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Unformatted text preview: 5E-10(pp 622-631) 1/18/06 9:18 AM Page 622 CHAPTER 10 By analyzing pairs of differential equations we gain insight into population cycles of predators and prey, such as the Canada lynx and snowshoe hare. W 150 100 R 3000 W R W 50 120 2000 80 0 1000 2000 3000 R 1000 40 0 t¡ t™ t£ Differential Equations t 5E-10(pp 622-631) 1/18/06 9:18 AM Page 623 Perhaps the most important of all the applications of calculus is to differential equations. When physical scientists or social scientists use calculus, more often than not it is to analyze a differential equation that has arisen in the process of modeling some phenomenon that they are studying. Although it is often impossible to find an explicit formula for the solution of a differential equation, we will see that graphical and numerical approaches provide the needed information. |||| 10.1 Modeling with Differential Equations |||| Now is a good time to read (or reread) the discussion of mathematical modeling on page 25. In describing the process of modeling in Section 1.2, we talked about formulating a mathematical model of a real-world problem either through intuitive reasoning about the phenomenon or from a physical law based on evidence from experiments. The mathematical model often takes the form of a differential equation, that is, an equation that contains an unknown function and some of its derivatives. This is not surprising because in a realworld problem we often notice that changes occur and we want to predict future behavior on the basis of how current values change. Let’s begin by examining several examples of how differential equations arise when we model physical phenomena. Models of Population Growth One model for the growth of a population is based on the assumption that the population grows at a rate proportional to the size of the population. That is a reasonable assumption for a population of bacteria or animals under ideal conditions (unlimited environment, adequate nutrition, absence of predators, immunity from disease). Let’s identify and name the variables in this model: t ෇ time ͑the independent variable͒ P ෇ the number of individuals in the population ͑the dependent variable͒ The rate of growth of the population is the derivative dP͞dt. So our assumption that the rate of growth of the population is proportional to the population size is written as the equation 1 dP ෇ kP dt where k is the proportionality constant. Equation 1 is our first model for population growth; it is a differential equation because it contains an unknown function P and its derivative dP͞dt. Having formulated a model, let’s look at its consequences. If we rule out a population of 0, then P͑t͒ Ͼ 0 for all t. So, if k Ͼ 0, then Equation 1 shows that PЈ͑t͒ Ͼ 0 for all t. This means that the population is always increasing. In fact, as P͑t͒ increases, Equation 1 shows that dP͞dt becomes larger. In other words, the growth rate increases as the population increases. 623 5E-10(pp 622-631) 624 ❙❙❙❙ 1/18/06 9:18 AM Page 624 CHAPTER 10 DIFFERENTIAL EQUATIONS P Let’s try to think of a solution of Equation 1. This equation asks us to find a function whose derivative is a constant multiple of itself. We know that exponential functions have that property. In fact, if we let P͑t͒ ෇ Ce kt, then PЈ͑t͒ ෇ C͑ke kt ͒ ෇ k͑Ce kt ͒ ෇ kP͑t͒ t Thus, any exponential function of the form P͑t͒ ෇ Ce kt is a solution of Equation 1. When we study this equation in detail in Section 10.4, we will see that there is no other solution. Allowing C to vary through all the real numbers, we get the family of solutions P͑t͒ ෇ Ce kt whose graphs are shown in Figure 1. But populations have only positive values and so we are interested only in the solutions with C Ͼ 0. And we are probably concerned only with values of t greater than the initial time t ෇ 0. Figure 2 shows the physically meaningful solutions. Putting t ෇ 0, we get P͑0͒ ෇ Ce k͑0͒ ෇ C, so the constant C turns out to be the initial population, P͑0͒. Equation 1 is appropriate for modeling population growth under ideal conditions, but we have to recognize that a more realistic model must reflect the fact that a given environment has limited resources. Many populations start by increasing in an exponential manner, but the population levels off when it approaches its carrying capacity K (or decreases toward K if it ever exceeds K). For a model to take into account both trends, we make two assumptions: FIGURE 1 The family of solutions of dP/dt=kP P 0 t ■ dP Ϸ kP if P is small (Initially, the growth rate is proportional to P.) dt ■ dP Ͻ 0 if P Ͼ K (P decreases if it ever exceeds K.) dt FIGURE 2 The family of solutions P(t)=Ce kt with C>0 and t˘0 A simple expression that incorporates both assumptions is given by the equation 2 P P =K equilibrium solutions P =0 0 FIGURE 3 Solutions of the logistic equation t ͩ ͪ dP P ෇ kP 1 Ϫ dt K Notice that if P is small compared with K, then P͞K is close to 0 and so dP͞dt Ϸ kP. If P Ͼ K , then 1 Ϫ P͞K is negative and so dP͞dt Ͻ 0. Equation 2 is called the logistic differential equation and was proposed by the Dutch mathematical biologist Pierre-François Verhulst in the 1840s as a model for world population growth. We will develop techniques that enable us to find explicit solutions of the logistic equation in Section 10.5, but for now we can deduce qualitative characteristics of the solutions directly from Equation 2. We first observe that the constant functions P͑t͒ ෇ 0 and P͑t͒ ෇ K are solutions because, in either case, one of the factors on the right side of Equation 2 is zero. (This certainly makes physical sense: If the population is ever either 0 or at the carrying capacity, it stays that way.) These two constant solutions are called equilibrium solutions. If the initial population P͑0͒ lies between 0 and K, then the right side of Equation 2 is positive, so dP͞dt Ͼ 0 and the population increases. But if the population exceeds the carrying capacity ͑P Ͼ K͒, then 1 Ϫ P͞K is negative, so dP͞dt Ͻ 0 and the population decreases. Notice that, in either case, if the population approaches the carrying capacity ͑P l K͒, then dP͞dt l 0, which means the population levels off. So we expect that the solutions of the logistic differential equation have graphs that look something like the ones in Figure 3. Notice that the graphs move away from the equilibrium solution P ෇ 0 and move toward the equilibrium solution P ෇ K . 5E-10(pp 622-631) 1/18/06 9:18 AM Page 625 SECTION 10.1 MODELING WITH DIFFERENTIAL EQUATIONS ❙❙❙❙ 625 A Model for the Motion of a Spring Let’s now look at an example of a model from the physical sciences. We consider the motion of an object with mass m at the end of a vertical spring (as in Figure 4). In Section 6.4 we discussed Hooke’s Law, which says that if the spring is stretched (or compressed) x units from its natural length, then it exerts a force that is proportional to x: m equilibrium position restoring force ෇ Ϫkx 0 x m where k is a positive constant (called the spring constant). If we ignore any external resisting forces (due to air resistance or friction) then, by Newton’s Second Law (force equals mass times acceleration), we have x FIGURE 4 m 3 d 2x ෇ Ϫkx dt 2 This is an example of what is called a second-order differential equation because it involves second derivatives. Let’s see what we can guess about the form of the solution directly from the equation. We can rewrite Equation 3 in the form d 2x k x 2 ෇ Ϫ dt m which says that the second derivative of x is proportional to x but has the opposite sign. We know two functions with this property, the sine and cosine functions. In fact, it turns out that all solutions of Equation 3 can be written as combinations of certain sine and cosine functions (see Exercise 3). This is not surprising; we expect the spring to oscillate about its equilibrium position and so it is natural to think that trigonometric functions are involved. General Differential Equations In general, a differential equation is an equation that contains an unknown function and one or more of its derivatives. The order of a differential equation is the order of the highest derivative that occurs in the equation. Thus, Equations 1 and 2 are first-order equations and Equation 3 is a second-order equation. In all three of those equations the independent variable is called t and represents time, but in general the independent variable doesn’t have to represent time. For example, when we consider the differential equation 4 yЈ ෇ xy it is understood that y is an unknown function of x. A function f is called a solution of a differential equation if the equation is satisfied when y ෇ f ͑x͒ and its derivatives are substituted into the equation. Thus, f is a solution of Equation 4 if f Ј͑x͒ ෇ xf ͑x͒ for all values of x in some interval. When we are asked to solve a differential equation we are expected to find all possible solutions of the equation. We have already solved some particularly simple differential 5E-10(pp 622-631) 626 ❙❙❙❙ 1/18/06 9:18 AM Page 626 CHAPTER 10 DIFFERENTIAL EQUATIONS equations, namely, those of the form yЈ ෇ f ͑x͒ For instance, we know that the general solution of the differential equation yЈ ෇ x 3 is given by y෇ x4 ϩC 4 where C is an arbitrary constant. But, in general, solving a differential equation is not an easy matter. There is no systematic technique that enables us to solve all differential equations. In Section 10.2, however, we will see how to draw rough graphs of solutions even when we have no explicit formula. We will also learn how to find numerical approximations to solutions. EXAMPLE 1 Show that every member of the family of functions y෇ 1 ϩ ce t 1 Ϫ ce t is a solution of the differential equation yЈ ෇ 1 ͑y 2 Ϫ 1͒. 2 SOLUTION We use the Quotient Rule to differentiate the expression for y: yЈ ෇ |||| Figure 5 shows graphs of seven members of the family in Example 1. The differential equation shows that if y Ϸ Ϯ1, then yЈ Ϸ 0. That is borne out by the flatness of the graphs near y ෇ 1 and y ෇ Ϫ1. 5 ෇ ͑y 2 Ϫ 1͒ ෇ 5 ෇ _5 FIGURE 5 ce t Ϫ c 2e 2t ϩ ce t ϩ c 2e 2t 2ce t ෇ ͑1 Ϫ ce t ͒2 ͑1 Ϫ ce t ͒2 The right side of the differential equation becomes 1 2 _5 ͑1 Ϫ ce t ͒͑ce t ͒ Ϫ ͑1 ϩ ce t ͒͑Ϫce t ͒ ͑1 Ϫ ce t ͒2 1 2 ͫͩ 1 ϩ ce t 1 Ϫ ce t ͪ ͬ ͫ 2 Ϫ1 ෇ 1 2 ͑1 ϩ ce t ͒2 Ϫ ͑1 Ϫ ce t ͒2 ͑1 Ϫ ce t ͒2 ͬ 1 4ce t 2ce t ෇ 2 ͑1 Ϫ ce t ͒2 ͑1 Ϫ ce t ͒2 Therefore, for every value of c, the given function is a solution of the differential equation. When applying differential equations, we are usually not as interested in finding a family of solutions (the general solution) as we are in finding a solution that satisfies some additional requirement. In many physical problems we need to find the particular solution that satisfies a condition of the form y͑t0 ͒ ෇ y0 . This is called an initial condition, and the problem of finding a solution of the differential equation that satisfies the initial condition is called an initial-value problem. Geometrically, when we impose an initial condition, we look at the family of solution curves and pick the one that passes through the point ͑t0 , y0 ͒. Physically, this corresponds to measuring the state of a system at time t0 and using the solution of the initial-value problem to predict the future behavior of the system. 5E-10(pp 622-631) 1/18/06 9:18 AM Page 627 SECTION 10.1 MODELING WITH DIFFERENTIAL EQUATIONS ❙❙❙❙ 627 EXAMPLE 2 Find a solution of the differential equation yЈ ෇ 2 ͑y 2 Ϫ 1͒ that satisfies the 1 initial condition y͑0͒ ෇ 2. SOLUTION Substituting the values t ෇ 0 and y ෇ 2 into the formula y෇ 1 ϩ ce t 1 Ϫ ce t from Example 1, we get 2෇ 1 ϩ ce 0 1ϩc 0 ෇ 1 Ϫ ce 1Ϫc Solving this equation for c, we get 2 Ϫ 2c ෇ 1 ϩ c, which gives c ෇ 1 . So the solution 3 of the initial-value problem is y෇ |||| 10.1 Exercises (d) Find a solution of the differential equation yЈ ෇ xy that satisfies the initial condition y͑1͒ ෇ 2. 1. Show that y ෇ x Ϫ x Ϫ1 is a solution of the differential equation xyЈ ϩ y ෇ 2x. 2. Verify that y ෇ sin x cos x Ϫ cos x is a solution of the initial- 7. (a) What can you say about a solution of the equation yЈ ෇ Ϫy 2 just by looking at the differential equation? (b) Verify that all members of the family y ෇ 1͑͞x ϩ C ͒ are solutions of the equation in part (a). (c) Can you think of a solution of the differential equation yЈ ෇ Ϫy 2 that is not a member of the family in part (b)? (d) Find a solution of the initial-value problem value problem yЈ ϩ ͑tan x͒y ෇ cos2 x y͑0͒ ෇ Ϫ1 on the interval Ϫ␲͞2 Ͻ x Ͻ ␲͞2. 3. (a) For what nonzero values of k does the function y ෇ sin kt satisfy the differential equation yЉ ϩ 9y ෇ 0 ? (b) For those values of k, verify that every member of the family of functions yЈ ෇ Ϫy 2 is also a solution. 4. For what values of r does the function y ෇ e rt satisfy the differential equation y Љ ϩ yЈ Ϫ 6y ෇ 0? 5. Which of the following functions are solutions of the differen- tial equation y Љ ϩ 2yЈ ϩ y ෇ 0? (a) y ෇ e t (b) y ෇ e Ϫt (c) y ෇ te Ϫt (d) y ෇ t 2e Ϫt 6. (a) Show that every member of the family of functions 2 y ෇ Ce x ͞2 is a solution of the differential equation yЈ ෇ xy. (b) Illustrate part (a) by graphing several members of the family of solutions on a common screen. (c) Find a solution of the differential equation yЈ ෇ xy that satisfies the initial condition y͑0͒ ෇ 5. y͑0͒ ෇ 0.5 8. (a) What can you say about the graph of a solution of the equa- y ෇ A sin kt ϩ B cos kt ; 1 ϩ 1et 3 ϩ et 3 ෇ 1 Ϫ 1et 3 Ϫ et 3 ; tion yЈ ෇ xy 3 when x is close to 0? What if x is large? (b) Verify that all members of the family y ෇ ͑c Ϫ x 2 ͒Ϫ1͞2 are solutions of the differential equation yЈ ෇ xy 3. (c) Graph several members of the family of solutions on a common screen. Do the graphs confirm what you predicted in part (a)? (d) Find a solution of the initial-value problem yЈ ෇ xy 3 y͑0͒ ෇ 2 9. A population is modeled by the differential equation ͩ dP P ෇ 1.2P 1 Ϫ dt 4200 ͪ (a) For what values of P is the population increasing? (b) For what values of P is the population decreasing? (c) What are the equilibrium solutions? 5E-10(pp 622-631) 628 ❙❙❙❙ 1/18/06 9:18 AM Page 628 CHAPTER 10 DIFFERENTIAL EQUATIONS 10. A function y͑t͒ satisfies the differential equation 13. Psychologists interested in learning theory study learning dy ෇ y 4 Ϫ 6y 3 ϩ 5y 2 dt (a) What are the constant solutions of the equation? (b) For what values of y is y increasing? (c) For what values of y is y decreasing? 11. Explain why the functions with the given graphs can’t be solu- tions of the differential equation dy ෇ e t͑ y Ϫ 1͒2 dt (a) y dP ෇ k͑M Ϫ P͒ dt (b) y 1 t 1 t 12. The function with the given graph is a solution of one of the following differential equations. Decide which is the correct equation and justify your answer. y 0 A. yЈ ෇ 1 ϩ xy |||| 10.2 B. yЈ ෇ Ϫ2xy k a positive constant is a reasonable model for learning. (c) Make a rough sketch of a possible solution of this differential equation. 1 1 curves. A learning curve is the graph of a function P͑t͒, the performance of someone learning a skill as a function of the training time t. The derivative dP͞dt represents the rate at which performance improves. (a) When do you think P increases most rapidly? What happens to dP͞dt as t increases? Explain. (b) If M is the maximum level of performance of which the learner is capable, explain why the differential equation x C. yЈ ෇ 1 Ϫ 2xy 14. Suppose you have just poured a cup of freshly brewed coffee with temperature 95ЊC in a room where the temperature is 20ЊC. (a) When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain. (b) Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton’s Law of Cooling for this particular situation. What is the initial condition? In view of your answer to part (a), do you think this differential equation is an appropriate model for cooling? (c) Make a rough sketch of the graph of the solution of the initial-value problem in part (b). Direction Fields and Euler’s Method Unfortunately, it’s impossible to solve most differential equations in the sense of obtaining an explicit formula for the solution. In this section we show that, despite the absence of an explicit solution, we can still learn a lot about the solution through a graphical approach (direction fields) or a numerical approach (Euler’s method). Direction Fields Suppose we are asked to sketch the graph of the solution of the initial-value problem yЈ ෇ x ϩ y y͑0͒ ෇ 1 We don’t know a formula for the solution, so how can we possibly sketch its graph? Let’s think about what the differential equation means. The equation yЈ ෇ x ϩ y tells us that the slope at any point ͑x, y͒ on the graph (called the solution curve) is equal to the sum of the 5E-10(pp 622-631) 1/18/06 9:18 AM Page 629 SECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD ❙❙❙❙ 629 x- and y-coordinates of the point (see Figure 1). In particular, because the curve passes through the point ͑0, 1͒, its slope there must be 0 ϩ 1 ෇ 1. So a small portion of the solution curve near the point ͑0, 1͒ looks like a short line segment through ͑0, 1͒ with slope 1 (see Figure 2). y y Slope at (¤, fi) is ¤+fi. Slope at (⁄, ›) is ⁄+›. (0, 1) 0 x Slope at (0, 1) is 0+1=1. 0 x FIGURE 1 FIGURE 2 A solution of yª=x+y Beginning of the solution curve through (0, 1) As a guide to sketching the rest of the curve, let’s draw short line segments at a number of points ͑x, y͒ with slope x ϩ y. The result is called a direction field and is shown in Figure 3. For instance, the line segment at the point ͑1, 2͒ has slope 1 ϩ 2 ෇ 3. The direction field allows us to visualize the general shape of the solution curves by indicating the direction in which the curves proceed at each point. y y (0, 1) 0 1 2 x 0 1 2 FIGURE 3 FIGURE 4 Direction field for yª=x+y x The solution curve through (0, 1) Now we can sketch the solution curve through the point ͑0, 1͒ by following the direction field as in Figure 4. Notice that we have drawn the curve so that it is parallel to nearby line segments. In general, suppose we have a first-order differential equation of the form yЈ ෇ F͑x, y͒ where F͑x, y͒ is some expression in x and y. The differential equation says that the slope of a solution curve at a point ͑x, y͒ on the curve is F͑x, y͒. If we draw short line segments with slope F͑x, y͒ at several points ͑x, y͒, the result is called a direction field (or slope field). These line segments indicate the direction in which a solution curve is heading, so the direction field helps us visualize the general shape of these curves. 5E-10(pp 622-631) 630 ❙❙❙❙ 1/18/06 9:18 AM Page 630 CHAPTER 10 DIFFERENTIAL EQUATIONS y EXAMPLE 1 2 (a) Sketch the direction field for the differential equation yЈ ෇ x 2 ϩ y 2 Ϫ 1. (b) Use part (a) to sketch the solution curve that passes through the origin. 1 SOLUTION _2 _1 0 1 2 (a) We start by computing the slope at several points in the following chart: x -1 yЈ ෇ x ϩ y Ϫ 1 2 FIGURE 5 2 1 0 1 2 0 1 2 Ϫ2 Ϫ1 0 1 2 ... 0 0 0 0 1 1 1 1 1 ... 3 0 Ϫ1 0 3 4 1 0 1 4 ... Now we draw short line segments with these slopes at these points. The result is the direction field shown in Figure 5. (b) We start at the origin and move to the right in the direction of the line segment (which has slope Ϫ1 ). We continue to draw the solution curve so that it moves parallel to the nearby line segments. The resulting solution curve is shown in Figure 6. Returning to the origin, we draw the solution curve to the left as well. y _1 2 Ϫ1 0 y _2 _2 Ϫ2 x x The more line segments we draw in a direction field, the clearer the picture becomes. Of course, it’s tedious to compute slopes and draw line segments for a huge number of points by hand, but computers are well suited for this task. Figure 7 shows a more detailed, computer-drawn direction field for the differential equation in Example 1. It enables us to draw, with reasonable accuracy, the solution curves shown in Figure 8 with y-intercepts Ϫ2, Ϫ1, 0, 1, and 2. -1 _2 FIGURE 6 3 3 Module 10.2A shows direction fields and solution curves for a variety of differential equations. _3 3 _3 _3 _3 FIGURE 7 R E L switch FIGURE 9 3 FIGURE 8 Now let’s see how direction fields give insight into physical situations. The simple electric circuit shown in Figure 9 contains an electromotive force (usually a battery or generator) that produces a voltage of E͑t͒ volts (V) and a current of I͑t͒ amperes (A) at time t. The circuit also contains a resistor with a resistance of R ohms ( ⍀ ) and an inductor with an inductance of L henries (H). Ohm’s Law gives the drop in voltage due to the resistor as RI. The voltage drop due to the inductor is L͑dI͞dt͒. One of Kirchhoff’s laws says that the sum of the voltage drops is equal to the supplied voltage E͑t͒. Thus, we have 1 L dI ϩ RI ෇ E͑t͒ dt which is a first-order differential equation that models the current I at time t . 5E-10(pp 622-631) 1/18/06 9:18 AM Page 631 SECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD ❙❙❙❙ 631 EXAMPLE 2 Suppose that in the simple circuit of Figure 9 the resistance is 12 ⍀, the inductance is 4 H, and a battery gives a constant voltage of 60 V. (a) Draw a direction field for Equation 1 with these values. (b) What can you say about the limiting value of the current? (c) Identify any equilibrium solutions. (d) If the switch is closed when t ෇ 0 so the current starts with I͑0͒ ෇ 0, use the direction field to sketch the solution curve. SOLUTION (a) If we put L ෇ 4, R ෇ 12, and E͑t͒ ෇ 60 in Equation 1, we get 4 dI ϩ 12I ෇ 60 dt or dI ෇ 15 Ϫ 3I dt The direction field for this differential equation is shown in Figure 10. I 6 4 2 0 1 2 3 t FIGURE 10 (b) It appears from the direction field that all solutions approach the value 5 A, that is, lim I͑t͒ ෇ 5 tlϱ (c) It appears that the constant function I͑t͒ ෇ 5 is an equilibrium solution. Indeed, we can verify this directly from the differential equation. If I͑t͒ ෇ 5, then the left side is dI͞dt ෇ 0 and the right side is 15 Ϫ 3͑5͒ ෇ 0. (d) We use the direction field to sketch the solution curve that passes through ͑0, 0͒, as shown in red in Figure 11. dI ෇ 15 Ϫ 3I dt I 6 4 2 0 FIGURE 11 1 2 3 t 5E-10(pp 632-641) 632 ❙❙❙❙ 1/18/06 9:20 AM Page 632 CHAPTER 10 DIFFERENTIAL EQUATIONS Notice from Figure 10 that the line segments along any horizontal line are parallel. That is because the independent variable t does not occur on the right side of the equation IЈ ෇ 15 Ϫ 3I . In general, a differential equation of the form yЈ ෇ f ͑y͒ in which the independent variable is missing from the right side, is called autonomous. For such an equation, the slopes corresponding to two different points with the same y-coordinate must be equal. This means that if we know one solution to an autonomous differential equation, then we can obtain infinitely many others just by shifting the graph of the known solution to the right or left. In Figure 11 we have shown the solutions that result from shifting the solution curve of Example 2 one and two time units (namely, seconds) to the right. They correspond to closing the switch when t ෇ 1 or t ෇ 2. Euler’s Method The basic idea behind direction fields can be used to find numerical approximations to solutions of differential equations. We illustrate the method on the initial-value problem that we used to introduce direction fields: yЈ ෇ x ϩ y y solution curve 1 y=L(x) 0 FIGURE 12 First Euler approximation 1 x y͑0͒ ෇ 1 The differential equation tells us that yЈ͑0͒ ෇ 0 ϩ 1 ෇ 1, so the solution curve has slope 1 at the point ͑0, 1͒. As a first approximation to the solution we could use the linear approximation L͑x͒ ෇ x ϩ 1. In other words, we could use the tangent line at ͑0, 1͒ as a rough approximation to the solution curve (see Figure 12). Euler’s idea was to improve on this approximation by proceeding only a short distance along this tangent line and then making a midcourse correction by changing direction as indicated by the direction field. Figure 13 shows what happens if we start out along the tangent line but stop when x ෇ 0.5. (This horizontal distance traveled is called the step size.) Since L͑0.5͒ ෇ 1.5, we have y͑0.5͒ Ϸ 1.5 and we take ͑0.5, 1.5͒ as the starting point for a new line segment. The differential equation tells us that yЈ͑0.5͒ ෇ 0.5 ϩ 1.5 ෇ 2, so we use the linear function y ෇ 1.5 ϩ 2͑x Ϫ 0.5͒ ෇ 2x ϩ 0.5 as an approximation to the solution for x Ͼ 0.5 (the orange segment in Figure 13). If we decrease the step size from 0.5 to 0.25, we get the better Euler approximation shown in Figure 14. y 1 0 y 1 1.5 0.5 1 x 0 0.25 1 x FIGURE 13 FIGURE 14 Euler approximation with step size 0.5 Euler approximation with step size 0.25 In general, Euler’s method says to start at the point given by the initial value and proceed in the direction indicated by the direction field. Stop after a short time, look at the slope at the new location, and proceed in that direction. Keep stopping and changing direc- 5E-10(pp 632-641) 1/18/06 9:20 AM Page 633 SECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD y slope=F(x¸, y¸) (⁄, ›) hF(x¸, y¸) h y¸ 0 ⁄ x FIGURE 15 633 tion according to the direction field. Euler’s method does not produce the exact solution to an initial-value problem—it gives approximations. But by decreasing the step size (and therefore increasing the number of midcourse corrections), we obtain successively better approximations to the exact solution. (Compare Figures 12, 13, and 14.) For the general first-order initial-value problem yЈ ෇ F͑x, y͒, y͑x 0͒ ෇ y0 , our aim is to find approximate values for the solution at equally spaced numbers x 0 , x 1 ෇ x 0 ϩ h, x 2 ෇ x 1 ϩ h, . . . , where h is the step size. The differential equation tells us that the slope at ͑x 0 , y0 ͒ is yЈ ෇ F͑x 0 , y0 ͒, so Figure 15 shows that the approximate value of the solution when x ෇ x 1 is y1 ෇ y0 ϩ hF͑x 0 , y0 ͒ Similarly, x¸ ❙❙❙❙ y2 ෇ y1 ϩ hF͑x 1, y1 ͒ In general, yn ෇ ynϪ1 ϩ hF͑x nϪ1, ynϪ1 ͒ EXAMPLE 3 Use Euler’s method with step size 0.1 to construct a table of approximate values for the solution of the initial-value problem yЈ ෇ x ϩ y y͑0͒ ෇ 1 SOLUTION We are given that h ෇ 0.1, x 0 ෇ 0, y0 ෇ 1, and F͑x, y͒ ෇ x ϩ y. So we have y1 ෇ y0 ϩ hF͑x 0 , y0 ͒ ෇ 1 ϩ 0.1͑0 ϩ 1͒ ෇ 1.1 y2 ෇ y1 ϩ hF͑x 1, y1 ͒ ෇ 1.1 ϩ 0.1͑0.1 ϩ 1.1͒ ෇ 1.22 y3 ෇ y2 ϩ hF͑x 2 , y2 ͒ ෇ 1.22 ϩ 0.1͑0.2 ϩ 1.22͒ ෇ 1.362 Module 10.2B shows how Euler’s method works numerically and visually for a variety of differential equations and step sizes. This means that if y͑x͒ is the exact solution, then y͑0.3͒ Ϸ 1.362. Proceeding with similar calculations, we get the values in the table: n xn yn n xn yn 1 2 3 4 5 0.1 0.2 0.3 0.4 0.5 1.100000 1.220000 1.362000 1.528200 1.721020 6 7 8 9 10 0.6 0.7 0.8 0.9 1.0 1.943122 2.197434 2.487178 2.815895 3.187485 For a more accurate table of values in Example 3 we could decrease the step size. But for a large number of small steps the amount of computation is considerable and so we need to program a calculator or computer to carry out these calculations. The following table shows the results of applying Euler’s method with decreasing step size to the initialvalue problem of Example 3. Step size |||| Computer software packages that produce numerical approximations to solutions of differential equations use methods that are refinements of Euler’s method. Although Euler’s method is simple and not as accurate, it is the basic idea on which the more accurate methods are based. Euler estimate of y͑0.5͒ Euler estimate of y͑1͒ 0.500 0.250 0.100 0.050 0.020 0.010 0.005 0.001 1.500000 1.625000 1.721020 1.757789 1.781212 1.789264 1.793337 1.796619 2.500000 2.882813 3.187485 3.306595 3.383176 3.409628 3.423034 3.433848 5E-10(pp 632-641) 634 ❙❙❙❙ 1/18/06 9:20 AM Page 634 CHAPTER 10 DIFFERENTIAL EQUATIONS Notice that the Euler estimates in the table seem to be approaching limits, namely, the true values of y͑0.5͒ and y͑1͒. Figure 16 shows graphs of the Euler approximations with step sizes 0.5, 0.25, 0.1, 0.05, 0.02, 0.01, and 0.005. They are approaching the exact solution curve as the step size h approaches 0. y 1 FIGURE 16 Euler approximations approaching the exact solution 0 0.5 1 x EXAMPLE 4 In Example 2 we discussed a simple electric circuit with resistance 12 ⍀, inductance 4 H, and a battery with voltage 60 V. If the switch is closed when t ෇ 0, we modeled the current I at time t by the initial-value problem dI ෇ 15 Ϫ 3I dt I͑0͒ ෇ 0 Estimate the current in the circuit half a second after the switch is closed. SOLUTION We use Euler’s method with F͑t, I͒ ෇ 15 Ϫ 3I, t0 ෇ 0, I0 ෇ 0, and step size h ෇ 0.1 second: I1 ෇ 0 ϩ 0.1͑15 Ϫ 3 ؒ 0͒ ෇ 1.5 I2 ෇ 1.5 ϩ 0.1͑15 Ϫ 3 ؒ 1.5͒ ෇ 2.55 I3 ෇ 2.55 ϩ 0.1͑15 Ϫ 3 ؒ 2.55͒ ෇ 3.285 I4 ෇ 3.285 ϩ 0.1͑15 Ϫ 3 ؒ 3.285͒ ෇ 3.7995 I5 ෇ 3.7995 ϩ 0.1͑15 Ϫ 3 ؒ 3.7995͒ ෇ 4.15965 So the current after 0.5 s is I͑0.5͒ Ϸ 4.16 A 5E-10(pp 632-641) 1/18/06 9:20 AM Page 635 SECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD |||| 10.2 ❙❙❙❙ 635 Exercises 1. A direction field for the differential equation yЈ ෇ y (1 Ϫ 4 y 2) 1 I (iii) y͑0͒ ෇ Ϫ3 II y y 2 is shown. (a) Sketch the graphs of the solutions that satisfy the given initial conditions. (i) y͑0͒ ෇ 1 (ii) y͑0͒ ෇ Ϫ1 (iv) y͑0͒ ෇ 3 2 2x _2 2x _2 (b) Find all the equilibrium solutions. y _2 _2 3 III IV y y 2 2 2 1 _3 _2 _1 0 1 2 3 x 2x _2 2x _2 _1 _2 _2 _3 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 7. Use the direction field labeled I (for Exercises 3–6) to 2. A direction field for the differential equation yЈ ෇ x sin y is sketch the graphs of the solutions that satisfy the given initial conditions. (a) y͑0͒ ෇ 1 (b) y͑0͒ ෇ 0 (c) y͑0͒ ෇ Ϫ1 shown. (a) Sketch the graphs of the solutions that satisfy the given initial conditions. (i) y͑0͒ ෇ 1 (ii) y͑0͒ ෇ 2 (iii) y͑0͒ ෇ ␲ (iv) y͑0͒ ෇ 4 ■ _2 8. Repeat Exercise 7 for the direction field labeled III. 9–10 |||| Sketch a direction field for the differential equation. Then use it to sketch three solution curves. (v) y͑0͒ ෇ 5 9. yЈ ෇ 1 ϩ y (b) Find all the equilibrium solutions. ■ y ■ ■ 10. yЈ ෇ x 2 Ϫ y 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ 11–14 |||| Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. 5 4 11. yЈ ෇ y Ϫ 2x, ͑1, 0͒ 13. yЈ ෇ y ϩ x y, 3 ■ ■ ■ 12. yЈ ෇ 1 Ϫ x y, 14. yЈ ෇ x Ϫ x y, ͑0, 1͒ ■ ■ ■ ͑0, 0͒ ͑1, 0͒ ■ ■ ■ ■ ■ ■ 2 CAS 1 _3 _2 _1 0 1 2 3 x 15–16 |||| Use a computer algebra system to draw a direction field for the given differential equation. Get a printout and sketch on it the solution curve that passes through ͑0, 1͒. Then use the CAS to draw the solution curve and compare it with your sketch. 15. yЈ ෇ y sin 2x ■ 3–6 |||| Match the differential equation with its direction field (labeled I–IV). Give reasons for your answer. 3. yЈ ෇ y Ϫ 1 5. yЈ ෇ y Ϫ x 2 4. yЈ ෇ y Ϫ x 2 6. yЈ ෇ y 3 Ϫ x 3 CAS ■ ■ 16. yЈ ෇ sin͑x ϩ y͒ ■ ■ ■ ■ ■ ■ ■ ■ ■ 17. Use a computer algebra system to draw a direction field for the differential equation yЈ ෇ y 3 Ϫ 4y. Get a printout and sketch on it solutions that satisfy the initial condition y͑0͒ ෇ c for various values of c. For what values of c does lim t l ϱ y͑t͒ exist? What are the possible values for this limit? 5E-10(pp 632-641) 636 ❙❙❙❙ 1/18/06 9:20 AM Page 636 CHAPTER 10 DIFFERENTIAL EQUATIONS 22. Use Euler’s method with step size 0.2 to estimate y͑1͒, where 18. Make a rough sketch of a direction field for the autonomous y͑x͒ is the solution of the initial-value problem yЈ ෇ 1 Ϫ x y, y͑0͒ ෇ 0. differential equation yЈ ෇ f ͑ y͒, where the graph of f is as shown. How does the limiting behavior of solutions depend on the value of y͑0͒? 23. Use Euler’s method with step size 0.1 to estimate y͑0.5͒, where y͑x͒ is the solution of the initial-value problem yЈ ෇ y ϩ x y, y͑0͒ ෇ 1. f(y) 24. (a) Use Euler’s method with step size 0.2 to estimate y͑1.4͒, _2 _1 0 1 2 where y͑x͒ is the solution of the initial-value problem yЈ ෇ x Ϫ x y, y͑1͒ ෇ 0. (b) Repeat part (a) with step size 0.1. y ; 25. (a) Program a calculator or computer to use Euler’s method to compute y͑1͒, where y͑x͒ is the solution of the initial-value problem 19. (a) Use Euler’s method with each of the following step sizes to estimate the value of y͑0.4͒, where y is the solution of the initial-value problem yЈ ෇ y, y͑0͒ ෇ 1. (i) h ෇ 0.4 (ii) h ෇ 0.2 (iii) h ෇ 0.1 (b) We know that the exact solution of the initial-value problem in part (a) is y ෇ e x. Draw, as accurately as you can, the graph of y ෇ e x, 0 ഛ x ഛ 0.4, together with the Euler approximations using the step sizes in part (a). (Your sketches should resemble Figures 12, 13, and 14.) Use your sketches to decide whether your estimates in part (a) are underestimates or overestimates. (c) The error in Euler’s method is the difference between the exact value and the approximate value. Find the errors made in part (a) in using Euler’s method to estimate the true value of y͑0.4͒, namely e 0.4. What happens to the error each time the step size is halved? 20. A direction field for a differential equation is shown. Draw, with a ruler, the graphs of the Euler approximations to the solution curve that passes through the origin. Use step sizes h ෇ 1 and h ෇ 0.5. Will the Euler estimates be underestimates or overestimates? Explain. y dy ϩ 3x 2 y ෇ 6x 2 dx (i) h ෇ 1 (iii) h ෇ 0.01 y͑0͒ ෇ 3 (ii) h ෇ 0.1 (iv) h ෇ 0.001 3 (b) Verify that y ෇ 2 ϩ eϪx is the exact solution of the differential equation. (c) Find the errors in using Euler’s method to compute y͑1͒ with the step sizes in part (a). What happens to the error when the step size is divided by 10? CAS 26. (a) Program your computer algebra system, using Euler’s method with step size 0.01, to calculate y͑2͒, where y is the solution of the initial-value problem yЈ ෇ x 3 Ϫ y 3 y͑0͒ ෇ 1 (b) Check your work by using the CAS to draw the solution curve. 27. The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of C farads (F), and a resistor with a resistance of R ohms (⍀). The voltage drop across the capacitor is Q͞C, where Q is the charge (in coulombs), so in this case Kirchhoff’s Law gives 2 RI ϩ Q ෇ E͑t͒ C But I ෇ dQ͞dt, so we have R 1 1 dQ ϩ Q ෇ E͑t͒ dt C Suppose the resistance is 5 ⍀, the capacitance is 0.05 F, and a battery gives a constant voltage of 60 V. (a) Draw a direction field for this differential equation. (b) What is the limiting value of the charge? C 0 1 2 x 21. Use Euler’s method with step size 0.5 to compute the approxi- mate y-values y1, y2 , y3 , and y4 of the solution of the initialvalue problem yЈ ෇ y Ϫ 2x, y͑1͒ ෇ 0. E R 5E-10(pp 632-641) 1/18/06 9:20 AM Page 637 SECTION 10.3 SEPARABLE EQUATIONS (c) Is there an equilibrium solution? (d) If the initial charge is Q͑0͒ ෇ 0 C, use the direction field to sketch the solution curve. (e) If the initial charge is Q͑0͒ ෇ 0 C, use Euler’s method with step size 0.1 to estimate the charge after half a second. 637 at a rate of 1ЊC per minute when its temperature is 70ЊC. (a) What does the differential equation become in this case? (b) Sketch a direction field and use it to sketch the solution curve for the initial-value problem. What is the limiting value of the temperature? (c) Use Euler’s method with step size h ෇ 2 minutes to estimate the temperature of the coffee after 10 minutes. 28. In Exercise 14 in Section 10.1 we considered a 95ЊC cup of cof- fee in a 20ЊC room. Suppose it is known that the coffee cools |||| 10.3 ❙❙❙❙ Separable Equations We have looked at first-order differential equations from a geometric point of view (direction fields) and from a numerical point of view (Euler’s method). What about the symbolic point of view? It would be nice to have an explicit formula for a solution of a differential equation. Unfortunately, that is not always possible. But in this section we examine a certain type of differential equation that can be solved explicitly. A separable equation is a first-order differential equation in which the expression for dy͞dx can be factored as a function of x times a function of y. In other words, it can be written in the form dy ෇ t͑x͒f ͑y͒ dx The name separable comes from the fact that the expression on the right side can be “separated” into a function of x and a function of y. Equivalently, if f ͑y͒ 0, we could write dy t͑x͒ ෇ dx h͑y͒ 1 where h͑y͒ ෇ 1͞f ͑y͒. To solve this equation we rewrite it in the differential form h͑y͒ dy ෇ t͑x͒ dx |||| The technique for solving separable differential equations was first used by James Bernoulli (in 1690) in solving a problem about pendulums and by Leibniz (in a letter to Huygens in 1691). John Bernoulli explained the general method in a paper published in 1694. so that all y’s are on one side of the equation and all x’s are on the other side. Then we integrate both sides of the equation: y h͑y͒ dy ෇ y t͑x͒ dx 2 Equation 2 defines y implicitly as a function of x. In some cases we may be able to solve for y in terms of x. We use the Chain Rule to justify this procedure: If h and t satisfy (2), then d dx so d dy ͩy and Thus, Equation 1 is satisfied. ͩy ͪ ͩy h͑y͒ dy ෇ ͪ h͑y͒ dy h͑y͒ d dx dy ෇ t͑x͒ dx dy ෇ t͑x͒ dx ͪ t͑x͒ dx 5E-10(pp 632-641) 638 ❙❙❙❙ 1/18/06 9:20 AM Page 638 CHAPTER 10 DIFFERENTIAL EQUATIONS EXAMPLE 1 dy x2 ෇ 2. dx y (b) Find the solution of this equation that satisfies the initial condition y͑0͒ ෇ 2. (a) Solve the differential equation |||| Figure 1 shows graphs of several members of the family of solutions of the differential equation in Example 1. The solution of the initialvalue problem in part (b) is shown in red. SOLUTION (a) We write the equation in terms of differentials and integrate both sides: y 2 dy ෇ x 2 dx 3 _3 3 2 dy ෇ y x 2 dx 1 3 yy y3 ෇ 1 x3 ϩ C 3 where C is an arbitrary constant. (We could have used a constant C1 on the left side and another constant C 2 on the right side. But then we could combine these constants by writing C ෇ C 2 Ϫ C1.) Solving for y, we get _3 3 y ෇ sx 3 ϩ 3C FIGURE 1 We could leave the solution like this or we could write it in the form 3 y ෇ sx 3 ϩ K where K ෇ 3C. (Since C is an arbitrary constant, so is K .) 3 (b) If we put x ෇ 0 in the general solution in part (a), we get y͑0͒ ෇ sK . To satisfy the 3 initial condition y͑0͒ ෇ 2, we must have sK ෇ 2 and so K ෇ 8. Thus, the solution of the initial-value problem is 3 y ෇ sx 3 ϩ 8 |||| Some computer algebra systems can plot curves defined by implicit equations. Figure 2 shows the graphs of several members of the family of solutions of the differential equation in Example 2. As we look at the curves from left to right, the values of C are 3, 2, 1, 0, Ϫ1, Ϫ2, and Ϫ3. EXAMPLE 2 Solve the differential equation SOLUTION Writing the equation in differential form and integrating both sides, we have ͑2y ϩ cos y͒dy ෇ 6x 2 dx y ͑2y ϩ cos y͒dy ෇ y 6x 4 3 _2 2 _4 FIGURE 2 dy 6x 2 ෇ . dx 2y ϩ cos y 2 dx y 2 ϩ sin y ෇ 2x 3 ϩ C where C is a constant. Equation 3 gives the general solution implicitly. In this case it’s impossible to solve the equation to express y explicitly as a function of x. EXAMPLE 3 Solve the equation yЈ ෇ x 2 y. SOLUTION First we rewrite the equation using Leibniz notation: dy ෇ x2y dx 5E-10(pp 632-641) 1/18/06 9:20 AM Page 639 SECTION 10.3 SEPARABLE EQUATIONS |||| If a solution y is a function that satisfies y͑x͒ 0 for some x, it follows from a uniqueness theorem for solutions of differential equations that y͑x͒ 0 for all x. If y ❙❙❙❙ 639 0, we can rewrite it in differential notation and integrate: dy ෇ x 2 dx y y y 0 dy ෇ y x 2 dx y Խ Խ ln y ෇ x3 ϩC 3 This equation defines y implicitly as a function of x. But in this case we can solve explicitly for y as follows: ԽyԽ ෇ e Խ Խ ෇ e ͑x ͞3͒ϩC ෇ e Ce x ͞3 3 ln y 3 3 y ෇ Ϯe Ce x ͞3 so We can easily verify that the function y ෇ 0 is also a solution of the given differential equation. So we can write the general solution in the form 3 y ෇ Ae x ͞3 where A is an arbitrary constant ( A ෇ e C, or A ෇ Ϫe C, or A ෇ 0). y 6 4 |||| Figure 3 shows a direction field for the differential equation in Example 3. Compare it with Figure 4, in which we use the equation 3 y ෇ Ae x / 3 to graph solutions for several values of A. If you use the direction field to sketch solution curves with y-intercepts 5, 2, 1, Ϫ1, and Ϫ2, they will resemble the curves in Figure 4. 6 2 _2 _1 0 1 2 x _2 _2 _4 _6 _6 FIGURE 4 FIGURE 3 R E FIGURE 5 EXAMPLE 4 In Section 10.2 we modeled the current I͑t͒ in the electric circuit shown in Figure 5 by the differential equation L switch 2 L dI ϩ RI ෇ E͑t͒ dt Find an expression for the current in a circuit where the resistance is 12 ⍀, the inductance is 4 H, a battery gives a constant voltage of 60 V, and the switch is turned on when t ෇ 0. What is the limiting value of the current? SOLUTION With L ෇ 4, R ෇ 12, and E͑t͒ ෇ 60, the equation becomes 4 dI ϩ 12I ෇ 60 dt 5E-10(pp 632-641) 640 ❙❙❙❙ 1/18/06 9:20 AM Page 640 CHAPTER 10 DIFFERENTIAL EQUATIONS dI ෇ 15 Ϫ 3I dt or and the initial-value problem is dI ෇ 15 Ϫ 3I dt I͑0͒ ෇ 0 We recognize this equation as being separable, and we solve it as follows: dI y 15 Ϫ 3I ෇ y dt ͑15 Ϫ 3I 0͒ Խ Խ Խ 15 Ϫ 3I Խ ෇ e Ϫ 1 ln 15 Ϫ 3I ෇ t ϩ C 3 |||| Figure 6 shows how the solution in Example 4 (the current) approaches its limiting value. Comparison with Figure 11 in Section 10.2 shows that we were able to draw a fairly accurate solution curve from the direction field. Ϫ3͑tϩC͒ 15 Ϫ 3I ෇ ϮeϪ3CeϪ3t ෇ AeϪ3t I ෇ 5 Ϫ 1 AeϪ3t 3 6 1 Since I͑0͒ ෇ 0, we have 5 Ϫ 3 A ෇ 0, so A ෇ 15 and the solution is y=5 I͑t͒ ෇ 5 Ϫ 5eϪ3t The limiting current, in amperes, is 0 lim I͑t͒ ෇ lim ͑5 Ϫ 5eϪ3t ͒ ෇ 5 Ϫ 5 lim eϪ3t 2.5 tlϱ tlϱ tlϱ ෇5Ϫ0෇5 FIGURE 6 Orthogonal Trajectories An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family orthogonally, that is, at right angles (see Figure 7). For instance, each member of the family y ෇ mx of straight lines through the origin is an orthogonal trajectory of the family x 2 ϩ y 2 ෇ r 2 of concentric circles with center the origin (see Figure 8). We say that the two families are orthogonal trajectories of each other. y x orthogonal trajectory FIGURE 7 FIGURE 8 EXAMPLE 5 Find the orthogonal trajectories of the family of curves x ෇ ky 2, where k is an arbitrary constant. SOLUTION The curves x ෇ ky 2 form a family of parabolas whose axis of symmetry is the x-axis. The first step is to find a single differential equation that is satisfied by all 5E-10(pp 632-641) 1/18/06 9:20 AM Page 641 SECTION 10.3 SEPARABLE EQUATIONS ❙❙❙❙ 641 members of the family. If we differentiate x ෇ ky 2, we get 1 ෇ 2ky dy dx dy 1 ෇ dx 2ky or This differential equation depends on k, but we need an equation that is valid for all values of k simultaneously. To eliminate k we note that, from the equation of the given general parabola x ෇ ky 2, we have k ෇ x͞y 2 and so the differential equation can be written as dy 1 ෇ ෇ dx 2ky 1 x 2 2 y y dy y ෇ dx 2x or This means that the slope of the tangent line at any point ͑x, y͒ on one of the parabolas is yЈ ෇ y͑͞2x͒. On an orthogonal trajectory the slope of the tangent line must be the negative reciprocal of this slope. Therefore, the orthogonal trajectories must satisfy the differential equation dy 2x ෇Ϫ dx y y This differential equation is separable, and we solve it as follows: y y dy ෇ Ϫy 2x dx y2 ෇ Ϫx 2 ϩ C 2 x x2 ϩ 4 FIGURE 9 y2 ෇C 2 where C is an arbitrary positive constant. Thus, the orthogonal trajectories are the family of ellipses given by Equation 4 and sketched in Figure 9. Orthogonal trajectories occur in various branches of physics. For example, in an electrostatic field the lines of force are orthogonal to the lines of constant potential. Also, the streamlines in aerodynamics are orthogonal trajectories of the velocity-equipotential curves. Mixing Problems A typical mixing problem involves a tank of fixed capacity filled with a thoroughly mixed solution of some substance, such as salt. A solution of a given concentration enters the tank at a fixed rate and the mixture, thoroughly stirred, leaves at a fixed rate, which may differ from the entering rate. If y͑t͒ denotes the amount of substance in the tank at time t, then yЈ͑t͒ is the rate at which the substance is being added minus the rate at which it is being removed. The mathematical description of this situation often leads to a first-order separable differential equation. We can use the same type of reasoning to model a variety of phenomena: chemical reactions, discharge of pollutants into a lake, injection of a drug into the bloodstream. 5E-10(pp 642-651) 642 ❙❙❙❙ 1/18/06 9:21 AM Page 642 CHAPTER 10 DIFFERENTIAL EQUATIONS EXAMPLE 6 A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that contains 0.03 kg of salt per liter of water enters the tank at a rate of 25 L͞min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt remains in the tank after half an hour? SOLUTION Let y͑t͒ be the amount of salt (in kilograms) after t minutes. We are given that y͑0͒ ෇ 20 and we want to find y͑30͒. We do this by finding a differential equation satisfied by y͑t͒. Note that dy͞dt is the rate of change of the amount of salt, so dy ෇ ͑rate in͒ Ϫ ͑rate out͒ dt 5 where (rate in) is the rate at which salt enters the tank and (rate out) is the rate at which salt leaves the tank. We have ͩ rate in ෇ 0.03 kg L ͪͩ 25 L min ͪ ෇ 0.75 kg min The tank always contains 5000 L of liquid, so the concentration at time t is y͑t͒͞5000 (measured in kilograms per liter). Since the brine flows out at a rate of 25 L͞min, we have rate out ෇ ͩ y͑t͒ kg 5000 L ͪͩ 25 L min ͪ ෇ y͑t͒ kg 200 min Thus, from Equation 5 we get dy y͑t͒ 150 Ϫ y͑t͒ ෇ 0.75 Ϫ ෇ dt 200 200 Solving this separable differential equation, we obtain dy y 150 Ϫ y ෇ y Խ Խ Ϫln 150 Ϫ y ෇ |||| Figure 10 shows the graph of the function y͑t͒ of Example 6. Notice that, as time goes by, the amount of salt approaches 150 kg. dt 200 t ϩC 200 Since y͑0͒ ෇ 20, we have Ϫln 130 ෇ C, so Խ t Ϫ ln 130 200 Խ Ϫln 150 Ϫ y ෇ y Խ 150 Ϫ y Խ ෇ 130e Ϫt͞200 150 Therefore 100 Since y͑t͒ is continuous and y͑0͒ ෇ 20 and the right side is never 0, we deduce that 150 y͑t͒ is always positive. Thus, 150 Ϫ y ෇ 150 Ϫ y and so Խ 50 0 FIGURE 10 Խ y͑t͒ ෇ 150 Ϫ 130eϪt͞200 200 400 t The amount of salt after 30 min is y͑30͒ ෇ 150 Ϫ 130eϪ30͞200 Ϸ 38.1 kg 5E-10(pp 642-651) 1/18/06 9:21 AM Page 643 SECTION 10.3 SEPARABLE EQUATIONS |||| 10.3 1–10 1. |||| Solve the differential equation. 2. 5. ͑1 ϩ tan y͒yЈ ෇ x ϩ 1 7. dy te t ෇ dt y s1 ϩ y 2 9. du ෇ 2 ϩ 2u ϩ t ϩ tu dt 8. yЈ ෇ ■ ■ 10. ■ members of the family of solutions (if your CAS does implicit plots). How does the solution curve change as the constant C varies? CAS xy 2 ln y dz ϩ e tϩz ෇ 0 dt ■ ■ ■ ■ ■ ■ y͑1͒ ෇ 0 12. dy y cos x , ෇ dx 1 ϩ y2 y͑0͒ ෇ 1 15. du 2t ϩ sec 2t ෇ , dt 2u 16. dy ෇ te y, dt ■ ■ y͑0͒ ෇ 0 ■ ■ ■ ■ ■ ■ ■ |||| Find the orthogonal trajectories of the family of curves. Use a graphing device to draw several members of each family on a common screen. 28. x 2 Ϫ y 2 ෇ k 30. y ෇ keϪx ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 32. In Exercise 28 in Section 10.2 we discussed a differential equa- u͑0͒ ෇ Ϫ5 tion that models the temperature of a 95ЊC cup of coffee in a 20ЊC room. Solve the differential equation to find an expression for the temperature of the coffee at time t. y͑1͒ ෇ 0 33. In Exercise 13 in Section 10.1 we formulated a model for learning in the form of the differential equation 0 Ͻ x Ͻ ␲͞2 y͑1͒ ෇ Ϫ1 ■ ■ ■ ■ ■ ■ ■ ■ 19. Find an equation of the curve that satisfies dy͞dx ෇ 4x y and whose y-intercept is 7. 20. Find an equation of the curve that passes through the point ͑1, 1͒ and whose slope at ͑x, y͒ is y 2͞x 3. 21. (a) Solve the differential equation yЈ ෇ 2x s1 Ϫ y 2. (b) Solve the initial-value problem yЈ ෇ 2x s1 Ϫ y 2, y͑0͒ ෇ 0, and graph the solution. (c) Does the initial-value problem yЈ ෇ 2x s1 Ϫ y 2, y͑0͒ ෇ 2, have a solution? Explain. Ϫy ; 22. Solve the equation e yЈ ϩ cos x ෇ 0 and graph several mem- bers of the family of solutions. How does the solution curve change as the constant C varies? CAS ■ to find an expression for the charge at time t. Find the limiting value of the charge. 3 ; ■ 31. Solve the initial-value problem in Exercise 27 in Section 10.2 P͑1͒ ෇ 2 ■ ■ 29. y ෇ ͑x ϩ k͒Ϫ1 17. yЈ tan x ෇ a ϩ y, y͑␲͞3͒ ෇ a, ■ ■ ; 27–30 ■ dP ෇ sPt, dt 18. x yЈ ϩ y ෇ y 2, ■ 26. yЈ ෇ x 2͞y 27. y ෇ kx 2 13. x cos x ෇ ͑2y ϩ e 3y ͒yЈ, 14. |||| 25. yЈ ෇ 1͞y |||| Find the solution of the differential equation that satisfies the given initial condition. dy ෇ y 2 ϩ 1, dx 25–26 (a) Use a computer algebra system to draw a direction field for the differential equation. Get a printout and use it to sketch some solution curves without solving the differential equation. (b) Solve the differential equation. (c) Use the CAS to draw several members of the family of solutions obtained in part (b). Compare with the curves from part (a). 11–18 11. 24. Solve the equation yЈ ෇ x sx 2 ϩ 1͑͞ ye y ͒ and graph several dy e 2x ෇ dx 4y 3 du 1 ϩ sr 6. ෇ dr 1 ϩ su 2 ■ CAS 4. yЈ ෇ y 2 sin x 3. ͑x 2 ϩ 1͒yЈ ෇ xy ■ 643 Exercises y dy ෇ dx x ■ ❙❙❙❙ 23. Solve the initial-value problem yЈ ෇ ͑sin x͒͞sin y, y͑0͒ ෇ ␲͞2, and graph the solution (if your CAS does implicit plots). dP ෇ k͑M Ϫ P͒ dt where P͑t͒ measures the performance of someone learning a skill after a training time t, M is the maximum level of performance, and k is a positive constant. Solve this differential equation to find an expression for P͑t͒. What is the limit of this expression? 34. In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C: A ϩ B l C. The law of mass action states that the rate of reaction is proportional to the product of the concentrations of A and B: d ͓C͔ ෇ k ͓A͔͓B͔ dt (See Example 4 in Section 3.4.) Thus, if the initial concentrations are ͓A͔ ෇ a moles͞L and ͓B͔ ෇ b moles͞L and we write 5E-10(pp 642-651) 644 ❙❙❙❙ 1/18/06 9:21 AM Page 644 CHAPTER 10 DIFFERENTIAL EQUATIONS x ෇ ͓C͔, then we have dx ෇ k͑a Ϫ x͒͑b Ϫ x͒ dt CAS (a) Assuming that a b, find x as a function of t. Use the fact that the initial concentration of C is 0. (b) Find x ͑t͒ assuming that a ෇ b. How does this expression for x ͑t͒ simplify if it is known that ͓C͔ ෇ a͞2 after 20 seconds? 35. In contrast to the situation of Exercise 34, experiments show that the reaction H 2 ϩ Br 2 l 2HBr satisfies the rate law d ͓HBr͔ ෇ k ͓H 2 ͔͓Br 2 ͔ 1͞2 dt and so for this reaction the differential equation becomes dx ෇ k͑a Ϫ x͒͑b Ϫ x͒1͞2 dt where x ෇ ͓HBr͔ and a and b are the initial concentrations of hydrogen and bromine. (a) Find x as a function of t in the case where a ෇ b. Use the fact that x͑0͒ ෇ 0. (b) If a Ͼ b, find t as a function of x. [Hint: In performing the integration, make the substitution u ෇ sb Ϫ x.] 36. A sphere with radius 1 m has temperature 15ЊC. It lies inside a concentric sphere with radius 2 m and temperature 25ЊC. The temperature T ͑r͒ at a distance r from the common center of the spheres satisfies the differential equation d 2T 2 dT ෇0 ϩ dr 2 r dr If we let S ෇ dT͞dr, then S satisfies a first-order differential equation. Solve it to find an expression for the temperature T ͑r͒ between the spheres. 37. A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at that time. Thus, a model for the concentration C ෇ C͑t͒ of the glucose solution in the bloodstream is dC ෇ r Ϫ kC dt where k is a positive constant. (a) Suppose that the concentration at time t ෇ 0 is C0. Determine the concentration at any time t by solving the differential equation. (b) Assuming that C0 Ͻ r͞k, find lim t l ϱ C͑t͒ and interpret your answer. 38. A certain small country has $10 billion in paper currency in circulation, and each day $50 million comes into the country’s banks. The government decides to introduce new currency by having the banks replace old bills with new ones whenever old currency comes into the banks. Let x ෇ x ͑t͒ denote the amount of new currency in circulation at time t, with x ͑0͒ ෇ 0. (a) Formulate a mathematical model in the form of an initialvalue problem that represents the “flow” of the new currency into circulation. (b) Solve the initial-value problem found in part (a). (c) How long will it take for the new bills to account for 90% of the currency in circulation? 39. A tank contains 1000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 L͞min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank (a) after t minutes and (b) after 20 minutes? 40. A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L͞min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L͞min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L͞min. How much salt is in the tank (a) after t minutes and (b) after one hour? 41. When a raindrop falls, it increases in size and so its mass at time t is a function of t, m͑t͒. The rate of growth of the mass is km͑t͒ for some positive constant k. When we apply Newton’s Law of Motion to the raindrop, we get ͑mv͒Ј ෇ tm, where v is the velocity of the raindrop (directed downward) and t is the acceleration due to gravity. The terminal velocity of the raindrop is lim t l ϱ v͑t͒. Find an expression for the terminal velocity in terms of t and k. 42. An object of mass m is moving horizontally through a medium which resists the motion with a force that is a function of the velocity; that is, m d 2s dv ෇m ෇ f ͑v͒ dt 2 dt where v ෇ v͑t͒ and s ෇ s͑t͒ represent the velocity and position of the object at time t, respectively. For example, think of a boat moving through the water. (a) Suppose that the resisting force is proportional to the velocity, that is, f ͑v͒ ෇ Ϫk v, k a positive constant. (This model is appropriate for small values of v.) Let v͑0͒ ෇ v0 and s͑0͒ ෇ s0 be the initial values of v and s. Determine v and s at any time t . What is the total distance that the object travels from time t ෇ 0? (b) For larger values of v a better model is obtained by supposing that the resisting force is proportional to the square of the velocity, that is, f ͑v͒ ෇ Ϫk v 2, k Ͼ 0. (This model was first proposed by Newton.) Let v0 and s0 be the initial values of v and s. Determine v and s at any time t . What is the total distance that the object travels in this case? 43. Let A͑t͒ be the area of a tissue culture at time t and let M be the final area of the tissue when growth is complete. Most cell divisions occur on the periphery of the tissue and the number of cells on the periphery is proportional to sA͑t͒. So a reason- 5E-10(pp 642-651) 1/18/06 9:21 AM Page 645 APPLIED PROJECT HOW FAST DOES A TANK DRAIN? CAS able model for the growth of tissue is obtained by assuming that the rate of growth of the area is jointly proportional to sA͑t͒ and M Ϫ A͑t͒. (a) Formulate a differential equation and use it to show that the tissue grows fastest when A͑t͒ ෇ M͞3. (b) Solve the differential equation to find an expression for A͑t͒. Use a computer algebra system to perform the integration. m dv mtR 2 ෇Ϫ dt ͑x ϩ R͒2 (a) Suppose a rocket is fired vertically upward with an initial velocity v0. Let h be the maximum height above the surface reached by the object. Show that gravitational force on an object of mass m that has been projected vertically upward from Earth’s surface is mtR 2 ͑x ϩ R͒2 where x ෇ x͑t͒ is the object’s distance above the surface at time t , R is Earth’s radius, and t is the acceleration due to gravity. Also, by Newton’s Second Law, F ෇ ma ෇ m ͑dv͞dt͒ 645 and so 44. According to Newton’s Law of Universal Gravitation, the F෇ ❙❙❙❙ v0 ෇ ͱ 2tRh Rϩh [Hint: By the Chain Rule, m ͑dv͞dt͒ ෇ mv ͑dv͞dx͒.] (b) Calculate ve ෇ lim h l ϱ v0. This limit is called the escape velocity for Earth. (c) Use R ෇ 3960 mi and t ෇ 32 ft͞s2 to calculate ve in feet per second and in miles per second. APPLIED PROJECT How Fast Does a Tank Drain? If water (or other liquid) drains from a tank, we expect that the flow will be greatest at first (when the water depth is greatest) and will gradually decrease as the water level decreases. But we need a more precise mathematical description of how the flow decreases in order to answer the kinds of questions that engineers ask: How long does it take for a tank to drain completely? How much water should a tank hold in order to guarantee a certain minimum water pressure for a sprinkler system? Let h͑t͒ and V͑t͒ be the height and volume of water in a tank at time t. If water leaks through a hole with area a at the bottom of the tank, then Torricelli’s Law says that dV ෇ Ϫa s2th dt 1 where t is the acceleration due to gravity. So the rate at which water flows from the tank is proportional to the square root of the water height. 1. (a) Suppose the tank is cylindrical with height 6 ft and radius 2 ft and the hole is circular with radius 1 in. If we take t ෇ 32 ft͞s2, show that y satisfies the differential equation 1 dh ෇Ϫ sh dt 72 (b) Solve this equation to find the height of the water at time t, assuming the tank is full at time t ෇ 0. (c) How long will it take for the water to drain completely? 2. Because of the rotation and viscosity of the liquid, the theoretical model given by Equation 1 isn’t quite accurate. Instead, the model 2 dh ෇ ksh dt 5E-10(pp 642-651) 646 ❙❙❙❙ 1/18/06 9:21 AM Page 646 CHAPTER 10 DIFFERENTIAL EQUATIONS |||| This part of the project is best done as a classroom demonstration or as a group project with three students in each group: a timekeeper to call out seconds, a bottle keeper to estimate the height every 10 seconds, and a record keeper to record these values. is often used and the constant k (which depends on the physical properties of the liquid) is determined from data concerning the draining of the tank. (a) Suppose that a hole is drilled in the side of a cylindrical bottle and the height h of the water (above the hole) decreases from 10 cm to 3 cm in 68 seconds. Use Equation 2 to find an expression for h͑t͒. Evaluate h͑t͒ for t ෇ 10, 20, 30, 40, 50, 60. (b) Drill a 4-mm hole near the bottom of the cylindrical part of a two-liter plastic soft-drink bottle. Attach a strip of masking tape marked in centimeters from 0 to 10, with 0 corresponding to the top of the hole. With one finger over the hole, fill the bottle with water to the 10-cm mark. Then take your finger off the hole and record the values of h͑t͒ for t ෇ 10, 20, 30, 40, 50, 60 seconds. (You will probably find that it takes 68 seconds for the level to decrease to h ෇ 3 cm.) Compare your data with the values of h͑t͒ from part (a). How well did the model predict the actual values? 3. In many parts of the world, the water for sprinkler systems in large hotels and hospitals is supplied by gravity from cylindrical tanks on or near the roofs of the buildings. Suppose such a tank has radius 10 ft and the diameter of the outlet is 2.5 inches. An engineer has to guarantee that the water pressure will be at least 2160 lb͞ft 2 for a period of 10 minutes. (When a fire happens, the electrical system might fail and it could take up to 10 minutes for the emergency generator and fire pump to be activated.) What height should the engineer specify for the tank in order to make such a guarantee? (Use the fact that the water pressure at a depth of d feet is P ෇ 62.5d. See Section 9.3.) 4. Not all water tanks are shaped like cylinders. Suppose a tank has cross-sectional area A͑h͒ at height h. Then the volume of water up to height h is V ෇ x0h A͑u͒ du and so the Fundamental Theorem of Calculus gives dV͞dh ෇ A͑h͒. It follows that dV dV dh dh ෇ ෇ A͑h͒ dt dh dt dt and so Torricelli’s Law becomes A͑h͒ dh ෇ Ϫa s2th dt (a) Suppose the tank has the shape of a sphere with radius 2 m and is initially half full of water. If the radius of the circular hole is 1 cm and we take t ෇ 10 m͞s2, show that h satisfies the differential equation ͑4h Ϫ h 2 ͒ dh ෇ Ϫ0.0001 s20h dt (b) How long will it take for the water to drain completely? APPLIED PROJECT Which Is Faster, Going Up or Coming Down? Suppose you throw a ball into the air. Do you think it takes longer to reach its maximum height or to fall back to Earth from its maximum height? We will solve the problem in this project but, before getting started, think about that situation and make a guess based on your physical intuition. 1. A ball with mass m is projected vertically upward from Earth’s surface with a positive initial velocity v0. We assume the forces acting on the ball are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude p v͑t͒ , where p is a positive constant and v͑t͒ is the velocity of the ball at time t. In both Խ Խ 5E-10(pp 642-651) 1/18/06 9:21 AM Page 647 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY |||| In modeling force due to air resistance, various functions have been used, depending on the physical characteristics and speed of the ball. Here we use a linear model, Ϫpv, but a quadratic model (Ϫpv 2 on the way up and pv 2 on the way down) is another possibility for higher speeds (see Exercise 42 in Section 10.3). For a golf ball, experiments have shown that a good model is Ϫpv 1.3 going up and p v 1.3 coming down. But no matter which force function Ϫf ͑v͒ is used [where f ͑v͒ Ͼ 0 for v Ͼ 0 and f ͑v͒ Ͻ 0 for v Ͻ 0], the answer to the question remains the same. See F. Brauer, “What Goes Up Must Come Down, Eventually,” Amer. Math. Monthly 108 (2001), pp. 437–440. ❙❙❙❙ 647 the ascent and the descent, the total force acting on the ball is Ϫpv Ϫ mt. [During ascent, v͑t͒ is positive and the resistance acts downward; during descent, v͑t͒ is negative and the resistance acts upward.] So, by Newton’s Second Law, the equation of motion is mvЈ ෇ Ϫpv Ϫ mt Solve this differential equation to show that the velocity is Խ Խ v͑t͒ ෇ ͩ ͪ mt Ϫpt͞m mt e Ϫ p p v0 ϩ 2. Show that the height of the ball, until it hits the ground, is ͩ y͑t͒ ෇ v0 ϩ mt p ͪ m mtt ͑1 Ϫ eϪpt͞m ͒ Ϫ p p 3. Let t1 be the time that the ball takes to reach its maximum height. Show that t1 ෇ ͩ m mt ϩ pv0 ln p mt ͪ Find this time for a ball with mass 1 kg and initial velocity 20 m͞s. Assume the air 1 resistance is 10 of the speed. ; 4. Let t2 be the time at which the ball falls back to Earth. For the particular ball in Problem 3, estimate t2 by using a graph of the height function y͑t͒. Which is faster, going up or coming down? 5. In general, it’s not easy to find t2 because it’s impossible to solve the equation y͑t͒ ෇ 0 explicitly. We can, however, use an indirect method to determine whether ascent or descent is faster; we determine whether y͑2t1 ͒ is positive or negative. Show that y͑2t1 ͒ ෇ m 2t p2 ͩ xϪ ͪ 1 Ϫ 2 ln x x where x ෇ e pt1͞m. Then show that x Ͼ 1 and the function f ͑x͒ ෇ x Ϫ 1 Ϫ 2 ln x x is increasing for x Ͼ 1. Use this result to decide whether y͑2t1 ͒ is positive or negative. What can you conclude? Is ascent or descent faster? |||| 10.4 Exponential Growth and Decay One of the models for population growth that we considered in Section 10.1 was based on the assumption that the population grows at a rate proportional to the size of the population: dP ෇ kP dt Is that a reasonable assumption? Suppose we have a population (of bacteria, for instance) 5E-10(pp 642-651) 648 ❙❙❙❙ 1/18/06 9:21 AM Page 648 CHAPTER 10 DIFFERENTIAL EQUATIONS with size P ෇ 1000 and at a certain time it is growing at a rate of PЈ ෇ 300 bacteria per hour. Now let’s take another 1000 bacteria of the same type and put them with the first population. Each half of the new population was growing at a rate of 300 bacteria per hour. We would expect the total population of 2000 to increase at a rate of 600 bacteria per hour initially (provided there’s enough room and nutrition). So if we double the size, we double the growth rate. In general, it seems reasonable that the growth rate should be proportional to the size. The same assumption applies in other situations as well. In nuclear physics, the mass of a radioactive substance decays at a rate proportional to the mass. In chemistry, the rate of a unimolecular first-order reaction is proportional to the concentration of the substance. In finance, the value of a savings account with continuously compounded interest increases at a rate proportional to that value. In general, if y͑t͒ is the value of a quantity y at time t and if the rate of change of y with respect to t is proportional to its size y͑t͒ at any time, then dy ෇ ky dt 1 where k is a constant. Equation 1 is sometimes called the law of natural growth (if k Ͼ 0) or the law of natural decay (if k Ͻ 0). Because it is a separable differential equation we can solve it by the methods of Section 10.3: y dy ෇ y k dt y Խ Խ ԽyԽ ෇ e ln y ෇ kt ϩ C ktϩC ෇ e Ce kt y ෇ Ae kt where A (෇ Ϯe C or 0) is an arbitrary constant. To see the significance of the constant A, we observe that y͑0͒ ෇ Ae k ؒ 0 ෇ A Therefore, A is the initial value of the function. Because Equation 1 occurs so frequently in nature, we summarize what we have just proved for future use. 2 The solution of the initial-value problem dy ෇ ky dt y͑0͒ ෇ y0 y͑t͒ ෇ y0 e kt is Population Growth What is the significance of the proportionality constant k? In the context of population growth, we can write 3 dP ෇ kP dt or 1 dP ෇k P dt 5E-10(pp 642-651) 1/18/06 9:21 AM Page 649 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 649 The quantity 1 dP P dt is the growth rate divided by the population size; it is called the relative growth rate. According to (3), instead of saying “the growth rate is proportional to population size” we could say “the relative growth rate is constant.” Then (2) says that a population with constant relative growth rate must grow exponentially. Notice that the relative growth rate k appears as the coefficient of t in the exponential function y0 e kt. For instance, if dP ෇ 0.02P dt and t is measured in years, then the relative growth rate is k ෇ 0.02 and the population grows at a rate of 2% per year. If the population at time 0 is P0 , then the expression for the population is P͑t͒ ෇ P0 e 0.02t TABLE 1 Year Population (millions) 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 1650 1750 1860 2070 2300 2560 3040 3710 4450 5280 6080 EXAMPLE 1 Assuming that the growth rate is proportional to population size, use the data in Table 1 to model the population of the world in the 20th century. What is the relative growth rate? How well does the model fit the data? SOLUTION We measure the time t in years and let t ෇ 0 in the year 1900. We measure the population P͑t͒ in millions of people. Then the initial condition is P͑0͒ ෇ 1650. We are assuming that the growth rate is proportional to population size, so the initial-value problem is dP ෇ kP dt P͑0͒ ෇ 1650 From (2) we know that the solution is P͑t͒ ෇ 1650e kt One way to estimate the relative growth rate k is to use the fact that the population in 1910 was 1750 million. Therefore P͑10͒ ෇ 1650e k͑10͒ ෇ 1750 We solve this equation for k: e 10k ෇ k෇ 1750 1650 1 1750 ln Ϸ 0.005884 10 1650 Thus, the relative growth rate is about 0.6% per year and the model becomes P͑t͒ ෇ 1650e 0.005884t 5E-10(pp 642-651) 650 ❙❙❙❙ 1/18/06 9:21 AM Page 650 CHAPTER 10 DIFFERENTIAL EQUATIONS Table 2 and Figure 1 allow us to compare the predictions of this model with the actual data. You can see that the predictions become quite inaccurate after about 30 years and they underestimate by a factor of more than 2 in 2000. TABLE 2 P Year Model Population 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 1650 1750 1856 1969 2088 2214 2349 2491 2642 2802 2972 1650 1750 1860 2070 2300 2560 3040 3710 4450 5280 6080 6000 Population (in millions) P=1650e 0.005884t 20 40 60 80 100 t Years since 1900 FIGURE 1 A possible model for world population growth Another possibility for estimating k would be to use the given population for 1950, for instance, instead of 1910. Then P͑50͒ ෇ 1650e 50k ෇ 2560 |||| In Sections 7.2 and 7.4* we modeled the same data with an exponential function, but there we used the method of least squares. k෇ 1 2560 ln Ϸ 0.0087846 50 1650 The estimate for the relative growth rate is now 0.88% per year and the model is P͑t͒ ෇ 1650e 0.0087846t The predictions with this second model are shown in Table 3 and Figure 2. This exponential model is more accurate over a longer period of time, but it too lags behind reality in recent years. TABLE 3 P Year Model Population 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 1650 1802 1967 2148 2345 2560 2795 3052 3332 3638 3972 1650 1750 1860 2070 2300 2560 3040 3710 4450 5280 6080 6000 Population (in millions) P=1650e 0.0087846t 20 40 60 80 Years since 1900 FIGURE 2 Another model for world population growth 100 t 5E-10(pp 642-651) 1/18/06 9:21 AM Page 651 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 651 EXAMPLE 2 Use the data in Table 1 to model the population of the world in the second half of the 20th century. Use the model to estimate the population in 1993 and to predict the population in the year 2010. SOLUTION Here we let t ෇ 0 in the year 1950. Then the initial-value problem is dP ෇ kP dt P͑0͒ ෇ 2560 and the solution is P͑t͒ ෇ 2560e kt Let’s estimate k by using the population in 1960: P͑10͒ ෇ 2560e 10k ෇ 3040 k෇ 1 3040 ln Ϸ 0.017185 10 2560 The relative growth rate is about 1.7% per year and the model is P͑t͒ ෇ 2560e 0.017185t We estimate that the world population in 1993 was P͑43͒ ෇ 2560e 0.017185͑43͒ Ϸ 5360 million The model predicts that the population in 2010 will be P͑60͒ ෇ 2560e 0.017185͑60͒ Ϸ 7179 million The graph in Figure 3 shows that the model is fairly accurate to date, so the estimate for 1993 is quite reliable. But the prediction for 2010 is riskier. P 6000 P=2560e 0.017185t Population (in millions) FIGURE 3 A model for world population growth in the second half of the 20th century 20 Years since 1950 40 t 5E-10(pp 652-661) 652 ❙❙❙❙ 1/18/06 9:23 AM Page 652 CHAPTER 10 DIFFERENTIAL EQUATIONS Radioactive Decay Radioactive substances decay by spontaneously emitting radiation. If m͑t͒ is the mass remaining from an initial mass m0 of the substance after time t, then the relative decay rate Ϫ 1 dm m dt has been found experimentally to be constant. (Since dm͞dt is negative, the relative decay rate is positive.) It follows that dm ෇ km dt where k is a negative constant. In other words, radioactive substances decay at a rate proportional to the remaining mass. This means that we can use (2) to show that the mass decays exponentially: m͑t͒ ෇ m0 e kt Physicists express the rate of decay in terms of half-life, the time required for half of any given quantity to decay. EXAMPLE 3 The half-life of radium-226 ( .226 Ra) is 1590 years. 88 (a) A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of .226 Ra 88 that remains after t years. (b) Find the mass after 1000 years correct to the nearest milligram. (c) When will the mass be reduced to 30 mg? SOLUTION (a) Let m͑t͒ be the mass of radium-226 (in milligrams) that remains after t years. Then dm͞dt ෇ km and y͑0͒ ෇ 100, so (2) gives m͑t͒ ෇ m͑0͒e kt ෇ 100e kt In order to determine the value of k, we use the fact that y͑1590͒ ෇ 2 ͑100͒. Thus 1 100e 1590k ෇ 50 so e 1590k ෇ 1 2 and 1590k ෇ ln 1 ෇ Ϫln 2 2 k෇Ϫ ln 2 1590 m͑t͒ ෇ 100eϪ͑ln 2͞1590͒t Therefore We could use the fact that e ln 2 ෇ 2 to write the expression for m͑t͒ in the alternative form m͑t͒ ෇ 100 ϫ 2 Ϫt͞1590 (b) The mass after 1000 years is m͑1000͒ ෇ 100eϪ͑ln 2͞1590͒1000 Ϸ 65 mg 5E-10(pp 652-661) 1/18/06 9:23 AM Page 653 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 653 (c) We want to find the value of t such that m͑t͒ ෇ 30, that is, 100eϪ͑ln 2͞1590͒t ෇ 30 or eϪ͑ln 2͞1590͒t ෇ 0.3 We solve this equation for t by taking the natural logarithm of both sides: 150 Ϫ ln 2 t ෇ ln 0.3 1590 m=100e _(ln 2)t/1590 t ෇ Ϫ1590 Thus m=30 0 FIGURE 4 4000 ln 0.3 Ϸ 2762 years ln 2 As a check on our work in Example 3, we use a graphing device to draw the graph of m͑t͒ in Figure 4 together with the horizontal line m ෇ 30. These curves intersect when t Ϸ 2800, and this agrees with the answer to part (c). Newton’s Law of Cooling Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. (This law also applies to warming.) If we let T͑t͒ be the temperature of the object at time t and Ts be the temperature of the surroundings, then we can formulate Newton’s Law of Cooling as a differential equation: dT ෇ k͑T Ϫ Ts͒ dt where k is a constant. We could solve this equation as a separable differential equation by the method of Section 10.3, but an easier method is to make the change of variable y͑t͒ ෇ T͑t͒ Ϫ Ts . Because Ts is constant, we have yЈ͑t͒ ෇ TЈ͑t͒ and so the equation becomes dy ෇ ky dt We can then use (2) to find an expression for y, from which we can find T . EXAMPLE 4 A bottle of soda pop at room temperature (72Њ F) is placed in a refrigerator where the temperature is 44Њ F. After half an hour the soda pop has cooled to 61Њ F. (a) What is the temperature of the soda pop after another half hour? (b) How long does it take for the soda pop to cool to 50Њ F? SOLUTION (a) Let T͑t͒ be the temperature of the soda after t minutes. The surrounding temperature is Ts ෇ 44Њ F, so Newton’s Law of Cooling states that dT ෇ k͑T Ϫ 44) dt If we let y ෇ T Ϫ 44, then y͑0͒ ෇ T͑0͒ Ϫ 44 ෇ 72 Ϫ 44 ෇ 28, so y is a solution of the initial-value problem dy y͑0͒ ෇ 28 ෇ ky dt 5E-10(pp 652-661) 654 ❙❙❙❙ 1/18/06 9:23 AM Page 654 CHAPTER 10 DIFFERENTIAL EQUATIONS and by (2) we have y͑t͒ ෇ y͑0͒e kt ෇ 28e kt We are given that T͑30͒ ෇ 61, so y͑30͒ ෇ 61 Ϫ 44 ෇ 17 and 28e 30k ෇ 17 e 30k ෇ 17 28 Taking logarithms, we have k෇ ln ( 17 ) 28 Ϸ Ϫ0.01663 30 Thus y͑t͒ ෇ 28e Ϫ0.01663t T͑t͒ ෇ 44 ϩ 28e Ϫ0.01663t T͑60͒ ෇ 44 ϩ 28e Ϫ0.01663͑60͒ Ϸ 54.3 So after another half hour the pop has cooled to about 54Њ F. (b) We have T͑t͒ ෇ 50 when 44 ϩ 28e Ϫ0.01663t ෇ 50 6 e Ϫ0.01663t ෇ 28 T 72 t෇ 44 6 ln ( 28 ) Ϸ 92.6 Ϫ0.01663 The pop cools to 50Њ F after about 1 hour 33 minutes. Notice that in Example 4, we have 0 FIGURE 5 30 60 90 t lim T͑t͒ ෇ lim ͑44 ϩ 28e Ϫ0.01663t ͒ ෇ 44 ϩ 28 ؒ 0 ෇ 44 tlϱ tlϱ which is to be expected. The graph of the temperature function is shown in Figure 5. Continuously Compounded Interest EXAMPLE 5 If $1000 is invested at 6% interest, compounded annually, then after 1 year the investment is worth $1000͑1.06͒ ෇ $1060, after 2 years it’s worth $͓1000͑1.06͔͒1.06 ෇ $1123.60, and after t years it’s worth $1000͑1.06͒t. In general, if an amount A0 is invested at an interest rate r ͑r ෇ 0.06 in this example), then after t years it’s worth A0͑1 ϩ r͒t. Usually, however, interest is compounded more frequently, say, n times a year. Then in each compounding period the interest rate is r͞n and there are nt compounding periods in t years, so the value of the investment is ͩ ͪ A0 1 ϩ r n nt 5E-10(pp 652-661) 1/18/06 9:23 AM Page 655 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 655 For instance, after 3 years at 6% interest a $1000 investment will be worth $1000͑1.06͒3 ෇ $1191.02 with annual compounding $1000͑1.03͒6 ෇ $1194.05 with semiannual compounding $1000͑1.015͒12 ෇ $1195.62 $1000͑1.005͒36 ෇ $1196.68 ͩ with quarterly compounding with monthly compounding $1000 1 ϩ 0.06 365 ͪ 365 ؒ 3 ෇ $1197.20 with daily compounding You can see that the interest paid increases as the number of compounding periods ͑n͒ increases. If we let n l ϱ, then we will be compounding the interest continuously and the value of the investment will be ͩ ͪ ͫͩ ͪ ͬ ͫ ͩ ͪͬ ͫ ͩ ͪͬ nt r n A͑t͒ ෇ lim A0 1 ϩ nlϱ ෇ lim A0 nlϱ ෇ A0 lim 1ϩ r n ෇ A0 lim 1ϩ 1 m nlϱ mlϱ 1ϩ n͞r n͞r rt rt m r n rt (where m ෇ n͞r) But the limit in this expression is equal to the number e (see Equation 7.4.9 or 7.4*.9). So with continuous compounding of interest at interest rate r, the amount after t years is A͑t͒ ෇ A0 e rt If we differentiate this equation, we get dA ෇ rA0 e rt ෇ rA͑t͒ dt which says that, with continuous compounding of interest, the rate of increase of an investment is proportional to its size. Returning to the example of $1000 invested for 3 years at 6% interest, we see that with continuous compounding of interest the value of the investment will be A͑3͒ ෇ $1000e ͑0.06͒3 ෇ $1000e 0.18 ෇ $1197.22 Notice how close this is to the amount we calculated for daily compounding, $1197.20. But the amount is easier to compute if we use continuous compounding. 5E-10(pp 652-661) 656 ❙❙❙❙ 1/18/06 9:23 AM Page 656 CHAPTER 10 DIFFERENTIAL EQUATIONS |||| 10.4 Exercises 1. A population of protozoa develops with a constant relative Year 2. A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 60 cells. (a) Find the relative growth rate. (b) Find an expression for the number of cells after t hours. (c) Find the number of cells after 8 hours. (d) Find the rate of growth after 8 hours. (e) When will the population reach 20,000 cells? 3. A bacteria culture starts with 500 bacteria and grows at a rate proportional to its size. After 3 hours there are 8000 bacteria. (a) Find an expression for the number of bacteria after t hours. (b) Find the number of bacteria after 4 hours. (c) Find the rate of growth after 4 hours. (d) When will the population reach 30,000? ; Population Year Population 1900 1910 1920 1930 1940 1950 growth rate of 0.7944 per member per day. On day zero the population consists of two members. Find the population size after six days. 76 92 106 123 131 150 1960 1970 1980 1990 2000 179 203 227 250 275 (a) Use the exponential model and the census figures for 1900 and 1910 to predict the population in 2000. Compare with the actual figure and try to explain the discrepancy. (b) Use the exponential model and the census figures for 1980 and 1990 to predict the population in 2000. Compare with the actual population. Then use this model to predict the population in the years 2010 and 2020. (c) Graph both of the exponential functions in parts (a) and (b) together with a plot of the actual population. Are these models reasonable ones? 7. Experiments show that if the chemical reaction N2O5 l 2NO 2 ϩ 1 O 2 2 4. A bacteria culture grows with constant relative growth rate. After 2 hours there are 600 bacteria and after 8 hours the count is 75,000. (a) Find the initial population. (b) Find an expression for the population after t hours. (c) Find the number of cells after 5 hours. (d) Find the rate of growth after 5 hours. (e) When will the population reach 200,000? 5. The table gives estimates of the world population, in millions, from 1750 to 2000: Year Population Year 790 980 1260 1900 1950 2000 1650 2560 6080 Ϫ d͓N2O5͔ ෇ 0.0005͓N2O5͔ dt (See Example 4 in Section 3.4.) (a) Find an expression for the concentration ͓N2O5͔ after t seconds if the initial concentration is C. (b) How long will the reaction take to reduce the concentration of N2O5 to 90% of its original value? Population 1750 1800 1850 takes place at 45ЊC, the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows: (a) Use the exponential model and the population figures for 1750 and 1800 to predict the world population in 1900 and 1950. Compare with the actual figures. (b) Use the exponential model and the population figures for 1850 and 1900 to predict the world population in 1950. Compare with the actual population. (c) Use the exponential model and the population figures for 1900 and 1950 to predict the world population in 2000. Compare with the actual population and try to explain the discrepancy. 6. The table gives the population of the United States, in millions, for the years 1900–2000. 8. Bismuth-210 has a half-life of 5.0 days. (a) A sample originally has a mass of 800 mg. Find a formula for the mass remaining after t days. (b) Find the mass remaining after 30 days. (c) When is the mass reduced to 1 mg? (d) Sketch the graph of the mass function. 9. The half-life of cesium-137 is 30 years. Suppose we have a 100-mg sample. (a) Find the mass that remains after t years. (b) How much of the sample remains after 100 years? (c) After how long will only 1 mg remain? 10. After 3 days a sample of radon-222 decayed to 58% of its orig- inal amount. (a) What is the half-life of radon-222? (b) How long would it take the sample to decay to 10% of its original amount? 5E-10(pp 652-661) 1/18/06 9:23 AM Page 657 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY 11. Scientists can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, 14 C, with a half-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates 14 C through food chains. When a plant or animal dies, it stops replacing its carbon and the amount of 14 C begins to decrease through radioactive decay. Therefore, the level of radioactivity must also decay exponentially. A parchment fragment was discovered that had about 74% as much 14 C radioactivity as does plant material on Earth today. Estimate the age of the parchment. 12. A curve passes through the point ͑0, 5͒ and has the property that the slope of the curve at every point P is twice the y-coordinate of P. What is the equation of the curve? 13. A roast turkey is taken from an oven when its temperature has reached 185ЊF and is placed on a table in a room where the temperature is 75ЊF. (a) If the temperature of the turkey is 150ЊF after half an hour, what is the temperature after 45 min? (b) When will the turkey have cooled to 100ЊF? 14. A thermometer is taken from a room where the temperature is 20ЊC to the outdoors, where the temperature is 5ЊC. After one minute the thermometer reads 12ЊC. (a) What will the reading on the thermometer be after one more minute? (b) When will the thermometer read 6ЊC? 15. When a cold drink is taken from a refrigerator, its temperature is 5ЊC. After 25 minutes in a 20ЊC room its temperature has increased to 10ЊC. (a) What is the temperature of the drink after 50 minutes? (b) When will its temperature be 15ЊC? 16. A freshly brewed cup of coffee has temperature 95ЊC in a 20ЊC room. When its temperature is 70ЊC, it is cooling at a rate of 1ЊC per minute. When does this occur? 17. The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At 15ЊC the pressure is 101.3 kPa at sea level and 87.14 kPa at h ෇ 1000 m. (a) What is the pressure at an altitude of 3000 m? (b) What is the pressure at the top of Mount McKinley, at an altitude of 6187 m? 18. (a) If $500 is borrowed at 14% interest, find the amounts due at the end of 2 years if the interest is compounded (i) annually, (ii) quarterly, (iii) monthly, (iv) daily, (v) hourly, and (vi) continuously. ; ❙❙❙❙ 657 (b) Suppose $500 is borrowed and the interest is compounded continuously. If A͑t͒ is the amount due after t years, where 0 ഛ t ഛ 2, graph A͑t͒ for each of the interest rates 14%, 10%, and 6% on a common screen. 19. (a) If $3000 is invested at 5% interest, find the value of the investment at the end of 5 years if the interest is compounded (i) annually, (ii) semiannually, (iii) monthly, (iv) weekly, (v) daily, and (vi) continuously. (b) If A͑t͒ is the amount of the investment at time t for the case of continuous compounding, write a differential equation and an initial condition satisfied by A͑t͒. 20. (a) How long will it take an investment to double in value if the interest rate is 6% compounded continuously? (b) What is the equivalent annual interest rate? 21. Consider a population P ෇ P͑t͒ with constant relative birth and death rates ␣ and ␤, respectively, and a constant emigration rate m, where ␣, ␤, and m are positive constants. Assume that ␣ Ͼ ␤. Then the rate of change of the population at time t is modeled by the differential equation dP ෇ kP Ϫ m dt where k ෇ ␣ Ϫ ␤ (a) Find the solution of this equation that satisfies the initial condition P͑0͒ ෇ P0. (b) What condition on m will lead to an exponential expansion of the population? (c) What condition on m will result in a constant population? A population decline? (d) In 1847, the population of Ireland was about 8 million and the difference between the relative birth and death rates was 1.6% of the population. Because of the potato famine in the 1840s and 1850s, about 210,000 inhabitants per year emigrated from Ireland. Was the population expanding or declining at that time? 22. Let c be a positive number. A differential equation of the form dy ෇ ky 1ϩc dt where k is a positive constant, is called a doomsday equation because the exponent in the expression ky 1ϩc is larger than that for natural growth (that is, ky). (a) Determine the solution that satisfies the initial condition y͑0͒ ෇ y0. (b) Show that there is a finite time t ෇ T (doomsday) such that lim t l T Ϫ y͑t͒ ෇ ϱ. (c) An especially prolific breed of rabbits has the growth term ky 1.01. If 2 such rabbits breed initially and the warren has 16 rabbits after three months, then when is doomsday? 5E-10(pp 652-661) 658 ❙❙❙❙ 1/18/06 9:23 AM Page 658 CHAPTER 10 DIFFERENTIAL EQUATIONS APPLIED PROJECT Calculus and Baseball In this project we explore three of the many applications of calculus to baseball. The physical interactions of the game, especially the collision of ball and bat, are quite complex and their models are discussed in detail in a book by Robert Adair, The Physics of Baseball, 3d ed. (New York: HarperPerennial, 2002). 1. It may surprise you to learn that the collision of baseball and bat lasts only about a thou- sandth of a second. Here we calculate the average force on the bat during this collision by first computing the change in the ball’s momentum. The momentum p of an object is the product of its mass m and its velocity v, that is, p ෇ mv. Suppose an object, moving along a straight line, is acted on by a force F ෇ F͑t͒ that is a continuous function of time. (a) Show that the change in momentum over a time interval ͓t0 , t1 ͔ is equal to the integral of F from t0 to t1; that is, show that t1 Batter’s box An overhead view of the position of a baseball bat, shown every fiftieth of a second during a typical swing. (Adapted from The Physics of Baseball) p͑t1 ͒ Ϫ p͑t0 ͒ ෇ y F͑t͒ dt t0 This integral is called the impulse of the force over the time interval. (b) A pitcher throws a 90-mi͞h fastball to a batter, who hits a line drive directly back to the pitcher. The ball is in contact with the bat for 0.001 s and leaves the bat with velocity 110 mi͞h. A baseball weighs 5 oz and, in U.S. Customary units, its mass is measured in slugs: m ෇ w͞t where t ෇ 32 ft͞s 2. (i) Find the change in the ball’s momentum. (ii) Find the average force on the bat. 2. In this problem we calculate the work required for a pitcher to throw a 90-mi͞h fastball by first considering kinetic energy. The kinetic energy K of an object of mass m and velocity v is given by K ෇ 1 mv 2. Sup2 pose an object of mass m, moving in a straight line, is acted on by a force F ෇ F͑s͒ that depends on its position s. According to Newton’s Second Law F͑s͒ ෇ ma ෇ m dv dt where a and v denote the acceleration and velocity of the object. (a) Show that the work done in moving the object from a position s0 to a position s1 is equal to the change in the object’s kinetic energy; that is, show that s1 2 2 W ෇ y F͑s͒ ds ෇ 1 mv1 Ϫ 1 mv 0 2 2 s0 where v0 ෇ v͑s0 ͒ and v1 ෇ v͑s1 ͒ are the velocities of the object at the positions s0 and s1. Hint: By the Chain Rule, dv dv ds dv m ෇m ෇ mv dt ds dt ds (b) How many foot-pounds of work does it take to throw a baseball at a speed of 90 mi͞h? 3. (a) An outfielder fields a baseball 280 ft away from home plate and throws it directly to the catcher with an initial velocity of 100 ft͞s. Assume that the velocity v͑t͒ of the ball after t seconds satisfies the differential equation dv͞dt ෇ Ϫv͞10 because of air resistance. How long does it take for the ball to reach home plate? (Ignore any vertical motion of the ball.) (b) The manager of the team wonders whether the ball will reach home plate sooner if it is relayed by an infielder. The shortstop can position himself directly between the outfielder and home plate, catch the ball thrown by the outfielder, turn, and throw the ball to 5E-10(pp 652-661) 1/18/06 9:23 AM Page 659 SECTION 10.5 THE LOGISTIC EQUATION ; |||| 10.5 ❙❙❙❙ 659 the catcher with an initial velocity of 105 ft͞s. The manager clocks the relay time of the shortstop (catching, turning, throwing) at half a second. How far from home plate should the shortstop position himself to minimize the total time for the ball to reach the plate? Should the manager encourage a direct throw or a relayed throw? What if the shortstop can throw at 115 ft͞s? (c) For what throwing velocity of the shortstop does a relayed throw take the same time as a direct throw? The Logistic Equation In this section we discuss in detail a model for population growth, the logistic model, that is more sophisticated than exponential growth. In doing so we use all the tools at our disposal—direction fields and Euler’s method from Section 10.2 and the explicit solution of separable differential equations from Section 10.3. In the exercises we investigate other possible models for population growth, some of which take into account harvesting and seasonal growth. The Logistic Model As we discussed in Section 10.1, a population often increases exponentially in its early stages but levels off eventually and approaches its carrying capacity because of limited resources. If P͑t͒ is the size of the population at time t, we assume that dP Ϸ kP dt if P is small This says that the growth rate is initially close to being proportional to size. In other words, the relative growth rate is almost constant when the population is small. But we also want to reflect the fact that the relative growth rate decreases as the population P increases and becomes negative if P ever exceeds its carrying capacity K, the maximum population that the environment is capable of sustaining in the long run. The simplest expression for the relative growth rate that incorporates these assumptions is ͩ ͪ 1 dP P ෇k 1Ϫ P dt K Multiplying by P, we obtain the model for population growth known as the logistic differential equation: 1 ͩ ͪ dP P ෇ kP 1 Ϫ dt K Notice from Equation 1 that if P is small compared with K, then P͞K is close to 0 and so dP͞dt Ϸ kP. However, if P l K (the population approaches its carrying capacity), then P͞K l 1, so dP͞dt l 0. We can deduce information about whether solutions increase or decrease directly from Equation 1. If the population P lies between 0 and K, then the right side of the equation is positive, so dP͞dt Ͼ 0 and the population increases. But if the population exceeds the carrying capacity ͑P Ͼ K͒, then 1 Ϫ P͞K is negative, so dP͞dt Ͻ 0 and the population decreases. 5E-10(pp 652-661) 660 ❙❙❙❙ 1/18/06 9:23 AM Page 660 CHAPTER 10 DIFFERENTIAL EQUATIONS Direction Fields Let’s start our more detailed analysis of the logistic differential equation by looking at a direction field. EXAMPLE 1 Draw a direction field for the logistic equation with k ෇ 0.08 and carrying capacity K ෇ 1000. What can you deduce about the solutions? SOLUTION In this case the logistic differential equation is ͩ dP P ෇ 0.08P 1 Ϫ dt 1000 ͪ A direction field for this equation is shown in Figure 1. We show only the first quadrant because negative populations aren’t meaningful and we are interested only in what happens after t ෇ 0. P 1400 1200 1000 800 600 400 200 FIGURE 1 Direction field for the logistic equation in Example 1 0 20 40 60 80 t The logistic equation is autonomous (dP͞dt depends only on P, not on t), so the slopes are the same along any horizontal line. As expected, the slopes are positive for 0 Ͻ P Ͻ 1000 and negative for P Ͼ 1000. The slopes are small when P is close to 0 or 1000 (the carrying capacity). Notice that the solutions move away from the equilibrium solution P ෇ 0 and move toward the equilibrium solution P ෇ 1000. In Figure 2 we use the direction field to sketch solution curves with initial populations P͑0͒ ෇ 100, P͑0͒ ෇ 400, and P͑0͒ ෇ 1300. Notice that solution curves that start below P ෇ 1000 are increasing and those that start above P ෇ 1000 are decreasing. The slopes are greatest when P Ϸ 500 and, therefore, the solution curves that start below P ෇ 1000 have inflection points when P Ϸ 500. In fact we can prove that all solution curves that start below P ෇ 500 have an inflection point when P is exactly 500 (see Exercise 9). P 1400 1200 1000 800 600 400 200 FIGURE 2 Solution curves for the logistic equation in Example 1 0 20 40 60 80 t 5E-10(pp 652-661) 1/18/06 9:23 AM Page 661 SECTION 10.5 THE LOGISTIC EQUATION ❙❙❙❙ 661 Euler’s Method Next let’s use Euler’s method to obtain numerical estimates for solutions of the logistic differential equation at specific times. EXAMPLE 2 Use Euler’s method with step sizes 20, 10, 5, 1, and 0.1 to estimate the population sizes P͑40͒ and P͑80͒, where P is the solution of the initial-value problem ͩ dP P ෇ 0.08P 1 Ϫ dt 1000 ͪ P͑0͒ ෇ 100 SOLUTION With step size h ෇ 20, t0 ෇ 0, P0 ෇ 100, and ͩ F͑t, P͒ ෇ 0.08P 1 Ϫ P 1000 ͪ we get, using the notation of Section 10.2, P1 ෇ 100 ϩ 20F͑0, 100͒ ෇ 244 P2 ෇ 244 ϩ 20F͑20, 244͒ Ϸ 539.14 P3 ෇ 539.14 ϩ 20F͑40, 539.14͒ Ϸ 936.69 P4 ෇ 936.69 ϩ 20F͑60, 936.69͒ Ϸ 1031.57 Thus, our estimates for the population sizes at times t ෇ 40 and t ෇ 80 are P͑40͒ Ϸ 539 P͑80͒ Ϸ 1032 For smaller step sizes we need to program a calculator or computer. The table gives the results. Step size Euler estimate of P͑40͒ Euler estimate of P͑80͒ 20 10 5 1 0.1 539 647 695 725 731 1032 997 991 986 985 Figure 3 shows a graph of the Euler approximations with step sizes h ෇ 10 and h ෇ 1. We see that the Euler approximation with h ෇ 1 looks very much like the lower solution curve that we drew using a direction field in Figure 2. P 1000 h=1 h=10 FIGURE 3 Euler approximations of the solution curve in Example 2 0 20 40 60 80 t 5E-10(pp 662-671) 662 ❙❙❙❙ 1/18/06 9:25 AM Page 662 CHAPTER 10 DIFFERENTIAL EQUATIONS The Analytic Solution The logistic equation (1) is separable and so we can solve it explicitly using the method of Section 10.3. Since ͩ ͪ dP P ෇ kP 1 Ϫ dt K we have dP y P͑1 Ϫ P͞K͒ ෇ y k dt 2 To evaluate the integral on the left side, we write 1 K ෇ P͑1 Ϫ P͞K͒ P͑K Ϫ P͒ Using partial fractions (see Section 8.4), we get K 1 1 ෇ ϩ P͑K Ϫ P͒ P KϪP This enables us to rewrite Equation 2: y ͩ 1 1 ϩ P KϪP Խ Խ ͪ Խ dP ෇ y k dt Խ ln P Ϫ ln K Ϫ P ෇ kt ϩ C ln Ϳ Ϳ Ϳ Ϳ KϪP ෇ Ϫkt Ϫ C P KϪP ෇ eϪktϪC ෇ eϪCeϪkt P KϪP ෇ AeϪkt P 3 where A ෇ ϮeϪC. Solving Equation 3 for P, we get K Ϫ 1 ෇ AeϪkt P so P෇ ? P 1 ෇ K 1 ϩ AeϪkt K 1 ϩ AeϪkt We find the value of A by putting t ෇ 0 in Equation 3. If t ෇ 0, then P ෇ P0 (the initial population), so K Ϫ P0 ෇ Ae 0 ෇ A P0 5E-10(pp 662-671) 1/18/06 9:25 AM Page 663 SECTION 10.5 THE LOGISTIC EQUATION ❙❙❙❙ 663 Thus, the solution to the logistic equation is P͑t͒ ෇ 4 K 1 ϩ AeϪkt K Ϫ P0 P0 where A ෇ Using the expression for P͑t͒ in Equation 4, we see that lim P͑t͒ ෇ K tlϱ which is to be expected. EXAMPLE 3 Write the solution of the initial-value problem ͩ dP P ෇ 0.08P 1 Ϫ dt 1000 ͪ P͑0͒ ෇ 100 and use it to find the population sizes P͑40͒ and P͑80͒. At what time does the population reach 900? SOLUTION The differential equation is a logistic equation with k ෇ 0.08, carrying capacity K ෇ 1000, and initial population P0 ෇ 100. So Equation 4 gives the population at time t as P͑t͒ ෇ 1000 1 ϩ AeϪ0.08t where A ෇ P͑t͒ ෇ Thus 1000 Ϫ 100 ෇9 100 1000 1 ϩ 9eϪ0.08t So the population sizes when t ෇ 40 and 80 are |||| Compare these values with the Euler estimates from Example 2: P͑40͒ Ϸ 731 P͑80͒ Ϸ 985 P͑40͒ ෇ 1000 Ϸ 731.6 1 ϩ 9eϪ3.2 P͑80͒ ෇ 1000 Ϸ 985.3 1 ϩ 9eϪ6.4 The population reaches 900 when 1000 ෇ 900 1 ϩ 9eϪ0.08t Solving this equation for t, we get |||| Compare the solution curve in Figure 4 with the lowest solution curve we drew from the direction field in Figure 2. 1000 P= FIGURE 4 1 eϪ0.08t ෇ 81 1 Ϫ0.08t ෇ ln 81 ෇ Ϫln 81 P=900 0 1 ϩ 9eϪ0.08t ෇ 10 9 t෇ 1000 1+9e _0.08t 80 ln 81 Ϸ 54.9 0.08 So the population reaches 900 when t is approximately 55. As a check on our work, we graph the population curve in Figure 4 and observe where it intersects the line P ෇ 900. The cursor indicates that t Ϸ 55. 5E-10(pp 662-671) 664 ❙❙❙❙ 1/18/06 9:25 AM Page 664 CHAPTER 10 DIFFERENTIAL EQUATIONS Comparison of the Natural Growth and Logistic Models In the 1930s the biologist G. F. Gause conducted an experiment with the protozoan Paramecium and used a logistic equation to model his data. The table gives his daily count of the population of protozoa. He estimated the initial relative growth rate to be 0.7944 and the carrying capacity to be 64. t (days) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 P (observed) 2 3 22 16 39 52 54 47 50 76 69 51 57 70 53 59 57 EXAMPLE 4 Find the exponential and logistic models for Gause’s data. Compare the predicted values with the observed values and comment on the fit. SOLUTION Given the relative growth rate k ෇ 0.7944 and the initial population P0 ෇ 2, the exponential model is P͑t͒ ෇ P0 e kt ෇ 2e 0.7944t Gause used the same value of k for his logistic model. [This is reasonable because P0 ෇ 2 is small compared with the carrying capacity (K ෇ 64). The equation 1 dP P0 dt Ϳ ͩ t෇0 ෇k 1Ϫ 2 64 ͪ Ϸk shows that the value of k for the logistic model is very close to the value for the exponential model.] Then the solution of the logistic equation in Equation 4 gives P͑t͒ ෇ A෇ where K 64 Ϫkt ෇ 1 ϩ Ae 1 ϩ AeϪ0.7944t K Ϫ P0 64 Ϫ 2 ෇ ෇ 31 P0 2 P͑t͒ ෇ So 64 1 ϩ 31e Ϫ0.7944t We use these equations to calculate the predicted values (rounded to the nearest integer) and compare them in the table. t (days) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 P (observed) 2 3 22 16 39 52 54 47 50 76 69 51 57 70 53 59 57 P (logistic model) 2 4 9 17 28 40 51 57 61 62 63 64 64 64 64 64 64 P (exponential model) 2 4 10 22 48 106 ... We notice from the table and from the graph in Figure 5 that for the first three or four days the exponential model gives results comparable to those of the more sophisticated logistic model. For t ജ 5, however, the exponential model is hopelessly inaccurate, but the logistic model fits the observations reasonably well. 5E-10(pp 662-671) 1/18/06 9:25 AM Page 665 SECTION 10.5 THE LOGISTIC EQUATION ❙❙❙❙ 665 P P=2e0.7944t 60 40 P= 20 64 1+31e _0.7944t FIGURE 5 0 The exponential and logistic models for the Paramecium data 4 8 16 t 12 Other Models for Population Growth The Law of Natural Growth and the logistic differential equation are not the only equations that have been proposed to model population growth. In Exercise 14 we look at the Gompertz growth function and in Exercises 15 and 16 we investigate seasonalgrowth models. Two of the other models are modifications of the logistic model. The differential equation dP P ෇ kP 1 Ϫ Ϫc dt K ͩ ͪ has been used to model populations that are subject to “harvesting” of one sort or another. (Think of a population of fish being caught at a constant rate). This equation is explored in Exercises 11 and 12. For some species there is a minimum population level m below which the species tends to become extinct. (Adults may not be able to find suitable mates.) Such populations have been modeled by the differential equation ͩ ͪͩ ͪ dP P ෇ kP 1 Ϫ dt K 1Ϫ m P where the extra factor, 1 Ϫ m͞P, takes into account the consequences of a sparse population (see Exercise 13). |||| 10.5 Exercises 1. Suppose that a population develops according to the logistic equation dP ෇ 0.05P Ϫ 0.0005P 2 dt where t is measured in weeks. (a) What is the carrying capacity? What is the value of k ? (b) A direction field for this equation is shown at the right. Where are the slopes close to 0? Where are they largest? Which solutions are increasing? Which solutions are decreasing? (c) Use the direction field to sketch solutions for initial populations of 20, 40, 60, 80, 120, and 140. What do these P 150 100 50 0 20 40 60 t 5E-10(pp 662-671) 666 ❙❙❙❙ 1/18/06 9:25 AM Page 666 CHAPTER 10 DIFFERENTIAL EQUATIONS solutions have in common? How do they differ? Which solutions have inflection points? At what population levels do they occur? (d) What are the equilibrium solutions? How are the other solutions related to these solutions? ; 2. Suppose that a population grows according to a logistic model with carrying capacity 6000 and k ෇ 0.0015 per year. (a) Write the logistic differential equation for these data. (b) Draw a direction field (either by hand or with a computer algebra system). What does it tell you about the solution curves? (c) Use the direction field to sketch the solution curves for initial populations of 1000, 2000, 4000, and 8000. What can you say about the concavity of these curves? What is the significance of the inflection points? (d) Program a calculator or computer to use Euler’s method with step size h ෇ 1 to estimate the population after 50 years if the initial population is 1000. (e) If the initial population is 1000, write a formula for the population after t years. Use it to find the population after 50 years and compare with your estimate in part (d). (f) Graph the solution in part (e) and compare with the solution curve you sketched in part (c). 3. The Pacific halibut fishery has been modeled by the differential equation ͩ ͪ y dy ෇ ky 1 Ϫ dt K 5. The population of the world was about 5.3 billion in 1990. Birth rates in the 1990s ranged from 35 to 40 million per year and death rates ranged from 15 to 20 million per year. Let’s assume that the carrying capacity for world population is 100 billion. (a) Write the logistic differential equation for these data. (Because the initial population is small compared to the carrying capacity, you can take k to be an estimate of the initial relative growth rate.) (b) Use the logistic model to estimate the world population in the year 2000 and compare with the actual population of 6.1 billion. (c) Use the logistic model to predict the world population in the years 2100 and 2500. (d) What are your predictions if the carrying capacity is 50 billion? 6. (a) Make a guess as to the carrying capacity for the U.S. population. Use it and the fact that the population was 250 million in 1990 to formulate a logistic model for the U.S. population. (b) Determine the value of k in your model by using the fact that the population in 2000 was 275 million. (c) Use your model to predict the U.S. population in the years 2100 and 2200. (d) Use your model to predict the year in which the U.S. population will exceed 300 million. 7. One model for the spread of a rumor is that the rate of spread where y͑t͒ is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be K ෇ 8 ϫ 10 7 kg, and k ෇ 0.71 per year. (a) If y͑0͒ ෇ 2 ϫ 10 7 kg, find the biomass a year later. (b) How long will it take for the biomass to reach 4 ϫ 10 7 kg? 4. The table gives the number of yeast cells in a new laboratory culture. Time (hours) Yeast cells Time (hours) Yeast cells 0 2 4 6 8 18 39 80 171 336 10 12 14 16 18 509 597 640 664 672 (a) Plot the data and use the plot to estimate the carrying capacity for the yeast population. (b) Use the data to estimate the initial relative growth rate. (c) Find both an exponential model and a logistic model for these data. (d) Compare the predicted values with the observed values, both in a table and with graphs. Comment on how well your models fit the data. (e) Use your logistic model to estimate the number of yeast cells after 7 hours. is proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisfied by y. (b) Solve the differential equation. (c) A small town has 1000 inhabitants. At 8 A.M., 80 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor? 8. Biologists stocked a lake with 400 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 10,000. The number of fish tripled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years. (b) How long will it take for the population to increase to 5000? 9. (a) Show that if P satisfies the logistic equation (1), then ͩ ͪͩ P d 2P ෇ k 2P 1 Ϫ dt 2 K 1Ϫ 2P K ͪ (b) Deduce that a population grows fastest when it reaches half its carrying capacity. ; 10. For a fixed value of K (say K ෇ 10), the family of logistic functions given by Equation 4 depends on the initial value P0 and the proportionality constant k. Graph several members of 5E-10(pp 662-671) 1/18/06 9:25 AM Page 667 SECTION 10.5 THE LOGISTIC EQUATION 11. Let’s modify the logistic differential equation of Example 1 as ͩ dP P ෇ 0.08P 1 Ϫ dt 1000 CAS CAS ͪ 14. Another model for a growth function for a limited population is given by the Gompertz function, which is a solution of the differential equation Ϫ 15 ͩͪ K dP P ෇ c ln dt P (a) Suppose P͑t͒ represents a fish population at time t, where t is measured in weeks. Explain the meaning of the term Ϫ15. (b) Draw a direction field for this differential equation. (c) What are the equilibrium solutions? (d) Use the direction field to sketch several solution curves. Describe what happens to the fish population for various initial populations. (e) Solve this differential equation explicitly, either by using partial fractions or with a computer algebra system. Use the initial populations 200 and 300. Graph the solutions and compare with your sketches in part (d). where c is a constant and K is the carrying capacity. (a) Solve this differential equation. (b) Compute lim t l ϱ P͑t͒. (c) Graph the Gompertz growth function for K ෇ 1000, P0 ෇ 100, and c ෇ 0.05, and compare it with the logistic function in Example 3. What are the similarities? What are the differences? (d) We know from Exercise 9 that the logistic function grows fastest when P ෇ K͞2. Use the Gompertz differential equation to show that the Gompertz function grows fastest when P ෇ K͞e. 12. Consider the differential equation ͩ P dP ෇ 0.08P 1 Ϫ dt 1000 ͪ 15. In a seasonal-growth model, a periodic function of time is Ϫc as a model for a fish population, where t is measured in weeks and c is a constant. (a) Use a CAS to draw direction fields for various values of c. (b) From your direction fields in part (a), determine the values of c for which there is at least one equilibrium solution. For what values of c does the fish population always die out? (c) Use the differential equation to prove what you discovered graphically in part (b). (d) What would you recommend for a limit to the weekly catch of this fish population? introduced to account for seasonal variations in the rate of growth. Such variations could, for example, be caused by seasonal changes in the availability of food. (a) Find the solution of the seasonal-growth model dP ෇ kP cos͑rt Ϫ ␾͒ dt ; 1Ϫ m P (a) Use the differential equation to show that any solution is increasing if m Ͻ P Ͻ K and decreasing if 0 Ͻ P Ͻ m. (b) For the case where k ෇ 0.08, K ෇ 1000, and m ෇ 200, draw a direction field and use it to sketch several solution curves. Describe what happens to the population for various initial populations. What are the equilibrium solutions? (c) Solve the differential equation explicitly, either by using partial fractions or with a computer algebra system. Use the initial population P0 . where k, r, and ␾ are positive constants. (b) By graphing the solution for several values of k, r, and ␾, explain how the values of k, r, and ␾ affect the solution. What can you say about lim t l ϱ P͑t͒? follows: some species there is a minimum population m such that the species will become extinct if the size of the population falls below m. This condition can be incorporated into the logistic equation by introducing the factor ͑1 Ϫ m͞P͒. Thus, the modified logistic model is given by the differential equation ͩ ͪͩ ͪ P͑0͒ ෇ P0 16. Suppose we alter the differential equation in Exercise 15 as 13. There is considerable evidence to support the theory that for P dP ෇ kP 1 Ϫ dt K 667 (d) Use the solution in part (c) to show that if P0 Ͻ m, then the species will become extinct. [Hint: Show that the numerator in your expression for P͑t͒ is 0 for some value of t.] this family. How does the graph change when P0 varies? How does it change when k varies? follows: ❙❙❙❙ dP ෇ kP cos 2͑rt Ϫ ␾͒ dt ; P͑0͒ ෇ P0 (a) Solve this differential equation with the help of a table of integrals or a CAS. (b) Graph the solution for several values of k, r, and ␾. How do the values of k, r, and ␾ affect the solution? What can you say about lim t l ϱ P͑t͒ in this case? 17. Graphs of logistic functions (Figures 2 and 4) look suspiciously similar to the graph of the hyperbolic tangent function (Figure 3 in Section 7.6). Explain the similarity by showing that the logistic function given by Equation 4 can be written as [ P͑t͒ ෇ 1 K 1 ϩ tanh ( 1 k͑ t Ϫ c͒) 2 2 where c ෇ ͑ln A͒͞ k. Thus, the logistic function is really just a shifted hyperbolic tangent. 5E-10(pp 662-671) 668 ❙❙❙❙ 1/18/06 9:25 AM Page 668 CHAPTER 10 DIFFERENTIAL EQUATIONS |||| 10.6 Linear Equations A first-order linear differential equation is one that can be put into the form dy ϩ P͑x͒y ෇ Q͑x͒ dx 1 where P and Q are continuous functions on a given interval. This type of equation occurs frequently in various sciences, as we will see. An example of a linear equation is xyЈ ϩ y ෇ 2x because, for x 0, it can be written in the form yЈ ϩ 2 1 y෇2 x Notice that this differential equation is not separable because it’s impossible to factor the expression for yЈ as a function of x times a function of y. But we can still solve the equation by noticing, by the Product Rule, that xyЈ ϩ y ෇ ͑xy͒Ј and so we can rewrite the equation as ͑xy͒Ј ෇ 2x If we now integrate both sides of this equation, we get xy ෇ x 2 ϩ C or C x y෇xϩ If we had been given the differential equation in the form of Equation 2, we would have had to take the preliminary step of multiplying each side of the equation by x. It turns out that every first-order linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function I͑x͒ called an integrating factor. We try to find I so that the left side of Equation 1, when multiplied by I͑x͒, becomes the derivative of the product I͑x͒y: 3 I͑x͒͑yЈ ϩ P͑x͒y͒ ෇ ͑I͑x͒y͒Ј If we can find such a function I , then Equation 1 becomes ͑I͑x͒y͒Ј ෇ I͑x͒Q͑x͒ Integrating both sides, we would have I͑x͒y ෇ y I͑x͒Q͑x͒ dx ϩ C so the solution would be 4 y͑x͒ ෇ 1 I͑x͒ ͫy ͬ I͑x͒Q͑x͒ dx ϩ C 5E-10(pp 662-671) 1/18/06 9:26 AM Page 669 SECTION 10.6 LINEAR EQUATIONS ❙❙❙❙ 669 To find such an I, we expand Equation 3 and cancel terms: I͑x͒yЈ ϩ I͑x͒P͑x͒y ෇ ͑I͑x͒y͒Ј ෇ IЈ͑x͒y ϩ I͑x͒yЈ I͑x͒P͑x͒ ෇ IЈ͑x͒ This is a separable differential equation for I , which we solve as follows: y dI ෇ y P͑x͒ dx I ԽԽ ln I ෇ y P͑x͒ dx I ෇ Ae x P͑x͒ dx where A ෇ Ϯe C. We are looking for a particular integrating factor, not the most general one, so we take A ෇ 1 and use I͑x͒ ෇ e x P͑x͒ dx 5 Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where I is given by Equation 5. Instead of memorizing this formula, however, we just remember the form of the integrating factor. To solve the linear differential equation yЈ ϩ P͑x͒y ෇ Q͑x͒, multiply both sides by the integrating factor I͑x͒ ෇ e x P͑x͒ dx and integrate both sides. EXAMPLE 1 Solve the differential equation dy ϩ 3x 2 y ෇ 6x 2. dx SOLUTION The given equation is linear since it has the form of Equation 1 with P͑x͒ ෇ 3x 2 and Q͑x͒ ෇ 6x 2. An integrating factor is I͑x͒ ෇ e x 3x 2 dx ෇ ex 3 3 Multiplying both sides of the differential equation by e x , we get |||| Figure 1 shows the graphs of several members of the family of solutions in Example 1. Notice that they all approach 2 as x l ϱ. ex 6 C=2 3 dy 3 3 ϩ 3x 2e x y ෇ 6x 2e x dx d x3 3 ͑e y͒ ෇ 6x 2e x dx or C=1 C=0 Integrating both sides, we have C=_1 _1.5 1.8 C=_2 3 3 3 e x y ෇ y 6x 2e x dx ෇ 2e x ϩ C _3 FIGURE 1 y ෇ 2 ϩ CeϪx 3 5E-10(pp 662-671) 670 ❙❙❙❙ 1/18/06 9:26 AM Page 670 CHAPTER 10 DIFFERENTIAL EQUATIONS EXAMPLE 2 Find the solution of the initial-value problem x 2 yЈ ϩ xy ෇ 1 xϾ0 y͑1͒ ෇ 2 SOLUTION We must first divide both sides by the coefficient of yЈ to put the differential equation into standard form: 6 yЈ ϩ 1 1 y෇ 2 x x xϾ0 The integrating factor is I͑x͒ ෇ e x ͑1͞x͒ dx ෇ e ln x ෇ x Multiplication of Equation 6 by x gives xyЈ ϩ y ෇ 5 1 x 1 dx ෇ ln x ϩ C x y෇ and so ͑xy͒Ј ෇ or xy ෇ y Then |||| The solution of the initial-value problem in Example 2 is shown in Figure 2. 1 x ln x ϩ C x Since y͑1͒ ෇ 2, we have (1, 2) 0 2෇ 4 ln 1 ϩ C ෇C 1 Therefore, the solution to the initial-value problem is _5 ln x ϩ 2 x y෇ FIGURE 2 EXAMPLE 3 Solve yЈ ϩ 2xy ෇ 1. SOLUTION The given equation is in the standard form for a linear equation. Multiplying by the integrating factor |||| Even though the solutions of the differential equation in Example 3 are expressed in terms of an integral, they can still be graphed by a computer algebra system (Figure 3). e x 2x dx ෇ e x we get 2 2 e x yЈ ϩ 2xe x y ෇ e x 2 (e x y)Ј ෇ e x or 2.5 C=2 2 2 2 2 e x y ෇ y e x dx ϩ C Therefore _2.5 2 2.5 2 C=_2 Recall from Section 8.5 that x e x dx can’t be expressed in terms of elementary functions. Nonetheless, it’s a perfectly good function and we can leave the answer as _2.5 FIGURE 3 y ෇ eϪx 2 ye x2 dx ϩ CeϪx 2 5E-10(pp 662-671) 1/18/06 9:26 AM Page 671 SECTION 10.6 LINEAR EQUATIONS ❙❙❙❙ 671 Another way of writing the solution is y ෇ eϪx 2 y x 0 2 e t dt ϩ CeϪx 2 (Any number can be chosen for the lower limit of integration.) Application to Electric Circuits R E L switch In Section 10.2 we considered the simple electric circuit shown in Figure 4: An electromotive force (usually a battery or generator) produces a voltage of E͑t͒ volts (V) and a current of I͑t͒ amperes (A) at time t . The circuit also contains a resistor with a resistance of R ohms (⍀) and an inductor with an inductance of L henries (H). Ohm’s Law gives the drop in voltage due to the resistor as RI . The voltage drop due to the inductor is L͑dI͞dt͒. One of Kirchhoff’s laws says that the sum of the voltage drops is equal to the supplied voltage E͑t͒. Thus, we have FIGURE 4 L 7 dI ϩ RI ෇ E͑t͒ dt which is a first-order linear differential equation. The solution gives the current I at time t . EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is 12 ⍀ and the inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closed when t ෇ 0 so the current starts with I͑0͒ ෇ 0, find (a) I͑t͒, (b) the current after 1 s, and (c) the limiting value of the current. SOLUTION |||| The differential equation in Example 4 is both linear and separable, so an alternative method is to solve it as a separable equation (Example 4 in Section 10.3). If we replace the battery by a generator, however, we get an equation that is linear but not separable (Example 5). (a) If we put L ෇ 4, R ෇ 12, and E͑t͒ ෇ 60 in Equation 7, we obtain the initial-value problem or dI ϩ 12I ෇ 60 dt I͑0͒ ෇ 0 dI ϩ 3I ෇ 15 dt 4 I͑0͒ ෇ 0 Multiplying by the integrating factor e x 3 dt ෇ e 3t, we get e 3t dI ϩ 3e 3tI ෇ 15e 3t dt d 3t ͑e I͒ ෇ 15e 3t dt e 3tI ෇ y 15e 3t dt ෇ 5e 3t ϩ C I͑t͒ ෇ 5 ϩ CeϪ3t Since I͑0͒ ෇ 0, we have 5 ϩ C ෇ 0, so C ෇ Ϫ5 and I͑t͒ ෇ 5͑1 Ϫ eϪ3t ͒ 5E-10(pp 672-681) ❙❙❙❙ 672 1/18/06 5:16 PM Page 672 CHAPTER 10 DIFFERENTIAL EQUATIONS |||| Figure 5 shows how the current in Example 4 approaches its limiting value. (b) After 1 second the current is I͑1͒ ෇ 5͑1 Ϫ eϪ3 ͒ Ϸ 4.75 A 6 lim I͑t͒ ෇ lim 5͑1 Ϫ eϪ3t ͒ (c) y=5 tlϱ tlϱ ෇ 5 Ϫ 5 lim eϪ3t tlϱ ෇5Ϫ0෇5 2.5 0 EXAMPLE 5 Suppose that the resistance and inductance remain as in Example 4 but, instead of the battery, we use a generator that produces a variable voltage of E͑t͒ ෇ 60 sin 30t volts. Find I͑t͒. FIGURE 5 SOLUTION This time the differential equation becomes 4 dI ϩ 12I ෇ 60 sin 30t dt dI ϩ 3I ෇ 15 sin 30t dt or The same integrating factor e 3t gives d 3t dI ͑e I͒ ෇ e 3t ϩ 3e 3tI ෇ 15e 3t sin 30t dt dt |||| Figure 6 shows the graph of the current when the battery is replaced by a generator. Using Formula 98 in the Table of Integrals, we have 2 e 3tI ෇ y 15e 3t sin 30t dt ෇ 15 0 e 3t ͑3 sin 30t Ϫ 30 cos 30t͒ ϩ C 909 5 I ෇ 101 ͑sin 30t Ϫ 10 cos 30t͒ ϩ CeϪ3t 2.5 Since I͑0͒ ෇ 0, we get 50 Ϫ 101 ϩ C ෇ 0 _2 |||| 10.6 1–4 5 50 I͑t͒ ෇ 101 ͑sin 30t Ϫ 10 cos 30t͒ ϩ 101 eϪ3t so FIGURE 6 Exercises Determine whether the differential equation is linear. |||| 1. yЈ ϩ e x y ෇ x 2 y 2 2. y ϩ sin x ෇ x 3yЈ 3. xyЈ ϩ ln x Ϫ x 2 y ෇ 0 12. 4. yЈ ϩ cos y ෇ tan x ■ ■ 5–14 |||| ■ ■ ■ ■ ■ ■ ■ ■ 13. ͑1 ϩ t͒ ■ Solve the differential equation. 5. yЈ ϩ 2y ෇ 2e x 6. yЈ ෇ x ϩ 5y 7. xyЈ Ϫ 2y ෇ x 2 8. x 2 yЈ ϩ 2xy ෇ cos 2 x 9. xyЈ ϩ y ෇ sx 10. 1 ϩ xy ෇ xyЈ 11. dy ϩ 2xy ෇ x 2 dx dy ෇ x sin 2x ϩ y tan x, dx Ϫ␲͞2 Ͻ x Ͻ ␲͞2 du ϩ u ෇ 1 ϩ t, t Ͼ 0 dt ■ 14. t ln t ■ ■ 15–20 dr ϩ r ෇ te t dt ■ |||| 17. ■ ■ ■ Solve the initial-value problem. 15. yЈ ෇ x ϩ y, 16. t ■ y͑0͒ ෇ 2 dy ϩ 2y ෇ t 3, dt t Ͼ 0, y͑1͒ ෇ 0 dv 2 Ϫ 2tv ෇ 3t 2e t , v͑0͒ ෇ 5 dt ■ ■ ■ ■ ■ 5E-10(pp 672-681) 1/18/06 5:17 PM Page 673 SECTION 10.6 LINEAR EQUATIONS 18. 2xyЈ ϩ y ෇ 6x, x Ͼ 0, 20. x ■ dy y Ϫ ෇ x, dx xϩ1 ■ ; 21–22 ■ ■ y͑4͒ ෇ 20 y͑1͒ ෇ 0, ■ ■ ■ xϾ0 ■ RI ϩ ■ ■ ■ But I ෇ dQ͞dt (see Example 3 in Section 3.4), so we have Solve the differential equation and use a graphing calculator or computer to graph several members of the family of solutions. How does the solution curve change as C varies? R xϾ0 22. yЈ ϩ ͑cos x͒y ෇ cos x ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 23. A Bernoulli differential equation (named after James 31. Let P͑t͒ be the performance level of someone learning a skill Observe that, if n ෇ 0 or 1, the Bernoulli equation is linear. For other values of n, show that the substitution u ෇ y 1Ϫn transforms the Bernoulli equation into the linear equation du ϩ ͑1 Ϫ n͒P͑x͒u ෇ ͑1 Ϫ n͒Q͑x͒ dx Use the method of Exercise 23 to solve the differential 25. yЈ ϩ as a reasonable model for learning, where k is a positive constant. Solve it as a linear differential equation and use your solution to graph the learning curve. y3 2 y෇ 2 x x 32. Two new workers were hired for an assembly line. Jim 26. yЈ ϩ y ෇ xy 3 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 27. In the circuit shown in Figure 4, a battery supplies a constant voltage of 40 V, the inductance is 2 H, the resistance is 10 ⍀, and I͑0͒ ෇ 0. (a) Find I͑t͒. (b) Find the current after 0.1 s. 28. In the circuit shown in Figure 4, a generator supplies a voltage ; as a function of the training time t. The graph of P is called a learning curve. In Exercise 13 in Section 10.1 we proposed the differential equation dP ෇ k͓M Ϫ P͑t͔͒ dt equation. 24. xyЈ ϩ y ෇ Ϫxy 2 Suppose the resistance is 5 ⍀, the capacitance is 0.05 F, a battery gives a constant voltage of 60 V, and the initial charge is Q͑0͒ ෇ 0 C. Find the charge and the current at time t. and E͑t͒ ෇ 10 sin 60t. Find the charge and the current at time t. dy ϩ P͑x͒y ෇ Q͑x͒y n dx |||| 1 dQ ϩ Q ෇ E͑t͒ dt C 30. In the circuit of Exercise 29, R ෇ 2 ⍀, C ෇ 0.01 F, Q͑0͒ ෇ 0, Bernoulli) is of the form 24–26 Q ෇ E͑t͒ C ■ |||| 21. xyЈ ϩ y ෇ x cos x, 673 capacitor is Q͞C, where Q is the charge (in coulombs), so in this case Kirchhoff’s Law gives y͑␲͒ ෇ 0 19. xyЈ ෇ y ϩ x sin x, 2 ❙❙❙❙ of E͑t͒ ෇ 40 sin 60t volts, the inductance is 1 H, the resistance is 20 ⍀, and I͑0͒ ෇ 1 A. (a) Find I͑t͒. (b) Find the current after 0.1 s. (c) Use a graphing device to draw the graph of the current function. 29. The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of C farads (F), and a resistor with a resistance of R ohms (⍀). The voltage drop across the processed 25 units during the first hour and 45 units during the second hour. Mark processed 35 units during the first hour and 50 units the second hour. Using the model of Exercise 31 and assuming that P͑0͒ ෇ 0, estimate the maximum number of units per hour that each worker is capable of processing. 33. In Section 10.3 we looked at mixing problems in which the volume of fluid remained constant and saw that such problems give rise to separable equations. (See Example 6 in that section.) If the rates of flow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable. A tank contains 100 L of water. A solution with a salt concentration of 0.4 kg͞L is added at a rate of 5 L͞min. The solution is kept mixed and is drained from the tank at a rate of 3 L͞min. If y͑t͒ is the amount of salt (in kilograms) after t minutes, show that y satisfies the differential equation dy 3y ෇2Ϫ dt 100 ϩ 2t C Solve this equation and find the concentration after 20 minutes. E R 34. A tank with a capacity of 400 L is full of a mixture of water and chlorine with a concentration of 0.05 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 4 L͞s. The mixture is kept 5E-10(pp 672-681) 674 ❙❙❙❙ 1/18/06 5:17 PM Page 674 CHAPTER 10 DIFFERENTIAL EQUATIONS stirred and is pumped out at a rate of 10 L͞s. Find the amount of chlorine in the tank as a function of time. (a) Solve this as a linear equation to show that mt v෇ ͑1 Ϫ eϪct͞m ͒ c 35. An object with mass m is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If s͑t͒ is the distance dropped after t seconds, then the speed is v ෇ sЈ͑t͒ and the acceleration is a ෇ vЈ͑t͒. If t is the acceleration due to gravity, then the downward force on the object is mt Ϫ cv, where c is a positive constant, and Newton’s Second Law gives m |||| 10.7 dv ෇ mt Ϫ cv dt (b) What is the limiting velocity? (c) Find the distance the object has fallen after t seconds. 36. If we ignore air resistance, we can conclude that heavier objects fall no faster than lighter objects. But if we take air resistance into account, our conclusion changes. Use the expression for the velocity of a falling object in Exercise 35(a) to find dv͞dm and show that heavier objects do fall faster than lighter ones. Predator-Prey Systems We have looked at a variety of models for the growth of a single species that lives alone in an environment. In this section we consider more realistic models that take into account the interaction of two species in the same habitat. We will see that these models take the form of a pair of linked differential equations. We first consider the situation in which one species, called the prey, has an ample food supply and the second species, called the predators, feeds on the prey. Examples of prey and predators include rabbits and wolves in an isolated forest, food fish and sharks, aphids and ladybugs, and bacteria and amoebas. Our model will have two dependent variables and both are functions of time. We let R͑t͒ be the number of prey (using R for rabbits) and W͑t͒ be the number of predators (with W for wolves) at time t. In the absence of predators, the ample food supply would support exponential growth of the prey, that is, dR ෇ kR dt where k is a positive constant In the absence of prey, we assume that the predator population would decline at a rate proportional to itself, that is, dW ෇ ϪrW dt where r is a positive constant With both species present, however, we assume that the principal cause of death among the prey is being eaten by a predator, and the birth and survival rates of the predators depend on their available food supply, namely, the prey. We also assume that the two species encounter each other at a rate that is proportional to both populations and is therefore proportional to the product RW. (The more there are of either population, the more encounters there are likely to be.) A system of two differential equations that incorporates these assumptions is as follows: W represents the predator. R represents the prey. 1 dR ෇ kR Ϫ aRW dt dW ෇ ϪrW ϩ bRW dt where k, r, a, and b are positive constants. Notice that the term ϪaRW decreases the natural growth rate of the prey and the term bRW increases the natural growth rate of the predators. 5E-10(pp 672-681) 1/18/06 5:17 PM Page 675 SECTION 10.7 PREDATOR-PREY SYSTEMS |||| The Lotka-Volterra equations were proposed as a model to explain the variations in the shark and food-fish populations in the Adriatic Sea by the Italian mathematician Vito Volterra (1860–1940). ❙❙❙❙ 675 The equations in (1) are known as the predator-prey equations, or the Lotka-Volterra equations. A solution of this system of equations is a pair of functions R͑t͒ and W͑t͒ that describe the populations of prey and predator as functions of time. Because the system is coupled (R and W occur in both equations), we can’t solve one equation and then the other; we have to solve them simultaneously. Unfortunately, it is usually impossible to find explicit formulas for R and W as functions of t. We can, however, use graphical methods to analyze the equations. EXAMPLE 1 Suppose that populations of rabbits and wolves are described by the LotkaVolterra equations (1) with k ෇ 0.08, a ෇ 0.001, r ෇ 0.02, and b ෇ 0.00002. The time t is measured in months. (a) Find the constant solutions (called the equilibrium solutions) and interpret the answer. (b) Use the system of differential equations to find an expression for dW͞dR. (c) Draw a direction field for the resulting differential equation in the RW-plane. Then use that direction field to sketch some solution curves. (d) Suppose that, at some point in time, there are 1000 rabbits and 40 wolves. Draw the corresponding solution curve and use it to describe the changes in both population levels. (e) Use part (d) to make sketches of R and W as functions of t. SOLUTION (a) With the given values of k, a, r, and b, the Lotka-Volterra equations become dR ෇ 0.08R Ϫ 0.001RW dt dW ෇ Ϫ0.02W ϩ 0.00002RW dt Both R and W will be constant if both derivatives are 0, that is, RЈ ෇ R͑0.08 Ϫ 0.001W͒ ෇ 0 WЈ ෇ W͑Ϫ0.02 ϩ 0.00002R͒ ෇ 0 One solution is given by R ෇ 0 and W ෇ 0. (This makes sense: If there are no rabbits or wolves, the populations are certainly not going to increase.) The other constant solution is W෇ 0.08 ෇ 80 0.001 R෇ 0.02 ෇ 1000 0.00002 So the equilibrium populations consist of 80 wolves and 1000 rabbits. This means that 1000 rabbits are just enough to support a constant wolf population of 80. There are neither too many wolves (which would result in fewer rabbits) nor too few wolves (which would result in more rabbits). (b) We use the Chain Rule to eliminate t: dW dW dR ෇ dt dR dt so dW dW dt Ϫ0.02W ϩ 0.00002RW ෇ ෇ dR dR 0.08R Ϫ 0.001RW dt 5E-10(pp 672-681) 676 ❙❙❙❙ 1/18/06 5:17 PM Page 676 CHAPTER 10 DIFFERENTIAL EQUATIONS (c) If we think of W as a function of R, we have the differential equation dW Ϫ0.02W ϩ 0.00002RW ෇ dR 0.08R Ϫ 0.001RW We draw the direction field for this differential equation in Figure 1 and we use it to sketch several solution curves in Figure 2. If we move along a solution curve, we observe how the relationship between R and W changes as time passes. Notice that the curves appear to be closed in the sense that if we travel along a curve, we always return to the same point. Notice also that the point (1000, 80) is inside all the solution curves. That point is called an equilibrium point because it corresponds to the equilibrium solution R ෇ 1000, W ෇ 80. W W 150 150 100 100 50 50 0 FIGURE 1 1000 2000 0 3000 R FIGURE 2 Direction field for the predator-prey system 3000 R 2000 1000 Phase portrait of the system When we represent solutions of a system of differential equations as in Figure 2, we refer to the RW-plane as the phase plane, and we call the solution curves phase trajectories. So a phase trajectory is a path traced out by solutions ͑R, W͒ as time goes by. A phase portrait consists of equilibrium points and typical phase trajectories, as shown in Figure 2. (d) Starting with 1000 rabbits and 40 wolves corresponds to drawing the solution curve through the point P0(1000, 40). Figure 3 shows this phase trajectory with the direction field removed. Starting at the point P0 at time t ෇ 0 and letting t increase, do we move clockwise or counterclockwise around the phase trajectory? If we put R ෇ 1000 and W P™ 140 120 100 80 P£ P¡ 60 40 P¸ (1000, 40) 20 FIGURE 3 Phase trajectory through (1000, 40) 0 500 1000 1500 2000 2500 3000 R 5E-10(pp 672-681) 1/18/06 5:18 PM Page 677 SECTION 10.7 PREDATOR-PREY SYSTEMS ❙❙❙❙ 677 W ෇ 40 in the first differential equation, we get dR ෇ 0.08͑1000͒ Ϫ 0.001͑1000͒͑40͒ ෇ 80 Ϫ 40 ෇ 40 dt Since dR͞dt Ͼ 0, we conclude that R is increasing at P0 and so we move counterclockwise around the phase trajectory. We see that at P0 there aren’t enough wolves to maintain a balance between the populations, so the rabbit population increases. That results in more wolves and eventually there are so many wolves that the rabbits have a hard time avoiding them. So the number of rabbits begins to decline (at P1 , where we estimate that R reaches its maximum population of about 2800). This means that at some later time the wolf population starts to fall (at P2 , where R ෇ 1000 and W Ϸ 140). But this benefits the rabbits, so their population later starts to increase (at P3 , where W ෇ 80 and R Ϸ 210). As a consequence, the wolf population eventually starts to increase as well. This happens when the populations return to their initial values of R ෇ 1000 and W ෇ 40, and the entire cycle begins again. (e) From the description in part (d) of how the rabbit and wolf populations rise and fall, we can sketch the graphs of R͑t͒ and W͑t͒. Suppose the points P1 , P2 , and P3 in Figure 3 are reached at times t1 , t2 , and t3 . Then we can sketch graphs of R and W as in Figure 4. R W 140 2500 120 2000 100 1500 80 60 1000 40 500 0 20 t¡ t™ 0 t t£ t¡ t™ t t£ FIGURE 4 Graphs of the rabbit and wolf populations as functions of time To make the graphs easier to compare, we draw the graphs on the same axes but with different scales for R and W, as in Figure 5. Notice that the rabbits reach their maximum populations about a quarter of a cycle before the wolves. R 3000 W R W 120 Number 2000 of rabbits 80 Number of wolves 1000 40 FIGURE 5 Comparison of the rabbit and wolf populations 0 t¡ t™ t£ t 5E-10(pp 672-681) 678 ❙❙❙❙ 1/18/06 5:18 PM Page 678 CHAPTER 10 DIFFERENTIAL EQUATIONS An important part of the modeling process, as we discussed in Section 1.2, is to interpret our mathematical conclusions as real-world predictions and to test the predictions against real data. The Hudson’s Bay Company, which started trading in animal furs in Canada in 1670, has kept records that date back to the 1840s. Figure 6 shows graphs of the number of pelts of the snowshoe hare and its predator, the Canada lynx, traded by the company over a 90-year period. You can see that the coupled oscillations in the hare and lynx populations predicted by the Lotka-Volterra model do actually occur and the period of these cycles is roughly 10 years. 160 hare 120 9 lynx Thousands 80 of hares 6 Thousands of lynx 40 3 FIGURE 6 Relative abundance of hare and lynx from Hudson’s Bay Company records 0 1850 1875 1900 1925 Although the relatively simple Lotka-Volterra model has had some success in explaining and predicting coupled populations, more sophisticated models have also been proposed. One way to modify the Lotka-Volterra equations is to assume that, in the absence of predators, the prey grow according to a logistic model with carrying capacity K. Then the Lotka-Volterra equations (1) are replaced by the system of differential equations ͩ ͪ dR R ෇ kR 1 Ϫ dt K Ϫ aRW dW ෇ ϪrW ϩ bRW dt This model is investigated in Exercises 9 and 10. Models have also been proposed to describe and predict population levels of two species that compete for the same resources or cooperate for mutual benefit. Such models are explored in Exercise 2. |||| 10.7 Exercises 1. For each predator-prey system, determine which of the vari- ables, x or y, represents the prey population and which represents the predator population. Is the growth of the prey restricted just by the predators or by other factors as well? Do the predators feed only on the prey or do they have additional food sources? Explain. dx (a) ෇ Ϫ0.05x ϩ 0.0001xy dt dy ෇ 0.1y Ϫ 0.005xy dt (b) dx ෇ 0.2x Ϫ 0.0002x 2 Ϫ 0.006xy dt dy ෇ Ϫ0.015y ϩ 0.00008xy dt 2. Each system of differential equations is a model for two species that either compete for the same resources or cooperate for mutual benefit (flowering plants and insect pollinators, for instance). Decide whether each system describes competition or cooperation and explain why it is a reasonable model. (Ask yourself what effect an increase in one species has on the growth rate of the other.) dx (a) ෇ 0.12x Ϫ 0.0006x 2 ϩ 0.00001xy dt dy ෇ 0.08x ϩ 0.00004xy dt dx (b) ෇ 0.15x Ϫ 0.0002x 2 Ϫ 0.0006xy dt dy ෇ 0.2y Ϫ 0.00008y 2 Ϫ 0.0002xy dt 5E-10(pp 672-681) 1/18/06 5:18 PM Page 679 ❙❙❙❙ SECTION 10.7 PREDATOR-PREY SYSTEMS |||| A phase trajectory is shown for populations of rabbits ͑R͒ and foxes ͑F͒. (a) Describe how each population changes as time goes by. (b) Use your description to make a rough sketch of the graphs of R and F as functions of time. y 6. 3–4 species 1 1200 1000 800 F 600 300 3. 679 400 species 2 200 0 200 ■ ■ ■ ■ 10 ■ ■ 15 ■ ■ ■ t ■ ■ 7. In Example 1(b) we showed that the rabbit and wolf popula- t=0 100 ■ 5 tions satisfy the differential equation dW Ϫ0.02W ϩ 0.00002RW ෇ dR 0.08R Ϫ 0.001RW 0 800 400 1200 1600 R 2000 By solving this separable differential equation, show that R 0.02W 0.08 e e 0.00002R 0.001W 4. F where C is a constant. It is impossible to solve this equation for W as an explicit function of R (or vice versa). If you have a computer algebra system that graphs implicitly defined curves, use this equation and your CAS to draw the solution curve that passes through the point ͑1000, 40͒ and compare with Figure 3. t=0 160 ෇C 120 8. Populations of aphids and ladybugs are modeled by the 80 equations dA ෇ 2A Ϫ 0.01AL dt 40 0 ■ ■ 400 ■ ■ 800 ■ ■ 1200 ■ ■ R 1600 ■ ■ ■ ■ 5–6 |||| Graphs of populations of two species are shown. Use them to sketch the corresponding phase trajectory. 5. y dL ෇ Ϫ0.5L ϩ 0.0001AL dt (a) Find the equilibrium solutions and explain their significance. (b) Find an expression for dL͞dA. (c) The direction field for the differential equation in part (b) is shown. Use it to sketch a phase portrait. What do the phase trajectories have in common? L species 1 200 400 species 2 300 100 200 100 0 1 t 0 5000 10000 15000 A 5E-10(pp 672-681) ❙❙❙❙ 680 1/18/06 5:18 PM Page 680 CHAPTER 10 DIFFERENTIAL EQUATIONS (d) Suppose that at time t ෇ 0 there are 1000 aphids and 200 ladybugs. Draw the corresponding phase trajectory and use it to describe how both populations change. (e) Use part (d) to make rough sketches of the aphid and ladybug populations as functions of t. How are the graphs related to each other? (b) Find all the equilibrium solutions and explain their significance. (c) The figure shows the phase trajectory that starts at the point ͑1000, 40͒. Describe what eventually happens to the rabbit and wolf populations. (d) Sketch graphs of the rabbit and wolf populations as functions of time. 9. In Example 1 we used Lotka-Volterra equations to model popu- lations of rabbits and wolves. Let’s modify those equations as follows: dR ෇ 0.08R͑1 Ϫ 0.0002R͒ Ϫ 0.001RW dt CAS 10. In Exercise 8 we modeled populations of aphids and ladybugs with a Lotka-Volterra system. Suppose we modify those equations as follows: dA ෇ 2A͑1 Ϫ 0.0001A͒ Ϫ 0.01AL dt dW ෇ Ϫ0.02W ϩ 0.00002RW dt (a) According to these equations, what happens to the rabbit population in the absence of wolves? dL ෇ Ϫ0.5L ϩ 0.0001AL dt W (a) In the absence of ladybugs, what does the model predict about the aphids? (b) Find the equilibrium solutions. (c) Find an expression for dL͞dA. (d) Use a computer algebra system to draw a direction field for the differential equation in part (c). Then use the direction field to sketch a phase portrait. What do the phase trajectories have in common? (e) Suppose that at time t ෇ 0 there are 1000 aphids and 200 ladybugs. Draw the corresponding phase trajectory and use it to describe how both populations change. (f) Use part (e) to make rough sketches of the aphid and ladybug populations as functions of t. How are the graphs related to each other? 70 60 50 40 30 600 |||| 800 1000 1200 1400 1600 10 Review R ■ CONCEPT CHECK 1. (a) What is a differential equation? (b) What is the order of a differential equation? (c) What is an initial condition? 2. What can you say about the solutions of the equation yЈ ෇ x 2 ϩ y 2 just by looking at the differential equation? 3. What is a direction field for the differential equation yЈ ෇ F͑x, y͒? 4. Explain how Euler’s method works. 5. What is a separable differential equation? How do you solve it? 6. What is a first-order linear differential equation? How do you solve it? ■ 7. (a) Write a differential equation that expresses the law of natural growth. What does it say in terms of relative growth rate? (b) Under what circumstances is this an appropriate model for population growth? (c) What are the solutions of this equation? 8. (a) Write the logistic equation. (b) Under what circumstances is this an appropriate model for population growth? 9. (a) Write Lotka-Volterra equations to model populations of food fish ͑F͒ and sharks ͑S͒. (b) What do these equations say about each population in the absence of the other? 5E-10(pp 672-681) 1/18/06 5:19 PM Page 681 CHAPTER 10 REVIEW ■ TRUE-FALSE QUIZ ❙❙❙❙ 681 ■ 5. The equation e x yЈ ෇ y is linear. Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 6. The equation yЈ ϩ xy ෇ e y is linear. 1. All solutions of the differential equation yЈ ෇ Ϫ1 Ϫ y 4 are 7. If y is the solution of the initial-value problem decreasing functions. ͩ ͪ 2. The function f ͑x͒ ෇ ͑ln x͒͞x is a solution of the differential dy y ෇ 2y 1 Ϫ dt 5 equation x 2 yЈ ϩ xy ෇ 1. 3. The equation yЈ ෇ x ϩ y is separable. 4. The equation yЈ ෇ 3y Ϫ 2x ϩ 6xy Ϫ 1 is separable. y͑0͒ ෇ 1 then lim t l ϱ y ෇ 5. ■ EXERCISES ■ y 3 1. (a) A direction field for the differential equation yЈ ෇ y͑ y Ϫ 2͒͑ y Ϫ 4͒ is shown. Sketch the graphs of the solutions that satisfy the given initial conditions. (i) y͑0͒ ෇ Ϫ0.3 (ii) y͑0͒ ෇ 1 (iii) y͑0͒ ෇ 3 (iv) y͑0͒ ෇ 4.3 (b) If the initial condition is y͑0͒ ෇ c, for what values of c is lim t l ϱ y͑t͒ finite? What are the equilibrium solutions? 2 1 _3 y 6 _2 0 _1 1 3 x 2 _1 _2 4 _3 2 0 1 2 (b) Use Euler’s method with step size 0.1 to estimate y͑0.3͒ where y͑x͒ is the solution of the initial-value problem in part (a). Compare with your estimate from part (a). (c) On what lines are the centers of the horizontal line segments of the direction field in part (a) located? What happens when a solution curve crosses these lines? x 4. (a) Use Euler’s method with step size 0.2 to estimate y͑0.4͒, 2. (a) Sketch a direction field for the differential equation where y͑x͒ is the solution of the initial-value problem yЈ ෇ x͞y. Then use it to sketch the four solutions that satisfy the initial conditions y͑0͒ ෇ 1, y͑0͒ ෇ Ϫ1, y͑2͒ ෇ 1, and y͑Ϫ2͒ ෇ 1. (b) Check your work in part (a) by solving the differential equation explicitly. What type of curve is each solution curve? 3. (a) A direction field for the differential equation yЈ ෇ x 2 Ϫ y 2 yЈ ෇ 2xy 2 (b) Repeat part (a) with step size 0.1. (c) Find the exact solution of the differential equation and compare the value at 0.4 with the approximations in parts (a) and (b). 5–8 is shown. Sketch the solution of the initial-value problem yЈ ෇ x Ϫ y 2 2 Solve the differential equation. |||| 5. yЈ ෇ xeϪsin x Ϫ y cos x y͑0͒ ෇ 1 Use your graph to estimate the value of y͑0.3͒. y͑0͒ ෇ 1 6. 8. x 2 yЈ Ϫ y ෇ 2 x 3e Ϫ1͞x 7. ͑3y 2 ϩ 2y͒yЈ ෇ x cos x ■ ■ ■ ■ ■ dx ෇ 1 Ϫ t ϩ x Ϫ tx dt ■ ■ ■ ■ ■ ■ ■ 5E-10(pp 682-683) 1/18/06 9:30 AM Page 682 682 ❙❙❙❙ CHAPTER 10 DIFFERENTIAL EQUATIONS 9–11 |||| Solve the initial-value problem. 9. xyyЈ ෇ ln x, y͑1͒ ෇ 2 10. 1 ϩ x ෇ 2xyyЈ, x Ͼ 0, 11. yЈ ϩ y ෇ sxeϪx, ■ ■ ■ 19. The von Bertalanffy growth model is used to predict the length ■ y͑1͒ ෇ Ϫ2 y͑0͒ ෇ 3 ■ ■ ■ ■ ■ ■ ■ ■ x ; 12. Solve the initial-value problem 2yyЈ ෇ xe , y͑0͒ ෇ 1, and graph the solution. 13–14 |||| ■ ■ 20. A tank contains 100 L of pure water. Brine that contains Find the orthogonal trajectories of the family of curves. ■ 0.1 kg of salt per liter enters the tank at a rate of 10 L͞min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank after 6 minutes? k 14. y ෇ 1 ϩ x2 13. kx 2 ϩ y 2 ෇ 1 ■ ■ ■ ■ ■ ■ ■ ■ L͑t͒ of a fish over a period of time. If L ϱ is the largest length for a species, then the hypothesis is that the rate of growth in length is proportional to L ϱ Ϫ L , the length yet to be achieved. (a) Formulate and solve a differential equation to find an expression for L͑t͒. (b) For the North Sea haddock it has been determined that L ϱ ෇ 53 cm, L͑0͒ ෇ 10 cm, and the constant of proportionality is 0.2. What does the expression for L͑t͒ become with these data? ■ 15. A bacteria culture starts with 1000 bacteria and the growth rate is proportional to the number of bacteria. After 2 hours the population is 9000. (a) Find an expression for the number of bacteria after t hours. (b) Find the number of bacteria after 3 h. (c) Find the rate of growth after 3 h. (d) How long does it take for the number of bacteria to double? 16. An isotope of strontium, 90Sr, has a half-life of 25 years. (a) Find the mass of 90Sr that remains from a sample of 18 mg after t years. (b) How long would it take for the mass to decay to 2 mg? 17. Let C͑t͒ be the concentration of a drug in the bloodstream. As the body eliminates the drug, C͑t͒ decreases at a rate that is proportional to the amount of the drug that is present at the time. Thus, CЈ͑t͒ ෇ ϪkC͑t͒, where k is a positive number called the elimination constant of the drug. (a) If C0 is the concentration at time t ෇ 0, find the concentration at time t. (b) If the body eliminates half the drug in 30 h, how long does it take to eliminate 90% of the drug? 18. (a) The population of the world was 5.28 billion in 1990 and 6.07 billion in 2000. Find an exponential model for these data and use the model to predict the world population in the year 2020. (b) According to the model in part (a), when will the world population exceed 10 billion? (c) Use the data in part (a) to find a logistic model for the population. Assume a carrying capacity of 100 billion. Then use the logistic model to predict the population in 2020. Compare with your prediction from the exponential model. (d) According to the logistic model, when will the world population exceed 10 billion? Compare with your prediction in part (b). 21. One model for the spread of an epidemic is that the rate of spread is jointly proportional to the number of infected people and the number of uninfected people. In an isolated town of 5000 inhabitants, 160 people have a disease at the beginning of the week and 1200 have it at the end of the week. How long does it take for 80% of the population to become infected? 22. The Brentano-Stevens Law in psychology models the way that a subject reacts to a stimulus. It states that if R represents the reaction to an amount S of stimulus, then the relative rates of increase are proportional: k dS 1 dR ෇ R dt S dt where k is a positive constant. Find R as a function of S. 23. The transport of a substance across a capillary wall in lung physiology has been modeled by the differential equation R dh ෇Ϫ dt V ͩ ͪ h kϩh where h is the hormone concentration in the bloodstream, t is time, R is the maximum transport rate, V is the volume of the capillary, and k is a positive constant that measures the affinity between the hormones and the enzymes that assist the process. Solve this differential equation to find a relationship between h and t. 24. Populations of birds and insects are modeled by the equations dx ෇ 0.4x Ϫ 0.002xy dt dy ෇ Ϫ0.2y ϩ 0.000008xy dt (a) Which of the variables, x or y, represents the bird population and which represents the insect population? Explain. 5E-10(pp 682-683) 1/18/06 9:30 AM Page 683 CHAPTER 10 REVIEW (b) Find the equilibrium solutions and explain their significance. (c) Find an expression for dy͞dx. (d) The direction field for the differential equation in part (c) is shown. Use it to sketch the phase trajectory corresponding to initial populations of 100 birds and 40,000 insects. Then use the phase trajectory to describe how both populations change. ❙❙❙❙ 683 y 260 240 220 200 180 160 140 120 100 y 400 10000 15000 35000 45000 x (d) Sketch graphs of the bird and insect populations as functions of time. 300 26. Barbara weighs 60 kg and is on a diet of 1600 calories per day, 200 100 0 25000 20000 40000 60000 x of which 850 are used automatically by basal metabolism. She spends about 15 cal͞kg͞day times her weight doing exercise. If 1 kg of fat contains 10,000 cal and we assume that the storage of calories in the form of fat is 100% efficient, formulate a differential equation and solve it to find her weight as a function of time. Does her weight ultimately approach an equilibrium weight? 27. When a flexible cable of uniform density is suspended between (e) Use part (d) to make rough sketches of the bird and insect populations as functions of time. How are these graphs related to each other? 25. Suppose the model of Exercise 24 is replaced by the equations dx ෇ 0.4x ͑1 Ϫ 0.000005x͒ Ϫ 0.002xy dt dy ෇ Ϫ0.2y ϩ 0.000008xy dt two fixed points and hangs of its own weight, the shape y ෇ f ͑x͒ of the cable must satisfy a differential equation of the form dy 2 d 2y 1ϩ 2 ෇ k dx dx ͱ ͩ ͪ where k is a positive constant. Consider the cable shown in the figure. (a) Let z ෇ dy͞dx in the differential equation. Solve the resulting first-order differential equation (in z ), and then integrate to find y. (b) Determine the length of the cable. y (a) According to these equations, what happens to the insect population in the absence of birds? (b) Find the equilibrium solutions and explain their significance. (c) The figure shows the phase trajectory that starts with 100 birds and 40,000 insects. Describe what eventually happens to the bird and insect populations. (b, h) (_b, h) (0, a) _b 0 b x 5E-10(pp 684-685) 1/18/06 9:30 AM Page 684 PROBLEMS PLUS 1. Find all functions f such that f Ј is continuous and x [ f ͑x͒] 2 ෇ 100 ϩ y ͕[ f ͑t͒] 2 ϩ [ f Ј͑t͒] 2 ͖ dt 0 for all real x 2. A student forgot the Product Rule for differentiation and made the mistake of thinking that ͑ ft͒Ј ෇ f ЈtЈ. However, he was lucky and got the correct answer. The function f that he 2 used was f ͑x͒ ෇ e x and the domain of his problem was the interval ( 1 , ϱ). What was the 2 function t ? 3. Let f be a function with the property that f ͑0͒ ෇ 1, f Ј͑0͒ ෇ 1, and f ͑a ϩ b͒ ෇ f ͑a͒ f ͑b͒ for all real numbers a and b. Show that f Ј͑x͒ ෇ f ͑x͒ for all x and deduce that f ͑x͒ ෇ e x. 4. Find all functions f that satisfy the equation ͩy ͪͩy f ͑x͒ dx ͪ 1 dx ෇ Ϫ1 f ͑x͒ 5. A peach pie is taken out of the oven at 5:00 P.M. At that time it is piping hot: 100ЊC. At 5:10 P.M. its temperature is 80ЊC; at 5:20 P.M. it is 65ЊC. What is the temperature of the room? 6. Snow began to fall during the morning of February 2 and continued steadily into the after- noon. At noon a snowplow began removing snow from a road at a constant rate. The plow traveled 6 km from noon to 1 P.M. but only 3 km from 1 P.M. to 2 P.M. When did the snow begin to fall? [Hints: To get started, let t be the time measured in hours after noon; let x ͑t͒ be the distance traveled by the plow at time t; then the speed of the plow is dx͞dt. Let b be the number of hours before noon that it began to snow. Find an expression for the height of the snow at time t. Then use the given information that the rate of removal R (in m3͞h) is constant.] y 7. A dog sees a rabbit running in a straight line across an open field and gives chase. In a rectan- gular coordinate system (as shown in the figure), assume: (i) The rabbit is at the origin and the dog is at the point ͑L, 0͒ at the instant the dog first sees the rabbit. (ii) The rabbit runs up the y-axis and the dog always runs straight for the rabbit. (iii) The dog runs at the same speed as the rabbit. (a) Show that the dog’s path is the graph of the function y ෇ f ͑x͒, where y satisfies the differential equation (x, y) x 0 FIGURE FOR PROBLEM 7 (L, 0) x d 2y ෇ dx 2 ͱ ͩ ͪ 1ϩ dy dx 2 (b) Determine the solution of the equation in part (a) that satisfies the initial conditions y ෇ yЈ ෇ 0 when x ෇ L. [Hint: Let z ෇ dy͞dx in the differential equation and solve the resulting first-order equation to find z; then integrate z to find y.] (c) Does the dog ever catch the rabbit? 8. (a) Suppose that the dog in Problem 7 runs twice as fast as the rabbit. Find a differential equation for the path of the dog. Then solve it to find the point where the dog catches the rabbit. (b) Suppose the dog runs half as fast as the rabbit. How close does the dog get to the rabbit? What are their positions when they are closest? 684 5E-10(pp 684-685) 1/18/06 9:30 AM Page 685 9. A planning engineer for a new alum plant must present some estimates to his company regard- ing the capacity of a silo designed to contain bauxite ore until it is processed into alum. The ore resembles pink talcum powder and is poured from a conveyor at the top of the silo. The silo is a cylinder 100 ft high with a radius of 200 ft. The conveyor carries 60,000␲ ft 3 and ͞h the ore maintains a conical shape whose radius is 1.5 times its height. (a) If, at a certain time t, the pile is 60 ft high, how long will it take for the pile to reach the top of the silo? (b) Management wants to know how much room will be left in the floor area of the silo when the pile is 60 ft high. How fast is the floor area of the pile growing at that height? (c) Suppose a loader starts removing the ore at the rate of 20,000␲ ft 3 when the height of ͞h the pile reaches 90 ft. Suppose, also, that the pile continues to maintain its shape. How long will it take for the pile to reach the top of the silo under these conditions? 10. Find the curve that passes through the point ͑3, 2͒ and has the property that if the tangent line is drawn at any point P on the curve, then the part of the tangent line that lies in the first quadrant is bisected at P. 11. Recall that the normal line to a curve at a point P on the curve is the line that passes through P and is perpendicular to the tangent line at P. Find the curve that passes through the point ͑3, 2͒ and has the property that if the normal line is drawn at any point on the curve, then the y-intercept of the normal line is always 6. 12. Find all curves with the property that if the normal line is drawn at any point P on the curve, then the part of the normal line between P and the x-axis is bisected by the y-axis. 685 ...
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