Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 105 pptx

31.4 Solving Separable First Order Differential Equations 1021 Let us assume P is greater than zero and less than the carrying capacity 500. Then P 500 − P = C 3 e 0.5t , C 3 > 0. Solve for P . P = 500C 3 e 0.5t − PC 3 e 0.5t P + PC 3 e 0.5t = 500C 3 e 0.5t P(1+C 3 e 0.5t ) = 500C 3 e 0.5t P = 500C 3 e 0.5t 1 + C 3 e 0.5t P(t)= 500 Ce −0.5t + 1 This, together with P = 0 and P = 500, is the solution to the differential equation P(0.5 − 0.001P)for P(0) between 0 and 500. Let’s check that the answer is reasonable. lim t→∞ P(t)= lim t→∞ 500 Ce −0.5t + 1 = 500 because lim t→∞ Ce −0.5t = 0. This makes sense. For 0 <P(0)<500, the solution curves are asymptotic to the constant solutions. Verify that lim t→−∞ P(t)=0. ◆ ◆ EXAMPLE 31.23 Is 2 dy dt + 3y = 4 separable? SOLUTION 2 dy dt + 3y = 4 2 dy dt =−3y+4 2dy = (−3y + 4)dt 2dy −3y + 4 = dt Yes, the differential equation is separable. It can be solved either by separation of variables or by using substitution to put it into the form du dt = ku. ◆ EXERCISE 31.6 Solve 2 dy dt + 3y = 4. Do it twice; use each of the methods suggested above. Answer y = 4 3 + Ce −3/2t 1022 CHAPTER 31 Differential Equations Taking Inventory If a differential equation is separable and we can compute the relevant integrals, then we can find a relationship that is a solution to the differential equation. Differential equations of the forms dy dt = f(t)and dy dt = f(y)are special cases of separable differential equations. If dy dt = f(t)then  dy =  f(t)dt so y =  f(t)dt.Solutions are vertical translates of one another. If dy dt = f(y)then  dy f(y) =  dt. Solutions are horizontal translates of one another. What Can We Do With a Nonseparable First Order Differential Equation? Suppose the differential equation dy dt = f(t,y) with initial condition f(t 0 )=y 0 is nonsep- arable. One approach to approximating the solution is to start at the point (t 0 , y 0 ) and to follow the flow line from this point by following the slope field for the differential equation. Imagine placing a bug in the ty-plane at the point (t 0 , y 0 ). Evaluating the differential equa- tion at (t 0 , y 0 ) we find the slope there. Let the bug take one step in the direction indicated, arriving at a new point (t 1 , y 1 ) where t 1 = t 0 + t and y 1 = y 0 +  dy dt   (t 0 ,y 0 )  · t, and t is small. Evaluating the differential equation at (t 1 , y 1 ) gives the bug its direction for its next step. t 2 = t 1 + t and y 2 = y 1 +  dy dt   (t 1 ,y 1 )  · t . And so the bug proceeds, one step at a time, consulting the differential equation after each step in order to get directions for the next step. t k+1 = t k + t and y k+1 = y k +  dy dt     (t k ,y k )  t This method of approximating a solution to a differential equation is known as Euler’s Method (pronounced “oiler’s” method). In order to use Euler’s Method, one must choose a “step size,” that is, a value for t. We expect the approximation to improve as the size of t decreases. While Euler’s Method is tedious to perform by hand, computers and programmable graphing calculators are well suited to carry out the task. You need to input the differential equation, the initial condition, and the step size. Drawbacks to Euler’s Method: Suppose we apply Euler’s Method to Example 31.21, dy dx =− x y , with initial condition t = 3, y = 4. From the work we did using separation of variables, we know the solution curve is a circle of radius 5 centered at the origin. Using Euler’s method, we step out along the tangent lines, so regardless of how small we program the step size, the path taken will spiral outward instead of giving the closed curve. To correct for such problems, more sophisticated variations of Euler’s Method have been developed, but the underlying line of reasoning is similar. Improvements upon Euler’s Method basically involve using more information about the slope in the vicinity of a particular point so that a more informed choice of direction can be made. One commonly used improvement upon Euler’s Method is called the Runge-Kutta Method. PROBLEMS FOR SECTION 31.4 In Problems 1 through 10, solve the given differential equation. 31.4 Solving Separable First Order Differential Equations 1023 1. dy dx = x 2y 2. dy dx = x 2 y 3. dy dx = xy 2 4. dy dx = y x 5. dy dx = x−1 2y+1 6. 2y  − y = 1 7. dy dx − y 2 = 1 8. dy dt = t cos 2 y 9. 2 dy dx − 3xy = 0 10. dy dx = cos x − sin y 11. Suppose that a freighter moving in a straight line with velocity v encounters water and air resistance proportional to its velocity. If the freighter is traveling with velocity v 0 when the motor is cut off, how far does it travel in time t?(Hint: Force = mass · acceleration. The only force acting on the freighter is the resistance from air and water.) 12. A very large container of juice contains four gallons of apple juice and one gallon of cranberry juice. Cranberry-apple juice (60% apple, 40% cranberry) is entering the container at a rate of three gallons per hour. The well-stirred mixture is leaving the container at three gallons per hour. (a) Write a differential equation whose solution is C(t), the number of gallons of cranberry juice in the container at time t. Solve, using the initial condition. (b) Write a differential equation whose solution is A(t), the number of gallons of apple juice in the container at time t. Solve the equation. 13. Suppose we change the previous problem so that the mixture is leaving the container at two gallons per hour. (a) Write a differential equation whose solution is C(t), the number of gallons of cranberry juice in the container at time t. (This equation should be good as long as the container does not overflow.) (b) Write a differential equation whose solution is A(t), the number of gallons of apple juice in the container at time t. (c) Are the differential equations you got in parts (a) and (b) separable? If so, solve them. 1024 CHAPTER 31 Differential Equations 31.5 SYSTEMS OF DIFFERENTIAL EQUATIONS Adjusting the simple exponential population growth model to reflect competition between the members of a population for limited resources led us to the logistic population growth model. In this section we will develop tools enabling us to deal with an added level of complexity, the relationship between two distinct populations. We can model competitive relationships, symbiotic relationships, and predator/prey relationships. These same tools enable us to develop alternative models for the spread of an epidemic, reflecting the nature of the disease and the way it is spread. Our mathematical focus in this section is on systems of differential equations. Epidemic Models Epidemic Model A: Contagious, Nonfatal Disease of Long Duration Earlier in this chapter we considered a model for the spread of disease assuming that people are being infected by the disease at a rate proportional to the product of the number of people who have already been infected and the number of those who have not. This could reflect a contagious disease where the sick are not isolated. Letting P(t)denote the number of infected people at time t and N denote the total population affected by the epidemic, we wrote dP dt = kP(N − P). In this model we assume that the population is fixed and that every member is susceptible to the disease. The disease is long in duration (there are no recoveries during the time period we are analyzing) but not fatal (no deaths during this period). A solution, with 0 <P 0 <N,issketched below. According to this model, eventually everyone will become infected. P P 0 N t Figure 31.15 REMARK This differential equation is logistic; it is of the same form as the fish-in-a-pond example. N is a limiting factor because P(t)≤N.AsP gets close to N , there are very few healthy people to infect, so dP dt will be small. Epidemic Model B: A Contagious, Fatal Disease Consider the situation in which a small group of people having a fatal infectious disease reside in a large population capable of catching the disease. Let us assume the disease has a negligibly short incubation time, so an infected individual can immediately infect others. We partition the population into two groups: the infected and the susceptible. 7 Assume that the population remains fixed during the time interval in question except for deaths due to 7 In mathematics, to partition a population into classes is to assign every member of the population to one and only one class. In this example, every person is either susceptible or infected. 31.5 Systems of Differential Equations 1025 the disease; assume there are no births and no deaths apart from those deaths due to the disease. Let I (t) = the number of infected people at time t. Let S(t) = the number of susceptible people at time t. Consider dS dt . The susceptible group is not gaining new members. It is losing members to the infected group at a rate proportional to interactions between infected and susceptible individuals. We can write dS dt =−rSI, where r is a positive constant. As the susceptible class loses members, the infected class gains members, the rate being rSI.The infected class loses people due to death from the disease. It is reasonable to suppose that the rate at which people die is proportional to the number of sick people. Then dI dt = rSI − kI, where r and k are positive constants. We must consider the system of equations  dS dt =−rSI dI dt = rSI − kI . This is the first time we have run into a system of differential equations, so we will take a step back to analyze some simple systems of equations before returning to this particular example. Analyzing Systems of Differential Equations: Some Examples 8 ◆ EXAMPLE 31.24 Newton’s Law of Cooling. Let y = y(t) = the temperature of a large room at time t. Let x = x(t) = the difference in temperature between the room and a small object placed in it. Let x be positive if the object is hotter than the room. The differential equations given below could reflect the room-object system. dx dt =−kx, k>0 dy dt = 0 For the sake of simplicity, let k = 1. (a) i. Solve for y in terms of t and sketch the solution in the ty-plane. ii. Solve for x in terms of t and sketch the solution in the tx-plane. (b) Graph the solutions in the xy-plane as follows: Pick any initial temperatures y(0) = y 0 and x(0) = x 0 for the room and the small object, respectively. Think about what will happen to x and y as t increases and sketch the appropriate path, drawing an arrow to indicate the direction the path is traveled. (t will not appear explicitly in the sketch.) 8 The three examples that follow were originally suggested to me by Stephen DiPippo when we taught together. 1026 CHAPTER 31 Differential Equations Select various other starting points and sketch each path traced out by x and y as t increases. All paths should have arrows to indicate how x and y change with time. (c) If the system is at equilibrium, then both x and y will remain constant as t varies, that is, dy dt = 0 and dx dt = 0. Under what conditions is the system at equilibrium? How do these equilibrium points relate to the picture drawn in part (a)? SOLUTION (a) Each differential equation can be solved independently. i. dy dt = 0 ⇒ y = C If we let y 0 denote y(0), then y(t) = y 0 .Inthety-plane, solutions can be repre- sented as shown in Figure 31.16. y t Figure 31.16 ii. dx dt =−x⇒x(t) = x 0 e −t In the tx-plane, solutions can be represented as shown in Figure 31.17. x t Figure 31.17 (b) Pick a point (x 0 , y 0 ) as a starting point. The value of y will remain constant. If x 0 is positive, then as t increases x will remain positive but tend toward zero; if x is negative, then as t increases x will remain negative but tend toward zero. (This information can be read from Figure 31.17.) On the following page is a sketch in the xy-plane. Various starting points (corresponding to t = 0) are chosen, each designated by an “X.” Arrows indicate the direction the path is traveled as t increases. 31.5 Systems of Differential Equations 1027 x y Figure 31.18 (c) If the system is at equilibrium, then dy dt = 0 and dx dt = 0. dy dt is identically 0. dx dt =−x,therefore dx dt = 0 if and only if x = 0. In other words, the system is at equilibrium as long as x, the temperature difference, is zero. When we represent solutions to the system of differential equations in the xy-plane (with no t -axis), an equilibrium solution is simply a point. As t varies, x and y remain constant. (Contrast this with an equilibrium solution in the ty-plane or the tx-plane in which an equilibrium solution is a horizontal line.) In this example, every point along the y-axis is an equilibrium point. Below representative solution curves are drawn. These curves, or trajectories, are traced by (x(t), y(t)) as t varies. Arrows indicate the direction the curve is traced as t increases. The xy-plane is often referred to as the phase-plane of the solutions. Initial points are usually not explicitly indicated. y x Figure 31.19 REMARKS 1. Trajectories on the right-hand side of the y-axis and the left-hand side of the y-axis are distinct. If x is initially positive, it will always be positive (see Figure 31.17). Likewise, if x is initially negative, it is always negative. In other words, if one starts at a point to the right of the y-axis, the trajectory through that point will never cross the y-axis. In fact, although the trajectory gets arbitrarily close to the y-axis, x never becomes zero 1028 CHAPTER 31 Differential Equations (again, refer to Figure 31.17). Each point on the y-axis is a distinct equilibrium point. (We can think of an equilibrium point as a degenerative trajectory.) Each horizontal line in Figure 31.19 consists of three distinct solution trajectories. Trajectories do not intersect, although they can get arbitrarily close to one another. 2. We determined the direction of the arrows on the trajectories by referring to Figure 31.17, but that information is directly accessible from the differential equations them- selves. dx dt =−x,therefore for x>0, dx dt < 0; x decreases as t increases. For x<0, dx dt > 0; x increases as t increases. We can determine the “shape” of the trajectories in the xy-plane by using the original differential equations and looking at the relative rates of change of y and x as t varies. The slope of the trajectory is dy dx = dy/dt dx/dt , where equality follows from the Chain Rule, provided dx dt = 0. Therefore dy dx = dy dt dx dt = 0 −x = 0 ⇒ dy dx = 0 ⇒ y = C. These are horizontal lines in the xy-plane. In Remark 1 we point out that each of these horizontal lines actually consists of three distinct trajectories, one to the right of the y-axis, one to the left of the y-axis, and an equilibrium point on the y-axis. ◆ Note: dy dx = dy/dt dx/dt provided that dx dt is nonzero. If, at a certain point P , dx dt = 0 and dy dt = 0, then neither x nor y is changing with t,soP is an equilibrium point. If, at a certain point Q, dx dt = 0 and dy dt = 0, then y is changing with t but x is not, so the trajectory is vertical at Q. ◆ EXAMPLE 31.25 Analyze the system of differential equations below, sketching solution trajectories in the phase-plane.  dx dt =−x dy dt =−y SOLUTION Each of these differential equations can be solved independently. They describe exponential decay. dx dt =−x⇒x(t) = x 0 e −t dy dt =−y⇒y(t) = y 0 e −t In the tx-plane and the ty-plane, solutions can be represented as shown in Figure 31.20. 31.5 Systems of Differential Equations 1029 y t x t Figure 31.20 Now consider the trajectories in the xy-plane. As t changes, (x(t), y(t)) traces out a curve in the xy-plane. Wherever we start out in the plane, as t increases, both the x- and y-coordinates go toward zero. At the equilibrium dx dt =−x=0 and dy dt =−y=0 simultaneously. So x = 0 and y = 0. The origin is the only equilibrium point. x y Figure 31.21 To determine the shape of the trajectories in the xy-plane, observe that for x 0 = 0 y(t) x(t) = y 0 e −t x 0 e −t = y 0 x 0 = constant. Thus, the trajectory starting at (x 0 , y 0 ) when t = 0 satisfies y x = k, where k is the constant y 0 x 0 . We obtain y = kx, the equation of a straight line through the origin. The coordinates of the starting point determine the slope of the line. In Figure 31.22 arrows indicate the direction to travel along these lines as t increases. We need to treat the case x 0 = 0 separately. If x 0 = 0 then x(t) = 0. For y 0 > 0, y(t) decreases as t increases; for y 0 < 0, y(t) increases as t increases. This gives two trajectories along the y-axis. If both x 0 and y 0 are zero, the system is at equilibrium. Putting everything together, we obtain the picture sketched on the following page. Apart from the origin, each trajectory is a ray approaching the origin. 1030 CHAPTER 31 Differential Equations x y Figure 31.22 ◆ EXERCISE 31.7 In Example 31.25, we can find the shape of the trajectories by looking at dy dx . dy dx = dy dt dx dt = −y −x ,or dy dx = y x . This is a separable differential equation. Solve for y, showing that y = kx, where k is a constant. Examples 31.24 and 31.25 are special in that we can solve each of the original dif- ferential equations independently to get x and y explicitly in terms of t and we can see precisely what the trajectories look like in the xy- phase-plane. In general, the situation is more complex. General Strategy Consider a system of differential equations of the form  dx dt = f(x,y) dy dt = g(x, y) where f and g are continuous functions. 9 Suppose we want to analyze the behavior of solution trajectories in the xy-phase-plane. The general direction of a trajectory at any point can be determined by knowing the signs of dx dt and dy dt . Therefore, to get a handle on the phase-plane picture we begin by determining where dx dt = 0 and where dy dt = 0. The curves along which dx dt = 0or dy dt = 0 are called nullclines. Nullclines are not, in general, trajectories. They are the set of points along which trajectories have either vertical or horizontal tangent lines. 9 We haven’tdefined continuity for functions of two variables, but if f(x,y) is continuous then for fixed y, f is a continuous function of x and for fixed x, f is a continuous function of y. The crucial consequence of continuity for our analysis is that if dx dt = f(x,y) where f is continuous, then dx dt can change sign only by passing through zero. . increases. We can determine the “shape” of the trajectories in the xy-plane by using the original differential equations and looking at the relative rates of change of y and x as t varies. The slope of the. y) where f and g are continuous functions. 9 Suppose we want to analyze the behavior of solution trajectories in the xy-phase-plane. The general direction of a trajectory at any point can be determined. is a constant. Examples 31.24 and 31.25 are special in that we can solve each of the original dif- ferential equations independently to get x and y explicitly in terms of t and we can see precisely