FIRSTORDER DIFFERENTIAL EQUATIONS


 Ruth Fields
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1 Chapter 16 FIRSTORDER DIFFERENTIAL EQUATIONS OVERVIEW In Section 4.7 we introduced differential euations of the form >d = ƒ(), where ƒ is given and is an unknown function of. When ƒ is continuous over some interval, we found the general solution () b integration, = 1 ƒ() d. In Section 7.5 we solved separable differential euations. Such euations arise when investigating eponential growth or deca, for eample. In this chapter we stu some other tpes of firstorder differential euations. The involve onl first derivatives of the unknown function Solutions, Slope Fields, and Picard s Theorem We begin this section b defining general differential euations involving first derivatives. We then look at slope fields, which give a geometric picture of the solutions to such euations. Finall we present Picard s Theorem, which gives conditions under which firstorder differential euations have eactl one solution. General FirstOrder Differential Euations and Solutions A firstorder differential euation is an euation d in which ƒ(, ) is a function of two variables defined on a region in the plane. The euation is of first order because it involves onl the first derivative > d (and not higherorder derivatives). We point out that the euations =ƒs, d and = ƒs, d are euivalent to Euation (1) and all three forms will be used interchangeabl in the tet. A solution of Euation (1) is a differentiable function = sd defined on an interval I of values (perhaps infinite) such that d sd = ƒs, sdd d d = ƒs, d d on that interval. That is, when () and its derivative sd are substituted into Euation (1), the resulting euation is true for all over the interval I. The general solution to a firstorder differential euation is a solution that contains all possible solutions. The general (1) 161
2 162 Chapter 16: FirstOrder Differential Euations solution alwas contains an arbitrar constant, but having this propert doesn t mean a solution is the general solution. That is, a solution ma contain an arbitrar constant without being the general solution. Establishing that a solution is the general solution ma reuire deeper results from the theor of differential euations and is best studied in a more advanced course. EXAMPLE 1 Show that ever member of the famil of functions = C + 2 is a solution of the firstorder differential euation d = 1 s2  d on the interval s0, d, where C is an constant. Solution Differentiating = C> + 2 gives d = C d d a1 b + 0 =C 2. Thus we need onl verif that for all H s0, d,  C 2 = 1 c2  ac + 2b d. This last euation follows immediatel b epanding the epression on the righthand side: 1 c2  a C + 2b d = 1 a C b =C 2. Therefore, for ever value of C, the function = C> + 2 is a solution of the differential euation. As was the case in finding antiderivatives, we often need a particular rather than the general solution to a firstorder differential euation =ƒs, d. The particular solution satisfing the initial condition s 0 d = 0 is the solution = sd whose value is 0 when = 0. Thus the graph of the particular solution passes through the point s 0, 0 d in the plane. A firstorder initial value problem is a differential euation =ƒs, d whose solution must satisf an initial condition s 0 d = 0. EXAMPLE 2 Show that the function is a solution to the firstorder initial value problem Solution The euation = s + 1d e d = , s0d = 2 3. d =  is a firstorder differential euation with ƒs, d = .
3 16.1 Solutions, Slope Fields, and Picard s Theorem , FIGURE ( 1) 1 e 3 Graph of the solution = s + 1d  1 to the differential 3 e euation >d = , with initial condition s0d = 2 (Eample 2). 3 On the left side of the euation: On the right side of the euation: The function satisfies the initial condition because The graph of the function is shown in Figure Slope Fields: Viewing Solution Curves d = d d a e b = e.  = s + 1d e  = e. s0d = cs + 1d e d = 11 = 0 3 = 2 3. Each time we specif an initial condition s 0 d = 0 for the solution of a differential euation =ƒs, d, the solution curve (graph of the solution) is reuired to pass through the point s 0, 0 d and to have slope ƒs 0, 0 d there. We can picture these slopes graphicall b drawing short line segments of slope ƒ(, ) at selected points (, ) in the region of the plane that constitutes the domain of ƒ. Each segment has the same slope as the solution curve through (, ) and so is tangent to the curve there. The resulting picture is called a slope field (or direction field) and gives a visualization of the general shape of the solution curves. Figure 16.2a shows a slope field, with a particular solution sketched into it in Figure 16.2b. We see how these line segments indicate the direction the solution curve takes at each point it passes through , (a) (b) FIGURE 16.2 (a) Slope field for (b) The particular solution d = . curve through the point a0, 2 (Eample 2). 3 b Figure 16.3 shows three slope fields and we see how the solution curves behave b following the tangent line segments in these fields.
4 164 Chapter 16: FirstOrder Differential Euations (a) ' 2 2 (b) ' 1 2 (c) ' (1 ) 2 FIGURE 16.3 Slope fields (top row) and selected solution curves (bottom row). In computer renditions, slope segments are sometimes portraed with arrows, as the are here. This is not to be taken as an indication that slopes have directions, however, for the do not. Constructing a slope field with pencil and paper can be uite tedious. All our eamples were generated b a computer. The Eistence of Solutions A basic uestion in the stu of firstorder initial value problems concerns whether a solution even eists. A second important uestion asks whether there can be more than one solution. Some conditions must be imposed to assure the eistence of eactl one solution, as illustrated in the net eample. EXAMPLE 3 The initial value problem d = 4>5, (0) = 0 5 (7, 5.4) has more than one solution. One solution is the constant function () = 0 for which the graph lies along the ais. A second solution is found b separating variables and integrating, as we did in Section 7.5. This leads to = a 5. 5 b (5, 1) 5 ( 5, 1) FIGURE 16.4 The graph of the solution = (>5) 5 to the initial value problem in Eample 3. Another solution is = 0. The two solutions = 0 and = (>5) 5 both satisf the initial condition (0) = 0 (Figure 16.4). We have found a differential euation with multiple solutions satisfing the same initial condition. This differential euation has even more solutions. For instance, two additional solutions are 0, for 0 = a 5, 5 b for 7 0
5 16.1 Solutions, Slope Fields, and Picard s Theorem 165 and 5 = a 5 b, for 0. 0, for 7 0 In man applications it is desirable to know that there is eactl one solution to an initial value problem. Such a solution is said to be uniue. Picard s Theorem gives conditions under which there is precisel one solution. It guarantees both the eistence and uniueness of a solution. THEOREM 1 Picard s Theorem Suppose that both ƒ(, ) and its partial derivative 0ƒ>0 are continuous on the interior of a rectangle R, and that ( 0, 0 ) is an interior point of R. Then the initial value problem d = ƒ(, ), ( 0 ) = 0 (2) has a uniue solution () for in some open interval containing 0. The differential euation in Eample 3 fails to satisf the conditions of Picard s Theorem. Although the function ƒ(, ) = 4>5 from Eample 3 is continuous in the entire plane, the partial derivative 0ƒ>0 = (4>5) 1>5 fails to be continuous at the point (0, 0) specified b the initial condition. Thus we found the possibilit of more than one solution to the given initial value problem. Moreover, the partial derivative 0ƒ>0 is not even defined where = 0. However, the initial value problem of Eample 3 does have uniue solutions whenever the initial condition ( 0 ) = 0 has 0 Z 0. Picard s Iteration Scheme Picard s Theorem is proved b appling Picard s iteration scheme, which we now introduce. We begin b noticing that an solution to the initial value problem of Euations (2) must also satisf the integral euation () = 0 + ƒ(t, (t)) L 0 (3) because L 0 = ()  ( 0). The converse is also true: If () satisfies Euation (3), then =ƒ(, ()) and ( 0 ) = 0. So Euations (2) ma be replaced b Euation (3). This sets the stage for Picard s interation
6 166 Chapter 16: FirstOrder Differential Euations method: In the integrand in Euation (3), replace (t) b the constant 0, then integrate and call the resulting righthand side of Euation (3) 1 (): 1 () = 0 + ƒ(t, 0 ). L 0 (4) This starts the process. To keep it going, we use the iterative formulas n + 1 () = 0 + ƒ(t, n (t)). L 0 (5) The proof of Picard s Theorem consists of showing that this process produces a seuence of functions { n ()} that converge to a function () that satisfies Euations (2) and (3) for values of sufficientl near 0. (The proof also shows that the solution is uniue; that is, no other method will lead to a different solution.) The following eamples illustrate the Picard iteration scheme, but in most practical cases the computations soon become too burdensome to continue. EXAMPLE 4 Illustrate the Picard iteration scheme for the initial value problem = , (0) = 1. Solution For the problem at hand, ƒ(, ) = , and Euation (4) becomes 1 () = 1 + (t  1) L 0 0 = 1 = If we now use Euation (5) with n = 1, we get 2 () = 1 + L 0 at t tb Substitute for in ƒ(t, ). 1 = The net iteration, with n = 2, gives 3 () = 1 + L 0 at t  t 2 + t 3 6 b Substitute for in ƒ(t, ). 2 = !. In this eample it is possible to find the eact solution because d + =
7 16.1 Solutions, Slope Fields, and Picard s Theorem 167 is a firstorder differential euation that is linear in. You will learn how to find the general solution in the net section. The solution of the initial value problem is then If we substitute the Maclaurin series for = Ce  = e . e  in this particular solution, we get = a !  3 3! + 4 4!  Á b = a4 4!  5 5! + Á b, and we see that the Picard scheme producing 3 () has given us the first four terms of this epansion. In the net eample we cannot find a solution in terms of elementar functions. The Picard scheme is one wa we could get an idea of how the solution behaves near the initial point. EXAMPLE 5 Find n () for n = 0, 1, 2, and 3 for the initial value problem = 2 + 2, (0) = 0. Solution B definition, 0 () = (0) = 0. The other functions n () are generated b the integral representation We successivel calculate n + 1 () = 0 + Ct 2 + ( n (t)) 2 D L = 3 1 () = 3 3, 3 + L0 2 () = , 0 ( n (t)) 2. 3 () = In Section 16.4 we introduce numerical methods for solving initial value problems like those in Eamples 4 and 5.
8 168 Chapter 16: FirstOrder Differential Euations EXERCISES 16.1 In Eercises 1 4, match the differential euations with their slope fields, graphed here = + 2. = = 4. = 22 In Eercises 5 and 6, cop the slope fields and sketch in some of the solution curves. 5. =( + 2)(  2) 4 2 (a) =( + 1)(  1) 4 4 (b) (c) In Eercises 7 10, write an euivalent firstorder differential euation and initial condition for = 1 + (t  (t)) L = L 1 1 t = 2  (1 + (t)) sin t L = 1 + (t) L (d) Use Picard s iteration scheme to find n () for n = 0, 1, 2, 3 in Eercises =, (1) = =, (0) = 1
9 16.2 FirstOrder Linear Euations =, (1) = 1 In Eercises 25 and 26, obtain a slope field and graph the particular 14. = +, (0) = 0 solution over the specified interval. Use our CAS DE solver to find the general solution of the differential euation. 15. = +, (0) = A logistic euation =s2  d, s0d = 1>2; 16. =2 , (1) = 1 0 4, Show that the solution of the initial value problem 26. =ssin dssin d, s0d = 2; 6 6, 6 6 is = +, ( 0 ) = 0 = ( ) e What integral euation is euivalent to the initial value problem =ƒ(), ( 0 ) = 0? COMPUTER EXPLORATIONS In Eercises 19 24, obtain a slope field and add to it graphs of the solution curves passing through the given points. 19. = with a. (0, 1) b. (0, 2) c. s0, 1d 20. =2s  4d with a. (0, 1) b. (0, 4) c. (0, 5) 21. =s + d with a. (0, 1) b. (0, 2) c. (0, 1>4) d. s 1, 1d 22. = 2 with a. (0, 1) b. (0, 2) c. s0, 1d d. (0, 0) 23. =s  1ds + 2d with a. s0, 1d b. (0, 1) c. (0, 3) d. (1, 1) 24. = with a. (0, 2) b. s0, 6d c. A 223, 4B Eercises 27 and 28 have no eplicit solution in terms of elementar functions. Use a CAS to eplore graphicall each of the differential euations. 27. =cos s2  d, s0d = 2; 0 5, A Gompertz euation =s1>2  ln d, s0d = 1>3; 0 4, Use a CAS to find the solutions of + = ƒsd subject to the initial condition s0d = 0, if ƒ() is a. 2 b. sin 2 c. 3e >2 d. 2e >2 cos 2. Graph all four solutions over the interval 2 6 to compare the results. 30. a. Use a CAS to plot the slope field of the differential euation = s  1d over the region 3 3 and b. Separate the variables and use a CAS integrator to find the general solution in implicit form. c. Using a CAS implicit function grapher, plot solution curves for the arbitrar constant values C = 6, 4, 2, 0, 2, 4, 6. d. Find and graph the solution that satisfies the initial condition s0d = FirstOrder Linear Euations A firstorder linear differential euation is one that can be written in the form d + Ps = Qsd, where P and Q are continuous functions of. Euation (1) is the linear euation s standard form. Since the eponential growth> deca euation >d = k (Section 7.5) can be put in the standard form d  k = 0, we see it is a linear euation with Psd = k and Qsd = 0. Euation (1) is linear (in ) because and its derivative > d occur onl to the first power, are not multiplied together, nor do the appear as the argument of a function Asuch as sin, e, or 2>dB. (1)
10 1610 Chapter 16: FirstOrder Differential Euations EXAMPLE 1 Put the following euation in standard form: d = 2 + 3, 7 0. Solution Divide b Notice that P() is 3>, not +3>. The standard form is + Ps = Qsd, so the minus sign is part of the formula for P(). Solving Linear Euations We solve the euation b multipling both sides b a positive function () that transforms the lefthand side into the derivative of the product sd #. We will show how to find in a moment, but first we want to show how, once found, it provides the solution we seek. Here is wh multipling b () works: sd d d d = d = + 3 d  3 = d + Ps = Qsd + Pss = sdqsd d d ssd # d = sdqsd + Ps = Qsd sd # = L sdqsd d Standard form with Psd = 3> and Qsd = Original euation is in standard form. Multipl b positive (). sd is chosen to make d + P = d d s # d. Integrate with respect to. (2) = 1 sd L sdqsd d (3) Euation (3) epresses the solution of Euation (2) in terms of the function () and Q(). We call () an integrating factor for Euation (2) because its presence makes the euation integrable. Wh doesn t the formula for P() appear in the solution as well? It does, but indirectl, in the construction of the positive function (). We have d sd = d d + P d + d = d + P d = P Condition imposed on Product Rule for derivatives The terms cancel. d
11 16.2 FirstOrder Linear Euations This last euation will hold if d = P = P d Variables separated, 7 0 = P d L L ln = P d L e ln = e 1 P d Integrate both sides. Since 7 0, we do not need absolute value signs in ln. Eponentiate both sides to solve for. = e 1 P d (4) Thus a formula for the general solution to Euation (1) is given b Euation (3), where () is given b Euation (4). However, rather than memorizing the formula, just remember how to find the integrating factor once ou have the standard form so P() is correctl identified. To solve the linear euation integrating factor sd = e 1 + Ps = Qsd, multipl both sides b the and integrate both sides. Psd d When ou integrate the lefthand side product in this procedure, ou alwas obtain the product () of the integrating factor and solution function because of the wa is defined. EXAMPLE 2 Solve the euation d = 2 + 3, 7 0. HISTORICAL BIOGRAPHY Adrien Marie Legendre ( ) Solution First we put the euation in standard form (Eample 1): so Psd = 3> is identified. The integrating factor is d  3 =, sd = e 1 Psd d = e 1 s3>d d = e 3 ln ƒ ƒ 3 ln = e Constant of integration is 0, so is as simple as possible. 7 0 = e ln 3 = 1 3.
12 1612 Chapter 16: FirstOrder Differential Euations Net we multipl both sides of the standard form b () and integrate: 1 # 3 a d  3 b = 1 # d = 1 2 d d a 1 3 b = = 1 L 2 d 1 3 =1 + C. Solving this last euation for gives the general solution: Lefthand side is d d s # d. Integrate both sides. = 3 a 1 + Cb = 2 + C 3, 7 0. EXAMPLE 3 Find the particular solution of 3  = ln + 1, 7 0, satisfing s1d = 2. Solution With 7 0, we write the euation in standard form: = ln Then the integrating factor is given b = e 1  d>3 = e s1>3dln = 1>3. Thus 1>3 = 1 sln + 1d 34>3 d. L 7 0 Lefthand side is. Integration b parts of the righthand side gives 1>3 =  1>3 sln + 1d + 4>3 d + C. L Therefore 1>3 =  1>3 sln + 1d  31>3 + C or, solving for, = sln + 4d + C 1>3. When = 1 and = 2 this last euation becomes 2 = s0 + 4d + C,
13 16.2 FirstOrder Linear Euations V a i R FIGURE 16.5 Eample 4. Switch b L The RL circuit in so Substitution into the euation for gives the particular solution In solving the linear euation in Eample 2, we integrated both sides of the euation after multipling each side b the integrating factor. However, we can shorten the amount of work, as in Eample 3, b remembering that the lefthand side alwas integrates into the product sd # of the integrating factor times the solution function. From Euation (3) this means that We need onl integrate the product of the integrating factor () with the righthand side Q() of Euation (1) and then euate the result with () to obtain the general solution. Nevertheless, to emphasize the role of () in the solution process, we sometimes follow the complete procedure as illustrated in Eample 2. Observe that if the function Q() is identicall zero in the standard form given b Euation (1), the linear euation is separable: Separating the variables We now present two applied problems modeled b a firstorder linear differential euation. RL Circuits d + Ps = Qsd d + Ps = 0 C = 2. = 2 1>3  ln  4. s = L sdqsd d. = Psd d Qsd K 0 The diagram in Figure 16.5 represents an electrical circuit whose total resistance is a constant R ohms and whose selfinductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. Ohm s Law, V = RI, has to be modified for such a circuit. The modified form is L di + Ri = V, (5) where i is the intensit of the current in amperes and t is the time in seconds. B solving this euation, we can predict how the current will flow after the switch is closed. EXAMPLE 4 The switch in the RL circuit in Figure 16.5 is closed at time t = 0. How will the current flow as a function of time? Solution Euation (5) is a firstorder linear differential euation for i as a function of t. Its standard form is di (6) + R L i = V L,
14 1614 Chapter 16: FirstOrder Differential Euations I V R 0 i I e L R L 2 R i V (1 e Rt/L ) R L 3 R L 4 R FIGURE 16.6 The growth of the current in the RL circuit in Eample 4. I is the current s steastate value. The number t = L>R is the time constant of the circuit. The current gets to within 5% of its steastate value in 3 time constants (Eercise 31). t and the corresponding solution, given that i = 0 when t = 0, is i = V R  V R e sr>l (Eercise 32). Since R and L are positive, sr>ld is negative and e sr>l : 0 as t :. Thus, lim i = lim av t: t: R  V R e sr>l b = V R  V # 0 = V R R. At an given time, the current is theoreticall less than V> R, but as time passes, the current approaches the steastate value VR. > According to the euation L di + Ri = V, I = V>R is the current that will flow in the circuit if either L = 0 (no inductance) or di> = 0 (stea current, i = constant) (Figure 16.6). Euation (7) epresses the solution of Euation (6) as the sum of two terms: a steastate solution VRand > a transient solution sv>rde sr>l that tends to zero as t :. (7) Miture Problems A chemical in a liuid solution (or dispersed in a gas) runs into a container holding the liuid (or the gas) with, possibl, a specified amount of the chemical dissolved as well. The miture is kept uniform b stirring and flows out of the container at a known rate. In this process, it is often important to know the concentration of the chemical in the container at an given time. The differential euation describing the process is based on the formula Rate of change of amount in container = rate at which rate at which chemical  chemical arrives departs. (8) If (t) is the amount of chemical in the container at time t and V(t) is the total volume of liuid in the container at time t, then the departure rate of the chemical at time t is Accordingl, Euation (8) becomes Departure rate = std Vstd # soutflow rated concentration in = acontainer at time t b # soutflow rated. = schemical s arrival rated  std Vstd # soutflow rated. If, sa, is measured in pounds, V in gallons, and t in minutes, the units in Euation (10) are pounds minutes = pounds minutes  pounds # gallons gallons minutes. (9) (10)
15 16.2 FirstOrder Linear Euations EXAMPLE 5 In an oil refiner, a storage tank contains 2000 gal of gasoline that initiall has 100 lb of an additive dissolved in it. In preparation for winter weather, gasoline containing 2 lb of additive per gallon is pumped into the tank at a rate of 40 gal> min. The wellmied solution is pumped out at a rate of 45 gal> min. How much of the additive is in the tank 20 min after the pumping process begins (Figure 16.7)? 40 gal/min containing 2 lb/gal 45 gal/min containing lb/gal V FIGURE 16.7 The storage tank in Eample 5 mies input liuid with stored liuid to produce an output liuid. Solution Let be the amount (in pounds) of additive in the tank at time t. We know that = 100 when t = 0. The number of gallons of gasoline and additive in solution in the tank at an time t is Therefore, Also, Vstd = 2000 gal + a40 gal min = s20005td gal. Rate out = std Vstd # outflow rate = a t b 45 = 45 lb t min. Rate in = a2 lb gal ba40 gal min b gal  45 b st mind min E. (9) Outflow rate is 45 gal> min and = t. The differential euation modeling the miture process is in pounds per minute. = 80 lb min. = t E. (10)
16 1616 Chapter 16: FirstOrder Differential Euations To solve this differential euation, we first write it in standard form: + Thus, Pstd = 45>s20005td and Qstd = 80. The integrating factor is std = e 1 P = e t t = ln s20005td = e t 7 0 Multipling both sides of the standard euation b (t) and integrating both sides gives s20005td 9 # a 9 s20005td = s20005td t b = 80s20005td s20005td 10 = 80s20005td 9 d Cs20005td9 D = 80s20005td 9 s20005td 9 = L 80s20005td 9 s20005td 9 = 80 # s20005td8 s 8ds 5d + C. The general solution is = 2s20005td + Cs20005td 9. Because = 100 when t = 0, we can determine the value of C: 100 = 2s20000d + Cs20000d 9 C = s2000d 9. The particular solution of the initial value problem is = 2s20005td s2000d 9 s20005td9. The amount of additive 20 min after the pumping begins is s20d = 2[20005s20d] s2000d 9 [20005s20d]9 L 1342 lb.
17 16.2 FirstOrder Linear Euations EXERCISES 16.2 Solve the differential euations in Eercises e d + 2e = 1 d + = e, = sin 2, 7 0 +stan = cos 2, p>2 6 6 p>2 d + 2 = 11, s1 + + = =e > e 2 +2e 2 = = 2 ln Solve the initial value problems in Eercises d = cos  2, 7 0 st  1d 3 ds + 4st  1d2 s = t + 1, t 7 1 st + 1d ds + 2s = 3st + 1d + 1 st + 1d 2, t 71 sin u dr + scos udr = tan u, 0 6 u 6 p>2 du tan u dr du + r = sin2 u, 0 6 u 6 p>2 t u du u du  2 = u3 sec u tan u, u 7 0, sp>3d = 2 s + 1d d  2s2 + = =, s0d = 6 d 21. Solve the eponential growth> deca initial value problem for as a function of t thinking of the differential euation as a firstorder linear euation with Psd = k and Qsd = 0: 22. Solve the following initial value problem for u as a function of t: du + 2 = 3, s0d = = t 3, t 7 0, s2d = 1 + = sin u, u 7 0, sp>2d = 1 e 2, 71, s0d = = k sk constantd, s0d = 0 + k m u = 0 sk and m positive constantsd, us0d = u 0 a. as a firstorder linear euation. b. as a separable euation. 23. Is either of the following euations correct? Give reasons for our answers. 1 1 a. d = ln ƒ ƒ + C b. d = ln ƒ ƒ + C L L 24. Is either of the following euations correct? Give reasons for our answers. a. b. 1 cos L cos d = tan + C 1 cos L cos d = tan + C cos 25. Salt miture A tank initiall contains 100 gal of brine in which 50 lb of salt are dissolved. A brine containing 2 lb> gal of salt runs into the tank at the rate of 5 gal> min. The miture is kept uniform b stirring and flows out of the tank at the rate of 4 gal> min. a. At what rate (pounds per minute) does salt enter the tank at time t? b. What is the volume of brine in the tank at time t? c. At what rate (pounds per minute) does salt leave the tank at time t? d. Write down and solve the initial value problem describing the miing process. e. Find the concentration of salt in the tank 25 min after the process starts. 26. Miture problem A 200gal tank is half full of distilled water. At time t = 0, a solution containing 0.5 lb> gal of concentrate enters the tank at the rate of 5 gal> min, and the wellstirred miture is withdrawn at the rate of 3 gal> min. a. At what time will the tank be full? b. At the time the tank is full, how man pounds of concentrate will it contain? 27. Fertilizer miture A tank contains 100 gal of fresh water. A solution containing 1 lb> gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal> min, and the miture is pumped out of the tank at the rate of 3 gal> min. Find the maimum amount of fertilizer in the tank and the time reuired to reach the maimum. 28. Carbon monoide pollution An eecutive conference room of a corporation contains 4500 ft 3 of air initiall free of carbon monoide. Starting at time t = 0, cigarette smoke containing 4% carbon monoide is blown into the room at the rate of 0.3 ft 3 >min. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of 0.3 ft 3 >min. Find the time when the concentration of carbon monoide in the room reaches 0.01%.
18 1618 Chapter 16: FirstOrder Differential Euations 29. Current in a closed RL circuit How man seconds after the switch in an RL circuit is closed will it take the current i to reach half of its steastate value? Notice that the time depends on R and L and not on how much voltage is applied. 30. Current in an open RL circuit If the switch is thrown open after the current in an RL circuit has built up to its steastate value I = V>R, the decaing current (see accompaning figure) obes the euation which is Euation (5) with V = 0. a. Solve the euation to epress i as a function of t. b. How long after the switch is thrown will it take the current to fall to half its original value? c. Show that the value of the current when t = L>R is I>e. (The significance of this time is eplained in the net eercise.) V R 0 i I e L di + Ri = 0, L R L 2 R 31. Time constants Engineers call the number L>R the time constant of the RL circuit in Figure The significance of the time constant is that the current will reach 95% of its final value within 3 time constants of the time the switch is closed (Figure 16.6). Thus, the time constant gives a builtin measure of how rapidl an individual circuit will reach euilibrium. a. Find the value of i in Euation (7) that corresponds to t = 3L>R and show that it is about 95% of the steastate value I = V>R. b. Approimatel what percentage of the steastate current will be flowing in the circuit 2 time constants after the switch is closed (i.e., when t = 2L>R)? 32. Derivation of Euation (7) in Eample 4 a. Show that the solution of the euation di + R L i = V L L 3 R t is i = V R + Ce sr>l. b. Then use the initial condition is0d = 0 to determine the value of C. This will complete the derivation of Euation (7). c. Show that i = V>R is a solution of Euation (6) and that i = Ce sr>l satisfies the euation HISTORICAL BIOGRAPHY James Bernoulli ( ) di + R L i = 0. A Bernoulli differential euation is of the form Observe that, if n = 0 or 1, the Bernoulli euation is linear. For other values of n, the substitution u = 1  n transforms the Bernoulli euation into the linear euation du d For eample, in the euation we have n = 2, so that u = 12 = 1 and du>d =  2 >d. Then >d =  2 du>d = u 2 du>d. Substitution into the original euation gives or, euivalentl, d + Ps = Qs n. + s1  ndpsdu = s1  ndqsd. d  = e  2 u 2 du d  u 1 = e  u 2 du d + u = e . This last euation is linear in the (unknown) dependent variable u. Solve the differential euations in Eercises = = = = 3
19 16.3 Applications Applications We now look at three applications of firstorder differential euations. The first application analzes an object moving along a straight line while subject to a force opposing its motion. The second is a model of population growth. The last application considers a curve or curves intersecting each curve in a second famil of curves orthogonall (that is, at right angles). Resistance Proportional to Velocit In some cases it is reasonable to assume that the resistance encountered b a moving object, such as a car coasting to a stop, is proportional to the object s velocit. The faster the object moves, the more its forward progress is resisted b the air through which it passes. Picture the object as a mass m moving along a coordinate line with position function s and velocit at time t. From Newton s second law of motion, the resisting force opposing the motion is If the resisting force is proportional to velocit, we have m Force = mass * acceleration = m. = k or =m k sk 7 0d. This is a separable differential euation representing eponential change. The solution to the euation with initial condition = 0 at t = 0 is (Section 7.5) = 0 e sk>m. (1) What can we learn from Euation (1)? For one thing, we can see that if m is something large, like the mass of a 20,000ton ore boat in Lake Erie, it will take a long time for the velocit to approach zero (because t must be large in the eponent of the euation in order to make kt> m large enough for to be small). We can learn even more if we integrate Euation (1) to find the position s as a function of time t. Suppose that a bo is coasting to a stop and the onl force acting on it is a resistance proportional to its speed. How far will it coast? To find out, we start with Euation (1) and solve the initial value problem ds = 0 e sk>m, ss0d = 0. Integrating with respect to t gives Substituting s = 0 when t = 0 gives 0 = 0 m k The bo s position at time t is therefore sstd = 0 m k s = 0 m k + C and C = 0 m k. e sk>m + 0 m k e sk>m + C. = 0 m k s1  e sk/m d. (2)
20 1620 Chapter 16: FirstOrder Differential Euations To find how far the bo will coast, we find the limit of s(t) as t :. Since sk>md 6 0, we know that e sk>m : 0 as t :, so that Thus, 0 m lim sstd = lim s1  e t: t: k sk>m d = 0 m k s10d = 0 m k. Distance coasted = 0 m k. The number 0 m>k is onl an upper bound (albeit a useful one). It is true to life in one respect, at least: if m is large, it will take a lot of energ to stop the bo. (3) In the English sstem, where weight is measured in pounds, mass is measured in slugs. Thus, Pounds = slugs * 32, assuming the gravitational constant is 32 ft sec 2. > EXAMPLE 1 For a 192lb ice skater, the k in Euation (1) is about 1> 3 slug> sec and m = 192>32 = 6 slugs. How long will it take the skater to coast from 11 ft> sec (7.5 mph) to 1 ft> sec? How far will the skater coast before coming to a complete stop? Solution We answer the first uestion b solving Euation (1) for t: 11e t>18 = 1 e t>18 = 1>11 t>18 = ln s1>11d = ln 11 t = 18 ln 11 L 43 sec. We answer the second uestion with Euation (3): Distance coasted = 0 m k = 198 ft. E. (1) with k = 1>3, m = 6, 0 = 11, = 1 = 11 # 6 1>3 Modeling Population Growth In Section 7.5 we modeled population growth with the Law of Eponential Change: dp = kp, Ps0d = P 0 where P is the population at time t, k 7 0 is a constant growth rate, and P 0 is the size of the population at time t = 0. In Section 7.5 we found the solution P = P 0 e kt to this model. To assess the model, notice that the eponential growth differential euation sas that dp> = k (4) P is constant. This rate is called the relative growth rate. Now, Table 16.1 gives the world population at miear for the ears 1980 to Taking = 1 and dp L P, we see from the table that the relative growth rate in Euation (4) is approimatel the constant Thus, based on the tabled data with t = 0 representing 1980, t = 1 representing 1981, and so forth, the world population could be modeled b the initial value problem, dp = 0.017P, Ps0d = 4454.
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