope field for the differential equation. [The slope fields are labeled (a), (b), (c), and (d).] To print an enlarged cop of the graph, go to the website www.mathgraphs.com. (a) (c) Differential Equation d 6 d 6 0 d 3 4 d 5 9 9 5 Initial Condition 0, 5 0, 3,, 0 (b) (d) 5 5.5 -value.5 9 60. The rate of change of with respect to is proportional to the difference between and 4. 6. The rate of change of with respect to is proportional to the product of and the difference between and 4. 6. The rate of change of with respect to is proportional to. CAS 63. Weight Gain A calf that weighs 60 pounds at birth gains weight at the rate dw dt k 00 w, where w is weight in pounds and t is time in ears. Solve the differential equation. (a) Use a computer algebra sstem to solve the differential equation for k 0.8, 0.9, and. Graph the three solutions. (b) If the animal is sold when its weight reaches 800 pounds, find the time of sale for each of the models in part (a). (c) What is the maimum weight of the animal for each of the models? 64. Weight Gain A calf that weighs w 0 pounds at birth gains weight at the rate dw dt 00 w, where w is weight in pounds and t is time in ears. Solve the differential equation. In Eercises 65 70, find the orthogonal trajectories of the famil. Use a graphing utilit to graph several members of each famil. 65. C 66. C 67. C 68. C 69. C 3 70. Ce In Eercises 7 74, match the logistic equation with its graph. [The graphs are labeled (a), (b), (c), and (d).] (a) (c) 4 0 8 6 4 4 0 8 6 4 6 4 4 6 8 0 (b) (d) 4 0 8 4 0 8 6 4 6 4 4 6 8 0 5 5 5 5.5 59. The rate of change of with respect to is proportional to the difference between and 4. 6 4 4 6 8 0 6 4 4 6 8 0 7. 7. e 3e 73. 74. e e

6.3 Separation of Variables and the Logistic Equation 433 In Eercises 75 and 76, the logistic equation models the growth of a population. Use the equation to (a) find the value of k, (b) find the carring capacit, (c) find the initial population, (d) determine when the population will reach 50% of its carring capacit, and (e) write a logistic differential equation that has the solution P t. 00 5000 75. P t 76. P t 9e 0.75t 39e 0.t CAS In Eercises 77 and 78, the logistic differential equation models the growth rate of a population. Use the equation to (a) find the value of k, (b) find the carring capacit, (c) graph a slope field using a computer algebra sstem, and (d) determine the value of P at which the population growth rate is the greatest. dp dp 77. 78. 0.P 0.0004P dt 3P 00 P dt In Eercises 79 8, find the logistic equation that satisfies the initial condition. 79. 80. 8. 8. Logistic Differential Equation dt 36 dt.8 0 4 dt 5 50 3 dt 0 600 Initial Condition 0, 4 0, 7 0, 8 0, 5 83. Endangered Species A conservation organization releases 5 Florida panthers into a game preserve. After ears, there are 39 panthers in the preserve. The Florida preserve has a carring capacit of 00 panthers. (a) Write a logistic equation that models the population of panthers in the preserve. (b) Find the population after 5 ears. (c) When will the population reach 00? (d) Write a logistic differential equation that models the growth rate of the panther population. Then repeat part (b) using Euler s Method with a step size of h. Compare the approimation with the eact answers. (e) At what time is the panther population growing most rapidl? Eplain. 84. Bacteria Growth At time t 0, a bacterial culture weighs gram. Two hours later, the culture weighs 4 grams. The maimum weight of the culture is 0 grams. (a) Write a logistic equation that models the weight of the bacterial culture. (b) Find the culture s weight after 5 hours. (c) When will the culture s weight reach 8 grams? (d) Write a logistic differential equation that models the growth rate of the culture s weight. Then repeat part (b) using Euler s Method with a step size of h. Compare the approimation with the eact answers. (e) At what time is the culture s weight increasing most rapidl? Eplain. WRITING ABOUT CONCEPTS 85. In our own words, describe how to recognize and solve differential equations that can be solved b separation of variables. 86. State the test for determining if a differential equation is homogeneous. Give an eample. 87. In our own words, describe the relationship between two families of curves that are mutuall orthogonal. CAPSTONE 88. Suppose the growth of a population is modeled b a logistic equation. As the population increases, its rate of growth decreases. What do ou think causes this to occur in real-life situations such as animal or human populations? 89. Show that if then k. be kt, dt 90. Sailing Ignoring resistance, a sailboat starting from rest accelerates dv dt at a rate proportional to the difference between the velocities of the wind and the boat. (a) The wind is blowing at 0 knots, and after half-hour the boat is moving at 0 knots. Write the velocit v as a function of time t. (b) Use the result of part (a) to write the distance traveled b the boat as a function of time. True or False? In Eercises 9 94, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 9. The function 0 is alwas a solution of a differential equation that can be solved b separation of variables. 9. The differential equation can be written in separated variables form. 93. The function f, 4 6 is homogeneous. 94. The families C and K are mutuall orthogonal. PUTNAM EXAM CHALLENGE 95. A not uncommon calculus mistake is to believe that the product rule for derivatives sas that fg f g. If f e, determine, with proof, whether there eists an open interval a, b and a nonzero function g defined on a, b such that this wrong product rule is true for in a, b. This problem was composed b the Committee on the Putnam Prize Competition. The Mathematical Association of America. All rights reserved.

434 Chapter 6 Differential Equations 6.4 First-Order Linear Differential Equations Solve a first-order linear differential equation. Use linear differential equations to solve applied problems. Solve a Bernoulli differential equation. First-Order Linear Differential Equations In this section, ou will see how to solve a ver important class of first-order differential equations first-order linear differential equations. DEFINITION OF FIRST-ORDER LINEAR DIFFERENTIAL EQUATION A first-order linear differential equation is an equation of the form P Q d where P and Q are continuous functions of. This first-order linear differential equation is said to be in standard form. NOTE It is instructive to see wh the integrating factor helps solve a linear differential equation of the form When both sides of the equation are multiplied b the integrating factor u e P d, the left-hand side becomes the derivative of a product. P Q. e P d P e P d Q e P d e P d Q e P d Integrating both sides of this second equation and dividing b u produces the general solution. To solve a linear differential equation, write it in standard form to identif the functions P and Q. Then integrate and form the epression P u e P d Integrating factor which is called an integrating factor. The general solution of the equation is u Q u d. General solution EXAMPLE Solving a Linear Differential Equation Find the general solution of e. Solution For this equation, P and Q e. So, the integrating factor is u e P d Integrating factor e d e. This implies that the general solution is u Q u d e e e d e e C e Ce. General solution

6.4 First-Order Linear Differential Equations 435 ANNA JOHNSON PELL WHEELER (883 966) Anna Johnson Pell Wheeler was awarded a master s degree from the Universit of Iowa for her thesis The Etension of Galois Theor to Linear Differential Equations in 904. Influenced b David Hilbert, she worked on integral equations while stuing infinite linear spaces. THEOREM 6.3 SOLUTION OF A FIRST-ORDER LINEAR DIFFERENTIAL EQUATION An integrating factor for the first-order linear differential equation P Q is u e P The solution of the differential equation is e P d d. Q e P d d C. STUDY TIP Rather than memorizing the formula in Theorem 6.3, just remember that multiplication b the integrating factor e P d converts the left side of the differential equation into the derivative of the product e P d. EXAMPLE Solving a First-Order Linear Differential Equation Find the general solution of Solution. The standard form of the given equation is P Q. So, P, and ou have P d d ln e P d e ln eln Standard form. Integrating factor Figure 6.9 C = 0 C = C = 4 C = 3 C = C = C = So, multipling each side of the standard form b ields 3 d d d ln C ln C. General solution Several solution curves for C,, 0,,, 3, and 4 are shown in Figure 6.9.

436 Chapter 6 Differential Equations EXAMPLE 3 Solving a First-Order Linear Differential Equation tan t, Find the general solution of < t <. Solution The equation is alrea in the standard form So, P t tan t, and P t dt tan t dt ln cos t. P t Q t. C = Because < t <, ou can drop the absolute value signs and conclude that the integrating factor is e P t dt e ln cos t cos t. Integrating factor π C = C = 0 C = π t So, multipling b cos t produces d cos t cos t dt tan t cos t cos t dt C = cos t sin t C tan t C sec t. General solution Figure 6.0 Several solution curves are shown in Figure 6.0. Applications One tpe of problem that can be described in terms of a differential equation involves chemical mitures, as illustrated in the net eample. EXAMPLE 4 A Miture Problem 4 gal/min Figure 6. 5 gal/min A tank contains 50 gallons of a solution composed of 90% water and 0% alcohol. A second solution containing 50% water and 50% alcohol is added to the tank at the rate of 4 gallons per minute. As the second solution is being added, the tank is being drained at a rate of 5 gallons per minute, as shown in Figure 6.. Assuming the solution in the tank is stirred constantl, how much alcohol is in the tank after 0 minutes? Solution Let be the number of gallons of alcohol in the tank at an time t. You know that 5 when t 0. Because the number of gallons of solution in the tank at an time is 50 t, and the tank loses 5 gallons of solution per minute, it must lose 5 50 t gallons of alcohol per minute. Furthermore, because the tank is gaining gallons of alcohol per minute, the rate of change of alcohol in the tank is given b dt 5 50 t To solve this linear equation, let P t 5 50 t and obtain P t dt 5 50 t dt 5 ln 50 t. Because t < 50, ou can drop the absolute value signs and conclude that e P t dt e 5 ln 50 t 50 t 5. dt 5. 50 t

6.4 First-Order Linear Differential Equations 437 So, the general solution is 50 t 5 50 dt t 5 50 t C 4 Because 5 when t 0, ou have 5 50 which means that the particular solution is Finall, when t 0, the amount of alcohol in the tank is 50 t 50 0 50 t C 50 5 C 50 t 5. 0 50 t 50 5. 0 50 5 C 0 50 0 50 5 3.45 gal which represents a solution containing 33.6% alcohol. In most falling-bo problems discussed so far in the tet, air resistance has been neglected. The net eample includes this factor. In the eample, the air resistance on the falling object is assumed to be proportional to its velocit v. If g is the gravitational constant, the downward force F on a falling object of mass m is given b the difference mg kv. But b Newton s Second Law of Motion, ou know that F ma m dv dt acceleration which ields the following differential equation. m dv mg kv dt EXAMPLE 5 A Falling Object with Air Resistance An object of mass m is dropped from a hovering helicopter. Find its velocit as a function of time t. Assume that the air resistance is proportional to the object s velocit. Solution The velocit v satisfies the equation dv dt kv g. m Letting b k m, ou can separate variables to obtain dv g bv dt dv g bv dt b ln g bv t C a dv dt kv m g g gravitational constant, k constant of proportionalit NOTE Notice in Eample 5 that the velocit approaches a limit of mg k as a result of the air resistance. For fallingbo problems in which air resistance is neglected, the velocit increases without bound. ln g bv bt bc g bv Ce bt. Because the object was dropped, v 0 when t 0; so g C, and it follows that bv g ge bt v C e bc g ge bt b mg k e kt m.

438 Chapter 6 Differential Equations E S R I A simple electric circuit consists of electric current I (in amperes), a resistance R (in ohms), an inductance L (in henrs), and a constant electromotive force E (in volts), as shown in Figure 6.. According to Kirchhoff s Second Law, if the switch S is closed when t 0, the applied electromotive force (voltage) is equal to the sum of the voltage drops in the rest of the circuit. This in turn means that the current I satisfies the differential equation L L di RI E. dt Figure 6. EXAMPLE 6 An Electric Circuit Problem Find the current I as a function of time t (in seconds), given that I satisfies the differential equation L di dt RI sin t where R and L are nonzero constants. Solution In standard form, the given linear equation is di dt R L I sin t. L Let P t R L, so that e P t dt e R L t, and, b Theorem 6.3, Ie R L t L e R L t sin t dt So the general solution is I e R L t 4L R e R L t R sin t L cos t C I 4L R e R L t R sin t L cos t C. 4L R R sin t L cos t Ce R L t. TECHNOLOGY The integral in Eample 6 was found using smbolic algebra software. If ou have access to Maple, Mathematica, or the TI-89, tr using it to integrate e L R L t sin t dt. In Chapter 8 ou will learn how to integrate functions of this tpe using integration b parts. Bernoulli Equation A well-known nonlinear equation that reduces to a linear one with an appropriate substitution is the Bernoulli equation, named after James Bernoulli (654 705). P Q n Bernoulli equation

6.4 First-Order Linear Differential Equations 439 This equation is linear if n 0, and has separable variables if n. So, in the following development, assume that n 0 and n. Begin b multipling b n and n to obtain n P n Q n n n P n n Q d d n n P n n Q which is a linear equation in the variable n. Letting z n produces the linear equation dz n P z n Q. d Finall, b Theorem 6.3, the general solution of the Bernoulli equation is n n P d e n Q e n P d d C. EXAMPLE 7 Solving a Bernoulli Equation Find the general solution of Solution For this Bernoulli equation, let n 3, and use the substitution z 4 Let z n 3. Differentiate. Multipling the original equation b 3 43 produces Write original equation. z 4 3. e 4 3 4 4 4e Multipl each side b 4 3. z 4z 4e. Linear equation: This equation is linear in z. Using P 4 produces P d 4 d which implies that factor produces e e 3. z P z Q is an integrating factor. Multipling the linear equation b this z 4z 4e z e 4ze 4e d 4e d ze ze 4e d ze Finall, substituting z 4, the general solution is 4 e Ce. e C z e Ce. Linear equation Multipl b integrating factor. Write left side as derivative. Integrate each side. Divide each side b e. General solution

440 Chapter 6 Differential Equations So far ou have studied several tpes of first-order differential equations. Of these, the separable variables case is usuall the simplest, and solution b an integrating factor is ordinaril used onl as a last resort. SUMMARY OF FIRST-ORDER DIFFERENTIAL EQUATIONS Method Form of Equation. Separable variables: M d N 0. Homogeneous: M, d N, 0, where M and N are nth-degree homogeneous functions 3. Linear: P Q 4. Bernoulli equation: P Q n 6.4 Eercises See www.calcchat.com for worked-out solutions to odd-numbered eercises. In Eercises 4, determine whether the differential equation is linear. Eplain our reasoning.. 3 e. ln sin 3. 4. In Eercises 5 4, solve the first-order linear differential equation. 5. 6 6. d 6 7. 8. 9. cos d 0 0. sin d 0.. 3. 4. Slope Fields In Eercises 5 and 6, (a) sketch an approimate solution of the differential equation satisfing the given initial condition b hand on the slope field, (b) find the particular solution that satisfies the given initial condition, and (c) use a graphing utilit to graph the particular solution. Compare the graph with the hand-drawn graph in part (a). To print an enlarged cop of the graph, go to the website www.mathgraphs.com. 5. 6. d e, 0, 5 4 4 3 5 d 3 5 0 3 e 3 3 e 3 tan sec, 0 sin, 4 4 4 4 In Eercises 7 4, find the particular solution of the differential equation that satisfies the boundar condition. 7. 8. 9. 0. Differential Equation cos 0 0 5 3 e e tan sec cos sec sec. 0. 3. d 4. 3 0 5. Population Growth When predicting population growth, demographers must consider birth and death rates as well as the net change caused b the difference between the rates of immigration and emigration. Let P be the population at time t and let N be the net increase per unit time resulting from the difference between immigration and emigration. So, the rate of growth of the population is given b dp kp N, dt N is constant. Solve this differential equation to find P as a function of time if at time t 0 the size of the population is P 0. 6. Investment Growth A large corporation starts at time t 0 to invest part of its receipts continuousl at a rate of P dollars per ear in a fund for future corporate epansion. Assume that the fund earns r percent interest per ear compounded continuousl. So, the rate of growth of the amount A in the fund is given b da ra P dt Boundar Condition 0 0 4 0 4 where A 0 when t 0. Solve this differential equation for A as a function of t.

6.4 First-Order Linear Differential Equations 44 Investment Growth Eercise 6. In Eercises 7 and 8, use the result of 7. Find A for the following. (a) P $75,000, r 8%, and t 0 ears (b) P $550,000, r 5.9%, and t 5 ears 8. Find t if the corporation needs $,000,000 and it can invest $5,000 per ear in a fund earning 8% interest compounded continuousl. 9. Intravenous Feeding Glucose is added intravenousl to the bloodstream at the rate of q units per minute, and the bo removes glucose from the bloodstream at a rate proportional to the amount present. Assume that Q t is the amount of glucose in the bloodstream at time t. (a) Determine the differential equation describing the rate of change of glucose in the bloodstream with respect to time. (b) Solve the differential equation from part (a), letting Q Q 0 when t 0. (c) Find the limit of Q t as t. 30. Learning Curve The management at a certain factor has found that the maimum number of units a worker can produce in a da is 75. The rate of increase in the number of units N produced with respect to time t in das b a new emploee is proportional to 75 N. (a) Determine the differential equation describing the rate of change of performance with respect to time. (b) Solve the differential equation from part (a). (c) Find the particular solution for a new emploee who produced 0 units on the first da at the factor and 35 units on the twentieth da. Miture In Eercises 3 35, consider a tank that at time t 0 contains v 0 gallons of a solution of which, b weight, q 0 pounds is soluble concentrate. Another solution containing q pounds of the concentrate per gallon is running into the tank at the rate of r gallons per minute. The solution in the tank is kept well stirred and is withdrawn at the rate of gallons per minute. 3. If Q is the amount of concentrate in the solution at an time t, show that dq dt r Q v 0 r r t q r. 3. If Q is the amount of concentrate in the solution at an time t, write the differential equation for the rate of change of Q with respect to t if r r r. 33. A 00-gallon tank is full of a solution containing 5 pounds of concentrate. Starting at time t 0, distilled water is admitted to the tank at a rate of 0 gallons per minute, and the well-stirred solution is withdrawn at the same rate. (a) Find the amount of concentrate Q in the solution as a function of t. (b) Find the time at which the amount of concentrate in the tank reaches 5 pounds. (c) Find the quantit of the concentrate in the solution as t. r 34. Repeat Eercise 33, assuming that the solution entering the tank contains 0.04 pound of concentrate per gallon. 35. A 00-gallon tank is half full of distilled water. At time t 0, a solution containing 0.5 pound of concentrate per gallon enters the tank at the rate of 5 gallons per minute, and the well-stirred miture is withdrawn at the rate of 3 gallons per minute. (a) At what time will the tank be full? (b) At the time the tank is full, how man pounds of concentrate will it contain? (c) Repeat parts (a) and (b), assuming that the solution entering the tank contains pound of concentrate per gallon. CAPSTONE 36. Suppose the epression u is an integrating factor for Which of the following is equal to u? Verif our answer. (a) P ) u (b) P ) u (c) Q u ) (d) Q u ) P Q. Falling Object In Eercises 37 and 38, consider an eight-pound object dropped from a height of 5000 feet, where the air resistance is proportional to the velocit. 37. Write the velocit of the object as a function of time if the velocit after 5 seconds is approimatel 0 feet per second. What is the limiting value of the velocit function? 38. Use the result of Eercise 37 to write the position of the object as a function of time. Approimate the velocit of the object when it reaches ground level. Electric Circuits In Eercises 39 and 40, use the differential equation for electric circuits given b L di RI E. dt In this equation, I is the current, R is the resistance, L is the inductance, and E is the electromotive force (voltage). 39. Solve the differential equation for the current given a constant voltage E 0. 40. Use the result of Eercise 39 to find the equation for the current if I 0 0, E 0 0 volts, R 600 ohms, and L 4 henrs. When does the current reach 90% of its limiting value? WRITING ABOUT CONCEPTS 4. Give the standard form of a first-order linear differential equation. What is its integrating factor? 4. Give the standard form of the Bernoulli equation. Describe how one reduces it to a linear equation.

44 Chapter 6 Differential Equations In Eercises 43 46, match the differential equation with its solution. 43. (a) Ce 44. (b) Ce 45. (c) C 46. (d) Ce In Eercises 47 54, solve the Bernoulli differential equation. 47. 48. 49. 50. 5. 5. 53. 54. Slope Fields In Eercises 55 58, (a) use a graphing utilit to graph the slope field for the differential equation, (b) find the particular solutions of the differential equation passing through the given points, and (c) use a graphing utilit to graph the particular solutions on the slope field. 55. 56. Differential Equation 0 0 0 3 3 3 3 e 3 e Differential Equation d d 43 3 Solution Points, 4,, 8 0, 7, 0, 57. 58. Differential Equation cot d d In Eercises 59 70, solve the first-order differential equation b an appropriate method. e 59. d e 3 60. d 4 6. cos cos d 0 6. 63. 3 4 d 0 64. d 0 65. e d 0 66. d 0 67. 4 d 3 3 0 68. d 3 4 0 69. 3 4 d 0 70. d e 0 True or False? In Eercises 7 and 7, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 7. is a first-order linear differential equation. 7. is a first-order linear differential equation. e Points,, 3, 0, 3, 0, SECTION PROJECT Weight Loss A person s weight depends on both the number of calories consumed and the energ used. Moreover, the amount of energ used depends on a person s weight the average amount of energ used b a person is 7.5 calories per pound per da. So, the more weight a person loses, the less energ a person uses (assuming that the person maintains a constant level of activit). An equation that can be used to model weight loss is dw dt C 7.5 3500 3500 w where w is the person s weight (in pounds), t is the time in das, and C is the constant dail calorie consumption. (a) Find the general solution of the differential equation. (b) Consider a person who weighs 80 pounds and begins a diet of 500 calories per da. How long will it take the person to lose 0 pounds? How long will it take the person to lose 35 pounds? (c) Use a graphing utilit to graph the solution. What is the limiting weight of the person? (d) Repeat parts (b) and (c) for a person who weighs 00 pounds when the diet is started. FOR FURTHER INFORMATION For more information on modeling weight loss, see the article A Linear Diet Model b Arthur C. Segal in The College Mathematics Journal.

Review Eercises 443 6 REVIEW EXERCISES See www.calcchat.com for worked-out solutions to odd-numbered eercises.. Determine whether the function 3 is a solution of the differential equation. Determine whether the function sin is a solution of the differential equation In Eercises 3 0, use integration to find a general solution of the differential equation. 3. d 4 7 4. 5. cos d 6. 7. 5 d 8. 9. e d 0. Slope Fields In Eercises and, a differential equation and its slope field are given. Determine the slopes (if possible) in the slope field at the points given in the table... d Slope Fields In Eercises 3 8, (a) sketch the slope field for the differential equation, and (b) use the slope field to sketch the solution that passes through the given point. Use a graphing utilit to verif our results. 3. 4. 5. 6. 7. 8. 4 8 4 Differential Equation 3 4 3 4 4 4 0 3. 8 8 0. 4 0 4 8 0 4 4 6 8 /d Point, 0, 0, 3, 0, 0, d 33 8 sin d 7 d 3e 3 d d sin 4 4 0 8 In Eercises 9 4, solve the differential equation. 9. 8 0. 8 d d. 3. 0 d d 3. 0 4. In Eercises 5 8, find the eponential function Ce kt that passes through the two points. 5. 6. 5 4 3 3 0, 4) 3 4 5 7. 8., 9, 6, 0, 5, 5, 6 9. Air Pressure Under ideal conditions, air pressure decreases continuousl with the height above sea level at a rate proportional to the pressure at that height. The barometer reads 30 inches at sea level and 5 inches at 8,000 feet. Find the barometric pressure at 35,000 feet. 30. Radioactive Deca Radioactive radium has a half-life of approimatel 599 ears. The initial quantit is 5 grams. How much remains after 750 ears? 3. Sales The sales S (in thousands of units) of a new product after it has been on the market for t ears is given b S Ce k t. (a) Find S as a function of t if 5000 units have been sold after ear and the saturation point for the market is 30,000 units that is, lim S 30. t (b) How man units will have been sold after 5 ears? (c) Use a graphing utilit to graph this sales function. 3. Sales The sales S (in thousands of units) of a new product after it has been on the market for t ears is given b S 5 e kt. (5, 5) t 3 4 5 (a) Find S as a function of t if 4000 units have been sold after ear. (b) How man units will saturate this market? (c) How man units will have been sold after 5 ears? (d) Use a graphing utilit to graph this sales function. 33. Population Growth A population grows continuousl at the rate of.85%. How long will it take the population to double? 5 4 3, 3 ) ) ) 0 (4, 5) t

444 Chapter 6 Differential Equations 34. Fuel Econom An automobile gets 8 miles per gallon of gasoline for speeds up to 50 miles per hour. Over 50 miles per hour, the number of miles per gallon drops at the rate of percent for each 0 miles per hour. (a) s is the speed and is the number of miles per gallon. Find as a function of s b solving the differential equation (b) Use the function in part (a) to complete the table. In Eercises 35 40, solve the differential equation. 35. 36. d e d 5 7 e 37. 38. 39. 3 40. d d 4. Verif that the general solution C C 3 satisfies the differential equation 3 3 0. Then find the particular solution that satisfies the initial condition 0 and when. 4. Vertical Motion A falling object encounters air resistance that is proportional to its velocit. The acceleration due to gravit is 9.8 meters per second per second. The net change in velocit is dv dt kv 9.8. (a) Find the velocit of the object as a function of time if the initial velocit is v 0. (b) Use the result of part (a) to find the limit of the velocit as t approaches infinit. (c) Integrate the velocit function found in part (a) to find the position function s. 4 Slope Fields In Eercises 43 and 44, sketch a few solutions of the differential equation on the slope field and then find the general solution analticall. To print an enlarged cop of the graph, go to the website www.mathgraphs.com. 43. 44. 3 d 4 d 4 ds 0.0, 6 0 4 4 s > 50. Speed 50 55 60 65 70 Miles per Gallon 4 e sin 0 4 4 4 4 In Eercises 45 and 46, the logistic equation models the growth of a population. Use the equation to (a) find the value of k, (b) find the carring capacit, (c) find the initial population, (d) determine when the population will reach 50% of its carring capacit, and (e) write a logistic differential equation that has the solution P t. 550 4800 45. P t 46. P t 34e 0.55t 4e 0.5t In Eercises 47 and 48, find the logistic equation that satisfies the initial condition. 47. 48. 49. Environment A conservation department releases 00 brook trout into a lake. It is estimated that the carring capacit of the lake for the species is 0,400. After the first ear, there are 000 brook trout in the lake. (a) Write a logistic equation that models the number of brook trout in the lake. (b) Find the number of brook trout in the lake after 8 ears. (c) When will the number of brook trout reach 0,000? 50. Environment Write a logistic differential equation that models the growth rate of the brook trout population in Eercise 49. Then repeat part (b) using Euler s Method with a step size of h. Compare the approimation with the eact answers. In Eercises 5 60, solve the first-order linear differential equation. 5. 5. e 4e 53. 54. 5 d 55. 56. 3 3 57. 3 sin d 0 58. 59. 60. In Eercises 6 64, solve the Bernoulli differential equation. 6. Hint: 6. 63. 64. Logistic Differential Equation dt 80 dt.76 8 0 4 e 4 tan e d 5 e 5 a b 4 e d e 3 Initial Condition 0, 8 0, 3 In Eercises 65 68, write an eample of the given differential equation. Then solve our equation. 65. Homogeneous 66. Logistic 67. First-order linear 68. Bernoulli

P.S. Problem Solving 445 P.S. PROBLEM SOLVING. The differential equation k dt where k and are positive constants, is called the doomsda equation. (a) Solve the doomsda equation.0 dt given that 0. Find the time T at which lim t. t T (b) Solve the doomsda equation k dt given that 0 0. Eplain wh this equation is called the doomsda equation.. A thermometer is taken from a room at 7 F to the outdoors, where the temperature is 0 F. The reading drops to 48 F after minute. Determine the reading on the thermometer after 5 minutes. 3. Let S represent sales of a new product (in thousands of units), let L represent the maimum level of sales (in thousands of units), and let t represent time (in months). The rate of change of S with respect to t varies jointl as the product of S and L S. (a) Write the differential equation for the sales model if L 00, S 0 when t 0, and S 0 when t. Verif that S (b) At what time is the growth in sales increasing most rapidl? (c) Use a graphing utilit to graph the sales function. (d) Sketch the solution from part (a) on the slope field shown in the figure below. To print an enlarged cop of the graph, go to the website www.mathgraphs.com. 40 0 00 80 60 40 0 L Ce kt. S 3 4 t (e) If the estimated maimum level of sales is correct, use the slope field to describe the shape of the solution curves for sales if, at some period of time, sales eceed L. 4. Another model that can be used to represent population growth is the Gompertz equation, which is the solution of the differential equation dt k ln L where k is a constant and L is the carring capacit. (a) Solve the differential equation. (b) Use a graphing utilit to graph the slope field for the differential equation when k 0.05 and L 000. (c) Describe the behavior of the graph as t. (d) Graph the equation ou found in part (a) for L 5000, 0 500, and k 0.0. Determine the concavit of the graph and how it compares with the general solution of the logistic differential equation. 5. Show that the logistic equation L be kt can be written as L tanh k ln b t k. What can ou conclude about the graph of the logistic equation? 6. Although it is true for some functions f and g, a common mistake in calculus is to believe that the Product Rule for derivatives is fg f g. (a) Given g, find f such that fg f g. (b) Given an arbitrar function g, find a function f such that fg f g. (c) Describe what happens if g e. 7. Torricelli s Law states that water will flow from an opening at the bottom of a tank with the same speed that it would attain falling from the surface of the water to the opening. One of the forms of Torricelli s Law is A h dh dt k gh where h is the height of the water in the tank, k is the area of the opening at the bottom of the tank, A h is the horizontal crosssectional area at height h, and g is the acceleration due to gravit g 3 feet per second per second. A hemispherical water tank has a radius of 6 feet. When the tank is full, a circular valve with a radius of inch is opened at the bottom, as shown in the figure. How long will it take for the tank to drain completel? 6 ft 6 h h

446 Chapter 6 Differential Equations 8. The clindrical water tank shown in the figure has a height of 8 feet. When the tank is full, a circular valve is opened at the bottom of the tank. After 30 minutes, the depth of the water is feet. r In Eercises 4, a medical researcher wants to determine the concentration C (in moles per liter) of a tracer drug injected into a moving fluid. Solve this problem b considering a singlecompartment dilution model (see figure). Assume that the fluid is continuousl mied and that the volume of the fluid in the compartment is constant. 8 ft h Tracer injected Flow R (pure) Volume V (a) How long will it take for the tank to drain completel? (b) What is the depth of the water in the tank after hour? 9. Suppose the tank in Eercise 8 has a height of 0 feet and a radius of 8 feet, and the valve is circular with a radius of inches. The tank is full when the valve is opened. How long will it take for the tank to drain completel? 0. In hill areas, radio reception ma be poor. Consider a situation in which an FM transmitter is located at the point, behind a hill modeled b the graph of and a radio receiver is on the opposite side of the hill. (Assume that the -ais represents ground level at the base of the hill.) (a) What is the closest position, 0 the radio can be to the hill so that reception is unobstructed? (b) Write the closest position, 0 of the radio with represented as a function of h if the transmitter is located at, h. (c) Use a graphing utilit to graph the function for in part (b). Determine the vertical asmptote of the function and interpret its meaning.. Biomass is a measure of the amount of living matter in an ecosstem. Suppose the biomass s t in a given ecosstem increases at a rate of about 3.5 tons per ear, and decreases b about.9% per ear. This situation can be modeled b the differential equation ds 3.5 0.09s. dt (a) Solve the differential equation. (b) Use a graphing utilit to graph the slope field for the differential equation. What do ou notice? (c) Eplain what happens as t. Figure for 4. If the tracer is injected instantaneousl at time t 0, then the concentration of the fluid in the compartment begins diluting according to the differential equation dc dt R V C, C C 0 when t 0. (a) Solve this differential equation to find the concentration C as a function of time t. (b) Find the limit of C as t. 3. Use the solution of the differential equation in Eercise to find the concentration C as a function of time t, and use a graphing utilit to graph the function. (a) V liters, R 0.5 liter per minute, and C 0 0.6 mole per liter (b) V liters, R.5 liters per minute, and C 0 0.6 mole per liter 4. In Eercises and 3, it was assumed that there was a single initial injection of the tracer drug into the compartment. Now consider the case in which the tracer is continuousl injected beginning at t 0 at the rate of Q moles per minute. Considering Q to be negligible compared with R, use the differential equation dc dt Q V R V C, Flow R (concentration C) C 0 when t 0. (a) Solve this differential equation to find the concentration C as a function of time t. (b) Find the limit of C as t.