AN EQUATION involving the derivative, or differential, of an

Transcription

1 59695_09_ch9_p qxd 9/18/09 10:43 AM Page 611 DIFFERENTIAL EQUATIONS 9 AN EQUATION involving the derivative, or differential, of an unknown function is called a differential equation. In this CO Photo to come chapter, we show how differential equations are used to solve problems involving the growth of an amount of mone earning interest compounded continuousl, the growth of a population of bacteria, the deca of radioactive material, and the rate at which a person learns a new subject, to name just a few. To come How fast is the flu spreading among the arm personnel at Fort MacArthur? In Example 3, page, we will see how a differential equation can be used to determine the number of arm personnel that will contract the flu after a given period of time.

2 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS 9.1 Differential Equations Models Involving Differential Equations We first encountered differential equations in Section 6.1. Recall that a differential equation is an equation that involves an unknown function and its derivative(s). Here are some examples of differential equations: xex dx 2 x2 dx Differential equations appear in practicall ever branch of applied mathematics, and the stu of these equations remains one of the most active areas of research in mathematics. As ou will see in the next few examples, models involving differential equations often arise from the mathematical formulation of practical problems. Unrestricted Growth Models The unrestricted growth model was first discussed in Chapter 5. There we saw that the size of a population at an time t, Q(t ), increases at a rate that is proportional to Q(t) itself. Thus, d a b t 8 0 kq where k is a constant of proportionalit. This is a differential equation involving the unknown function Q and its derivative Q. (1) C A Q(t) = C Ae kt FIGURE 1 Q(t) describes a learning curve. = C C 1 + A Q(t) = C 1 + Ae Ckt = C FIGURE 2 Two solutions of the logistic equation t t Restricted Growth Models In man applications the quantit Q(t) does not exhibit unrestricted growth but approaches some definite upper bound. The learning curves and logistic functions we discussed in Chapter 5 are examples of restricted growth models. Let s derive the mathematical models that lead to these functions. Suppose Q(t) does not exceed some number C, called the carring capacit of the environment. Furthermore, suppose the rate of growth of this quantit is proportional to the difference between its upper bound and its current size. The resulting differential equation is where k is a constant of proportionalit. Observe that if the initial population is small relative to C, then the rate of growth of Q is relativel large. But as Q(t) approaches C, the difference C Q(t) approaches zero, as does the rate of growth of Q. In Section 9.3, ou will see that the solution of the differential Equation (2) is a function that describes a learning curve (Figure 1). Next, let s consider a restricted growth model in which the rate of growth of a quantit Q(t) is jointl proportional to its current size and the difference between its upper bound and its current size; that is, k1c Q2 kq1c Q2 where k is a constant of proportionalit. Observe that when Q(t) is small relative to C, the rate of growth of Q is approximatel proportional to Q. But as Q(t) approaches C, the growth rate slows down to zero. If Q C, then / 0 and the quantit is decreasing with time, with the deca rate slowing down as Q approaches C. We will show later that the solution of the differential Equation (3) is just the logistic function we discussed in Chapter 5. Its graph is shown in Figure 2. (2) (3)

3 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS 613 ED: Cap to match manuscript? S 0 FIGURE 3 The solution to the differential equation (4) describes the response to a stimulus. FIGURE 4 The rate of change of the amount of salt at time t (rate of salt flowing in) (rate of salt flowing out) R S Stimulus Response In the quantitative theor of pscholog, one model that describes the relationship between a stimulus S and the resulting response R is the Weber Fechner law. This law asserts that the rate of change of a reaction R is inversel proportional to the stimulus S. Mathematicall, this law ma be expressed as where k is a constant of proportionalit. Furthermore, suppose that the threshold level, the lowest level of stimulation at which sensation is detected, is S 0. Then we have the condition R 0 when S S 0 ; that is, R(S 0 ) 0. The graph of R versus S is shown in Figure 3. Mixture Problems Our next example is a tpical mixture problem. Suppose a tank initiall contains 10 gallons of pure water. Brine containing 3 pounds of salt per gallon flows into the tank at a rate of 2 gallons per minute, and the well-stirred mixture flows out of the tank at the same rate. How much salt is in the tank at an given time? Let s formulate this problem mathematicall. Suppose A(t) denotes the amount of salt in the tank at an time t. Then the derivative da/, the rate of change of the amount of salt at an time t, must satisf the condition da dr ds k S (Rate of salt flowing in) (Rate of salt flowing out) (Figure 4). But the rate at which salt flows into the tank is given b (2 gal/min)(3 lb/gal) (Rate of flow) (Concentration) or 6 pounds per minute. Since the rate at which the solution leaves the tank is the same as the rate at which the brine is poured into it, that tank contains 10 gallons of the mixture at an time t. Since the salt content at an time t is A pounds, the concentration of the salt in the mixture is (A/10) pounds per gallon. Therefore, the rate at which salt flows out of the tank is given b 12 gal/min2a A 10 lb/gal b or (A/5) pounds per minute. Therefore, we are led to the differential equation da 6 A 5 An additional condition arises from the fact that initiall there is no salt in the solution. This condition ma be expressed mathematicall as A 0 when t 0 or, more concisel, A(0) 0. We will solve each of the differential equations we have introduced here in Section 9.3. Solutions of Differential Equations Suppose we are given a differential equation involving the derivative(s) of a function. Recall that a solution to a differential equation is an function f(x) that satisfies the differential equation. Thus, f(x) is a solution of the differential equation, provided that the replacement of and its derivative(s) b the function f(x) and its corresponding derivatives reduces the given differential equation to an identit for all values of x. (4) (5)

4 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS EXAMPLE 1 Show that the function f(x) e x x 1 is a solution of the differential equation x Solution Let f(x) e x x 1 so that f (x) e x 1 Substituting these equations into the left side of the given differential equation ields ( e x 1) (e x x 1) e x 1 e x x 1 x which is equal to the right side of the given equation for all values of x. Therefore, f(x) e x x 1 is a solution of the given differential equation. In Example 1, we verified that e x x 1 is a solution of the differential equation x. This is b no means the onl solution of this differential equation, as the next example shows. EXAMPLE 2 Show that an function of the form f(x) ce x x 1, where c is a constant, is a solution of the differential equation x Solution Let f(x) ce x x 1 so that f (x) ce x 1 Substituting these equations into the left side of the given differential equation ields ce x 1 ce x x 1 x and we have verified the assertion. It can be shown that ever solution of the differential equation x must have the form ce x x 1, where c is a constant; therefore, this is the general solution of the differential equation x. Figure 5 shows a famil of solutions of this differential equation for selected values of c.

5 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS 615 c = 1 c = 2 c = x 2 FIGURE 5 Some solutions of x c = 0 c = 1 c = 2 Recall that the solution obtained b assigning a specific value to the constant c is called a particular solution of the differential equation. For example, the particular solution e x x 1 of Example 1 is obtained from the general solution b taking c 1. In practice, a particular solution of a differential equation is obtained from the general solution of the differential equation b requiring that the solution and/or its derivative(s) satisf certain conditions at one or more values of x. EXAMPLE 3 Use the results of Example 2 to find the particular solution of the equation x that satisfies the condition (0) 0; that is, f(0) 0, where f denotes the solution. Solution From the results of Example 2, we see that the general solution of the given differential equation is given b Using the given condition, we see that f(x) ce x x 1 f(0) ce c 1 0 or c 1 Therefore, the required particular solution is e x x 1. Explore & Discuss Consider the differential equation /dx F(x, ) and suppose f(x) is a solution of the differential equation. 1. If (a, b) is a point in the domain of F, explain wh F(a, b) gives the slope of f at x a. 2. For the differential equation /dx x/, compute F(x, ) for selected integral values of x and. (For example, tr x 0, 1, 2 and 1, 2, 3.) Verif that if ou draw (continued)

6 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS a lineal element (a tin line segment) having slope F(x, ) through each point (x, ), ou obtain a direction field similar to the one shown in the figure: x 3. The direction field associated with the differential equation hints at the solution curves for the differential equation. Sketch a few solution curves for the differential equation. [You will be asked to verif our answer to part (3) in the next section. 9.1 Self-Check Exercises 1. Consider the differential equation x 2 4x 2 a. Show that x 2 (c/x 2 ) is the general solution of the differential equation. b. Find the particular solution of the differential equation that satisfies (1) The population of a certain species grows at a rate directl proportional to the square root of its size. If the initial population is N 0, find the population at an time t. Formulate but do not solve the problem. Solutions to Self-Check Exercises 9.1 can be found on page. 9.1 Concept Questions 1. Define the following terms in our own words. a. A differential equation. b. The general solution of a differential equation. c. A particular solution of a differential equation. 2. Give a differential equation that describes the following situation. a. The size of a population at an time t, Q(t), increases at a rate that is proportional to Q(t) itself. b. The size of a population at an time t, Q(t), does not exceed some number C, and the rate of growth of Q(t) is proportional to the difference between its upper bound and its current size. 3. a. Does the equation of Question 2(a) describe restricted or unrestricted growth? b. Does the equation of Question 2(b) describe restricted or unrestricted growth? 4. Given a tpical mixture problem where A(t) denotes the amount of salt in the tank at an time t, what condition must the derivative da/, the rate of change of the amount of salt at an time t, satisf?

7 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS Exercises In Exercises 1 12, verif that is a solution of the differential equation. 1. x 2 ; x 3x 2 2. e x ; ce x, c an constant; 2x x 2 4. Ce kx, C an constant; k dx 5. e 2x ; C 1 e x C 2 e 2x ; C 1 e 2x C 2 xe 2x ; C 1 C 2 x 1/3 ; 3x C 1 x 2 3x 0 x C ln x 2 x ; 10. C 1 e x C 2 xe x C 3 x 2 e x ; C Ae kt, A and C constants; k(c ) C 12. A and C constants; k(c ) 1 Ae Ckt, In Exercises 13 18, verif that is a general solution of the differential equation. Then find a particular solution of the differential equation that satisfies the side condition. 13. Cx 2 2x; 2 a 2; (1) 10 x b 14. Ce x2 ; 2x; (0) 0 C 15. a 1 b 0; (1) 1 x ; x 16. Ce 2x 2x 1; 2 4x 0; (0) 3 Ce x 17. a 1 x 1 b e x ; (1) x 1 2 xex ; x 2 e 18. C 1 x 3 C 2 x 2 ; x 2 4x 6 0; (2) 0 and (2) RADIOACTIVE DECAY A radioactive substance decas at a rate directl proportional to the amount present. If the substance is present in the amount of Q 0 g initiall (t 0), find the amount present at an time t. Formulate the problem in terms of a differential equation with a side condition. Do not solve it. 20. SUPPLY AND DEMAND Let S(t) denote the suppl of a certain commodit as a function of time t. Suppose the rate of change of the suppl is proportional to the difference between the demand D(t) and the suppl. Find a differential equation that describes this situation. 21. NET INVESTMENT The management of a compan has decided that the level of investment should not exceed C dollars. Furthermore, management has decided that the rate of net investment (the rate of change of the total capital invested) should be proportional to the difference between C and the total capital invested. Formulate but do not solve the problem in terms of a differential equation. 22. LAMBERT S LAW OF ABSORPTION Lambert s law of absorption states that the percentage of incident light L, absorbed in passing through a thin laer of material x, is proportional to the thickness of the material. If, for a certain material, x 0 in. of the material reduces the light to half its intensit, how much additional material is needed to reduce the intensit to a quarter of its initial value? Formulate but do not solve the problem in terms of a differential equation with a side condition. 23. CONCENTRATION OF A DRUG IN THE BLOODSTREAM The rate at which the concentration of a drug in the bloodstream decreases is proportional to the concentration at an time t. Initiall, the concentration of the drug in the bloodstream is C 0 g/ml. What is the concentration of the drug in the bloodstream at an time t? Formulate but do not solve the problem in terms of a differential equation with a side condition. 24. AMOUNT OF GLUCOSE IN THE BLOODSTREAM Suppose glucose is infused into the bloodstream at a constant rate of C g/min and, at the same time, the glucose is converted and removed from the bloodstream at a rate proportional to the amount of glucose present. Show that the amount of glucose A(t) present in the bloodstream at an time t is governed b the differential equation. where k is a constant. A C ka 25. NEWTON S LAW OF COOLING Newton s law of cooling states that the temperature of a bo drops at a rate that is proportional to the difference between the temperature of the bo and the constant temperature C of the surrounding medium (assume that the temperature of the bo is initiall greater than C). Show that Newton s law of cooling ma be expressed as the differential equation k1 C where 0 denotes the temperature of the bo before immersion in the medium.

8 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS 26. FISK S LAW Suppose a cell of volume V cc is surrounded b a homogeneous chemical solution of concentration C g/cc. Let denote the concentration of the solute inside the cell at an time t and suppose that, initiall, the concentration is 0. Fisk s law, named after the German phsiologist Adolf Fisk ( ), states that the rate of change of the concentration of solute inside the cell at an time t is proportional to the difference between the concentration of the solute outside the cell and the concentration inside the cell and inversel proportional to the volume of the cell. Show that Fisk s law ma be expressed as the differential equation where k is a constant. (Note: The constant of proportionalit k depends on the area and permeabilit of the cell membrane.) 27. ALLOMETRIC LAWS Allometr is the stu of the relative growth of a part of an organism in relation to the growth of an entire organism. Suppose x(t) denotes the weight of an animal s organ at time t and t(t) denotes the size of another organ in the same animal at the same time t. An allometric law states that the relative growth rate of one organ, (dx/)/x, is proportional to the relative growth rate of the other, (/)/. Show that this allometric law ma be stated in terms of the differential equation where k is a constant. k V 1C dx x k GOMPERTZ GROWTH CURVE Suppose a quantit Q(t) does not exceed some number C; that is, Q(t) C for all t. Sup- pose further that the rate of growth of Q(t) is jointl proportional to its current size and the difference between its upper bound and the natural logarithm of its current size. What is the size of the quantit Q(t) at an time t? Show that the mathematical formulation of this problem leads to the differential equation kq1c ln Q2 3Q102 Q 0 4 where Q 0 denotes the size of the quantit present initiall. The graph of Q(t) is called the Gompertz growth curve. This model, like the ones leading to the learning curve and the logistic curve, describes restricted growth. In Exercises 29 34, determine whether the statement is true or false. If it is true, explain wh it is true. If it is false, give an example to show wh it is false. 4 e3x The function f(x) x 2 2x is a solution of the differential equation x 3x 2 4x. x The function f(x) ce x is a solution of the differential equation e 3x. ce x3 31. The function f(x) 2 is a solution of the differential equation 3x 2 x The function f(x) 1 cx 2 is a solution of the differential equation x If f(x) is a solution of a first-order differential equation, then Cf(x) is also a solution. 34. If f(x) is a solution of a first-order differential equation, then f(x) C is also a solution. 9.1 Solutions to Self-Check Exercises 1. a. We compute 2x 2c x 3 and the required particular solution is x 2 3 x 2 Substituting this into the left side of the given differential equation gives x a 2x 2c x b 2 a 3 x2 c x b 2 2x2 2c x 2 2x2 2c 4x2 2 x which equals the expression on the right side of the differential equation, and this verifies the assertion. b. Using the given condition, we have 2. Let N denote the size of the population at an time t. Then the required differential equation is dn kn1/2 and the initial condition is N(0) N c or c 3 2 1

9 59695_09_ch9_p qxd 9/18/09 10:43 AM Page SEPARATION OF VARIABLES Separation of Variables The Method of Separation of Variables Differential equations are classified according to their basic form. A compelling reason for this categorization is that different methods are used to solve different tpes of equations. The order of a differential equation is the order of the highest derivative of the unknown function appearing in the equation. A differential equation ma be classified b the order of its derivative. For example, the differential equations xe x and 2 x 2 are first-order equations, whereas the differential equation d a b t 8 0 is a second-order equation. For the remainder of this chapter, we restrict our stu to first-order differential equations. In this section we describe a method for solving an important class of first-order differential equations that can be written in the form where f(x) is a function of x onl and t() is a function of onl. Such differential equations are said to be separable because the variables can be separated. Equations (1) through (5) are first-order separable differential equations. As another example, the equation f 1x2t12 dx kq1c Q2 has the form / f(t)t(q), where f(t) k and t(q) Q(C Q), and so it is separable. On the other hand, the differential equation dx x2 2 is not separable. Separable first-order equations can be solved using the method of separation of variables. Method of Separation of Variables Suppose we are given a first-order separable differential equation in the form Step 1 Step 2 Write Equation (6) in the form f 1x2t12 dx f 1x2 dx t12 When written in this form, the variables in (7) are said to be separated. Integrate each side of Equation (7) with respect to the appropriate variable. (6) (7)

10 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS We will justif this method at the end of this section. Solving Separable Differential Equations EXAMPLE 1 Find the general solution of the first-order differential equation Solution x x 2 1 Step 1 Step 2 Observe that the given differential equation has the form where f(x) x/(x 2 1) and t(), and is therefore separable. Separating the variables, we obtain Integrating each side of the last equation with respect to the appropriate variable, we have or dx a x x 2 b f 1x2t12 1 a x x 2 1 b dx x x 2 1 dx ln 0 0 C ln1x2 12 C 2 ln ln1x2 12 C 2 C 1 where C 1 and C 2 are arbitrar constants of integration. Letting C denote the constant such that C 2 C 1 ln 0 C 0, we have ln ln1x2 12 ln 0 C 0 so the general solution is ln2x 2 1 ln 0 C 0 ln 0 C 2x ln A ln B ln AB C 2x 2 1 Exploring with TECHNOLOGY Refer to Example 1, where it was shown that the general solution of the given differential equation is C 2x 2 1. Use a graphing utilit to plot the graphs of the members of this famil of solutions corresponding to C 3, 2, 1, 0, 1, 2, and 3. Use the standard viewing window.

11 59695_09_ch9_p qxd 9/18/09 10:43 AM Page SEPARATION OF VARIABLES 621 EXAMPLE 2 Find the particular solution of the differential equation that satisfies the condition (0) 1. Solution Step 1 e x ( 2 1) 0 Writing the given differential equation in the form e x and separating the variables, we obtain dx 0 or dx ex 2 1 e x dx Step 2 Integrating each side of this equation with respect to the appropriate variable, we have 2 1 e x dx a 1 b ex dx ln 0 0 e x C 1 Using the condition (0) 1, we have Therefore, the required solution is 2 ln 2 2e x C 1C 2C ln 1 2 C or C 3 2 ln 2 2e x 3 Example 2 is an initial value problem. In general, an initial value problem consists of a differential equation with one or more side conditions specified at a point. Also observe that the solution of Example 2 appeared as an implicit equation involving x and. This often happens when we solve separable differential equations. EXAMPLE 3 Find an equation describing f given that (1) the slope of the tangent line to the graph of f at an point P(x, ) is given b the expression x/(2) and (2) the graph of f passes through the point P(1, 2). Solution The slope of the tangent line to the graph of f at an point P(x, ) is given b the derivative dx x 2 which is a separable first-order differential equation. Separating the variables, we obtain 2 xdx

12 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS which, upon integration, ields x2 C x or x C (C 2C 1 ) 3 FIGURE 6 The graph of x where C is an arbitrar constant. To evaluate C, we use the second condition, which implies that when x 1, 2. This gives 1 2 2(2 2 ) C or C 9 Hence, the required equation is x The graph of the equation f appears in Figure 6. Explore & Discuss Refer to the Explore & Discuss problem on page. Use the method of separation of variables to solve the differential equation /dx x/ and thus verif that our solution to part 3 in the problem is indeed correct. Justification of the Method of Separation of Variables To justif the method of separation of variables, let s consider the separable Equation (6) in its general form: f 1x2t12 dx If g() 0, we ma rewrite the equation in the form 1 f 1x2 0 t12 dx Now, suppose that G is an antiderivative of 1/g and F is an antiderivative of f. Using the chain rule, we see that d 1 3G12 F1x24 G 12 F 1x2 f 1x2 dx dx t12 dx Therefore, and so d 3G12 F1x24 0 dx G12 F1x2 C C, a constant But the last equation is equivalent to G12 F1x2 C or t12 f 1x2 dx which is precisel the result of step 2 in the method of separation of variables.

13 59695_09_ch9_p qxd 9/18/09 10:43 AM Page SEPARATION OF VARIABLES Self-Check Exercise Find the solution of the differential equation 2x 2 2x 2 that satisfies the condition (0) 0. The solution to Self-Check Exercise 9.2 can be found on page. 9.2 Concept Questions 1. a. What is the order of a differential equation? Give an example. b. What is a separable equation? Give an example of a differential equation that is separable and one that is not separable. 2. Explain how ou would solve a separable equation. 3. What is an initial value problem? 9.2 Exercises In Exercises 1 16, find the general solution of the firstorder differential equation b separating variables. 1. x ex ( 1) 7. x (3 4) x x2 x 2 x xex 2 x B x 21 x 2 1x ln x 16. x x In Exercises 17 28, find the solution of the initial value problem x ; xe ; (0) ; (0) ; x 21. 3x 2x; (0) xe x2 ; x ; x 2 1/2 ; (1) 1 x xe x ; (1) xe ; (0) 1 x x 2 e ; (0) ; x Find an equation defining a function f given that (1) the slope of the tangent line to the graph of f at an point P(x, ) is given b the expression /dx (3x 2 )/(2) and (2) the graph of f passes through the point (1, 3). 30. Find a function f given that (1) the slope of the tangent line to the graph of f at an point P(x, ) is given b the expression /dx 3x and (2) the graph of f passes through the point (0, 2). 31. EXPONENTIAL DECAY Use separation of variables to solve the differential equation where k and Q 0 are positive constants, describing exponential deca. 32. FISK S LAW Refer to Exercise 26, Section 9.1. Use separation of variables to solve the differential equation where k, V, C, and 0 are constants with C 0. Find and interpret our result. lim ts kq 3Q102 Q 0 4 k V 1C CONCENTRATION OF GLUCOSE IN THE BLOODSTREAM Refer to Exercise 24, Section 9.1. Use separation of variables to solve the differential equation A C ka, where C and k are positive constants. Hint: Rewrite the given differential equation in the form da k a C k A b

14 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS 34. ALLOMETRIC LAWS Refer to Exercise 27, Section 9.1. Use separation of variables to solve the differential equation where k is a constant. 1 dx x k SUPPLY AND DEMAND Assume that the rate of change of the suppl of a commodit is proportional to the difference between the demand and the suppl so that ds k1d S2 where k is a constant of proportionalit. Suppose that D is constant and S(0) S 0. Find a formula for S(t). 36. SUPPLY AND DEMAND Assume that the rate of change of the unit price of a commodit is proportional to the difference between the demand and the suppl so that dp k1d S2 In Exercises 37 42, determine whether the statement is true or false. If it is true, explain wh it is true. If it is false, give an example to show wh it is false. 37. The differential equation x 2x 2 is separable. 38. The differential equation x 2 2 is separable. 39. If the differentiable function M(x, ) dx N(x, ) 0 can be written so that M(x, ) f(x)t() and N(x, ) F(x)G() for functions f, t, F, and G, then it is separable. 40. The differential equation (x 2 2) dx (2x 4x) 0 is separable. 41. The differential equation dx ( x 2 ) 0is separable. 42. The differential equation is separable. dx f 1x2t12 F1x2 G12 where k is a constant of proportionalit. Suppose that D 50 2p, S 5 3p, and p(0) 4. Find a formula for p(t). 9.2 Solutions to Self-Check Exercises Writing the differential equation in the form dx 2x and separating variables, we obtain 1 2x2 dx Integrating each side of the last equation with respect to the appropriate variable, we have 1 2x2 dx or ln x3 C Using the initial condition (0) 0, we have ln 1 C or C 0 Therefore, ln x3 1 e 12/32x3 e 12/32x3 1

15 59695_09_ch9_p qxd 9/18/09 10:43 AM Page APPLICATIONS OF SEPARABLE DIFFERENTIAL EQUATIONS Applications of Separable Differential Equations In this section, we look at some applications of first-order separable differential equations. We begin b reexamining some of the applications discussed in Section 9.1. Unrestricted Growth Models The differential equation describing an unrestricted growth model is given b where Q(t) represents the size of a certain population at an time t and k is a positive constant. Separating the variables in this differential equation, we have which, upon integration, ields Q kq k Q k ln 0 Q 0 kt C 1 Q e kt C 1 Cekt where C e C 1 is an arbitrar positive constant. Thus, we ma write the solution as Q(t) Ce kt Q 0 Q(t) = Q 0 e kt FIGURE 7 An unrestricted growth model t Observe that if the quantit present initiall is denoted b Q 0, then Q(0) Q 0. Appling this condition ields the equation Ce 0 Q 0 or C Q 0 Therefore, the model for unrestricted exponential growth with initial population Q 0 is given b (Figure 7). Q(t) Q 0 e kt (8) APPLIED EXAMPLE 1 Growth of Bacteria Under ideal laborator conditions, the rate of growth of bacteria in a culture is proportional to the size of the culture at an time t. Suppose that 10,000 bacteria are present initiall in a culture and 60,000 are present 2 hours later. How man bacteria will there be in the culture at the end of 4 hours? Solution Let Q(t) denote the number of bacteria present in the culture at time t. Then kq

16 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS where k is a constant of proportionalit. Solving this separable first-order differential equation, we obtain Q(t) Q 0 e kt Equation (8) where Q 0 denotes the initial bacteria population. Since Q 0 10,000, we have Q(t) 10,000e kt Next, the condition that 60,000 bacteria are present 2 hours later translates into Q(2) 60,000, or 60,000 10,000e 2k e 2k 6 e k 6 1/2 Thus, the number of bacteria present at an time t is given b Q1t2 10,000e kt 10,0001e k 2 t 110,00026 t/2 In particular, the number of bacteria present in the culture at the end of 4 hours is given b Q142 10, / ,000 Restricted Growth Models From Section 9.1, we see that a differential equation describing a restricted growth model is given b k1c Q2 (9) where both k and C are positive constants. To solve this separable first-order differential equation, we first separate the variables, obtaining k C Q Integrating each side with respect to the appropriate variable ields = C C Q k C A Q(t) = C Ae kt ln 0 C Q 0 kt d ln 0 C Q 0 kt d C Q e kt d e kt e d d, an arbitrar constant (10) t Q1t2 C Ae kt FIGURE 8 A restricted exponential growth model where we have denoted the constant e d b A. This is the equation of the learning curve (Figure 8) studied in Chapter 5.

17 59695_09_ch9_p qxd 9/18/09 10:43 AM Page APPLICATIONS OF SEPARABLE DIFFERENTIAL EQUATIONS 627 APPLIED EXAMPLE 2 Yield of a Wheat Field In an experiment conducted b researchers of the Agriculture Department of a midwestern universit, it was found that the maximum ield of wheat in the universit s experimental field station was 150 bushels per acre. Furthermore, the researchers discovered that the rate at which the ield of wheat increased was governed b the differential equation dx k1150 Q2 where Q(x) denotes the ield in bushels per acre and x is the amount in pounds of an experimental fertilizer used per acre of land. Data obtained in the experiment indicated that 10 pounds of fertilizer per acre of land would result in a ield of 80 bushels of wheat per acre, whereas 20 pounds of fertilizer per acre of land would result in a ield of 120 bushels of wheat per acre. Determine the ield if 30 pounds of fertilizer were used per acre. Solution The given differential equation has the same form as Equation (9) with C 150. Solving it directl or using the result obtained in the solution of Equation (9), we see that the ield per acre is given b Q(x) 150 Ae kx The first of the given conditions implies that Q(10) 80; that is, 150 Ae 10k 80 or A 70e 10k. Therefore, Q1x e 10k e kx e k1x 102 The second of the given conditions implies that Q(20) 120, or e k e 10k 30 e 10k 3 7 Taking the logarithm of each side of the equation, we find ln e 10k ln a 3 7 b 10k ln 3 ln k = 150 Therefore, Q(x) e 0.085(x 10) 100 In particular, when x 30, we have 50 Q(x) = e 0.085(x 10) FIGURE 9 Q(x) is a function relating crop ield to the amount of fertilizer used. x Q e e So the ield would be 137 bushels per acre if 30 pounds of fertilizer were used per acre. The graph of Q(x) is shown in Figure 9. Note that it was not necessar to find the specific value of k in order to solve this problem. (Wh?)

18 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS Next, let s consider a differential equation describing another tpe of restricted growth: where k and C are positive constants. Separating variables leads to Integrating each side of this equation with respect to the appropriate variable, we have As it stands, the integrand on the left side of this equation is not in a form that can be easil integrated. However, observe that as ou ma verif b adding the terms between the brackets on the right-hand side. Making use of this identit, we have and or ln 0 Q 0 ln 0 C Q 0 Ckt b b, an arbitrar constant Q ln ` ` Ckt b C Q Q (B e b ) C Q eckt b e b e Ckt Be Ckt 11 Be Ckt 2Q CBe Ckt Q1t2 kq1c Q2 k Q1C Q2 1 Q1C Q2 k 1 Q1C Q2 1 C c 1 Q 1 C Q d 1 C c 1 Q 1 C Q d k Q C Q Ck Q CBe Ckt QBe Ckt CBeCkt Q 1 Be Ckt C 1 Ae Ckt a A 1 B b (see Figure 2, page ). In its final form, this function is equivalent to the logistic function encountered in Chapter 5. APPLIED EXAMPLE 3 Spread of a Flu Epidemic During a flu epidemic, 5% of the 5000 arm personnel stationed at Fort MacArthur had contracted influenza at time t 0. Furthermore, the rate at which the were contracting influenza was jointl proportional to the number of personnel who had alrea contracted the disease and the noninfected population. If 20% of the per- (11)

19 59695_09_ch9_p qxd 9/18/09 10:43 AM Page APPLICATIONS OF SEPARABLE DIFFERENTIAL EQUATIONS 629 sonnel had contracted the flu b the 10th da, find the number of personnel who had contracted the flu b the 13th da. Solution Let Q(t) denote the number of arm personnel who had contracted the flu after t das. Then We ma solve this separable differential equation directl, or we ma use the result for the more general problem obtained earlier. Opting for the latter, we use Equation (11) with C 5000 to obtain The condition that 5% of the population had contracted influenza at time t 0 implies that from which we see that A 19. Therefore, Next, the condition that 20% of the population had contracted influenza b the tenth da implies that or kq15000 Q Q1t2 1 Ae 5000kt Q A 5000 Q1t2 1 19e 5000kt 5000 Q ,000k 1 19e 1 19e 50,000k 5 and Therefore, In particular, the number of arm personnel who had contracted the flu b the 13th da is given b Q1132 e 50,000k ,000k ln 4 ln 19 k 1 1ln 4 ln , Q1t e 0.156t e e or approximatel 29%. The graph of Q(t) is shown in Figure 10.

20 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS 5000 = Q(t) = e 0.156t 1000 FIGURE 10 An epidemic model t Explore & Discuss Consider the model for restricted growth described b the differential equation / kq(c Q) with the solution given in Equation (11). 1. Show that the rate of growth of Q is greatest at t (ln A)/kC. Hint: Use both the differential equation and Equation (11). 2. Refer to Example 3. At what time is the number of influenza cases increasing at the greatest rate? APPLIED EXAMPLE 4 Weber Fechner Law Derive the Weber Fechner law describing the relationship between a stimulus S and the resulting response R b solving differential equation (4) subject to the condition R 0 when S S 0, where S 0 is the threshold level. Solution The differential equation under consideration is and is separable. Separating the variables, we have which, upon integration, ields dr ds k S ds dr k S dr k ds S R k ln S C where C is an arbitrar constant. Using the condition R 0 when S S 0 gives 0 k ln S 0 C C k ln S 0 R = k ln S0 S Substituting this value of C in the expression for R leads to S 0 S R k ln S k ln S 0 k ln S S 0 FIGURE 11 The Weber Fechner law the required relationship between R and S. The graph of R is shown in Figure 11.

21 59695_09_ch9_p qxd 9/18/09 10:43 AM Page APPLICATIONS OF SEPARABLE DIFFERENTIAL EQUATIONS 631 APPLIED EXAMPLE 5 A Mixture Problem A tank initiall contains 10 gallons of pure water. Brine containing 3 pounds of salt per gallon flows into the tank at a rate of 2 gallons per minute, and the well-stirred mixture flows out of the tank at the same rate. How much salt is present at the end of 10 minutes? How much salt is present in the long run? Solution The problem was formulated mathematicall on page, and we were led to the differential equation subject to the condition A(0) 0. The differential equation is separable. Separating the variables and integrating, we obtain da 6 A 5 da 6 A 30 A 5 5 da 30 A 1 5 ln 0 30 A t b b, a constant A(t) = 30(1 e t/5 ) FIGURE 12 The solution of the differential equation da 6 A 5 = 30 t The condition A(0) 0 implies that giving C 30, and so 0 30 C A(t) 30(1 e t/5 ) The amount of salt present after 10 minutes is given b A(10) 30(1 e 2 ) or pounds. The amount of salt present in the long run is given b or 30 pounds (Figure 12). 1 ln 0 30 A 0 5 t b 30 A e b e t/5 lim ts A 30 Ce t/5 1C e b 2 A1t2 lim 3011 e t/ ts 9.3 Self-Check Exercise Newton s law of cooling states that the temperature of an object drops at a rate that is proportional to the difference in the temperature between the object and that of the surrounding medium. Suppose that an apple pie is taken out of the oven at a temperature of 200 F and placed on the counter in a room where the temperature is 70 F. If the temperature of the apple pie is 150 F after 5 min, find its temperature (t) as a function of time t. The solution to Self-Check Exercise 9.3 can be found on page.

22 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS 9.3 Concept Questions 1. Consider the differential equation kq 1k 02 describing unrestricted growth. a. What does the equation sa about the rate of growth of Q with respect to t? What happens to the rate of growth as t approaches infinit? b. Reconcile our answer to part (a) with the solution Q(t) Q 0 e kt of the differential equation. dr 2. Consider the differential equation ds k 1k 02 S describing the relationship between a stimulus S and the resulting response R (the Weber Fechner law). a. What can ou sa about the rate of change of R with respect to S? What happens to the rate of change as S approaches infinit? b. Reconcile our answers to part (a) with the solution R k ln S of the differential equation. S Exercises 1. CHEMICAL DECOMPOSITION The rate of decomposition of a certain chemical substance is directl proportional to the amount present at an time t. If 0 g of the chemical are present at time t 0, find an expression for the amount present at an time t. 2. GROWTH OF BACTERIA Under ideal laborator conditions, the rate of growth of bacteria in a culture is proportional to the size of the culture at an time t. Suppose that 2000 bacteria are present initiall in the culture and 5000 are present 1 hr later. How man bacteria will be in the culture at the end of 2 hr? 3. WORLD POPULATION GROWTH The world population at the beginning of 1980 was 4.5 billion. Assuming that the population continues to grow at its present rate of approximatel 2%/ear, find a function Q(t) that expresses the world population (in billions) as a function of time t (in ears). What will be the world population at the beginning of 2012? 4. POPULATION GROWTH The population of a certain communit is increasing at a rate directl proportional to the population at an time t. In the last 3 r, the population has doubled. How long will it take for the population to triple? 5. LAMBERT S LAW OF ABSORPTION According to Lambert s law of absorption, the percentage of incident light L, absorbed in passing through a thin laer of material x, is proportional to the 1 thickness of the material. If 2 in. of a certain material reduces the light to half of its intensit, how much additional material is needed to reduce the intensit to one-fourth of its initial value? 6. SAVINGS ACCOUNTS An amount of mone deposited in a savings account grows at a rate proportional to the amount present. (It can be shown that an amount of mone grows in this manner if it earns interest compounded continuousl.) Suppose $10,000 is deposited in a fixed account earning interest at the rate of 10%/ear compounded continuousl. a. What is the accumulated amount after 5 r? b. How long does it take for the original deposit to double in value? 7. CHEMICAL REACTIONS In a certain chemical reaction, a substance is converted into another substance at a rate proportional to the square of the amount of the first substance present at an time t. Initiall (t 0) 50 g of the first substance was present; 1 hr later, onl 10 g of it remained. Find an expression that gives the amount of the first substance present at an time t. What is the amount present after 2 hr? 8. NEWTON S LAW OF COOLING Newton s law of cooling states that the rate at which the temperature of an object changes is directl proportional to the difference in temperature between the object and that of the surrounding medium. A horseshoe heated to a temperature of 100 C is immersed in a large tank of water at a (constant) temperature of 30 C at time t 0. Three minutes later the temperature of the horseshoe is reduced to 70 C. Derive an expression that gives the temperature of the horseshoe at an time t. What is the temperature of the horseshoe 5 min after it has been immersed in the water? 9. NEWTON S LAW OF COOLING Newton s law of cooling states that the rate at which the temperature of an object changes is directl proportional to the difference in temperature between the object and that of the surrounding medium. A cup of coffee is prepared with boiling water (212 F) and left to cool on the counter in a room where the temperature is 72 F. If the temperature of the coffee is 140 F after 2 min, determine when the coffee will be cool enough to drink (sa, 110 F).

23 59695_09_ch9_p qxd 9/18/09 10:43 AM Page APPLICATIONS OF SEPARABLE DIFFERENTIAL EQUATIONS LEARNING CURVES The American Stenographic Institute finds that the average student taking Elementar Shorthand will progress at a rate given b k180 Q2 in a 20-wk course, where Q(t) measures the number of words of dictation a student can take per minute after t wk in the course. If the average student can take 50 words of dictation per minute after 10 wk in the course, how man words per minute can the average student take after completing the course? 11. TRAINING PERSONNEL The personnel manager of Gibraltar Insurance Compan estimates that the number of insurance claims an experienced clerk can process in a da is 40. Furthermore, the rate at which a clerk can process insurance claims during the tth wk of training is proportional to the difference between the maximum number possible (40) and the number he or she can process in the tth wk. If the number of claims the average trainee can process after 2 wk on the job is 10/da, determine how man claims the average trainee can process after 6 wk on the job. 12. EFFECT OF IMMIGRATION ON POPULATION GROWTH Suppose a countr s population at an time t grows in accordance with the rule dp kp I where P denotes the population at an time t, k is a positive constant reflecting the natural growth rate of the population, and I is a constant giving the (constant) rate of immigration into the countr. If the total population of the countr at time t 0 is P 0, find an expression for the population at an time t. 13. EFFECT OF IMMIGRATION ON POPULATION GROWTH Refer to Exercise 12. The population of the United States in the ear 1980 (t 0) was million. Suppose the natural growth rate is 0.8% annuall (k 0.008) and net immigration is allowed at the rate of.5 million people/ear (I 0.5). What will be the U.S. population in 2012? 14. SINKING FUNDS The proprietor of Carson Hardware Store has decided to set up a sinking fund for the purpose of purchasing a computer server 2 r from now. It is expected that the purchase will involve a sum of $30,000. The fund grows at the rate of da ra P where A denotes the size of the fund at an time t, r is the annual interest rate earned b the fund compounded continuousl, and P is the amount (in dollars) paid into the fund b the proprietor per ear (assume this is done on a frequent basis in small deposits over the ear so that it is essentiall continuous). If the fund earns 10% interest per ear compounded continuousl, determine the size of the earl investment the proprietor should pa into the fund. 15. SPREAD OF A RUMOR The rate at which a rumor spreads through an Alpine village of 400 residents is jointl proportional to the number of residents who have heard it and the number who have not. Initiall, 10 residents heard the rumor, but 2 das later this number increased to 80. Find the number of people who will have heard the rumor after 1 wk. 16. GROWTH OF A FRUIT FLY COLONY A biologist has determined that the maximum number of fruit flies that can be sustained in a carefull controlled environment (with a limited suppl of space and food) is 400. Suppose that the rate at which the population of the colon increases obes the rule kq1c Q2 where C is the carring capacit (400) and Q denotes the number of fruit flies in the colon at an time t. If the initial population of fruit flies in the experiment is 10 and it grows to 45 after 10 das, determine the population of the colon of fruit flies on the 20th da. 17. DISCHARGING WATER FROM A TANK A container that has a constant cross section A is filled with water to height H. The water is discharged through an opening of cross section B at the base of the container. B using Torricelli s Law, it can be shown that the height of the water at time t satisfies the initialvalue problem dh B A 22th a. Find an expression for h. b. Find the time T it takes for the tank to empt. c. Find T if A 4 (ft 2 ), B 1 (in. 2 ), H 16 (ft), and t 32 (ft/sec 2 ). B ft 2 A ft GOMPERTZ GROWTH CURVES Refer to Exercise 28, Section 9.1. Consider the differential equation kq1c ln Q2 h102 H H ft with the side condition Q(0) Q 0. The solution Q(t) describes restricted growth and has a graph known as the Gompertz curve. Using separation of variables, solve this differential equation.

24 59695_09_ch9_p qxd 9/18/09 10:43 AM Page DIFFERENTIAL EQUATIONS 19. GOMPERTZ GROWTH CURVES Consider the Gompertz differential equation where c is a positive constant and L is the carring capacit of the environment. a. Solve the differential equation. b. Find lim P(t). ts c. Show that P(t) is increasing most rapidl when P L/e. d. Show that P(t) is increasing most rapidl when 20. CONCENTRATION OF A DRUG IN THE BLOODSTREAM Suppose that the rate at which the concentration of a drug in the bloodstream decreases is proportional to the concentration at time t. Initiall, there is no drug in the bloodstream. At time t 0 a drug having a concentration of C 0 g/ml is introduced into the bloodstream. a. What is the concentration of drug in the bloodstream at the end of T hr? b. If at time T another dosage having the concentration of C 0 g/ml is infused into the bloodstream, what is the concentration of the drug at the end of 2T hr? c. If the process were continued, what would the concentration of the drug be at the end of NT hr? d. Find the concentration of the drug in the bloodstream in the long run. Hint: Evaluate lim (NT), where (NT) denotes the concentration NS of the drug at the end of NT hr. 21. SPREAD OF DISEASE A simple mathematical model in epidemiolog for the spread of a disease assumes that the rate at which the disease spreads is jointl proportional to the number of infected people and the number of uninfected people. Suppose that there are a total of N people in the population, of whom N 0 are infected initiall. Show that the number of infected people after t weeks, x(t), is given b x1t2 dp cp ln a L P b t where k is a positive constant. ln ln a L P 0 b 22. SPREAD OF DISEASE Refer to Exercise 21. Suppose there are 8000 students in a college and 400 students had contracted the flu at the beginning of the week. a. If 1200 had contracted the flu at the end of the week, how man will have contracted the flu at the end of 2, 3, and 4 wk? b. How long does it take for 80% of the student population to become infected? c. Plot the graph of the function x(t). c N 1 a N N 0 b e knt N VON BERTALANFFY GROWTH MODEL The von Bertalanff growth model is used to predict the length of commercial fish. The model is described b the differential equation dx k1l x2 where x(t) is the length of the fish at time, t, k is a positive constant called the von Bertalanff growth rate, and L is the maximum length of the fish. a. Find x(t) given that the length of the fish at t 0 is x 0. b. At the time the larvae hatch, the North Sea haddock are about 0.4 cm long, and the average haddock grows to a length of 10 cm after 1 r. Find an expression for the length of the North Sea haddock at time t. c. Plot the graph of x. Take L 100 (cm). d. On average, the haddock that are caught toda are between 40 cm and 60 cm long. What are the ages of the haddock that are caught? 24. CHEMICAL REACTION RATES Two chemical solutions, one containing N molecules of chemical A and another containing M molecules of chemical B, are mixed together at time t 0. The molecules from the two chemicals combine to form another chemical solution containing (AB) molecules. The rate at which the AB molecules are formed, /, is called the reaction rate and is jointl proportional to (N ) and (M ). Thus, k1n 21M 2 where k is a constant. (We assume the temperature of the chemical mixture remains constant during the interaction.) Solve this differential equation with the side condition (0) 0 assuming that N 0 and M 0. Hint: Use the identit 1 1N 21M 2 1 M N a 1 N 1 M b 25. MIXTURE PROBLEMS A tank initiall contains 20 gal of pure water. Brine containing 2 lb of salt per gallon flows into the tank at a rate of 3 gal/min, and the well-stirred mixture flows out of the tank at the same rate. How much salt is present in the tank at an time t? How much salt is present at the end of 20 min? How much salt is present in the long run? 26. MIXTURE PROBLEMS A tank initiall contains 50 gal of brine, in which 10 lb of salt is dissolved. Brine containing 2 lb of dissolved salt per gallon flows into the tank at the rate of 2 gal/min, and the well-stirred mixture flows out of the tank at the same rate. How much salt is present in the tank at the end of 10 min?