31.2 Solutions to Differential Equations: An Introduction 991 6. Let’s suppose that the population in a certain country has a growth rate of 2% and a population of 9 million at a time we’ll designate as t = 0. Due to the political and economic situation, there is a massive rearrangement of populations in the region. The immigration and emigration rates are both constant, with people entering the country at a rate of 100,000 per year and leaving at a rate of 300,000 per year. Let P = P(t)be the population in millions at time t. (a) Write a differential equation reflecting the situation. Keep in mind that P is in millions. (b) If this situation goes on indefinitely, what will happen to the country’s population? (c) What initial population would support a net emigration of 200,000 per year? 7. In the beginning of a chemical reaction there are 600 moles of substance A and none of substance B. Over the course of the reaction, the 600 moles of substance A are converted to 600 moles of substance B. (Each molecule of A is converted to a molecule of B via the reaction.) Suppose the rate at which A is turning into B is proportional to the product of the number of moles of A and the number of moles of B. (a) Let N = N(t) be the number of moles of substance A at time t. Translate the statement above into mathematical language. (Note: The number of moles of substance B should be expressed in terms of the number of moles of substance A.) (b) Using your answer to part (a), find d 2 N dt 2 . Your answer will involve the proportion- ality constant used in part (a). (c) N(t) is a decreasing function. The rate at which N is changing is a function of N, the number of moles of substance A. When the rate at which A is being converted to B is highest, how many moles are there of substance A? 8. There are many places in the world where populations are changing and immigration and/or emigration play a big role. People may move to find food, or to find jobs, or to flee political or religious persecution. Pick a situation that interests you. You could look at the number of Tibetans in Tibet, or the number of Tibetans in India, or the number of lions in the Serengeti, or the number of tourists in Nepal. Get some data and try to model the population dynamics using a differential equation. What simplifying assumptions have you made? 31.2 SOLUTIONS TO DIFFERENTIAL EQUATIONS: AN INTRODUCTION Although knowing about the rate of change of a quantity is useful, often what we really want to know about is the actual amount of that quantity. After all, it’s nice to know that the money in your bank account is growing at an instantaneous rate of 5%, but what you really want to know is when you’ll finally have enough money to buy that new motorcycle or whatever it is that you’re saving for. In this section we turn our attention to the solutions of differential equations. 992 CHAPTER 31 Differential Equations What Does It Mean to Be a Solution to a Differential Equation? We have said that a function f is a solution to a differential equation if it satisfies the differential equation. By this we mean that when the function and its derivative(s) are substituted in the appropriate places in the differential equation, the two sides of the equation are equal. We review work presented in Section 15.2. ◆ EXAMPLE 31.10 Is y = x 3 a solution to the differential equation dy dx = 3y x ? SOLUTION To determine whether y = x 3 is a solution, we need to substitute it into the differential equation. dy dx = 3y x Replace y by x 3 wherever it appears. d dx x 3 ? = 3x 3 x 3x 2 ? = 3x 3 x 3x 2 = 3x 2 True. y = x 3 is a solution to dy dx = 3y x because it satisfies the equation. ◆ ◆ EXAMPLE 31.11 Is y = xe 3x a solution to the differential equation dy dx = 3y x ? SOLUTION dy dx = 3y x Replace y by xe 3x wherever it appears. d dx [xe 3x ] ? = 3xe 3x x 3xe 3x + e 3x ? = 3e 3x e 3x [3x + 1] = e 3x (3) The equation is not satisfied, so y = xe 3x is not a solution to the differential equation. ◆ Differential equations have families of solutions. In Section 15.2 we looked at the family of solutions to each of the three differential equations given below. The discussion is summarized here. i. dy dt = 2 ii. dy dt = 2t iii. dy dt = 2y Graphical perspective: Solving a differential equation that involves dy dt means finding y as a function of t; therefore, on our graph we will label the vertical axis y and the horizontal axis t. At any point P in the ty-plane we can use the differential equation to find the slope of the tangent line to the solution curve through P.We’ll draw a short line segment through P indicating the slope of the solution curve there. The resulting diagram is called a slope field. (See Figure 31.1 on the following page.) Observations In part (i), dy dt = 2, the slope is independent of the point P chosen. 31.2 Solutions to Differential Equations: An Introduction 993 In part (ii), dy dt = 2t, the slope depends only on the t-coordinate of P . This is true whenever we have a differential equation of the form dy dt = f(t). In part (iii), dy dt = 2y, the slope depends only on the y-coordinate of P . This is true whenever we have a differential equation of the form dy dt = f(y). y t y t y t (i) dy dt = 2 (ii) dy dt = 2t (iii) dy dt = 2y Figure 31.1 In Figure 31.1 (ii) the slope is positive whenever t is positive, negative whenever t is negative, and zero at t = 0. At (2, 3) the slope is 4; at (1, 5) the slope is 2. In Figure 31.1 (iii) the slope is positive whenever y is positive, negative whenever y is negative, and zero at y = 0. At (2, 3) the slope is 6; at (1, 5) the slope is 10. By following the slope fields, we can get a rough idea of the shapes of the solution curves. (See Figure 31.2.) y t y t y t C>O C<O (i) dy dt = 2 (ii) dy dt = 2t (iii) dy dt = 2y y = Ce 2t y = t 2 + Cy = 2t + C Figure 31.2 We state, without proof, the following fact (known as the Existence and Uniqueness Theorem). 4 4 The statements you’ll see in differential equations texts are actually much stronger than this. 994 CHAPTER 31 Differential Equations Existence and Uniqueness Theorem (Weak form) Let (a, b) be a point in the plane. Any differential equation of the form dy dt = g(t) where g is continuous, or of the form dy dt = f(y)where f and f are continuous, has a solution passing through (a, b). The solution exists, and it is unique. In particular the Existence and Uniqueness Theorem tells us given any point P = (a, b) in the plane, a differential equation of the form dy dt = k, dy dt = kt,or dy dt = ky where k is constant has exactly one solution passing through the point P . This makes sense in our examples above; because the slope at P is completely determined by the coordinates of P ,notwo solution curves can cross. We can combine this graphical analysis with our knowledge of analytic solutions to these differential equations. 5 y = 2t + C is a solution to part (i) for any constant C. y = t 2 + C is a solution to part (ii) for any constant C. y = Ce 2t is a solution to part (iii) for any constant C. The Existence and Uniqueness Theorem tells us that we have written general solutions to each of these differential equations. In other words, any solution to dy dt = 2 can be expressed in the form y = 2t + C; any solution to dy dt = 2t can be expressed in the form y = t 2 + C; any solution to dy dt = 2y can be expressed in the form y = Ce 2t . EXERCISE 31.1 Let k be an arbitrary constant. Show that for any point P in the plane there is a unique value of C such that the curve y = Ce kt passes through P . EXERCISE 31.2 Let C be an arbitrary constant. Show that for any point P in the plane there is a unique value of C such that the curve y = t 2 + C passes through P . Suppose we know the general solution of a first order differential equation. We can determine a particular solution if an initial condition is specified. Geometrically this is equivalent to knowing one point through which the solution curve passes. The conclusions drawn from these specific examples can be generalized. 1. Suppose a differential equations is of the form dy dt = f(t). . The slope of the solution curve at P is determined completely by the t-coordinate. . If F is an antiderivative of f , then F(t)+ C is the general solution to dy dt = f(t). Solving dy dt = f(t)is equivalent to finding f(t)dt. . The solutions to dy dt = f(t)are vertical translates of one another. (They differ from one another by an additive constant.) 2. Suppose a differential equation is of the form dy dt = f(y). . The slope of the solution curve at P is determined completely by the y-coordinate. . Solutions to dy dt = f(y)are horizontal translates of one another. 5 See Section 15.2 for a review of solutions to differential equations of the form dy dt = ky. 31.2 Solutions to Differential Equations: An Introduction 995 Solving Differential Equations: Analytic Solutions Solving a differential equation can be quite difficult. However, there are several types of differential equations that we can already solve. In this section we will look at differential equations of the form dy dt = f(t) and dy dt = ky. In Section 31.3 we’ll look qualitatively at solutions to differential equations of the form dy dt = f(y). Then in Section 31.4 we will look at solutions to a larger class of differential equations, differential equations of the form dy dt = f(y)g(t).The type of differential equations we look at here are special cases of those we will look at in Section 31.4. Differential Equations of the Form dy dt = f(t) Differential equations of the form dy dt = f(t) can be solved by integration. To solve such an equation is to find a function y whose derivative is f(t),i.e., to find an antiderivative of f(t).If dy dt = f(t),then y = f(t)dt. ◆ EXAMPLE 31.12 The differential equation governing the path of a projectile can be solved by finding antiderivatives. In Example 31.7 we considered an object falling through the air. Ignoring air resistance, we say that the object undergoes a constant downward acceleration of 32 feet per second due to the force of gravity. The object’s initial vertical velocity, v(0),is denoted by the constant v 0 and its initial height, s(0),bys 0 .Solve the differential equations modeling this situation. SOLUTION Let v(t) be the vertical component of the velocity of the object at time t. dv dt =−32 v(t) = −32 dt v(t) =−32t + C 1 We solve for C 1 using v(0) = v 0 to get v 0 =−32(0) + C 1 ,soC 1 =v 0 v(t) =−32t + v 0 . We know that v(t) = ds dt , where s(t) is the height (vertical position) of the object at time t. ds dt =−32t + v 0 s(t) = (−32t + v 0 )dt Then s(t) =−32 t 2 2 + v 0 t + C 2 . Using s(0) = s 0 , we can solve for C 2 to get s 0 =−32 · 0 2 + v 0 · 0 + C 2 ,soC 2 =s 0 . s(t) =−16t 2 + v 0 t + s 0 . ◆ 996 CHAPTER 31 Differential Equations Review: Differential Equations of the Form dy dt = ky In Section 15.2 we looked at the differential equation dy dt = ky, an equation that models any situation in which a quantity grows or decays at a rate proportional to the amount of the quantity itself and conclude the following. The general solution to dy dt = ky is y(t) = Ce kt , where C is an arbitrary constant. The general solution to dy dt = ky, together with an initial condition, enables us to determine a particular solution. Using substitution, we are able to get quite a bit of mileage out of knowing how to find a solution to differential equations of this form. ◆ EXAMPLE 31.13 Solve the following differential equations. (a) dP dt =−2P (b) d dt (700 − F)=−0.01(700 − F) (c) dT dt =−k(T − 65) SOLUTION Our strategy is to put each of these equations in the form dy dt = ky. Once in this form we know y(t) = Ce kt . (a) dP dt =−2P.The general solution is P(t)= Ce −2t . (b) d dt (700 − F)=−0.01(700 − F) 700 − F(t)plays the role of the dependent variable, y; the constant of proportionality is −0.01. The general solution is 700 − F(t)= Ce −0.01t . We can write F(t)=700 − Ce 0.01t . (c) dT dt =−k(T − 65). We need to rewrite this to get it into the form dy dt = ky. Let y = T − 65. Then dy dt = dT dt and the original equation becomes dy dt =−ky. y = Ce kt The general solution is T − 65 = Ce −kt ,orT(t)=65 + Ce −kt . ◆ We see that knowing how to solve any differential equation of the form dy dt = ky allows us to solve dT dt = kT − b because we can use substitution to transform it into the form dy dt = ky. Using substitution, we can solve several of the other differential equations that arose from examples in Section 31.1, namely, (a) dC dt = k(C − N), where k and N are constants (Example 31.2); (b) dA dt = 5 − kA, where k is a constant (Example 31.5); and (c) dM dt = 0.05M − 4000 (Example 31.6). We solve by using substitution to alter the form to be dy dt = ky. 31.2 Solutions to Differential Equations: An Introduction 997 (a) dC dt = k(C − N) can be solved using the substitution y = C − N, where N is constant. (b) dA dt = 5 − kA can be solved either by writing dA dt =−kA + 5 =−k A− 5 k and using the substitution y = A − 5 k or simply by using the substitution y = 5 − kA. (c) dM dt = 0.05M − 4000 can be solved by writing dM dt = 0.05 M − 4000 0.05 and using the substitution y = M − 4000 0.05 or by letting y = 0.05M − 4000. We can solve any differential equation of the form dP dt = aP + b either by rewriting it first as dP dt = a P + b a and then using the substitution y = P + b a to get it into the form dy dt = ay or by using the substitution u(t) = aP + b, P = 1 a [u(t) − b]. EXERCISE 31.3 Solve the differential equation dM dt = 0.05M − 4000 with the initial condition M(0) = 10,000. (This is the differential equation from Example 31.6.) Check your answer to make sure it works. The differential equation dP dt = aP + b is a special case of differential equations of the form dy dt = f(y).Differential equations of the form dy dt = f(y)are called autonomous differential equations. The word autonomous means “not controlled by outside forces”; in an autonomous differential equation the rate of change of y is controlled (determined) only by y, not by other outside “forces.” In other words, dy dt can be described in terms of y only, without reference to t or any other variables. In the next section we will study the behavior of solutions to autonomous differential equations (such as dS dt = kS(300 − S)) from a qualitative perspective. PROBLEMS FOR SECTION 31.2 1. Which one of the following is a solution to the differential equation y (t) =−5y? (a) y = e 5t (b) y = t 2 (c) y = e −5t (d) y = sin(5t) 2. Which one of the following is a solution to the differential equation y (t) =−25y? (a) y = e 5t (b) y = t 2 (c) y = e −5t (d) y = sin(5t) 3. (a) Verify that y(t) = Ce kt is a solution to the differential equation dy dt = ky. (b) Verify that y = ke t is not a solution to dy dt = ky. (c) Verify that y = e kt + C is not a solution to dy dt = ky. 4. Which of the following is a solution to dy dx = −y x + y 2 (ln x) 2 ? (a) y = x + 1 (b) y = 1 + 1/x (c) y = ln x x (d) y = ln x 5. Determine which of the following functions are solutions to each of the differential equations below. (A given differential equation may have more than one solution.) Differential Equations: i. dy dt = t ii. dy dt = y iii. dy dt = e t iv. d 2 y dt 2 = 4y 998 CHAPTER 31 Differential Equations Solution choices: (a) y = t 2 2 (b) y = t 2 2 + 5 (c) y = e −2t (d) y = e t + 5 (e) y = 2e t (f) y = e 2t (g) y = 5e 2t (h) y = e 2t + 5 6. Which of the following is a solution to the differential equation y − y − 6y = 0? (a) y = Ce t (b) y = sin 2t (c) y = 5e 3t + e −2t (d) y = e 3t − 2 7. Which of the following is a solution to the differential equation y + 9y = 0? (a) y = e 3t + e −3t (b) y = Ce t − t (c) y = C(t 2 + t) (d) y = sin 3t + 6 (e) y = 5 cos 3t 8. Which of the following is a solution to the differential equation dy dt = y + 1? (a) y = Ce t (b) y = Ce t − t (c) y = C(t 2 + t) (d) y = Ce t − 1 (e) y = Ce −t + 1 9. Is y = xe x 2 + e x 3x a solution to the differential equation x dy dx + (1 − x)y = xe x ? Justify your answer. 10. Solve the following differential equations by using the method of substitution to put them into the form dy dt = ky. (a) dP dt = 0.3(1000 − P) (b) dM dt = 0.4M − 2000 11. Solve the following. (a) dx dt = 6 − 2x. Do this using substitution in two ways. i. Factor out a −2 from the right-hand side and let u = x − 3. Then solve. ii. Let v = 6 − 2x. Express dx dt in terms of v and then convert the equation dx dt = 6 − 2x to an equation in v and dv dt and solve. (b) dx dt = 3x − 7 (c) dy dt = ky + B 12. When a population has unlimited resources and is free from disease and strife, the rate at which the population grows in often modeled as being proportional to the population. Assume that both the bee and the mosquito populations described below behave according to this model. In both scenarios described below you are given enough information to find the proportionality constant k. In one case the information allows you to find k solely using the differential equation, without requiring that you solve it. In the other scenario you must actually solve the differential equation in order to find k. 31.2 Solutions to Differential Equations: An Introduction 999 (a) Let M = M(t) be the mosquito population at time t, t in weeks. At t = 0 there are 1000 mosquitoes. Suppose that when there are 5000 mosquitoes the population is growing at a rate of 250 mosquitoes per week. Write a differential equation reflecting the situation. Include a value for k, the proportionality constant. (b) Let B = B(t) be the bee population at time t , t in weeks. At t = 0 there are 600 bees. When t = 10 there are 800 bees. Write a differential equation reflecting the situation. Include a value for k, the proportionality constant. 13. The population in a certain country grows at a rate proportional to the population at time t, with a proportionality constant of 0.03. Due to political turmoil, people are leaving the country at a constant rate of 6000 people per year. Assume that there is no immigration into the country. Let P = P(t)denote the population at time t. (a) Write a differential equation reflecting the situation. (b) Solve the differential equation for P(t) given the information that at time t = 0 there are 3 million people in the country. In other words, find P(t),the number of people in the country at time t. 14. A population of otters is declining. New otters are born at a rate proportional to the population with constant of proportionality 0.04, but otters die at a rate proportional to the population with constant 0.09. Today, the population is 1000. A group of people wants to try to prevent the otter population from dying out, so they plan to bring in otters from elsewhere at a rate of 40 otters per year. We’ll model the situation with continuous functions. Let P(t)be the population of the otters t years after today. (a) Write a differential equation whose solution is P(t). (b) Solve this differential equation. Your answer should include no unknown con- stants. (c) According to this model, will the attempt to save the otter population work? Explain your answer. If it won’t work, at what rate must otters be brought in to ensure the population’s survival? If it will work, for how many years must the importation of otters continue? 15. (a) Suppose a hot object is placed in a room whose temperature is kept fixed at F degrees. Let T(t)be the temperature of the object. Newton’s law of cooling says that the hot object will cool at a rate proportional to the difference in temperature between the object and its environment. Write a differential equation reflecting this statement and involving T . Explain why this differential equation is a special case of the differential equation in Example 31.3 part (a). (b) What is the sign of the constant of proportionality in the equation you wrote above? Explain. (c) Suppose that we are interested in the temperature of a cold cup of lemonade as it warms up to room temperature. Let L(t) represent the temperature of the lemonade at time t and assume that it sits in a room that is kept at 65 degrees. At time t = 0 the lemonade is at 40 degrees. Fifteen minutes later it has warmed to 50 degrees. i. Sketch a graph of L(t) using your intuition and the information given. ii. Is L(t) increasing at an increasing rate or a decreasing rate? iii. Write a differential equation reflecting the situation. Indicate the sign of the proportionality constant. 1000 CHAPTER 31 Differential Equations iv. Find L(t). Your final answer should have no undetermined constants and should be consistent with the answer to part (i). v. How long will it take the lemonade to reach a temperature of 55 degrees? 16. Money in a bank account is earning interest at a nominal rate of 4% per year com- pounded continuously. Withdrawals are made at a rate of $8000 per year. Assume that withdrawals are made continuously. (a) Write a differential equation modeling the situation. (b) Depending on the initial deposit, the amount of money in the account will either in- crease, decrease, or remain constant. Explain this in words; refer to the differential equation. (c) Suppose the money in the account remains constant. What was the initial deposit? For what initial deposits will the amount of money in the account actually continue to grow? (d) Show that M(t) = M 0 e 0.04t − 8000t is not a solution to the differential equation you got in part (a). 17. Suppose a population is changing according to the equation dP dt = kP − E, where E is the rate at which people are emigrating from the country. As established in part (d) of the previous problem, P(t)=P 0 e kt − Et is not a solution to this differential equation. (a) Use substitution to solve dP dt = kP − E. (Your answer ought to agree with that given in part (b).) (b) Verify that P(t)=Ce kt + E k , where C is a constant, is a solution to the differential equation dP dt = kP − E. 18. P(t)= Ce kt + E k , where C is a constant, is the general solution to the differential equation dP dt = kP − E. Below is the slope field for dP dt = 2P − 6. P t 5 4 3 2 1 –1 12345 –1 –2–3 (a) i. Find the particular solution that corresponds to the initial condition P(0)=2. ii. Sketch the solution curve through (0, 2). (b) i. Find the particular solution that corresponds to the initial condition P(0)=3. ii. Sketch the solution curve through (0, 3). (c) i. Find the particular solution that corresponds to the initial condition P(0)=4. ii. Sketch the solution curve through (0, 4). . A is turning into B is proportional to the product of the number of moles of A and the number of moles of B. (a) Let N = N(t) be the number of moles of substance A at time t. Translate the statement. the statement above into mathematical language. (Note: The number of moles of substance B should be expressed in terms of the number of moles of substance A.) (b) Using your answer to part (a), find d 2 N dt 2 solution to part (ii) for any constant C. y = Ce 2t is a solution to part (iii) for any constant C. The Existence and Uniqueness Theorem tells us that we have written general solutions to each of these