31.3 Qualitative Analysis of dy dt = f(y) 1011 dP dt = 0.45P(t)− 0.0005[P(t)] 2 d dt dP dt = 0.45 dP dt − 0.001P dP dt d 2 P dt 2 = (0.45 − 0.001P) dP dt We can replace dP dt by 0.45P − 0.0005P 2 ,orP(0.45 − 0.0005P),obtaining d 2 P dt 2 = (0.45 − 0.001P)P(0.45 − 0.0005P)· P d 2 P dt 2 is continuous, so it can change sign only on either side of a zero. The zeros are where 0.45 − 0.001P = 0, P = 0, and 0.45 − 0.0005P = 0. P = 450, P = 0, and P = 900. P = 450 is what we’re interested in. For P ∈ (0, 450) d 2 P dt 2 > 0; for P ∈ (450, 900) d 2 P dt 2 < 0. Therefore the point of inflection is at P = 450, halfway between the two equilibrium solutions. The fish population is growing most rapidly when there are 450 fish. ◆ Compartmental Analysis ◆ EXAMPLE 31.19 A 20-quart juice dispenser in a cafeteria is filled with a juice mixture that is 10% mango juice and 90% cranberry juice. An orange-mango blend is entering the dispenser at a rate of 4 quarts an hour and the well-stirred mixture leaves at the same rate. The orange-mango blend is 50% orange and 50% mango. (a) Model the situation with a differential equation whose solution is M = M(t), the number of quarts of mango juice in the container at time t. Include the initial condition. Sketch the solution curve, indicating the long-run behavior of M. (b) Model the situation with a differential equation whose solution is C(t), the number of quarts of cranberry juice in the container at time t. Include an initial condition. (c) Suppose that once the percentage of mango juice in the mixture reaches 30%, high enough to be irresistible, the rate of juice consumption increases to 5 quarts per hour while the rate at which the container is being filled remains at 4 quarts an hour. Revise the differential equation in part (a) to reflect the situation. Reset the clock to denote by t = 0 the moment at which the consumption rate increases. 1012 CHAPTER 31 Differential Equations 4 quarts/hr 4 quarts/hr 20 quart mixture Figure 31.13 SOLUTION (a) The basic structure for our model is rate of change = rate in − rate out. More specifically, rate of change of mango juice = rate at which mango juice enters − rate at which mango juice leaves . The left-hand side of the equation is dM dt with units quarts of mango juice hour . The rate in and rate out must also be measured in quarts of mango juice hour . Rate in: A mixture is entering at a rate of 4 quarts of mixture hour . Half of this is mango juice, so mango juice is entering at a rate of 2 quarts of mango juice hour . quarts of mixture hour · quarts of mango juice quarts of mixture = quarts of mango juice hour Rate out: The mixture is leaving at a rate of 4 quarts of mixture hour , but what fraction of this exiting juice is mango juice? Because the juice in the container is well mixed, the fraction is the same as the ratio of mango juice to mixture in the container. The amount of mango juice in the container changes with time; it is given by M(t). The amount of mixture in the dispenser remains constant at 20 quarts (because the rate at which juice is entering is equal to the rate that juice is leaving). rate out = 4 quarts of mixture hour · M(t) 20 quarts of mango juice quarts of mixture = 4M(t) 20 quarts of mango juice hour Now we can set up the differential equation. rate of change of mango juice = rate in of mango juice − rate out of mango juice dM dt = 2 − 4 · M 20 dM dt = 2 − 1 5 M where M(0) = 2 The equilibrium solution is the value of M for which dM dt = 0. 31.3 Qualitative Analysis of dy dt = f(y) 1013 2 − 1 5 M = 0 1 5 M = 2 M = 10 MM t M t (+) (–) sign of dM dt 10 2 M =10 M =10 Figure 31.14 We know that the solution curve through (0, 2) is concave down because its slope, given by 2 − 1 5 M, decreases for M ∈ (0, 10). In the long run the amount of mango juice in the container will approach 10 quarts. This makes sense, because the juice in the container will eventually look like the incoming juice; it will eventually be half mango juice and half orange juice. (b) Let C = C(t) be the number of quarts of cranberry juice in the container at time t. rate of change = rate in − rate out rate of change of cranberry juice = rate in of cranberry juice − rate out of cranberry juice dC dt = 0 − 4 quarts of mixture hour · C(t) quarts of cranberry juice 20 quarts of mixture dC dt =−4 C(t) 20 dC dt =− 1 5 C(t) (c) Notice that from t = 0 on the number of quarts of juice in the dispenser goes down by 1 quart every hour. rate of change = rate in − rate out The rate in remains at 2 quarts of mango juice hour but the rate out must be recalculated. Rate out: 5 quarts of mixture hour · M(t) 20−t quarts of mango juice quarts of mixture = 5 M(t) 20−t Therefore dM dt = 2 − 5M 20−t . 6 The initial condition is M(0) = .3(20) = 6. ◆ While the amount of mango juice in a juice dispenser is not itself an issue of vital importance, the ideas involved in solving problems of this type are directly applicable to interesting prob- lems in chemistry, ecology, and geology. For example, a lake may be fed by an underground 6 This differential equation gives dM dt in terms of both M and t . We have not discussed how to solve it. 1014 CHAPTER 31 Differential Equations spring and other tributaries and may itself feed an outlet stream. The water quality may vary; assuming a well-mixed lake, an environmentalist can analyze the spring/lake/stream system and the level of pollutants over time. Similarly, an ecologist looking at Mono Lake, where the diversion of water sources has led to a change in salination levels affecting brine shrimp populations, can analyze the salt concentration in terms of rate of change = rate in − rate out. Geologists trying to date events across millennia use the same setup. The time spans they are dealing with can be so enormous that the oceans themselves are considered “well-mixed.” PROBLEMS FOR SECTION 31.3 1. Do a qualitative analysis of the family of solutions to each of the differential equations below. Then, in another color pen or pencil, highlight the graphs of the solutions cor- responding to the given initial conditions. (If the solution is asymptotic to a horizontal line, draw and label that line.) (a) dy dt = 4y − 8; y(0) = 0, y(0) =−1, y(0) = 3 (b) dy dt = y 2 − 4; y(0) =−1, y(0) =−3, y(0) = 4 (c) dy dt = (y − 1)(y − 2)(y + 1); y(0) = 0, y(0) = 3 (d) dy dt = y 2 + 5y − 6; y(0) =−5, y(0) =−7, y(0) = 2 2. For each of the following families of graphs, write a differential equation for which the graphs drawn could be solutions. (There are infinitely many correct answers; produce one.) y (a) t 3 y (b) t 3 y (c) t 2 y (d) t 2 –2 3. In each of Problems 2 (a)–(d), identify the equilibria and classify them as stable or unstable. 4. Consider the differential equation dP dt = kP(L − P), where k and L are positive constants. 31.3 Qualitative Analysis of dy dt = f(y) 1015 (a) For what values of P is dP dt zero? (b) Show that P(t)= L 1+Ce −kLt , where C is a constant, is a solution of the logistic equation above. 5. The population of a town in the south of Bangladesh has been growing exponentially. However, recent flooding has alarmed residents and people are leaving the town at a rate of N thousand people per year, where N is a constant. The rate of change of the population of the town can be modeled by the differential equation dP dt = 0.02P − N, where P = P(t)is the number of people in the town in thousands. (a) If P(0)= 100, what is the largest yearly exodus rate the town can support in the long run? (b) How big must the population of the town be in order to support the loss of 1000 people per year? 6. Which of the graphs below could be the graph of a solution to dy dt = y 2 − 1, where y(0) = 0? Explain why you eliminated each of the other options. y t (a) y t (b) y t (c) y t (e) 1 –1 y t (d) 7. Which of the graphs on the following page could be a solution to the differential equation dx dt = x 2 (2 − x) with initial condition x(0)= 0.5? Explain why you eliminated each of the other options. 1016 CHAPTER 31 Differential Equations x t 0.5 x t 0.5 x t 0.5 x t 0.5 2 8. Essay question: One of your classmates is puzzled by what it means to solve a differ- ential equation. He has two questions for you. Answer them in plain English. (a) How can you tell if something is a solution to a differential equation? (b) What is the difference between a general and a particular solution? Is the difference always that the general solution just has a “plus C” at the end? Are there some cases where that’s the only difference? 9. Let P(t)be the number of crocodiles in a mud hole at time t. Suppose dP dt = 0.01P − 0.0025P 2 . (a) What is the carrying capacity of the mud hole? (b) Find d 2 P dt 2 . Remember: You are differentiating with respect to t, so the derivative of P is not 1. (c) Use your answer to part (b) to determine how many crocodiles are in the mud hole when the number of crocodiles is increasing most rapidly. (d) Sketch a solution curve if the number of crocodiles in the mud hole at time t = 0 is 3. (Label the vertical axis. You need not calibrate the t-axis.) 10. Sketch a representative family of solutions to the following differential equations. You need not take a second derivative. (a) dy dt = y(y 2 − 4) (b) dy dt = y 2 (y − 2) In Problems 11 through 16 sketch a representative family of solutions for each of the following differential equations. 11. (a) dy dt = 2y − 6 (b) dy dt = 6 − 2y 12. (a) dy dt = sin t (b) dy dt = sin y 13. dy dt = tan y 14. (a) dy dt = t 2 (b) dy dt = y 2 15. (a) dy dt = t 2 − 1 (b) dy dt = y 2 − 1 16. dx dt = (1 − x)(x + 2)(x − 3) 31.3 Qualitative Analysis of dy dt = f(y) 1017 17. Give an example of a differential equation with constant solutions at y =−1and y = 4 with the characteristics specified. (a) The equilibrium at y =−1isstable; the equilibrium at y = 4 is unstable. (b) The equilibrium at y =−1isunstable; the equilibrium at y = 4 is stable. (c) Neither equilibrium solution is stable. 18. Give an example of a differential equation of the form dv dt = f(v)and whose solutions depend upon v(0) as described below. If v(0)>5, then v(t) is increasing. If v(0) = 5, then v(t) = 5. If 2 <v(0)<5, then v(t) is decreasing. If v(0) = 2, then v(t) = 2. If v(0)<2, then v(t) is increasing. 19. A canister contains 10 liters of blue paint. Paint is being used at a rate of 2 liters per hour and the canister is being replenished at a rate of 2 liters per hour by a pale blue paint that is 80% blue and 20% white. Assuming the canister is well-mixed, write a differential equation whose solution is w(t), the amount of white paint in the canister at time t. Specify the initial condition. 20. A 5-gallon urn is filled with chai, a milky spicy tea. The chai in the urn is 90% tea and 10% milk. Chai is being consumed at a rate of 1/2 gallon per hour and the urn is kept full by adding a mixture that is 80% tea and 20% milk. Assume that the chai is well-mixed. (a) Write a differential equation whose solution is M(t),the number of gallons of milk in the urn at time t. Specify the initial condition. (b) Use qualitative analysis to sketch the solution to the differential equation in part (a). (c) How much milk is in the urn after 2 hours? 21. The population of wildebeest in the Serengeti was decimated by a rinderpest plague in the 1950s. In 1961 the Serengeti supported a population of a quarter of a million wildebeest. By 1978 the wildebeest population was 1.5 million and by 1991 it had reached 2 million. (Craig Packer Into Africa, Chicago, The University of Chicago Press, 1996 p. 250.) Given this data, would you be more inclined to model the growth of the wildebeest population using an exponential growth model or using a logistic growth model? Explain your reasoning. 22. A lake contains 10 10 liters of water. Acid rain containing 0.02 milligrams of pollutant per liter of rain falls into the lake at a rate of 10 3 liters per week. An outlet stream drains away 10 3 liters of water per week. Assume that the pollutant is always evenly distributed throughout the lake, so the runoff into the stream has the same concentration of pollutant as the lake as a whole. The volume of the lake stays constant at 10 10 liters because the water lost from the runoff balances exactly the water gained from the rain. 1018 CHAPTER 31 Differential Equations (a) Write a differential equation whose solution is P(t),the number of milligrams of pollutant in the lake as a function of t measured in weeks. (b) Find any equilibrium solutions. (c) Sketch some representative solution curves. (d) How would you alter the differential equation if there was a dry spell and rain was falling into the lake at a rate of only 10 2 liters per week. 23. Challenge Problem: (a) You plan to save money starting today at a rate of $4000 per year over the next 30 years. You will deposit this money at a nearly continuous rate (a constant amount each day) into a bank account that earns 5% interest compounded continuously. Let B(t) be the balance of money in the account t years from now, where 0 ≤ t ≤ 30. i. Write a differential equation whose solution is B(t). ii. Write an integral that is equal to B(30), the amount in the account at the end of 30 years. (b) Now assume that instead of making deposits continuously, you decide to make a deposit of $4000 once a year, starting today and continuing until you have made a total of 30 deposits. Suppose the bank account pays 5% interest compounded annually. i. Write a geometric sum equal tothebalanceimmediatelyafter the final deposit. ii. Find a closed form expression (no +···+,nosummation notation) for this sum. 31.4 SOLVING SEPARABLE FIRST ORDER DIFFERENTIAL EQUATIONS Differential equations of the forms dy dt = g(t) and dy dt = f(y)are special cases of a larger class of first order differential equations called separable differential equations. Separable differential equations can be written in the form dy dt = g(t)f (y). Equivalently, we can write dy dt = g(t) h(y) .Wecan“separate” variables as follows: h(y) dy dt = g(t) Integrating both sides with respect to t gives h(y) dy dt dt = g(t) dt or h(y) dy = g(t) dt. Generally this is accomplished by separating the variables into the differential form 31.4 Solving Separable First Order Differential Equations 1019 h(y) dy = g(t) dt and integrating both sides. Let’s use this method on a familiar example. ◆ EXAMPLE 31.20 Solve dy dt = ky using separation of variables. SOLUTION dy y = kdt Assuming y = 0, we can divide by y. dy y = kdt ln |y|=kt + C |y|=e kt+C Solve for y explicitly in terms of t. |y|=e C ·e kt = Ae kt y =±Ae kt where A>0. In other words, y = C 1 e kt , C 1 any constant. There’s one missing detail here. According to what we’ve done, C 1 shouldn’t be zero because the separation of variables assumed y = 0. Treating the case y = 0 independently, we see y = 0 is a solution. Hence C 1 may be zero. ◆ ◆ EXAMPLE 31.21 Solve dy dx = −x y . SOLUTION ydy=−xdx ydy= −xdx 1 2 y 2 =− 1 2 x 2 + C 1 x 2 + y 2 = C Observations y is not a function of x. The solution is a relationship between x and y. You might remember this particular problem from Chapter 17. We began with the equation of the circle and, using implicit differentiation, arrived at dy dx = −x y . ◆ In the previous section we discussed the logistic growth model, both in thecontext of lions on a Tanzanian savannah and fish in a lake. Using separation of variables and partial fractions, we can solve the logistic differential equation. ◆ EXAMPLE 31.22 Solve dP dt = 0.5P − 0.001P 2 . SOLUTION dP dt = P(0.5 − 0.001P)=0.001P(500 − P) Separating variables, we obtain 1020 CHAPTER 31 Differential Equations dP P(500 − P) =0.001 dt for P = 500, P = 0. Integrating gives dP P(500 − P) =0.001t + C 1 . (31.1) To integrate the left-hand side, we algebraically split the fraction into the sum of two fractions each of which is easier to integrate. We look for constants A and B such that 1 P(500 − P) = A P + B 500 − P . Clearing denominators gives 1 = A(500 − P)+BP and evaluating at P = 0 yields A = 0.002. If A = 0.002, then 1 = 1 − (0.002)P + BP,soB=0.002. Therefore 1 P(500 − P) = 0.002 P + 0.002 500 − P , i.e., 1 P(500 − P) = 1 500 1 P + 1 500 − P . Hence 1 P(500 − P) dP = 1 500 1 P + 1 500 − P dP. Returning to Equation (31.1) we can now write 1 500 1 P + 1 500 − P dP = 0.001t + C 1 . Then 1 500 [ln |P |−ln |500 − P |] = 0.001t + C 1 or ln P 500 − P = 0.5t + C 2 . Exponentiating, we obtain P 500 − P = C 3 e 0.5t , C 3 > 0. . 10% mango juice and 90% cranberry juice. An orange-mango blend is entering the dispenser at a rate of 4 quarts an hour and the well-stirred mixture leaves at the same rate. The orange-mango blend. mango juice and half orange juice. (b) Let C = C(t) be the number of quarts of cranberry juice in the container at time t. rate of change = rate in − rate out rate of change of cranberry juice = rate. logistic growth model, both in thecontext of lions on a Tanzanian savannah and fish in a lake. Using separation of variables and partial fractions, we can solve the logistic differential equation. ◆ EXAMPLE