FIRST-ORDER DIFFERENTIAL EQUATIONS

Chapter 16 FIRST-ORDER 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 first-order differential euations. The involve onl first derivatives of the unknown function. 16.1 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 first-order differential euations have eactl one solution. General First-Order Differential Euations and Solutions A first-order 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 higher-order 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) 16-1

16-2 Chapter 16: First-Order 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 first-order 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 right-hand 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 first-order 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 first-order 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 first-order initial value problem Solution The euation = s + 1d - 1 3 e d = -, s0d = 2 3. d = - is a first-order differential euation with ƒs, d = -.

16.1 Solutions, Slope Fields, and Picard s Theorem 16-3 0, 2 3 2 1 4 2 2 4 1 FIGURE 16.1 2 3 4 ( 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 16.1. Slope Fields: Viewing Solution Curves d = d d a + 1-1 3 e b = 1-1 3 e. - = s + 1d - 1 3 e - = 1-1 3 e. s0d = cs + 1d - 1 3 e d = 1-1 = 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. 4 4 0, 2 3 2 2 4 2 0 2 4 4 2 0 2 4 2 2 4 4 (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.

16-4 Chapter 16: First-Order 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 first-order 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) 1 5 1 5 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

16.1 Solutions, Slope Fields, and Picard s Theorem 16-5 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

16-6 Chapter 16: First-Order Differential Euations method: In the integrand in Euation (3), replace (t) b the constant 0, then integrate and call the resulting right-hand 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 = 1 + 2 2 -. If we now use Euation (5) with n = 1, we get 2 () = 1 + L 0 at - 1 - t 2 2 + tb Substitute for in ƒ(t, ). 1 = 1 - + 2-3 6. The net iteration, with n = 2, gives 3 () = 1 + L 0 at - 1 + t - t 2 + t 3 6 b Substitute for in ƒ(t, ). 2 = 1 - + 2-3 3 + 4 4!. In this eample it is possible to find the eact solution because d + =

16.1 Solutions, Slope Fields, and Picard s Theorem 16-7 is a first-order 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 = - 1 + Ce - = - 1 + 2e -. e - in this particular solution, we get = - 1 + 2 a1 - + 2 2! - 3 3! + 4 4! - Á b = 1 - + 2-3 3 + 2 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 () = 3 3 + 7 63, 0 ( n (t)) 2. 3 () = 3 3 + 7 63 + 211 2079 + 15 59535. In Section 16.4 we introduce numerical methods for solving initial value problems like those in Eamples 4 and 5.

16-8 Chapter 16: First-Order Differential Euations EXERCISES 16.1 In Eercises 1 4, match the differential euations with their slope fields, graphed here. 4 2 4 2 2 4 2 4 1. = + 2. = + 1 3. =- 4. = 2-2 In Eercises 5 and 6, cop the slope fields and sketch in some of the solution curves. 5. =( + 2)( - 2) 4 2 (a) 4 2 2 4 4 2 2 4 4 2 2 4 2 6. =( + 1)( - 1) 4 4 (b) 2 4 4 2 2 4 2 2 4 2 2 4 4 4 2 4 (c) 4 2 2 2 4 In Eercises 7 10, write an euivalent first-order differential euation and initial condition for. 7. 8. 9. 10. = -1 + (t - (t)) L = L 1 1 t = 2 - (1 + (t)) sin t L = 1 + (t) L 0 0 1 2 4 (d) Use Picard s iteration scheme to find n () for n = 0, 1, 2, 3 in Eercises 11 16. 11. =, (1) = 2 12. =, (0) = 1

16.2 First-Order Linear Euations 16-9 13. =, (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) = 1 25. A logistic euation =s2 - d, s0d = 1>2; 16. =2 -, (-1) = 1 0 4, 0 3 17. Show that the solution of the initial value problem 26. =ssin dssin d, s0d = 2; -6 6, -6 6 is = +, ( 0 ) = 0 = -1 - + (1 + 0 + 0 ) e - 0. 18. 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 2 + 4 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, 0 5 28. A Gompertz euation =s1>2 - ln d, s0d = 1>3; 0 4, 0 3 29. 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 = 32 + 4 + 2 2s - 1d over the region -3 3 and -3 3. 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 = -1. 16.2 First-Order Linear Euations A first-order 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)

16-10 Chapter 16: First-Order 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 left-hand 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 = 2 + 3 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

16.2 First-Order Linear Euations 16-11 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 left-hand 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 (1752 1833) 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 s-3>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.

16-12 Chapter 16: First-Order Differential Euations Net we multipl both sides of the standard form b () and integrate: 1 # 3 a d - 3 b = 1 # 3 1 3 d - 3 4 = 1 2 d d a 1 3 b = 1 2 1 3 = 1 L 2 d 1 3 =-1 + C. Solving this last euation for gives the general solution: Left-hand 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: - 1 3 = ln + 1. 3 Then the integrating factor is given b = e 1 - d>3 = e s-1>3dln = -1>3. Thus -1>3 = 1 sln + 1d 3-4>3 d. L 7 0 Left-hand side is. Integration b parts of the right-hand side gives -1>3 = - -1>3 sln + 1d + -4>3 d + C. L Therefore -1>3 = - -1>3 sln + 1d - 3-1>3 + C or, solving for, = -sln + 4d + C 1>3. When = 1 and = -2 this last euation becomes -2 = -s0 + 4d + C,

16.2 First-Order Linear Euations 16-13 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 left-hand 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 right-hand 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 first-order 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 self-inductance, 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 first-order linear differential euation for i as a function of t. Its standard form is di (6) + R L i = V L,

16-14 Chapter 16: First-Order 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 stea-state value. The number t = L>R is the time constant of the circuit. The current gets to within 5% of its stea-state 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 stea-state 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 stea-state 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)

16.2 First-Order Linear Euations 16-15 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 well-mied 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 = s2000-5td gal. Rate out = std Vstd # outflow rate = a 2000-5t b 45 = 45 lb 2000-5t 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 = 2000-5t. The differential euation modeling the miture process is in pounds per minute. = 80 lb min. = 80-45 2000-5t E. (10)

16-16 Chapter 16: First-Order Differential Euations To solve this differential euation, we first write it in standard form: + Thus, Pstd = 45>s2000-5td and Qstd = 80. The integrating factor is std = e 1 P = e 1 45 2000-5t 45 2000-5t = 80. -9 ln s2000-5td = e 2000-5t 7 0 Multipling both sides of the standard euation b (t) and integrating both sides gives s2000-5td -9 # a -9 s2000-5td = s2000-5td -9. + 45 2000-5t b = 80s2000-5td-9 + 45s2000-5td -10 = 80s2000-5td -9 d Cs2000-5td-9 D = 80s2000-5td -9 s2000-5td -9 = L 80s2000-5td -9 s2000-5td -9 = 80 # s2000-5td-8 s -8ds -5d + C. The general solution is = 2s2000-5td + Cs2000-5td 9. Because = 100 when t = 0, we can determine the value of C: 100 = 2s2000-0d + Cs2000-0d 9 C =- 3900 s2000d 9. The particular solution of the initial value problem is = 2s2000-5td - 3900 s2000d 9 s2000-5td9. The amount of additive 20 min after the pumping begins is s20d = 2[2000-5s20d] - 3900 s2000d 9 [2000-5s20d]9 L 1342 lb.

16.2 First-Order Linear Euations 16-17 EXERCISES 16.2 Solve the differential euations in Eercises 1 14. 1. 2. e d + 2e = 1 d + = e, 7 0 3. 4. 5. +3 = sin 2, 7 0 +stan = cos 2, -p>2 6 6 p>2 d + 2 = 1-1, 7 0 6. s1 + + = 2 7. 2 =e >2 + 8. e 2 +2e 2 = 2 9. - = 2 ln 10. 11. 12. 13. 14. Solve the initial value problems in Eercises 15 20. 15. 16. 17. 18. 19. 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 7-1 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 + = 20. + =, 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 first-order 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 = 1 + 2 = t 3, t 7 0, s2d = 1 + = sin u, u 7 0, sp>2d = 1 e 2, 7-1, s0d = 5 + 1 = k sk constantd, s0d = 0 + k m u = 0 sk and m positive constantsd, us0d = u 0 a. as a first-order 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 200-gal 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 well-stirred 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%.

16-18 Chapter 16: First-Order 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 stea-state 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 stea-state 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 16.6. 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 built-in 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 stea-state value I = V>R. b. Approimatel what percentage of the stea-state 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 (1654 1705) 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 = 1-2 = -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 33 36. 33. - = - 2 34. - = 2 35. + = -2 36. 2 +2 = 3

16.3 Applications 16-19 16.3 Applications We now look at three applications of first-order 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 20,000-ton 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)

16-20 Chapter 16: First-Order 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 s1-0d = 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 192-lb 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 1989. Taking = 1 and dp L P, we see from the table that the relative growth rate in Euation (4) is approimatel the constant 0.017. 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.

16.3 Applications 16-21 6000 5000 P World population (1980 99) P 4454e 0.017t 4000 0 10 20 FIGURE 16.8 Notice that the value of the solution P = 4454e 0.017t is 6152.16 when t = 19, which is slightl higher than the actual population in 1999. t TABLE 16.1 World population (miear) Population Year (millions) P>P 1980 4454 1981 4530 1982 4610 1983 4690 1984 4770 1985 4851 1986 4933 1987 5018 1988 5105 1989 5190 76>4454 L 0.0171 80>4530 L 0.0177 80>4610 L 0.0174 80>4690 L 0.0171 81>4770 L 0.0170 82>4851 L 0.0169 85>4933 L 0.0172 87>5018 L 0.0173 85>5105 L 0.0167 Orthogonal trajector Source: U.S. Bureau of the Census (Sept., 1999): www.census.gov> ipc> www> worldpop.html. FIGURE 16.9 An orthogonal trajector intersects the famil of curves at right angles, or orthogonall. FIGURE 16.10 Ever straight line through the origin is orthogonal to the famil of circles centered at the origin. The solution to this initial value problem gives the population function P = 4454e 0.017t. In ear 1999 (so t = 19), the solution predicts the world population in miear to be about 6152 million, or 6.15 billion (Figure 16.8), which is more than the actual population of 6001 million from the U.S. Bureau of the Census. A more realistic model would consider environmental factors affecting the growth rate. Orthogonal Trajectories An orthogonal trajector of a famil of curves is a curve that intersects each curve of the famil at right angles, or orthogonall (Figure 16.9). For instance, each straight line through the origin is an orthogonal trajector of the famil of circles 2 + 2 = a 2, centered at the origin (Figure 16.10). Such mutuall orthogonal sstems of curves are of particular importance in phsical problems related to electrical potential, where the curves in one famil correspond to flow of electric current and those in the other famil correspond to curves of constant potential. The also occur in hdronamics and heat-flow problems. EXAMPLE 2 Find the orthogonal trajectories of the famil of curves = a, where a Z 0 is an arbitrar constant. Solution The curves = a form a famil of hperbolas with asmptotes = ;. First we find the slopes of each curve in this famil, or their > d values. Differentiating = a implicitl gives d + = 0 or d =-. Thus the slope of the tangent line at an point (, ) on one of the hperbolas = a is =->. On an orthogonal trajector the slope of the tangent line at this same point

16-22 Chapter 16: First-Order Differential Euations must be the negative reciprocal, or >. Therefore, the orthogonal trajectories must satisf the differential euation d =. 0 2 2 b b 0 a, a 0 This differential euation is separable and we solve it as in Section 7.5: = d L = L d 1 2 2 = 1 2 2 + C Separate variables. Integrate both sides. FIGURE 16.11 Each curve is orthogonal to ever curve it meets in the other famil (Eample 2). 2-2 = b, (5) where b = 2C is an arbitrar constant. The orthogonal trajectories are the famil of hperbolas given b Euation (5) and sketched in Figure 16.11. EXERCISES 16.3 1. Coasting biccle A 66-kg cclist on a 7-kg biccle starts coasting on level ground at 9 m> sec. The k in Euation (1) is about 3.9 kg> sec. a. About how far will the cclist coast before reaching a complete stop? b. How long will it take the cclist s speed to drop to 1 m> sec? 2. Coasting battleship Suppose that an Iowa class battleship has mass around 51,000 metric tons (51,000,000 kg) and a k value in Euation (1) of about 59,000 kg> sec. Assume that the ship loses power when it is moving at a speed of 9 m> sec. a. About how far will the ship coast before it is dead in the water? b. About how long will it take the ship s speed to drop to 1 m> sec? 3. The data in Table 16.2 were collected with a motion detector and a CBL b Valerie Sharritts, a mathematics teacher at St. Francis DeSales High School in Columbus, Ohio. The table shows the distance s (meters) coasted on in-line skates in t sec b her daughter Ashle when she was 10 ears old. Find a model for Ashle s position given b the data in Table 16.2 in the form of Euation (2). Her initial velocit was 0 = 2.75 m>sec, her mass m = 39.92 kg (she weighed 88 lb), and her total coasting distance was 4.91 m. 4. Coasting to a stop Table 16.3 shows the distance s (meters) coasted on in-line skates in terms of time t (seconds) b Kell Schmitzer. Find a model for her position in the form of Euation (2). Her initial velocit was 0 = 0.80 m>sec, her mass m = 49.90 kg (110 lb), and her total coasting distance was 1.32 m. TABLE 16.2 Ashle Sharritts skating data t (sec) s (m) t (sec) s (m) t (sec) s (m) 0 0 2.24 3.05 4.48 4.77 0.16 0.31 2.40 3.22 4.64 4.82 0.32 0.57 2.56 3.38 4.80 4.84 0.48 0.80 2.72 3.52 4.96 4.86 0.64 1.05 2.88 3.67 5.12 4.88 0.80 1.28 3.04 3.82 5.28 4.89 0.96 1.50 3.20 3.96 5.44 4.90 1.12 1.72 3.36 4.08 5.60 4.90 1.28 1.93 3.52 4.18 5.76 4.91 1.44 2.09 3.68 4.31 5.92 4.90 1.60 2.30 3.84 4.41 6.08 4.91 1.76 2.53 4.00 4.52 6.24 4.90 1.92 2.73 4.16 4.63 6.40 4.91 2.08 2.89 4.32 4.69 6.56 4.91

16.4 Euler s Method 16-23 TABLE 16.3 Kell Schmitzer skating data t (sec) s (m) t (sec) s (m) t (sec) s (m) 0 0 1.5 0.89 3.1 1.30 0.1 0.07 1.7 0.97 3.3 1.31 0.3 0.22 1.9 1.05 3.5 1.32 0.5 0.36 2.1 1.11 3.7 1.32 0.7 0.49 2.3 1.17 3.9 1.32 0.9 0.60 2.5 1.22 4.1 1.32 1.1 0.71 2.7 1.25 4.3 1.32 1.3 0.81 2.9 1.28 4.5 1.32 In Eercises 5 10, find the orthogonal trajectories of the famil of curves. Sketch several members of each famil. 5. = m 6. = c 2 7. k 2 + 2 = 1 8. 2 2 + 2 = c 2 9. = ce - 10. = e k 11. Show that the curves 2 2 + 3 2 = 5 and 2 = 3 are orthogonal. 12. Find the famil of solutions of the given differential euation and the famil of orthogonal trajectories. Sketch both families. a. d + = 0 b. - 2 d = 0 13. Suppose a and b are positive numbers. Sketch the parabolas 2 = 4a 2-4a and 2 = 4b 2 + 4b in the same diagram. Show that the intersect at Aa - b, ;22abB, and that each a-parabola is orthogonal to ever b-parabola. 16.4 Euler s Method HISTORICAL BIOGRAPHY Leonhard Euler (1703 1783) 0 0 0 L() 0 f ( 0, 0 )( 0 ) ( 0, 0 ) () FIGURE 16.12 The linearization L() of = sd at = 0. 0 ( 0, 0 ) ( 1, L( 1 )) 0 ( 1, ( 1 )) d 1 0 d () FIGURE 16.13 The first Euler step approimates s 1 d with 1 = Ls 1 d. If we do not reuire or cannot immediatel find an eact solution for an initial value problem =ƒs, d, s 0 d = 0, we can often use a computer to generate a table of approimate numerical values of for values of in an appropriate interval. Such a table is called a numerical solution of the problem, and the method b which we generate the table is called a numerical method. Numerical methods are generall fast and accurate, and the are often the methods of choice when eact formulas are unnecessar, unavailable, or overl complicated. In this section we stu one such method, called Euler s method, upon which man other numerical methods are based. Euler s Method Given a differential euation >d = ƒs, d and an initial condition s 0 d = 0, we can approimate the solution = sd b its linearization Lsd = s 0 d + s 0 ds - 0 d or Lsd = 0 + ƒs 0, 0 ds - 0 d. The function L() gives a good approimation to the solution () in a short interval about 0 (Figure 16.12). The basis of Euler s method is to patch together a string of linearizations to approimate the curve over a longer stretch. Here is how the method works. We know the point s 0, 0 d lies on the solution curve. Suppose that we specif a new value for the independent variable to be 1 = 0 + d. (Recall that d = in the definition of differentials.) If the increment d is small, then 1 = Ls 1 d = 0 + ƒs 0, 0 d d is a good approimation to the eact solution value = s 1 d. So from the point s 0, 0 d, which lies eactl on the solution curve, we have obtained the point s 1, 1 d, which lies ver close to the point s 1, s 1 dd on the solution curve (Figure 16.13). Using the point s 1, 1 d and the slope ƒs 1, 1 d of the solution curve through s 1, 1 d, we take a second step. Setting 2 = 1 + d, we use the linearization of the solution curve through s 1, 1 d to calculate 2 = 1 + ƒs 1, 1 d d.

16-24 Chapter 16: First-Order Differential Euations 0 Euler approimation ( 0, 0 ) ( 1, 1 ) ( 2, 2 ) Error True solution curve () d d d 0 1 2 3 ( 3, 3 ) FIGURE 16.14 Three steps in the Euler approimation to the solution of the initial value problem =ƒs, d, s 0 d = 0. As we take more steps, the errors involved usuall accumulate, but not in the eaggerated wa shown here. This gives the net approimation s 2, 2 d to values along the solution curve = sd (Figure 16.14). Continuing in this fashion, we take a third step from the point s 2, 2 d with slope ƒs 2, 2 d to obtain the third approimation and so on. We are literall building an approimation to one of the solutions b following the direction of the slope field of the differential euation. The steps in Figure 16.14 are drawn large to illustrate the construction process, so the approimation looks crude. In practice, d would be small enough to make the red curve hug the blue one and give a good approimation throughout. EXAMPLE 1 Find the first three approimations 1, 2, 3 using Euler s method for the initial value problem starting at 0 = 0 with d = 0.1. 3 = 2 + ƒs 2, 2 d d, =1 +, s0d = 1, Solution We have 0 = 0, 0 = 1, 1 = 0 + d = 0.1, 2 = 0 + 2 d = 0.2, and 3 = 0 + 3 d = 0.3. First: Second: Third: 1 = 0 + ƒs 0, 0 d d = 0 + s1 + 0 d d = 1 + s1 + 1ds0.1d = 1.2 2 = 1 + ƒs 1, 1 d d = 1 + s1 + 1 d d = 1.2 + s1 + 1.2ds0.1d = 1.42 3 = 2 + ƒs 2, 2 d d = 2 + s1 + 2 d d = 1.42 + s1 + 1.42ds0.1d = 1.662 The step-b-step process used in Eample 1 can be continued easil. Using euall spaced values for the independent variable in the table and generating n of them, set Then calculate the approimations to the solution, 1 = 0 + ƒs 0, 0 d d 2 = 1 + ƒs 1, 1 d d o 1 = 0 + d 2 = 1 + d o n = n - 1 + d. n = n - 1 + ƒs n - 1, n - 1 d d. The number of steps n can be as large as we like, but errors can accumulate if n is too large.

16.4 Euler s Method 16-25 Euler s method is eas to implement on a computer or calculator. A computer program generates a table of numerical solutions to an initial value problem, allowing us to input 0 and 0, the number of steps n, and the step size d. It then calculates the approimate solution values 1, 2, Á, n in iterative fashion, as just described. Solving the separable euation in Eample 1, we find that the eact solution to the initial value problem is = 2e - 1. We use this information in Eample 2. EXAMPLE 2 Use Euler s method to solve on the interval 0 1, starting at 0 = 0 and taking (a) d = 0.1 and (b) d = 0.05. Compare the approimations with the values of the eact solution = 2e - 1. Solution =1 +, s0d = 1, (a) We used a computer to generate the approimate values in Table 16.4. The error column is obtained b subtracting the unrounded Euler values from the unrounded values found using the eact solution. All entries are then rounded to four decimal places. TABLE 16.4 Euler solution of =1 +, s0d = 1, step size d = 0.1 (Euler) (eact) Error 4 3 0 1 1 0 0.1 1.2 1.2103 0.0103 0.2 1.42 1.4428 0.0228 0.3 1.662 1.6997 0.0377 0.4 1.9282 1.9836 0.0554 0.5 2.2210 2.2974 0.0764 0.6 2.5431 2.6442 0.1011 0.7 2.8974 3.0275 0.1301 0.8 3.2872 3.4511 0.1639 0.9 3.7159 3.9192 0.2033 1.0 4.1875 4.4366 0.2491 2 2e 1 1 0 FIGURE 16.15 The graph of = 2e - 1 superimposed on a scatterplot of the Euler approimations shown in Table 16.4 (Eample 2). 1 B the time we reach = 1 (after 10 steps), the error is about 5.6% of the eact solution. A plot of the eact solution curve with the scatterplot of Euler solution points from Table 16.4 is shown in Figure 16.15. (b) One wa to tr to reduce the error is to decrease the step size. Table 16.5 shows the results and their comparisons with the eact solutions when we decrease the step size to 0.05, doubling the number of steps to 20. As in Table 16.4, all computations are performed before rounding. This time when we reach = 1, the relative error is onl about 2.9%.

16-26 Chapter 16: First-Order Differential Euations TABLE 16.5 Euler solution of =1 +, s0d = 1, step size d = 0.05 (Euler) (eact) Error 0 1 1 0 0.05 1.1 1.1025 0.0025 0.10 1.205 1.2103 0.0053 0.15 1.3153 1.3237 0.0084 0.20 1.4310 1.4428 0.0118 0.25 1.5526 1.5681 0.0155 0.30 1.6802 1.6997 0.0195 0.35 1.8142 1.8381 0.0239 0.40 1.9549 1.9836 0.0287 0.45 2.1027 2.1366 0.0340 0.50 2.2578 2.2974 0.0397 0.55 2.4207 2.4665 0.0458 0.60 2.5917 2.6442 0.0525 0.65 2.7713 2.8311 0.0598 0.70 2.9599 3.0275 0.0676 0.75 3.1579 3.2340 0.0761 0.80 3.3657 3.4511 0.0853 0.85 3.5840 3.6793 0.0953 0.90 3.8132 3.9192 0.1060 0.95 4.0539 4.1714 0.1175 1.00 4.3066 4.4366 0.1300 It might be tempting to reduce the step size even further in Eample 2 to obtain greater accurac. Each additional calculation, however, not onl reuires additional computer time but more importantl adds to the buildup of round-off errors due to the approimate representations of numbers inside the computer. The analsis of error and the investigation of methods to reduce it when making numerical calculations are important but are appropriate for a more advanced course. There are numerical methods more accurate than Euler s method, as ou can see in a further stu of differential euations. We stu one improvement here. HISTORICAL BIOGRAPHY Carl Runge (1856 1927) Improved Euler s Method We can improve on Euler s method b taking an average of two slopes. We first estimate n as in the original Euler method, but denote it b z n. We then take the average of ƒs n - 1, n - 1 d and ƒs n, z n d in place of ƒs n - 1, n - 1 d in the net step. Thus, we calculate the net approimation using n z n = n - 1 + ƒs n - 1, n - 1 d d n = n - 1 + c ƒs n - 1, n - 1 d + ƒs n, z n d d d. 2

16.4 Euler s Method 16-27 EXAMPLE 3 Use the improved Euler s method to solve =1 +, s0d = 1, on the interval 0 1, starting at 0 = 0 and taking d = 0.1. Compare the approimations with the values of the eact solution = 2e - 1. Solution We used a computer to generate the approimate values in Table 16.6. The error column is obtained b subtracting the unrounded improved Euler values from the unrounded values found using the eact solution. All entries are then rounded to four decimal places. TABLE 16.6 Improved Euler solution of =1 +, s0d = 1, step size d = 0.1 (improved Euler) (eact) Error 0 1 1 0 0.1 1.21 1.2103 0.0003 0.2 1.4421 1.4428 0.0008 0.3 1.6985 1.6997 0.0013 0.4 1.9818 1.9836 0.0018 0.5 2.2949 2.2974 0.0025 0.6 2.6409 2.6442 0.0034 0.7 3.0231 3.0275 0.0044 0.8 3.4456 3.4511 0.0055 0.9 3.9124 3.9192 0.0068 1.0 4.4282 4.4366 0.0084 B the time we reach = 1 (after 10 steps), the relative error is about 0.19%. B comparing Tables 16.4 and 16.6, we see that the improved Euler s method is considerabl more accurate than the regular Euler s method, at least for the initial value problem =1 +, s0d = 1. EXERCISES 16.4 In Eercises 1 6, use Euler s method to calculate the first three approimations to the given initial value problem for the specified increment size. Calculate the eact solution and investigate the accurac of our approimations. Round our results to four decimal places. 1. =1 -, s2d = -1, d = 0.5 2. =s1 - d, s1d = 0, d = 0.2 T T 3. =2 + 2, s0d = 3, d = 0.2 4. = 2 s1 + 2d, s -1d = 1, d = 0.5 5. =2e 2, s0d = 2, d = 0.1 6. = + e - 2, s0d = 2, d = 0.5 7. Use the Euler method with d = 0.2 to estimate (1) if and s0d = 1. What is the eact value of (1)? =

16-28 Chapter 16: First-Order Differential Euations 8. Use the Euler method with d = 0.2 to estimate (2) if => and s1d = 2. What is the eact value of (2)? 19. =2 2 s - 1d, s2d = -1>2, 0 = 2, * = 3 (See Eercise 17 for the eact solution.) 9. Use the Euler method with d = 0.5 to estimate (5) if 20. = - 1, s0d = 3, 0 = 0, * = 1 = 2 > 2 and s1d = -1. What is the eact value of (5)? (See Eercise 18 for the eact solution.) and s0d = 1. What is the eact value of (2)? Use a CAS to eplore graphicall each of the differential euations in 10. Use the Euler method with d = 1>3 to estimate (2) if Eercises 21 24. Perform the following steps to help with our eplorations. In Eercises 11 and 12, use the improved Euler s method to calculate the first three approimations to the given initial value problem. Compare the approimations with the values of the eact solution. 11. =2s + 1d, s0d = 3, d = 0.2 (See Eercise 3 for the eact solution.) 12. =s1 - d, s1d = 0, d = 0.2 (See Eercise 2 for the eact solution.) COMPUTER EXPLORATIONS In Eercises 13 16, use Euler s method with the specified step size to estimate the value of the solution at the given point *. Find the value of the eact solution at *. 13. =2e 2, s0d = 2, d = 0.1, * = 1 14. = + e - 2, s0d = 2, d = 0.5, * = 2 15. =2>, 7 0, s0d = 1, d = 0.1, * = 1 16. =1 + 2, s0d = 0, d = 0.1, * = 1 In Eercises 17 and 18, (a) find the eact solution of the initial value problem. Then compare the accurac of the approimation with s * d using Euler s method starting at 0 with step size (b) 0.2, (c) 0.1, and (d) 0.05. 17. =2 2 s - 1d, s2d = -1>2, 0 = 2, * = 3 18. = - 1, s0d = 3, 0 = 0, * = 1 In Eercises 19 and 20, compare the accurac of the approimation with s * d using the improved Euler s method starting at 0 with step size a. 0.2 b. 0.1 c. 0.05 d. Describe what happens to the error as the step size decreases. a. Plot a slope field for the differential euation in the given -window. b. Find the general solution of the differential euation using our CAS DE solver. c. Graph the solutions for the values of the arbitrar constant C = -2, -1, 0, 1, 2 superimposed on our slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval [0, b]. e. Find the Euler numerical approimation to the solution of the initial value problem with 4 subintervals of the -interval and plot the Euler approimation superimposed on the graph produced in part (d). f. Repeat part (e) for 8, 16, and 32 subintervals. Plot these three Euler approimations superimposed on the graph from part (e). g. Find the error s seactd - seulerdd at the specified point = b for each of our four Euler approimations. Discuss the improvement in the percentage error. 21. = +, s0d = -7>10; -4 4, -4 4; b = 1 22. =->, s0d = 2; -3 3, -3 3; b = 2 23. A logistic euation =s2 - d, s0d = 1>2; 0 4, 0 3; b = 3 24. =ssin dssin d, s0d = 2; -6 6, -6 6; b = 3p>2 16.5 Graphical Solutions of Autonomous Euations In Chapter 4 we learned that the sign of the first derivative tells where the graph of a function is increasing and where it is decreasing. The sign of the second derivative tells the concavit of the graph. We can build on our knowledge of how derivatives determine the shape of a graph to solve differential euations graphicall. The starting ideas for doing so are the notions of phase line and euilibrium value. We arrive at these notions b investigating what happens when the derivative of a differentiable function is zero from a point of view different from that studied in Chapter 4.

16.5 Graphical Solutions of Autonomous Euations 16-29 Euilibrium Values and Phase Lines When we differentiate implicitl the euation we obtain 1 ln s5-15d = + 1, 5 1 5 a 5 5-15 b d = 1. Solving for =>d we find =5-15 = 5s - 3d. In this case the derivative is a function of onl (the dependent variable) and is zero when = 3. A differential euation for which > d is a function of onl is called an autonomous differential euation. Let s investigate what happens when the derivative in an autonomous euation euals zero. We assume an derivatives are continuous. DEFINITION If >d = gs d is an autonomous differential euation, then the values of for which >d = 0 are called euilibrium values or rest points. Thus, euilibrium values are those at which no change occurs in the dependent variable, so is at rest. The emphasis is on the value of where >d = 0, not the value of, as we studied in Chapter 4. For eample, the euilibrium values for the autonomous differential euation d = s + 1ds - 2d are = -1 and = 2. To construct a graphical solution to an autonomous differential euation, we first make a phase line for the euation, a plot on the -ais that shows the euation s euilibrium values along with the intervals where > d and d 2 >d 2 are positive and negative. Then we know where the solutions are increasing and decreasing, and the concavit of the solution curves. These are the essential features we found in Section 4.4, so we can determine the shapes of the solution curves without having to find formulas for them. EXAMPLE 1 Draw a phase line for the euation d = s + 1ds - 2d and use it to sketch solutions to the euation. Solution 1. Draw a number line for and mark the euilibrium values = -1 and = 2, where >d = 0. 1 2

16-30 Chapter 16: First-Order Differential Euations 2. Identif and label the intervals where 70 and 60. This step resembles what we did in Section 4.3, onl now we are marking the -ais instead of the -ais. ' 0 ' 0 ' 0 1 2 We can encapsulate the information about the sign of on the phase line itself. Since 70 on the interval to the left of = -1, a solution of the differential euation with a -value less than -1 will increase from there toward = -1. We displa this information b drawing an arrow on the interval pointing to -1. 1 2 Similarl, 60 between = -1 and = 2, so an solution with a value in this interval will decrease toward = -1. For 7 2, we have 70, so a solution with a -value greater than 2 will increase from there without bound. In short, solution curves below the horizontal line = -1 in the -plane rise toward = -1. Solution curves between the lines = -1 and = 2 fall awa from = 2 toward = -1. Solution curves above = 2 rise awa from = 2 and keep going up. 3. Calculate and mark the intervals where 70 and 60. To find, we differentiate with respect to, using implicit differentiation. =s + 1ds - 2d = 2 - - 2 Formula for... = d d s d = d d s 2 - - 2d ' 0 '' 0 = 2 - = s2-1 Differentiated implicitl with respect to. 2 1 2 0 1 ' 0 '' 0 ' 0 '' 0 ' 0 '' 0 = s2-1ds + 1ds - 2d. From this formula, we see that changes sign at = -1, = 1>2, and = 2. We add the sign information to the phase line. ' 0 ' 0 ' 0 ' 0 '' 0 '' 0 '' 0 '' 0 1 1 2 2 FIGURE 16.16 Graphical solutions from Eample 1 include the horizontal lines = -1 and = 2 through the euilibrium values. From Theorem 1, no two solution curves will ever cross or touch each other. 4. Sketch an assortment of solution curves in the -plane. The horizontal lines = -1, = 1>2, and = 2 partition the plane into horizontal bands in which we know the signs of and. In each band, this information tells us whether the solution curves rise or fall and how the bend as increases (Figure 16.16). The euilibrium lines = -1 and = 2 are also solution curves. (The constant functions = -1 and = 2 satisf the differential euation.) Solution curves

16.5 Graphical Solutions of Autonomous Euations 16-31 that cross the line = 1>2 have an inflection point there. The concavit changes from concave down (above the line) to concave up (below the line). As predicted in Step 2, solutions in the middle and lower bands approach the euilibrium value = -1 as increases. Solutions in the upper band rise steadil awa from the value = 2. Stable and Unstable Euilibria Look at Figure 16.16 once more, in particular at the behavior of the solution curves near the euilibrium values. Once a solution curve has a value near = -1, it tends steadil toward that value; = -1 is a stable euilibrium. The behavior near = 2 is just the opposite: all solutions ecept the euilibrium solution = 2 itself move awa from it as increases. We call = 2 an unstable euilibrium. If the solution is at that value, it stas, but if it is off b an amount, no matter how small, it moves awa. (Sometimes an euilibrium value is unstable because a solution moves awa from it onl on one side of the point.) Now that we know what to look for, we can alrea see this behavior on the initial phase line. The arrows lead awa from = 2 and, once to the left of = 2, toward = -1. We now present several applied eamples for which we can sketch a famil of solution curves to the differential euation models using the method in Eample 1. In Section 7.5 we solved analticall the differential euation dh = -ksh - H S d, k 7 0 modeling Newton s law of cooling. Here H is the temperature (amount of heat) of an object at time t and H S is the constant temperature of the surrounding medium. Our first eample uses a phase line analsis to understand the graphical behavior of this temperature model over time. EXAMPLE 2 What happens to the temperature of the soup when a cup of hot soup is placed on a table in a room? We know the soup cools down, but what does a tpical temperature curve look like as a function of time? Solution Suppose that the surrounding medium has a constant Celsius temperature of 15 C. We can then epress the difference in temperature as Hstd - 15. Assuming H is a differentiable function of time t, b Newton s law of cooling, there is a constant of proportionalit k 7 0 such that dh dh 0 0 15 FIGURE 16.17 First step in constructing the phase line for Newton s law of cooling in Eample 2. The temperature tends towards the euilibrium (surroundingmedium) value in the long run. H dh = -ksh - 15d (minus k to give a negative derivative when H 7 15). Since dh> = 0 at H = 15, the temperature 15 C is an euilibrium value. If H 7 15, Euation (1) tells us that sh - 15d 7 0 and dh> 6 0. If the object is hotter than the room, it will get cooler. Similarl, if H 6 15, then sh - 15d 6 0 and dh> 7 0. An object cooler than the room will warm up. Thus, the behavior described b Euation (1) agrees with our intuition of how temperature should behave. These observations are captured in the initial phase line diagram in Figure 16.17. The value H = 15 is a stable euilibrium. (1)

16-32 Chapter 16: First-Order Differential Euations dh 0 d 2 H 0 15 dh 0 d 2 H 2 0 2 FIGURE 16.18 The complete phase line for Newton s law of cooling (Eample 2). 15 H Initial temperature Initial temperature Temperature of surrounding medium FIGURE 16.19 Temperature versus time. Regardless of initial temperature, the object s temperature H(t) tends toward 15 C, the temperature of the surrounding medium. t H We determine the concavit of the solution curves b differentiating both sides of Euation (1) with respect to t: d adh b = d s -ksh - 15dd d 2 H 2 = -k dh. Since -k is negative, we see that d 2 H> 2 is positive when dh> 6 0 and negative when dh> 7 0. Figure 16.18 adds this information to the phase line. The completed phase line shows that if the temperature of the object is above the euilibrium value of 15 C, the graph of H(t) will be decreasing and concave upward. If the temperature is below 15 C (the temperature of the surrounding medium), the graph of H(t) will be increasing and concave downward. We use this information to sketch tpical solution curves (Figure 16.19). From the upper solution curve in Figure 16.19, we see that as the object cools down, the rate at which it cools slows down because dh> approaches zero. This observation is implicit in Newton s law of cooling and contained in the differential euation, but the flattening of the graph as time advances gives an immediate visual representation of the phenomenon. The abilit to discern phsical behavior from graphs is a powerful tool in understanding real-world sstems. EXAMPLE 3 Galileo and Newton both observed that the rate of change in momentum encountered b a moving object is eual to the net force applied to it. In mathematical terms, F = d smd (2) where F is the force and m and the object s mass and velocit. If m varies with time, as it will if the object is a rocket burning fuel, the right-hand side of Euation (2) epands to m + dm F r k m F p mg positive 0 FIGURE 16.20 An object falling under the influence of gravit with a resistive force assumed to be proportional to the velocit. using the Product Rule. In man situations, however, m is constant, dm> = 0, and Euation (2) takes the simpler form F = m or F = ma, known as Newton s second law of motion (see Section 16.3). In free fall, the constant acceleration due to gravit is denoted b g and the one force acting downward on the falling bo is F p = mg, the propulsion due to gravit. If, however, we think of a real bo falling through the air sa, a penn from a great height or a parachutist from an even greater height we know that at some point air resistance is a factor in the speed of the fall. A more realistic model of free fall would include air resistance, shown as a force F r in the schematic diagram in Figure 16.20. (3)

16.5 Graphical Solutions of Autonomous Euations 16-33 FIGURE 16.21 Eample 3. 0 0 mg k Initial phase line for 0 0 d 2 d 0 2 2 0 2 FIGURE 16.22 for Eample 3. mg k Initial velocit Initial velocit mg k The completed phase line mg k FIGURE 16.23 Tpical velocit curves in Eample 3. The value = mg>k is the terminal velocit. t For low speeds well below the speed of sound, phsical eperiments have shown that F r is approimatel proportional to the bo s velocit. The net force on the falling bo is therefore giving We can use a phase line to analze the velocit functions that solve this differential euation. The euilibrium point, obtained b setting the right-hand side of Euation (4) eual to zero, is If the bo is initiall moving faster than this, > is negative and the bo slows down. If the bo is moving at a velocit below mg>k, then > 7 0 and the bo speeds up. These observations are captured in the initial phase line diagram in Figure 16.21. We determine the concavit of the solution curves b differentiating both sides of Euation (4) with respect to t: d 2 2 F = F p - F r, m = mg - k = g - k m. = mg k. = d ag - k m b =-k m. We see that d 2 > 2 6 0 when 6 mg>k and d 2 > 2 7 0 when 7 mg>k. Figure 16.22 adds this information to the phase line. Notice the similarit to the phase line for Newton s law of cooling (Figure 16.18). The solution curves are similar as well (Figure 16.23). Figure 16.23 shows two tpical solution curves. Regardless of the initial velocit, we see the bo s velocit tending toward the limiting value = mg>k. This value, a stable euilibrium point, is called the bo s terminal velocit. Skdivers can var their terminal velocit from 95 mph to 180 mph b changing the amount of bo area opposing the fall. (4) EXAMPLE 4 In Section 16.3 we eamined population growth using the model of eponential change. That is, if P represents the number of individuals and we neglect departures and arrivals, then dp = kp, (5) where k 7 0 is the birthrate minus the death rate per individual per unit time. Because the natural environment has onl a limited number of resources to sustain life, it is reasonable to assume that onl a maimum population M can be accommodated. As the population approaches this limiting population or carring capacit, resources become less abundant and the growth rate k decreases. A simple relationship ehibiting this behavior is k = rsm - Pd,

16-34 Chapter 16: First-Order Differential Euations where r 7 0 is a constant. Notice that k decreases as P increases toward M and that k is negative if P is greater than M. Substituting rsm - Pd for k in Euation (5) gives the differential euation dp 0 M FIGURE 16.24 Euation 6. dp 0 0 dp 0 0 M M 2 P The initial phase line for dp 0 d 2 P d 0 2 P d 2 P 2 0 0 2 2 FIGURE 16.25 The completed phase line for logistic growth (Euation 6). P dp = rsm - PdP = rmp - rp 2. The model given b Euation (6) is referred to as logistic growth. We can forecast the behavior of the population over time b analzing the phase line for Euation (6). The euilibrium values are P = M and P = 0, and we can see that dp> 7 0 if 0 6 P 6 M and dp> 6 0 if P 7 M. These observations are recorded on the phase line in Figure 16.24. We determine the concavit of the population curves b differentiating both sides of Euation (6) with respect to t: d 2 P 2 = d srmp - rp2 d = rm dp - 2rP dp = rsm - 2Pd dp. If P = M>2, then d 2 P> 2 = 0. If P 6 M>2, then sm - 2Pd and dp> are positive and d 2 P> 2 7 0. If M>2 6 P 6 M, then sm - 2Pd 6 0, dp> 7 0, and d 2 P> 2 6 0. If P 7 M, then sm - 2Pd and dp> are both negative and d 2 P> 2 7 0. We add this information to the phase line (Figure 16.25). The lines P = M>2 and P = M divide the first uadrant of the tp-plane into horizontal bands in which we know the signs of both dp> and d 2 P> 2. In each band, we know how the solution curves rise and fall, and how the bend as time passes. The euilibrium lines P = 0 and P = M are both population curves. Population curves crossing the line P = M>2 have an inflection point there, giving them a sigmoid shape (curved in two directions like a letter S). Figure 16.26 displas tpical population curves. (6) (7) P Population M M 2 Limiting population Time t FIGURE 16.26 Population curves in Eample 4.

16.5 Graphical Solutions of Autonomous Euations 16-35 EXERCISES 16.5 In Eercises 1 8, a. Identif the euilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identif the signs of and. c. Sketch several solution curves. 1. = s + 2ds - 3d 2. d = 2-4 d 3. 4. d = 2-2 d = 3-5. =2, 7 0 6. = - 2, 7 0 7. =s - 1ds - 2ds - 3d 8. = 3-2 The autonomous differential euations in Eercises 9 12 represent models for population growth. For each eercise, use a phase line analsis to sketch solution curves for P(t), selecting different starting values P(0) (as in Eample 4). Which euilibria are stable, and which are unstable? dp dp 9. = 1-2P 10. = Ps1-2Pd dp dp 11. = 2PsP - 3d 12. = 3Ps1 - Pd ap - 1 2 b 13. Catastrophic continuation of Eample 4 Suppose that a health population of some species is growing in a limited environment and that the current population P 0 is fairl close to the carring capacit M 0. You might imagine a population of fish living in a freshwater lake in a wilderness area. Suddenl a catastrophe such as the Mount St. Helens volcanic eruption contaminates the lake and destros a significant part of the food and ogen on which the fish depend. The result is a new environment with a carring capacit M 1 considerabl less than M 0 and, in fact, less than the current population P 0. Starting at some time before the catastrophe, sketch a before-and-after curve that shows how the fish population responds to the change in environment. 14. Controlling a population The fish and game department in a certain state is planning to issue hunting permits to control the deer population (one deer per permit). It is known that if the deer population falls below a certain level m, the deer will become etinct. It is also known that if the deer population rises above the carring capacit M, the population will decrease back to M through disease and malnutrition. a. Discuss the reasonableness of the following model for the growth rate of the deer population as a function of time: dp = rpsm - PdsP - md, where P is the population of the deer and r is a positive constant of proportionalit. Include a phase line. b. Eplain how this model differs from the logistic model dp> = rpsm - Pd. Is it better or worse than the logistic model? c. Show that if P 7 M for all t, then lim t: Pstd = M. d. What happens if P 6 m for all t? e. Discuss the solutions to the differential euation. What are the euilibrium points of the model? Eplain the dependence of the stea-state value of P on the initial values of P. About how man permits should be issued? 15. Skdiving If a bo of mass m falling from rest under the action of gravit encounters an air resistance proportional to the suare of velocit, then the bo s velocit t seconds into the fall satisfies the euation m = mg - k 2, k 7 0 where k is a constant that depends on the bo s aeronamic properties and the densit of the air. (We assume that the fall is too short to be affected b changes in the air s densit.) a. Draw a phase line for the euation. b. Sketch a tpical velocit curve. c. For a 160-lb skdiver smg = 160d and with time in seconds and distance in feet, a tpical value of k is 0.005. What is the diver s terminal velocit? 16. Resistance proportional to 2Y A bo of mass m is projected verticall downward with initial velocit 0. Assume that the resisting force is proportional to the suare root of the velocit and find the terminal velocit from a graphical analsis. 17. Sailing A sailboat is running along a straight course with the wind providing a constant forward force of 50 lb. The onl other force acting on the boat is resistance as the boat moves through the water. The resisting force is numericall eual to five times the boat s speed, and the initial velocit is 1 ft> sec. What is the maimum velocit in feet per second of the boat under this wind? 18. The spread of information Sociologists recognize a phenomenon called social diffusion,which is the spreading of a piece of information, technological innovation, or cultural fad among a population. The members of the population can be divided into two classes: those who have the information and those who do not. In a fied population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the information times the number et to receive it. If X denotes the number of individuals who have the information in a population of N people, then a mathematical model for social diffusion is given b dx = kxsn - X d, where t represents time in das and k is a positive constant.

16-36 Chapter 16: First-Order Differential Euations a. Discuss the reasonableness of the model. b. Construct a phase line identifing the signs of X and X. c. Sketch representative solution curves. d. Predict the value of X for which the information is spreading most rapidl. How man people eventuall receive the information? 19. Current in an RL-circuit The accompaning diagram represents an electrical circuit whose total resistance is a constant R ohms and whose self-inductance, 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. From Section 16.2, we have L di + Ri = V, where i is the intensit of the current in amperes and t is the time in seconds. V Switch Use a phase line analsis to sketch the solution curve assuming that the switch in the RL-circuit is closed at time t = 0. What happens to the current as t :? This value is called the stea-state solution. 20. A pearl in shampoo Suppose that a pearl is sinking in a thick fluid, like shampoo, subject to a frictional force opposing its fall and proportional to its velocit. Suppose that there is also a resistive buoant force eerted b the shampoo. According to Archimedes principle, the buoant force euals the weight of the fluid displaced b the pearl. Using m for the mass of the pearl and P for the mass of the shampoo displaced b the pearl as it descends, complete the following steps. a. Draw a schematic diagram showing the forces acting on the pearl as it sinks, as in Figure 16.20. b. Using (t) for the pearl s velocit as a function of time t, write a differential euation modeling the velocit of the pearl as a falling bo. c. Construct a phase line displaing the signs of and. d. Sketch tpical solution curves. e. What is the terminal velocit of the pearl? i a b R L 16.6 Sstems of Euations and Phase Planes In some situations we are led to consider not one, but several first-order differential euations. Such a collection is called a sstem of differential euations. In this section we present an approach to understanding sstems through a graphical procedure known as a phase-plane analsis. We present this analsis in the contet of modeling the populations of trout and bass living in a common pond. Phase Planes A general sstem of two first-order differential euations ma take the form d = F(, ), = G(, ). Such a sstem of euations is called autonomous because d> and > do not depend on the independent variable time t, but onl on the dependent variables and. A solution

16.6 Sstems of Euations and Phase Planes 16-37 of such a sstem consists of a pair of functions (t) and (t) that satisfies both of the differential euations simultaneousl for ever t over some time interval (finite or infinite). We cannot look at just one of these euations in isolation to find solutions (t) or (t) since each derivative depends on both and. To gain insight into the solutions, we look at both dependent variables together b plotting the points ((t), (t)) in the -plane starting at some specified point. Therefore the solution functions are considered as parametric euations (with parameter t), and a corresponding solution curve through the specified point is called a trajector of the sstem. The -plane itself, in which these trajectories reside, is referred to as the phase plane. Thus we consider both solutions together and stu the behavior of all the solution trajectories in the phase plane. It can be proved that two trajectories can never cross or touch each other. A Competitive-Hunter Model Imagine two species of fish, sa trout and bass, competing for the same limited resources in a certain pond. We let (t) represent the number of trout and (t) the number of bass living in the pond at time t. In realit (t) and (t) are alwas integer valued, but we will approimate them with real-valued differentiable functions. This allows us to appl the methods of differential euations. Several factors affect the rates of change of these populations. As time passes, each species breeds, so we assume its population increases proportionall to its size. Taken b itself, this would lead to eponential growth in each of the two populations. However, there is a countervailing effect from the fact that the two species are in competition. A large number of bass tends to cause a decrease in the number of trout, and vice-versa. Our model takes the size of this effect to be proportional to the freuenc with which the two species interact, which in turn is proportional to, the product of the two populations. These considerations lead to the following model for the growth of the trout and bass in the pond: d = (a - b), (1a) = (m - n). (1b) Here (t) represents the trout population, (t) the bass population, and a, b, m, n are positive constants. A solution of this sstem then consists of a pair of functions (t) and (t) that gives the population of each fish species at time t. Each euation in (1) contains both of the unknown functions and, so we are unable to solve them individuall. Instead, we will use a graphical analsis to stu the solution trajectories of this competitive-hunter model. We now eamine the nature of the phase plane in the trout-bass population model. We will be interested in the 1st uadrant of the -plane, where Ú 0 and Ú 0, since populations cannot be negative. First, we determine where the bass and trout populations are both constant. Noting that the ((t), (t)) values remain unchanged when d> = 0 and > = 0, Euations (1a and 1b) then become (a - b) = 0, (m - n) = 0. This pair of simultaneous euations has two solutions: (, ) = (0, 0) and (, ) = (m>n, a>b). At these (, ) values, called euilibrium or rest points, the two populations remain at constant values over all time. The point (0, 0) represents a pond containing no members of either fish species; the point (m>n, a>b) corresponds to a pond with an unchanging number of each fish species.

16-38 Chapter 16: First-Order Differential Euations Bass Net, we note that if = a>b, then Euation (1a) implies d> = 0, so the trout population (t) is constant. Similarl, if = m>n, then Euation (1b) implies > = 0, and the bass population (t) is constant. This information is recorded in Figure 16.27. m n Trout FIGURE 16.28 To the left of the line = m>n the trajectories move upward, and to the right the move downward. a b Bass d = 0 (a) Trout Bass m n (b) = 0 Trout a b Bass m n m a ( n, b ) (c) Trout Bass FIGURE 16.27 Rest points in the competitive-hunter model given b Euations (1a and 1b). a b Trout FIGURE 16.29 Above the line = a>b the trajectories move to the left, and below it the move to the right. a b Bass (0, 0) A C m n FIGURE 16.30 Composite graphical analsis of the trajector directions in the four regions determined b = m>n and = a>b. B D Trout In setting up our competitive-hunter model, precise values of the constants a, b, m, n will not generall be known. Nonetheless, we can analze the sstem of Euations (1) to learn the nature of its solution trajectories. We begin b determining the signs of d> and > throughout the phase plane. Although (t) represents the number of trout and (t) the number of bass at time t, we are thinking of the pair of values ((t), (t)) as a point tracing out a trajector curve in the phase plane. When d> is positive, (t) is increasing and the point is moving to the right in the phase plane. If d> is negative, the point is moving to the left. Likewise, the point is moving upward where > is positive and downward where > is negative. We saw that > = 0 along the vertical line = m>n. To the left of this line, > is positive since > = (m - n) and 6 m>n. So the trajectories on this side of the line are directed upward. To the right of this line, > is negative and the trajectories point downward. The directions of the associated trajectories are indicated in Figure 16.28. Similarl, above the horizontal line = a>b, we have d> 6 0 and the trajectories head leftward; below this line the head rightward, as shown in Figure 16.29. Combining this information gives four distinct regions in the plane A, B, C, D, with their respective trajector directions shown in Figure 16.30. Net, we eamine what happens near the two euilibrium points. The trajectories near (0, 0) point awa from it, upward and to the right. The behavior near the euilibrium point (m>n, a>b) depends on the region in which a trajector begins. If it starts in region B, for instance, then it will move downward and leftward towards the euilibrium point. Depending on where the trajector begins, it ma move downward into region D, leftward into region A, or perhaps straight into the euilibrium point. If it enters into regions A or D, then it will continue to move awa from the rest point. We sa that both rest points are unstable, meaning (in this setting) there are trajectories near each point that head awa from them. These features are indicated in Figure 16.31. It turns out that in each of the half-planes above and below the line = a>b, there is eactl one trajector approaching the euilibrium point (m>n, a>b) (see Eercise 7). Above these two trajectories the bass population increases and below them it decreases. The two trajectories approaching the euilibrium point are suggested in Figure 16.32.

16.6 Sstems of Euations and Phase Planes 16-39 a b Bass (0, 0) FIGURE 16.31 Motion along the trajectories near the rest points (0, 0) and (m>n, a>b). Bass (0, 0) A C Bass win m n FIGURE 16.32 Qualitative results of analzing the competitive-hunter model. There are eactl two trajectories approaching the point (m>n, a>b). B D m a ( n, b) Trout win Trout Trout Our graphical analsis leads us to conclude that, under the assumptions of the competitive-hunter model, it is unlikel that both species will reach euilibrium levels. This is because it would be almost impossible for the fish populations to move eactl along one of the two approaching trajectories for all time. Furthermore, the initial populations point ( 0, 0 ) determines which of the two species is likel to survive over time, and mutual coeistence of the species is highl improbable. Limitations of the Phase-Plane Analsis Method Unlike the situation for the competitive-hunter model, it is not alwas possible to determine the behavior of trajectories near a rest point. For eample, suppose we know that the trajectories near a rest point, chosen here to be the origin (0, 0), behave as in Figure 16.33. The information provided b Figure 16.33 is not sufficient to distinguish between the three possible trajectories shown in Figure 16.34. Even if we could determine that a trajector near an euilibrium point resembles that of Figure 16.34c, we would still not know how the other trajectories behave. It could happen that a trajector closer to the origin behaves like the motions displaed in Figure 16.34a or 16.34b. The spiraling trajector in Figure 16.34b can never actuall reach the rest point in a finite time period. ( 0, 0 ) ( 0, 0 ) ( 0, 0 ) (a) (b) (c) FIGURE 16.33 Trajector direction near the rest point (0, 0). 2 + 2 = 1 ( 1, 1 ) Another Tpe of Behavior The sstem FIGURE 16.34 Three possible trajector motions: (a) periodic motion, (b) motion toward an asmptoticall stable rest point, and (c) motion near an unstable rest point. d = + - ( 2 + 2 ), = - + - ( 2 + 2 ) (2a) (2b) FIGURE 16.35 The solution 2 + 2 = 1 is a limit ccle. ( 0, 0 ) can be shown to have onl one euilibrium point at (0, 0). Yet an trajector starting on the unit circle traverses it clockwise because, when 2 + 2 = 1, we have >d = -> (see Eercise 2). If a trajector starts inside the unit circle, it spirals outward, asmptoticall approaching the circle as t :. If a trajector starts outside the unit circle, it spirals inward, again asmptoticall approaching the circle as t :. The circle 2 + 2 = 1 is called a limit ccle of the sstem (Figure 16.35). In this sstem, the values of and eventuall become periodic.

16-40 Chapter 16: First-Order Differential Euations EXERCISES 16.6 1. List three important considerations that are ignored in the competitive-hunter model as presented in the tet. 2. For the sstem (2a and 2b), show that an trajector starting on the unit circle 2 + 2 = 1 will traverse the unit circle in a periodic solution. First introduce polar coordinates and rewrite the sstem as dr> = r(1 - r 2 ) and du> = 1. 3. Develop a model for the growth of trout and bass assuming that in isolation trout demonstrate eponential deca [so that a 6 0 in Euations (1a and 1b)] and that the bass population grows logisticall with a population limit M. Analze graphicall the motion in the vicinit of the rest points in our model. Is coeistence possible? 4. How might the competitive-hunter model be validated? Include a discussion of how the various constants a, b, m, and n might be estimated. How could state conservation authorities use the model to ensure the survival of both species? 5. Consider another competitive-hunter model defined b d = a a1 - k 1 b - b, = m a1 - k 2 b - n, where and represent trout and bass populations, respectivel. a. What assumptions are implicitl being made about the growth of trout and bass in the absence of competition? b. Interpret the constants a, b, m, n, k 1, and k 2 in terms of the phsical problem. c. Perform a graphical analsis: i) Find the possible euilibrium levels. ii) Determine whether coeistence is possible. iii) Pick several tpical starting points and sketch tpical trajectories in the phase plane. iv) Interpret the outcomes predicted b our graphical analsis in terms of the constants a, b, m, n, k 1, and k 2. Note: When ou get to part (iii), ou should realize that five cases eist. You will need to analze all five cases. 6. Consider the following economic model. Let P be the price of a single item on the market. Let Q be the uantit of the item available on the market. Both P and Q are functions of time. If one considers price and uantit as two interacting species, the following model might be proposed: dp dq = ap a b Q - Pb, = cq(ƒp - Q), where a, b, c, and ƒ are positive constants. Justif and discuss the adeuac of the model. a. If a = 1, b = 20,000, c = 1, and ƒ = 30, find the euilibrium points of this sstem. If possible, classif each euilibrium point with respect to its stabilit. If a point cannot be readil classified, give some eplanation. b. Perform a graphical stabilit analsis to determine what will happen to the levels of P and Q as time increases. c. Give an economic interpretation of the curves that determine the euilibrium points. 7. Show that the two trajectories leading to (m>n, a>b) shown in Figure 16.32 are uniue b carring out the following steps. a. From sstem (1a and 1b) derive the following euation: (m - n) = d (a - b). b. Separate variables, integrate, and eponentiate to obtain where K is a constant of integration. c. Let ƒ() = a >e b and g() = m >e n. Show that ƒ() has a uniue maimum of M = (a>eb) a when = a>b as shown in Figure 16.36. Similarl, show that g() has a uniue maimum M = (m>en) m when = m>n, also shown in Figure 16.36. M M f() g() a e -b = K m e -n a b m n a e b m e n FIGURE 16.36 Graphs of the functions ƒ() = a >e b and g() = m >e n. d. Consider what happens as (, ) approaches (m>n, a>b). Take limits in part (b) as : m>n and : a>b to show that either

16.6 Sstems of Euations and Phase Planes 16-41 or M >M = K. Thus an solution trajector that approaches (m>n, a>b) must satisf a e. Show that onl one trajector can approach (m>n, a>b) from below the line = a>b. Pick 0 6 a>b. From Figure 16.36 ou can see that ƒ( 0 ) 6 M, which implies that This in turn implies that a lim ca baen :m>n b e m bd = K : a>b e b = am ba m M e n b. M M a m e n b = 0 a >e b0 6 M. m e n 6 M. Figure 16.36 tells ou that for g() there is a uniue value 0 6 m>n satisfing this last ineualit. That is, for each 6 a>b there is a uniue value of satisfing the euation in part (d). Thus there can eist onl one trajector solution approaching (m>n, a>b) from below, as shown in Figure 16.37. f. Use a similar argument to show that the solution trajector leading to (m>n, a>b) is uniue if 0 7 a>b. a b 0 Bass Uniue 0 8. Show that the second-order differential euation =F(,, ) can be reduced to a sstem of two first-order differential euations m n d = z, dz = F(,, z). d Trout FIGURE 16.37 For an 6 a>b onl one solution trajector leads to the rest point (m>n, a>b). Can something similar be done to the nth-order differential euation (n) = FA,,,, Á, (n - 1) B?

Chapter 17 SECOND-ORDER DIFFERENTIAL EQUATIONS OVERVIEW In this chapter we etend our stu of differential euations to those of second order. Second-order differential euations arise in man applications in the sciences and engineering. For instance, the can be applied to the stu of vibrating springs and electric circuits. You will learn how to solve such differential euations b several methods in this chapter. 17.1 Second-Order Linear Euations An euation of the form P() () + Q() () + R()() = G(), which is linear in and its derivatives, is called a second-order linear differential euation. We assume that the functions P, Q, R, and G are continuous throughout some open interval I. If G() is identicall zero on I, the euation is said to be homogeneous; otherwise it is called nonhomogeneous. Therefore, the form of a second-order linear homogeneous differential euation is P() + Q() + R() = 0. We also assume that P() is never zero for an H I. Two fundamental results are important to solving Euation (2). The first of these sas that if we know two solutions 1 and 2 of the linear homogeneous euation, then an linear combination = c 1 1 + c 2 2 is also a solution for an constants and. c 1 c 2 (1) (2) THEOREM 1 The Superposition Principle If 1 () and 2 () are two solutions to the linear homogeneous euation (2), then for an constants c 1 and c 2, the function () = c 1 1 () + c 2 2 () is also a solution to Euation (2). 17-1

17-2 Chapter 17: Second-Order Differential Euations Proof Substituting into Euation (2), we have P() +Q() +R() = P()(c 1 1 + c 2 2 ) +Q()(c 1 1 + c 2 2 ) +R()(c 1 1 + c 2 2 ) = P()(c 1 1 +c 2 2 ) + Q()(c 1 1 +c 2 2 ) + R()(c 1 1 + c 2 2 ) = c 1 (P() 1 +Q() 1 +R() 1 ) + c 2 (P() 2 +Q() 2 +R() 2 ) 144442444443 144442444443 = 0, 1 is a solution 0, 2 is a solution = c 1 (0) + c 2 (0) = 0. Therefore, = c 1 1 + c 2 2 is a solution of Euation (2). Theorem 1 immediatel establishes the following facts concerning solutions to the linear homogeneous euation. 1. A sum of two solutions 1 + 2 to Euation (2) is also a solution. (Choose c 1 = c 2 = 1.) 2. A constant multiple k 1 of an solution 1 to Euation (2) is also a solution. (Choose c 1 = k and c 2 = 0.) 3. The trivial solution () K 0 is alwas a solution to the linear homogeneous euation. (Choose c 1 = c 2 = 0.) The second fundamental result about solutions to the linear homogeneous euation concerns its general solution or solution containing all solutions. This result sas that there are two solutions 1 and 2 such that an solution is some linear combination of them for suitable values of the constants c 1 and c 2. However, not just an pair of solutions will do. The solutions must be linearl independent, which means that neither 1 nor 2 is a constant multiple of the other. For eample, the functions ƒ() = e and g() = e are linearl independent, whereas ƒ() = 2 and g() = 7 2 are not (so the are linearl dependent). These results on linear independence and the following theorem are proved in more advanced courses. THEOREM 2 If P, Q, and R are continuous over the open interval I and P() is never zero on I, then the linear homogeneous euation (2) has two linearl independent solutions 1 and 2 on I. Moreover, if 1 and 2 are an two linearl independent solutions of Euation (2), then the general solution is given b where c 1 and c 2 are arbitrar constants. () = c 1 1 () + c 2 2 (), We now turn our attention to finding two linearl independent solutions to the special case of Euation (2), where P, Q, and R are constant functions. Constant-Coefficient Homogeneous Euations Suppose we wish to solve the second-order homogeneous differential euation a +b +c = 0, (3)

17.1 Second-Order Linear Euations 17-3 where a, b, and c are constants. To solve Euation (3), we seek a function which when multiplied b a constant and added to a constant times its first derivative plus a constant times its second derivative sums identicall to zero. One function that behaves this wa is the eponential function = e r, when r is a constant. Two differentiations of this eponential function give =re r and =r 2 e r, which are just constant multiples of the original eponential. If we substitute = e r into Euation (3), we obtain ar 2 e r + bre r + ce r = 0. Since the eponential function is never zero, we can divide this last euation through b e r. Thus, = e r is a solution to Euation (3) if and onl if r is a solution to the algebraic euation ar 2 + br + c = 0. (4) Euation (4) is called the auiliar euation (or characteristic euation) of the differential euation a +b +c = 0. The auiliar euation is a uadratic euation with roots r 1 = -b + 2b 2-4ac 2a and r 2 = -b - 2b 2-4ac. 2a There are three cases to consider which depend on the value of the discriminant b 2-4ac. Case 1: b 2 4ac>0. In this case the auiliar euation has two real and uneual roots and. Then and 2 = e r 2 1 = e r 1 r 1 r 2 are two linearl independent solutions to Euation (3) because is not a constant multiple of e r 1 e r 2 (see Eercise 61). From Theorem 2 we conclude the following result. THEOREM 3 If r 1 and r 2 are two real and uneual roots to the auiliar euation ar 2 + br + c = 0, then = c 1 e r 1 + c 2 e r 2 is the general solution to a +b +c = 0. EXAMPLE 1 Find the general solution of the differential euation - -6 = 0. Solution Substitution of = e r into the differential euation ields the auiliar euation r 2 - r - 6 = 0, which factors as (r - 3)(r + 2) = 0. The roots are r 1 = 3 and r 2 = -2. Thus, the general solution is = c 1 e 3 + c 2 e -2.

17-4 Chapter 17: Second-Order Differential Euations Case 2: b 2 4ac 0. In this case r 1 = r 2 = -b>2a. To simplif the notation, let r = -b>2a. Then we have one solution 1 = e r with 2ar + b = 0. Since multiplication of e r b a constant fails to produce a second linearl independent solution, suppose we tr multipling b a function instead. The simplest such function would be u() =, so let s see if 2 = e r is also a solution. Substituting into the differential euation gives a 2 +b 2 +c 2 = a(2re r + r 2 e r ) + b(e r + re r ) + ce r = (2ar + b)e r + (ar 2 + br + c)e r = 0(e r ) + (0)e r = 0. The first term is zero because r = -b>2a; the second term is zero because r solves the auiliar euation. The functions and 2 = e r 1 = e r are linearl independent (see Eercise 62). From Theorem 2 we conclude the following result. 2 THEOREM 4 If r is the onl (repeated) real root to the auiliar euation ar 2 + br + c = 0, then = c 1 e r + c 2 e r is the general solution to a +b +c = 0. EXAMPLE 2 Find the general solution to +4 +4 = 0. Solution which factors into The auiliar euation is r 2 + 4r + 4 = 0, (r + 2) 2 = 0. Thus, r = -2 is a double root. Therefore, the general solution is = c 1 e -2 + c 2 e -2. Case 3: b 2 4ac<0. In this case the auiliar euation has two comple roots r and, where and are real numbers and i 2 1 = a + ib r 2 = a - ib a b = -1. (These real numbers are a = -b>2a and b = 24ac - b 2 >2a.) These two comple roots then give rise to two linearl independent solutions and 2 = e (a - ib) = e a 1 = e (a + ib) = e a (cos b + i sin b) (cos b - i sin b). (The epressions involving the sine and cosine terms follow from Euler s identit in Section 9.9.) However, the solutions 1 and 2 are comple valued rather than real valued. Nevertheless, because of the superposition principle (Theorem 1), we can obtain from them the two real-valued solutions 3 = 1 2 1 + 1 2 2 = e a cos b and 4 = 1 2i 1-1 2i 2 = e a sin b. The functions 3 and 4 are linearl independent (see Eercise 63). From Theorem 2 we conclude the following result.

17.1 Second-Order Linear Euations 17-5 THEOREM 5 If r 1 = a + ib and r 2 = a - ib are two comple roots to the auiliar euation ar 2 + br + c = 0, then = e a (c 1 cos b + c 2 sin b) is the general solution to a +b +c = 0. EXAMPLE 3 Find the general solution to the differential euation -4 +5 = 0. Solution The auiliar euation is r 2-4r + 5 = 0. The roots are the comple pair r = (4 ; 216-20)>2 or r 1 = 2 + i and r 2 = 2 - i. Thus, a = 2 and b = 1 give the general solution = e 2 (c 1 cos + c 2 sin ). Initial Value and Boundar Value Problems To determine a uniue solution to a first-order linear differential euation, it was sufficient to specif the value of the solution at a single point. Since the general solution to a secondorder euation contains two arbitrar constants, it is necessar to specif two conditions. One wa of doing this is to specif the value of the solution function and the value of its derivative at a single point: ( 0 ) = 0 and ( 0 ) = 1. These conditions are called initial conditions. The following result is proved in more advanced tets and guarantees the eistence of a uniue solution for both homogeneous and nonhomogeneous second-order linear initial value problems. THEOREM 6 If P, Q, R, and G are continuous throughout an open interval I, then there eists one and onl one function () satisfing both the differential euation P() () + Q() () + R()() = G() on the interval I, and the initial conditions at the specified point 0 H I. ( 0 ) = 0 and ( 0 ) = 1 It is important to realize that an real values can be assigned to 0 and 1 and Theorem 6 applies. Here is an eample of an initial value problem for a homogeneous euation.

17-6 Chapter 17: Second-Order Differential Euations EXAMPLE 4 Find the particular solution to the initial value problem -2 + = 0, (0) = 1, (0) = -1. 4 3 2 1 0 1 2 = e 2e 4 6 8 FIGURE 17.1 Particular solution curve for Eample 4. Solution The auiliar euation is r 2-2r + 1 = (r - 1) 2 = 0. The repeated real root is r = 1, giving the general solution = c 1 e + c 2 e. Then, =c 1 e + c 2 ( + 1)e. From the initial conditions we have 1 = c 1 + c 2 # 0 and -1 = c1 + c 2 # 1. Thus, c 1 = 1 and c 2 = -2. The uniue solution satisfing the initial conditions is = e - 2e. The solution curve is shown in Figure 17.1. Another approach to determine the values of the two arbitrar constants in the general solution to a second-order differential euation is to specif the values of the solution function at two different points in the interval I. That is, we solve the differential euation subject to the boundar values ( 1 ) = 1 and ( 2 ) = 2, where 1 and 2 both belong to I. Here again the values for 1 and 2 can be an real numbers. The differential euation together with specified boundar values is called a boundar value problem. Unlike the result stated in Theorem 6, boundar value problems do not alwas possess a solution or more than one solution ma eist (see Eercise 65). These problems are studied in more advanced tets, but here is an eample for which there is a uniue solution. EXAMPLE 5 Solve the boundar value problem +4 = 0, (0) = 0, a p. 12 b = 1 Solution The auiliar euation is r 2 + 4 = 0, which has the comple roots r = ;2i. The general solution to the differential euation is = c 1 cos 2 + c 2 sin 2. The boundar conditions are satisfied if (0) = c 1 # 1 + c 2 # 0 = 0 a p. 12 b = c 1 cos a p 6 b + c 2 sin a p 6 b = 1 It follows that c 1 = 0 and c 2 = 2. The solution to the boundar value problem is = 2 sin 2.

17.1 Second-Order Linear Euations 17-7 EXERCISES 17.1 In Eercises 1 30, find the general solution of the given euation. 1. - -12 = 0 2. 3 - =0 3. +3-4 = 0 4. -9 = 0 5. -4 = 0 6. -64 = 0 7. 2 - -3 = 0 8. 9 - = 0 9. 8-10 -3 = 0 10. 3-20 +12 = 0 11. +9 = 0 12. +4 +5 = 0 13. +25 = 0 14. + = 0 15. -2 +5 = 0 16. +16 = 0 17. +2 +4 = 0 18. -2 +3 = 0 19. +4 +9 = 0 20. 4-4 +13 = 0 21. =0 22. +8 +16 = 0 23. d 2 d 2 24. d 2-6 d 2 + 4 d + 4 = 0 d + 9 = 0 25. d 2 d 2 + 6 d + 9 = 0 26. 4 d 2-12 2 d d + 9 = 0 27. 4 d 2 d 2 + 4 d + = 0 28. 4 d 2 d 2-4 d + = 0 29. 9 d 2 d 2 + 6 d + = 0 30. 9 d 2-12 2 d d + 4 = 0 In Eercises 31 40, find the uniue solution of the second-order initial value problem. 31. 32. 33. 34. 35. 36. 37. 38. 39. +6 +5 = 0, (0) = 0, (0) = 3 +16 = 0, (0) = 2, (0) = -2 +12 = 0, (0) = 0, (0) = 1 12 +5-2 = 0, (0) = 1, (0) = -1 +8 = 0, (0) = -1, (0) = 2 +4 +4 = 0, (0) = 0, (0) = 1-4 +4 = 0, (0) = 1, (0) = 0 4-4 + = 0, (0) = 4, (0) = 4 4 d 2 + 12 + 9 = 0, (0) = 2, 2 d d d (0) = 1 40. 9 d 2-12 + 4 = 0, (0) = -1, 2 d d d (0) = 1 In Eercises 41 55, find the general solution. 41. -2-3 = 0 42. 6 - - = 0 43. 4 +4 + = 0 44. 9 +12 +4 = 0 45. 4 +20 = 0 46. +2 +2 = 0 47. 25 +10 + = 0 48. 6 +13-5 = 0 49. 4 +4 +5 = 0 50. +4 +6 = 0 51. 16-24 +9 = 0 52. 6-5 -6 = 0 53. 9 +24 +16 = 0 54. 4 +16 +52 = 0 55. 6-5 -4 = 0 In Eercises 56 60, solve the initial value problem. 56. -2 +2 = 0, (0) = 0, (0) = 2 57. +2 + = 0, (0) = 1, (0) = 1 58. 4-4 + = 0, (0) = -1, (0) = 2 59. 3 + -14 = 0, (0) = 2, (0) = -1 60. 4 +4 +5 = 0, (p) = 1, (p) = 0 61. Prove that the two solution functions in Theorem 3 are linearl independent. 62. Prove that the two solution functions in Theorem 4 are linearl independent. 63. Prove that the two solution functions in Theorem 5 are linearl independent. 64. Prove that if 1 and 2 are linearl independent solutions to the homogeneous euation (2), then the functions 3 = 1 + 2 and 4 = 1-2 are also linearl independent solutions. 65. a. Show that there is no solution to the boundar value problem +4 = 0, (0) = 0, (p) = 1. b. Show that there are infinitel man solutions to the boundar value problem +4 = 0, (0) = 0, (p) = 0. 66. Show that if a, b, and c are positive constants, then all solutions of the homogeneous differential euation approach zero as :. a +b +c = 0

17-8 Chapter 17: Second-Order Differential Euations 17.2 Nonhomogeneous Linear Euations In this section we stu two methods for solving second-order linear nonhomogeneous differential euations with constant coefficients. These are the methods of undetermined coefficients and variation of parameters. We begin b considering the form of the general solution. Form of the General Solution Suppose we wish to solve the nonhomogeneous euation where a, b, and c are constants and G is continuous over some open interval I. Let c = c 1 1 + c 2 2 be the general solution to the associated complementar euation a +b +c = 0. (2) (We learned how to find c in Section 17.1.) Now suppose we could somehow come up with a particular function that solves the nonhomogeneous euation (1). Then the sum p also solves the nonhomogeneous euation (1) because a( c + p ) +b( c + p ) +c( c + p ) Moreover, if = () is the general solution to the nonhomogeneous euation (1), it must have the form of Euation (3). The reason for this last statement follows from the observation that for an function satisfing Euation (1), we have p = (a c +b c +c c ) + (a p +b p +c p ) = 0 + G() = G(). a +b +c = G(), = c + p a( - p ) +b( - p ) +c( - p ) = (a +b +c) - (a p +b p +c p ) = G() - G() = 0. c solves E. (2) and p solves E. (1) Thus, c = - p is the general solution to the homogeneous euation (2). We have established the following result. (1) (3) THEOREM 7 The general solution = () to the nonhomogeneous differential euation (1) has the form = c + p, where the complementar solution c is the general solution to the associated homogeneous euation (2) and p is an particular solution to the nonhomogeneous euation (1).

17.2 Nonhomogeneous Linear Euations 17-9 The Method of Undetermined Coefficients This method for finding a particular solution p to the nonhomogeneous euation (1) applies to special cases for which G() is a sum of terms of various polnomials p() multipling an eponential with possibl sine or cosine factors. That is, G() is a sum of terms of the following forms: p 1 ()e r, p 2 ()e a cos b, p 3 ()e a sin b. For instance, 1 -, e 2, e, cos, and 5e - sin 2 represent functions in this categor. (Essentiall these are functions solving homogeneous linear differential euations with constant coefficients, but the euations ma be of order higher than two.) We now present several eamples illustrating the method. EXAMPLE 1 Solve the nonhomogeneous euation -2-3 = 1-2. Solution The auiliar euation for the complementar euation -2-3 = 0 is It has the roots r = -1 and r = 3 giving the complementar solution c = c 1 e - + c 2 e 3. Now G() = 1-2 is a polnomial of degree 2. It would be reasonable to assume that a particular solution to the given nonhomogeneous euation is also a polnomial of degree 2 because if is a polnomial of degree 2, then -2-3 is also a polnomial of degree 2. So we seek a particular solution of the form We need to determine the unknown coefficients A, B, and C. When we substitute the polnomial and its derivatives into the given nonhomogeneous euation, we obtain p 2A - 2(2A + B) - 3(A 2 + B + C) = 1-2 or, collecting terms with like powers of, r 2-2r - 3 = (r + 1)(r - 3) = 0. p = A 2 + B + C. -3A 2 + (-4A - 3B) + (2A - 2B - 3C) = 1-2. This last euation holds for all values of if its two sides are identical polnomials of degree 2. Thus, we euate corresponding powers of to get -3A = -1, -4A - 3B = 0, and 2A - 2B - 3C = 1. These euations impl in turn that A 1>3, B -4>9, and C 5>27. Substituting these values into the uadratic epression for our particular solution gives p = 1 3 2-4 9 + 5 27. B Theorem 7, the general solution to the nonhomogeneous euation is = c + p = c 1e - + c 2 e 3 + 1 3 2-4 9 + 5 27.

17-10 Chapter 17: Second-Order Differential Euations EXAMPLE 2 Find a particular solution of - =2 sin. Solution If we tr to find a particular solution of the form and substitute the derivatives of euation p p = A sin in the given euation, we find that A must satisf the -A sin + A cos = 2 sin for all values of. Since this reuires A to eual both -2 and 0 at the same time, we conclude that the nonhomogeneous differential euation has no solution of the form A sin. It turns out that the reuired form is the sum p = A sin + B cos. The result of substituting the derivatives of this new trial solution into the differential euation is -A sin - B cos - (A cos - B sin ) = 2 sin or (B - A) sin - (A + B) cos = 2 sin. This last euation must be an identit. Euating the coefficients for like terms on each side then gives B - A = 2 and A + B = 0. Simultaneous solution of these two euations gives A = -1 and B = 1. Our particular solution is p = cos - sin. EXAMPLE 3 Find a particular solution of -3 +2 = 5e. Solution If we substitute p = Ae and its derivatives in the differential euation, we find that Ae - 3Ae + 2Ae = 5e or 0 = 5e. However, the eponential function is never zero. The trouble can be traced to the fact that = e is alrea a solution of the related homogeneous euation -3 +2 = 0. The auiliar euation is r 2-3r + 2 = (r - 1)(r - 2) = 0, which has r = 1 as a root. So we would epect Ae to become zero when substituted into the left-hand side of the differential euation. The appropriate wa to modif the trial solution in this case is to multipl Ae b. Thus, our new trial solution is p = Ae.

17.2 Nonhomogeneous Linear Euations 17-11 The result of substituting the derivatives of this new candidate into the differential euation is (Ae + 2Ae ) - 3(Ae + Ae ) + 2Ae = 5e or -Ae = 5e. Thus, A = -5 gives our sought-after particular solution p = -5e. EXAMPLE 4 Find a particular solution of -6 +9 = e 3. Solution The auiliar euation for the complementar euation r 2-6r + 9 = (r - 3) 2 = 0 has r = 3 as a repeated root. The appropriate choice for p in this case is neither nor Ae 3 because the complementar solution contains both of those terms alrea. Thus, we choose a term containing the net higher power of as a factor. When we substitute p = A 2 e 3 and its derivatives in the given differential euation, we get (9A 2 e 3 + 12Ae 3 + 2Ae 3 ) - 6(3A 2 e 3 + 2Ae 3 ) + 9A 2 e 3 = e 3 or 2Ae 3 = e 3. Thus, A = 1>2, and the particular solution is p = 1. 2 2 e 3 When we wish to find a particular solution of Euation (1) and the function G() is the sum of two or more terms, we choose a trial function for each term in G() and add them. EXAMPLE 5 Find the general solution to - =5e - sin 2. Solution We first check the auiliar euation r 2 - r = 0. Its roots are r = 1 and r = 0. Therefore, the complementar solution to the associated homogeneous euation is c = c 1 e + c 2. We now seek a particular solution p. That is, we seek a function that will produce 5e - sin 2 when substituted into the left-hand side of the given differential euation. One part of is to produce 5e p, the other -sin 2. Since an function of the form c 1 e is a solution of the associated homogeneous euation, we choose our trial solution p to be the sum p = Ae + B cos 2 + C sin 2, Ae 3 including e where we might otherwise have included onl e. When the derivatives of are substituted into the differential euation, the resulting euation is (Ae + 2Ae - 4B cos 2-4C sin 2) - (Ae + Ae - 2B sin 2 + 2C cos 2) = 5e - sin 2 p

17-12 Chapter 17: Second-Order Differential Euations or Ae - (4B + 2C) cos 2 + (2B - 4C) sin 2 = 5e - sin 2. This euation will hold if A = 5, 4B + 2C = 0, 2B - 4C = -1, or A = 5, B = -1>10, and C = 1>5. Our particular solution is p = 5e - 1. 10 cos 2 + 1 sin 2 5 The general solution to the differential euation is = c + p = c 1e + c 2 + 5e - 1. 10 cos 2 + 1 sin 2 5 You ma find the following table helpful in solving the problems at the end of this section. TABLE 17.1 The method of undetermined coefficients for selected euations of the form a + b + c = G(). If G() has a term Then include this that is a constant epression in the multiple of... And if trial function for p. e r r is not a root of the auiliar euation r is a single root of the auiliar euation r is a double root of the auiliar euation Ae r Ae r A 2 e r sin k, cos k p 2 + + m ki is not a root of the auiliar euation 0 is not a root of the auiliar euation 0 is a single root of the auiliar euation 0 is a double root of the auiliar euation B cos k + C sin k D 2 + E + F D 3 + E 2 + F D 4 + E 3 + F 2 The Method of Variation of Parameters This is a general method for finding a particular solution of the nonhomogeneous euation (1) once the general solution of the associated homogeneous euation is known. The method consists of replacing the constants c 1 and c 2 in the complementar solution b functions 1 = 1 () and 2 = 2 () and reuiring (in a wa to be eplained) that the

17.2 Nonhomogeneous Linear Euations 17-13 resulting epression satisf the nonhomogeneous euation (1). There are two functions to be determined, and reuiring that Euation (1) be satisfied is onl one condition. As a second condition, we also reuire that 1 1 + 2 2 = 0. (4) Then we have = 1 1 + 2 2, = 1 1 + 2 2, = 1 1 + 2 2 + 1 1 + 2 2. If we substitute these epressions into the left-hand side of Euation (1), we obtain 1 (a 1 +b 1 +c 1 ) + 2 (a 2 +b 2 +c 2 ) + a( 1 1 + 2 2 ) = G(). The first two parenthetical terms are zero since and are solutions of the associated homogeneous euation (2). So the nonhomogeneous euation (1) is satisfied if, in addition to Euation (4), we reuire that a( 1 1 + 2 2 ) = G(). (5) Euations (4) and (5) can be solved together as a pair 1 1 + 2 2 = 0, 1 1 + 2 2 = G() a for the unknown functions 1 and 2. The usual procedure for solving this simple sstem is to use the method of determinants (also known as Cramer s Rule), which will be demonstrated in the eamples to follow. Once the derivative functions 1 and 2 are known, the two functions 1 = 1 () and 2 = 2 () can be found b integration. Here is a summar of the method. 1 2 Variation of Parameters Procedure To use the method of variation of parameters to find a particular solution to the nonhomogeneous euation a + b + c = G(), we can work directl with Euations (4) and (5). It is not necessar to rederive them. The steps are as follows. 1. Solve the associated homogeneous euation a +b +c = 0 to find the functions 1 and 2. 2. Solve the euations 1 1 + 2 2 = 0, 1 1 + 2 2 = G() a simultaneousl for the derivative functions 1 and 2. 3. Integrate 1 and 2 to find the functions 1 = 1 () and 2 = 2 (). 4. Write down the particular solution to nonhomogeneous euation (1) as p = 1 1 + 2 2.

17-14 Chapter 17: Second-Order Differential Euations EXAMPLE 6 Find the general solution to the euation + = tan. Solution The solution of the homogeneous euation + = 0 is given b c = c 1 cos + c 2 sin. Since 1 () = cos and 2 () = sin, the conditions to be satisfied in Euations (4) and (5) are Solution of this sstem gives Likewise, 1 = 1 cos + 2 sin = 0, - 1 sin + 2 cos = tan. 0 sin ` tan cos ` = cos sin ` -sin cos ` 2 = After integrating 1 and 2, we have cos 0 ` -sin tan ` = sin. cos sin ` -sin cos ` 1 () = L -sin 2 cos d a = 1 -tan sin cos 2 + sin 2 = -sin2 cos. = - L (sec - cos ) d and = -ln ƒ sec + tan ƒ + sin, 2 () = sin d = -cos. L Note that we have omitted the constants of integration in determining 1 and 2. The would merel be absorbed into the arbitrar constants in the complementar solution. Substituting and into the epression for in Step 4 gives 1 2 p = [-ln ƒ sec + tan ƒ + sin ] cos + (-cos ) sin = (-cos ) ln ƒ sec + tan ƒ. The general solution is = c 1 cos + c 2 sin - (cos ) ln ƒ sec + tan ƒ. p

17.2 Nonhomogeneous Linear Euations 17-15 EXAMPLE 7 Solve the nonhomogeneous euation + -2 = e. Solution The auiliar euation is giving the complementar solution The conditions to be satisfied in Euations (4) and (5) are Solving the above sstem for 1 and 2 gives Likewise, 1 = r 2 + r - 2 = (r + 2)(r - 1) = 0 2 = 0 e ` e e ` e - 2 Integrating to obtain the parameter functions, we have ` c = c 1 e - 2 + c 2 e. 1 e - 2 + 2 e = 0, -2 1 e - 2 + 2 e = e. e ` -2e - 2 e ` e - 2 0-2e - 2 e ` 3e - = 1 () = L - 1 3 e3 d a = 1 = -e2 3e - = - 1 3 e3. e - 3e - = 3. = - 1 3 ae3 3 - L e 3 3 db = 1 27 (1-3)e3, and Therefore, 2 2 () = d = L 3 6. (1-3)e3 p = c de 27-2 + a 2 6 be = 1 27 e - 1 9 e + 1 6 2 e. The general solution to the differential euation is = c 1 e -2 + c 2 e - 1 9 e + 1 6 2 e, where the term (1>27)e in has been absorbed into the term c 2 e p in the complementar solution.

17-16 Chapter 17: Second-Order Differential Euations EXERCISES 17.2 Solve the euations in Eercises 1 16 b the method of undetermined coefficients. 1. -3-10 = -3 2. -3-10 = 2-3 3. - =sin 4. +2 + = 2 5. + = cos 3 6. + = e 2 7. - -2 = 20 cos 8. + = 2 + 3e 9. - = e + 2 10. +2 + = 6 sin 2 11. 12. - -6 = e - - 7 cos +3 +2 = e - + e -2-13. d 2 d 2 + 5 d = 152 14. 15. d 2 d 2-3 d = e3-12 16. Solve the euations in Eercises 17 28 b variation of parameters. 17. + = 18. 19. + = tan, + = sin - p 2 6 6 p 2 20. 21. +2 + = e - 22. - = 23. - = e 24. - = sin 25. +4 +5 = 10 26. - =2 27. 28. d 2 d 2 + = sec, - p 2 6 6 p 2 d 2 d 2 - d = e cos, 7 0 d 2 d 2 - d = -8 + 3 d 2 d 2 + 7 d = 422 + 5 + 1 In each of Eercises 29 32, the given differential euation has a particular solution p of the form given. Determine the coefficients in p. Then solve the differential euation. 29. -5 =e 5, p = A 2 e 5 + Be 5 30. - =cos + sin, p = A cos + B sin 31. + = 2 cos + sin, p = A cos + B sin 32. + - 2 = e, p = A 2 e + Be In Eercises 33 36, solve the given differential euations (a) b variation of parameters and (b) b the method of undetermined coefficients. d 2 d 2 33. 34. d 2-4 + 4 = 2e2 d 2 - d = e + e - d d 2 d 2 35. 36. d 2-9 d 2-4 d - 5 = e + 4 d = 9e9 Solve the differential euations in Eercises 37 46. Some of the euations can be solved b the method of undetermined coefficients, but others cannot. 37. 38. + = cot, 0 6 6 p + = csc, 0 6 6 p 39. -8 =e 8 40. +4 = sin 41. - = 3 42. +4 +5 = + 2 43. +2 = 2 - e 44. +9 = 9 - cos 45. 46. + = sec tan, - p 2 6 6 p 2-3 +2 = e - e 2 The method of undetermined coefficients can sometimes be used to solve first-order ordinar differential euations. Use the method to solve the euations in Eercises 47 50. 47. -3 = e 48. +4 = 49. -3 = 5e 3 50. + = sin Solve the differential euations in Eercises 51 and 52 subject to the given initial conditions. 51. 52. d 2 d 2 + = sec2, - p 2 6 6 p ; (0) = (0) = 1 2 d 2 d 2 + = e2 ; (0) = 0, (0) = 2 5 In Eercises 53 58, verif that the given function is a particular solution to the specified nonhomogeneous euation. Find the general solution and evaluate its arbitrar constants to find the uniue solution satisfing the euation and the given initial conditions. 53. 54. 55. + =, p = 2 2 + =, p = 2 sin +, (0) = 0, (0) = 0 1 2 + + = 4e (cos - sin ), p = 2e cos, (0) = 0, (0) = 1 56. - -2 = 1-2, p = - 1, (0) = 0, (0) = 1 57. 58. -2 + = 2e, p = 2 e, (0) = 1, (0) = 0-2 + = -1 e, 7 0, p = e ln, (1) = e, (1) = 0 -, (0) = 0, (0) = 0 In Eercises 59 and 60, two linearl independent solutions and are given to the associated homogeneous euation of the variablecoefficient nonhomogeneous euation. Use the method of variation of parameters to find a particular solution to the nonhomogeneous euation. Assume 7 0 in each eercise. 59. 2 +2-2 = 2, 1 = - 2, 2 = 60. 2 + - =, 1 = - 1, 2 = 1 2

17.3 Applications 17-17 17.3 Applications In this section we appl second-order differential euations to the stu of vibrating springs and electric circuits. 0 s mass m at euilibrium FIGURE 17.2 Mass m stretches a spring b length s to the euilibrium position at = 0. 0 0 s F s a position after release start position F p F r FIGURE 17.3 The propulsion force (weight) F p pulls the mass downward, but the spring restoring force F s and frictional force F r pull the mass upward. The motion starts at = 0 with the mass vibrating up and down. Vibrations A spring has its upper end fastened to a rigid support, as shown in Figure 17.2. An object of mass m is suspended from the spring and stretches it a length s when the spring comes to rest in an euilibrium position. According to Hooke s Law (Section 6.5), the tension force in the spring is ks, where k is the spring constant. The force due to gravit pulling down on the spring is mg, and euilibrium reuires that ks = mg. (1) Suppose that the object is pulled down an additional amount 0 beond the euilibrium position and then released. We want to stu the object s motion, that is, the vertical position of its center of mass at an future time. Let, with positive direction downward, denote the displacement position of the object awa from the euilibrium position = 0 at an time t after the motion has started. Then the forces acting on the object are (see Figure 17.3) F p = mg, the propulsion force due to gravit, F s = k(s + ), the restoring force of the spring s tension, F r = d a frictional force assumed proportional to velocit., The frictional force tends to retard the motion of the object. The resultant of these forces is F = F p - F s - F r, and b Newton s second law F = ma, we must then have m d 2. 2 = mg - ks - k - d B Euation (1), mg - ks = 0, so this last euation becomes subject to the initial conditions (0) = 0 and (0) = 0. (Here we use the prime notation to denote differentiation with respect to time t.) You might epect that the motion predicted b Euation (2) will be oscillator about the euilibrium position = 0 and eventuall damp to zero because of the retarding frictional force. This is indeed the case, and we will show how the constants m, d, and k determine the nature of the damping. You will also see that if there is no friction (so d = 0), then the object will simpl oscillate indefinitel. Simple Harmonic Motion m d 2 2 + d + k = 0, Suppose first that there is no retarding frictional force. Then d = 0 and there is no damping. If we substitute v = 2k>m to simplif our calculations, then the second-order euation (2) becomes +v 2 = 0, with (0) = 0 and (0) = 0. (2)

17-18 Chapter 17: Second-Order Differential Euations C = c 1 2 + c 2 2 c 1 The auiliar euation is having the imaginar roots r = ;vi. The general solution to the differential euation in (2) is To fit the initial conditions, we compute r 2 + v 2 = 0, = c 1 cos vt + c 2 sin vt. =-c 1 v sin vt + c 2 v cos vt and then substitute the conditions. This ields c 1 = 0 and c 2 = 0. The particular solution = 0 cos vt describes the motion of the object. Euation (4) represents simple harmonic motion of amplitude 0 and period T = 2p>v. The general solution given b Euation (3) can be combined into a single term b using the trigonometric identit sin (vt + f) = cos vt sin f + sin vt cos f. To appl the identit, we take (see Figure 17.4) c 1 = C sin f and c 2 = C cos f, where (3) (4) c 2 FIGURE 17.4 c 2 = C cos f. c 1 = C sin f and C = 2c 1 2 + c 2 2 Then the general solution in Euation (3) can be written in the alternative form = C sin (vt + f). (5) Here C and f ma be taken as two new arbitrar constants, replacing the two constants c 1 and c 2. Euation (5) represents simple harmonic motion of amplitude C and period T = 2p>v. The angle vt + f is called the phase angle, and f ma be interpreted as its initial value. A graph of the simple harmonic motion represented b Euation (5) is given in Figure 17.5. and f = tan -1 c 1 c 2. C C sin Period T = 2 0 t C = C sin( t + ) FIGURE 17.5 Simple harmonic motion of amplitude C and period T with initial phase angle f (Euation 5).

17.3 Applications 17-19 Damped Motion Assume now that there is friction in the spring sstem, so v = 2k>m and 2b = d>m, then the differential euation (2) is The auiliar euation is +2b +v 2 = 0. d Z 0. If we substitute r 2 + 2br + v2 = 0, with roots r = -b ; 2b 2 - v 2. Three cases now present themselves, depending upon the relative sizes of b and v. Case 1: b V. The double root of the auiliar euation is real and euals r = v. The general solution to Euation (6) is = (c 1 + c 2 t)e - vt. This situation of motion is called critical damping and is not oscillator. Figure 17.6a shows an eample of this kind of damped motion. Case 2: b>v. The roots of the auiliar euation are real and uneual, given b r and r 2 = -b - 2b 2 - v 2 1 = -b + 2b 2 - v 2. The general solution to Euation (6) is given b = c 1 e A - b + 2b2 - v 2 Bt + c 2 e A - b - 2b2 - v 2 Bt. Here again the motion is not oscillator and both r 1 and r 2 are negative. Thus approaches zero as time goes on. This motion is referred to as overdamping (see Figure 17.6b). Case 3: b<v. The roots to the auiliar euation are comple and given b r = -b ; i2v 2 - b 2. The general solution to Euation (6) is given b = e - bt Ac 1 cos2v 2 - b 2 t + c 2 sin2v 2 - b 2 tb. This situation, called underdamping, represents damped oscillator motion. It is analogous to simple harmonic motion of period T = 2p> 2v 2 - b 2 ecept that the amplitude is not constant but damped b the factor e - bt. Therefore, the motion tends to zero as t increases, so the vibrations tend to die out as time goes on. Notice that the period T = 2p> 2v 2 - b 2 is larger than the period T 0 = 2p>v in the friction-free sstem. Moreover, the larger the value of b = d>2m in the eponential damping factor, the more uickl the vibrations tend to become unnoticeable. A curve illustrating underdamped motion is shown in Figure 17.6c. (6) 0 t 0 t 0 t = (1 + t)e t = 2e 2t e t = e t sin(5t + /4) (a) Critical damping (b) Overdamping (c) Underdamping FIGURE 17.6 Three eamples of damped vibrator motion for a spring sstem with friction, so d Z 0.

17-20 Chapter 17: Second-Order Differential Euations An eternal force F(t) can also be added to the spring sstem modeled b Euation (2). The forcing function ma represent an eternal disturbance on the sstem. For instance, if the euation models an automobile suspension sstem, the forcing function might represent periodic bumps or potholes in the road affecting the performance of the suspension sstem; or it might represent the effects of winds when modeling the vertical motion of a suspension bridge. Inclusion of a forcing function results in the second-order nonhomogeneous euation m d 2 2 + d + k = F(t). We leave the stu of such spring sstems to a more advanced course. (7) Electric Circuits The basic uantit in electricit is the charge (analogous to the idea of mass). In an electric field we use the flow of charge, or current I = d>, as we might use velocit in a gravitational field. There are man similarities between motion in a gravitational field and the flow of electrons (the carriers of charge) in an electric field. Consider the electric circuit shown in Figure 17.7. It consists of four components: voltage source, resistor, inductor, and capacitor. Think of electrical flow as being like a fluid flow, where the voltage source is the pump and the resistor, inductor, and capacitor tend to block the flow. A batter or generator is an eample of a source, producing a voltage that causes the current to flow through the circuit when the switch is closed. An electric light bulb or appliance would provide resistance. The inductance is due to a magnetic field that opposes an change in the current as it flows through a coil. The capacitance is normall created b two metal plates that alternate charges and thus reverse the current flow. The following smbols specif the uantities relevant to the circuit: : charge at a cross section of a conductor measured in coulombs (abbreviated c); I: current or rate of change of charge d/ (flow of electrons) at a cross section of a conductor measured in amperes (abbreviated A); E: electric (potential) source measured in volts (abbreviated V); V: difference in potential between two points along the conductor measured in volts (V). R, Resistor Voltage source E L, Inductor C, Capacitor FIGURE 17.7 An electric circuit. Ohm observed that the current I flowing through a resistor, caused b a potential difference across it, is (approimatel) proportional to the potential difference (voltage drop). He named his constant of proportionalit 1>R and called R the resistance. So Ohm s law is I = 1 R V.

17.3 Applications 17-21 Similarl, it is known from phsics that the voltage drops across an inductor and a capacitor are and where L is the inductance and C is the capacitance (with the charge on the capacitor). The German phsicist Gustav R. Kirchhoff (1824 1887) formulated the law that the sum of the voltage drops in a closed circuit is eual to the supplied voltage E(t). Smbolicall, this sas that Since I = d>, Kirchhoff s law becomes The second-order differential euation (8), which models an electric circuit, has eactl the same form as Euation (7) modeling vibrator motion. Both models can be solved using the methods developed in Section 17.2. Summar L d 2 2 L di + R d The following chart summarizes our analogies for the phsics of motion of an object in a spring sstem versus the flow of charged particles in an electrical circuit. C, RI + L di + C = E(t). + 1 C = E(t). (8) Linear Second-Order Constant-Coefficient Models Mechanical Sstem Electrical Sstem m + +k = F(t) L +R + 1 C = E(t) : displacement : charge : velocit : current : acceleration : change in current m: mass L: inductance d: damping constant R: resistance k: spring constant 1> C: where C is the capacitance F(t): forcing function E(t): voltage source EXERCISES 17.3 1. A 16-lb weight is attached to the lower end of a coil spring suspended from the ceiling and having a spring constant of 1 lb> ft. The resistance in the spring mass sstem is numericall eual to the instantaneous velocit. At t = 0 the weight is set in motion from a position 2 ft below its euilibrium position b giving it a downward velocit of 2 ft> sec. Write an initial value problem that models the given situation. 2. An 8-lb weight stretches a spring 4 ft. The spring mass sstem resides in a medium offering a resistance to the motion that is numericall eual to 1.5 times the instantaneous velocit. If the weight is released at a position 2 ft above its euilibrium position with a downward velocit of 3 ft> sec, write an initial value problem modeling the given situation.

17-22 Chapter 17: Second-Order Differential Euations 3. A 20-lb weight is hung on an 18-in. spring and stretches it 6 in. The weight is pulled down 5 in. and 5 lb are added to the weight. If the weight is now released with a downward velocit of 0 in. > sec, write an initial value problem modeling the vertical displacement. 4. A 10-lb weight is suspended b a spring that is stretched 2 in. b the weight. Assume a resistance whose magnitude is 20> 1g lb times the instantaneous velocit in feet per second. If the weight is pulled down 3 in. below its euilibrium position and released, formulate an initial value problem modeling the behavior of the spring mass sstem. 5. An (open) electrical circuit consists of an inductor, a resistor, and a capacitor. There is an initial charge of 2 coulombs on the capacitor. At the instant the circuit is closed, a current of 3 amperes is present and a voltage of Estd = 20 cos t is applied. In this circuit the voltage drop across the resistor is 4 times the instantaneous change in the charge, the voltage drop across the capacitor is 10 times the charge, and the voltage drop across the inductor is 2 times the instantaneous change in the current. Write an initial value problem to model the circuit. 6. An inductor of 2 henrs is connected in series with a resistor of 12 ohms, a capacitor of 1> 16 farad, and a 300 volt batter. Initiall, the charge on the capacitor is zero and the current is zero. Formulate an initial value problem modeling this electrical circuit. Mechanical units in the British and metric sstems ma be helpful in doing the following problems. Unit British Sstem MKS Sstem Distance Feet (ft) Meters (m) Mass Slugs Kilograms (kg) Time Seconds (sec) Seconds (sec) Force Pounds (lb) Newtons (N) g(earth) 32 ft> sec 2 9.81 m> sec 2 7. A 16-lb weight is attached to the lower end of a coil spring suspended from the ceiling and having a spring constant of 1 lb> ft. The resistance in the spring mass sstem is numericall eual to the instantaneous velocit. At t = 0 the weight is set in motion from a position 2 ft below its euilibrium position b giving it a downward velocit of 2 ft> sec. At the end of p sec, determine whether the mass is above or below the euilibrium position and b what distance. 8. An 8-lb weight stretches a spring 4 ft. The spring mass sstem resides in a medium offering a resistance to the motion eual to 1.5 times the instantaneous velocit. If the weight is released at a position 2 ft above its euilibrium position with a downward velocit of 3 ft> sec, find its position relative to the euilibrium position 2 sec later. 9. A 20-lb weight is hung on an 18-in. spring stretching it 6 in. The weight is pulled down 5 in. and 5 lb are added to the weight. If the weight is now released with a downward velocit of 0 in. > sec, find the position of mass relative to the euilibrium in terms of 0 and valid for an time t Ú 0. 10. A mass of 1 slug is attached to a spring whose constant is 25> 4 lb> ft. Initiall the mass is released 1 ft above the euilibrium position with a downward velocit of 3 ft> sec, and the subseuent motion takes place in a medium that offers a damping force numericall eual to 3 times the instantaneous velocit. An eternal force ƒ(t) is driving the sstem, but assume that initiall ƒ(t) K 0. Formulate and solve an initial value problem that models the given sstem. Interpret our results. 11. A 10-lb weight is suspended b a spring that is stretched 2 in. b the weight. Assume a resistance whose magnitude is 40> 1g lb times the instantaneous velocit in feet per second. If the weight is pulled down 3 in. below its euilibrium position and released, find the time reuired to reach the euilibrium position for the first time. 12. A weight stretches a spring 6 in. It is set in motion at a point 2 in. below its euilibrium position with a downward velocit of 2 in. > sec. a. When does the weight return to its starting position? b. When does it reach its highest point? c. Show that the maimum velocit is 212g + 1 in. > sec. 13. A weight of 10 lb stretches a spring 10 in. The weight is drawn down 2 in. below its euilibrium position and given an initial velocit of 4 in. > sec. An identical spring has a different weight attached to it. This second weight is drawn down from its euilibrium position a distance eual to the amplitude of the first motion and then given an initial velocit of 2 ft> sec. If the amplitude of the second motion is twice that of the first, what weight is attached to the second spring? 14. A weight stretches one spring 3 in. and a second weight stretches another spring 9 in. If both weights are simultaneousl pulled down 1 in. below their respective euilibrium positions and then released, find the first time after t = 0 when their velocities are eual. 15. A weight of 16 lb stretches a spring 4 ft. The weight is pulled down 5 ft below the euilibrium position and then released. What initial velocit 0 given to the weight would have the effect of doubling the amplitude of the vibration? 16. A mass weighing 8 lb stretches a spring 3 in. The spring mass sstem resides in a medium with a damping constant of 2 lb-sec> ft. If the mass is released from its euilibrium position with a velocit of 4 in. > sec in the downward direction, find the time reuired for the mass to return to its euilibrium position for the first time. 17. A weight suspended from a spring eecutes damped vibrations with a period of 2 sec. If the damping factor decreases b 90% in 10 sec, find the acceleration of the weight when it is 3 in. below its euilibrium position and is moving upward with a speed of 2 ft> sec. 18. A 10-lb weight stretches a spring 2 ft. If the weight is pulled down 6 in. below its euilibrium position and released, find the highest point reached b the weight. Assume the spring mass sstem resides in a medium offering a resistance of 10> 1g lb times the instantaneous velocit in feet per second.

17.4 Euler Euations 17-23 19. An LRC circuit is set up with an inductance of 1> 5 henr, a resistance position thereb stretching the spring 1.96 m. The mass is in a of 1 ohm, and a capacitance of 5> 6 farad. Assuming the initial charge is 2 coulombs and the initial current is 4 amperes, find the solution function describing the charge on the capacitor at an time. What is the charge on the capacitor after a long period of time? viscous medium that offers a resistance in newtons numericall eual to 4 times the instantaneous velocit measured in meters per second. The mass is then pulled down 2 m below its euilibrium position and released with a downward velocit of 3 m> sec. 20. An (open) electrical circuit consists of an inductor, a resistor, and At this same instant an eternal force given b ƒ(t) = 20 cos t (in a capacitor. There is an initial charge of 2 coulombs on the capacitor. At the instant the circuit is closed, a current of 3 amperes is if the mass is above or below its euilibrium position and b how newtons) is applied to the sstem. At the end of p sec determine present but no eternal voltage is being applied. In this circuit the much. voltage drops at three points are numericall related as follows: across the capacitor, 10 times the charge; across the resistor, 4 times the instantaneous change in the charge; and across the inductor, 2 times the instantaneous change in the current. Find the charge on the capacitor as a function of time. 24. An 8-lb weight stretches a spring 4 ft. The spring mass sstem resides in a medium offering a resistance to the motion eual to 1.5 times the instantaneous velocit, and an eternal force given b ƒ(t) = 6 + e - t (in pounds) is being applied. If the weight is released at a position 2 ft above its euilibrium position with downward velocit of 3 ft> sec, find its position relative to the euilib- 21. A 16-lb weight stretches a spring 4 ft. This spring mass sstem is in a medium with a damping constant of 4.5 lb-sec> ft, and an eternal rium after 2 sec have elapsed. force given b ƒ(t) = 4 + e - 2t (in pounds) is being ap- 25. Suppose L = 10 henrs, R = 10 ohms, C = 1>500 farads, plied. What is the solution function describing the position of the mass at an time if the mass is released from 2 ft below the euilibrium position with an initial velocit of 4 ft> sec downward? 22. A 10-kg mass is attached to a spring having a spring constant of 140 N> m. The mass is started in motion from the euilibrium position with an initial velocit of 1 m> sec in the upward direction and with an applied eternal force given b ƒ(t) = 5 sin t (in newtons). The mass is in a viscous medium with a coefficient of resistance eual to 90 N-sec> m. Formulate an initial value problem that models the given sstem; solve the model and interpret the results. E = 100 volts, (0) = 10 coulombs, and (0) = i(0) = 0. Formulate and solve an initial value problem that models the given LRC circuit. Interpret our results. 26. A series circuit consisting of an inductor, a resistor, and a capacitor is open. There is an initial charge of 2 coulombs on the capacitor, and 3 amperes of current is present in the circuit at the instant the circuit is closed. A voltage given b E(t) = 20 cos t is applied. In this circuit the voltage drops are numericall eual to the following: across the resistor to 4 times the instantaneous change in the charge, across the capacitor to 10 times the charge, and across the inductor to 2 times the instantaneous change in the current. Find the charge on the capacitor as a function of time. Deter- 23. A 2-kg mass is attached to the lower end of a coil spring suspended from the ceiling. The mass comes to rest in its euilibrium mine the charge on the capacitor and the current at time t = 10. 17.4 Euler Euations In Section 17.1 we introduced the second-order linear homogeneous differential euation P() () + Q() () + R()() = 0 and showed how to solve this euation when the coefficients P, Q, and R are constants. If the coefficients are not constant, we cannot generall solve this differential euation in terms of elementar functions we have studied in calculus. In this section ou will learn how to solve the euation when the coefficients have the special forms P() = a 2, Q() = b, and R() = c, where a, b, and c are constants. These special tpes of euations are called Euler euations, in honor of Leonhard Euler who studied them and showed how to solve them. Such euations arise in the stu of mechanical vibrations. The General Solution of Euler Euations Consider the Euler euation a 2 +b +c = 0, 7 0. (1)

17-24 Chapter 17: Second-Order Differential Euations To solve Euation (1), we first make the change of variables z = ln and () = Y(z). We net use the chain rule to find the derivatives () and () : and () = d d Y(z) = d dz Y(z) dz d = Y (z) 1 () = d d () = d d Y (z) 1 = - 1 2 Y (z) + 1 dz Y (z) d = - 1 2 Y (z) + 1 2 Y (z). Substituting these two derivatives into the left-hand side of Euation (1), we find a 2 +b +c = a 2 a- 1 2 Y (z) + 1 2 Y (z)b + b a1 Y (z)b + cy(z) = ay (z) + (b - a)y (z) + cy(z). Therefore, the substitutions give us the second-order linear differential euation with constant coefficients ay (z) + (b - a)y (z) + cy(z) = 0. We can solve Euation (2) using the method of Section 17.1. That is, we find the roots to the associated auiliar euation ar 2 + (b - a)r + c = 0 to find the general solution for Y(z). After finding Y(z), we can determine () from the substitution z = ln. EXAMPLE 1 Find the general solution of the euation 2 +2-2 = 0. Solution This is an Euler euation with a = 1, b = 2, and c = -2. The auiliar euation (3) for Y(z) is r 2 + (2-1)r - 2 = (r - 1)(r + 2) = 0, with roots r = -2 and r = 1. The solution for Y(z) is given b Y(z) = c 1 e - 2z + c 2 e z. Substituting z = ln gives the general solution for () : () = c 1 e -2ln + c 2 e ln = c 1-2 + c 2. (2) (3) EXAMPLE 2 Solve the Euler euation 2-5 +9 = 0. Solution Since a = 1, b = -5, and c = 9, the auiliar euation (3) for Y(z) is r 2 + (-5-1)r + 9 = (r - 3) 2 = 0. The auiliar euation has the double root r = 3 giving Y(z) = c 1 e 3z + c 2 ze 3z. Substituting z = ln into this epression gives the general solution () = c 1 e 3ln + c 2 ln e 3ln = c 1 3 + c 2 3 ln.

17.4 Euler Euations 17-25 EXAMPLE 3 Find the particular solution to 2-3 +68 = 0 that satisfies the initial conditions (1) = 0 and (1) = 1. 10 5 0 2 4 6 8 10 5 10 FIGURE 17.8 Eample 3. = 2 sin(8ln) 8 Graph of the solution to Solution Here a = 1, b = -3, and c = 68 substituted into the auiliar euation (3) gives r 2-4r + 68 = 0. The roots are r = 2 + 8i and r = 2-8i giving the solution Y(z) = e 2z (c 1 cos 8z + c 2 sin 8z). Substituting z = ln into this epression gives () = e 2ln Ac 1 cos (8 ln ) + c 2 sin (8 ln )B. From the initial condition (1) = 0, we see that c 1 = 0 and () = c 2 2 sin (8 ln ). To fit the second initial condition, we need the derivative () = c 2 A8 cos (8 ln ) + 2 sin (8 ln )B. Since (1) = 1, we immediatel obtain c 2 = 1>8. Therefore, the particular solution satisfing both initial conditions is () = 1. 8 2 sin (8 ln ) Since -1 sin (8 ln ) 1, the solution satisfies - 2 2 (). 8 8 A graph of the solution is shown in Figure 17.8. EXERCISES 17.4 In Eercises 1 24, find the general solution to the given Euler euation. Assume 7 0 throughout. 1. 2 +2-2 = 0 2. 2 + -4 = 0 3. 2-6 = 0 4. 2 + - = 0 5. 2-5 +8 = 0 6. 2 2 +7 +2 = 0 7. 3 2 +4 =0 8. 2 +6 +4 = 0 9. 2 - + = 0 10. 2 - +2 = 0 11. 2 - +5 = 0 12. 2 +7 +13 = 0 13. 2 +3 +10 = 0 14. 2-5 +10 = 0 15. 4 2 +8 +5 = 0 16. 4 2-4 +5 = 0 17. 2 +3 + = 0 18. 2-3 +9 = 0 19. 2 + =0 20. 4 2 + = 0 21. 9 2 +15 + = 0 22. 16 2-8 +9 = 0 23. 16 2 +56 +25 = 0 24. 4 2-16 +25 = 0 In Eercises 25 30, solve the given initial value problem. 25. 2 +3-3 = 0, (1) = 1, (1) = -1 26. 6 2 +7-2 = 0, (1) = 0, (1) = 1 27. 2 - + = 0, (1) = 1, (1) = 1 28. 2 +7 +9 = 0, (1) = 1, (1) = 0 29. 2 - +2 = 0, (1) = -1, (1) = 1 30. 2 +3 +5 = 0, (1) = 1, (1) = 0

17-26 Chapter 17: Second-Order Differential Euations 17.5 Power-Series Solutions In this section we etend our stu of second-order linear homogeneous euations with variable coefficients. With the Euler euations in Section 17.4, the power of the variable in the nonconstant coefficient had to match the order of the derivative with which it was paired: 2 with, 1 with, and 0 (=1) with. Here we drop that reuirement so we can solve more general euations. Method of Solution The power-series method for solving a second-order homogeneous differential euation consists of finding the coefficients of a power series () = a c n n = c 0 + c 1 + c 2 2 + Á (1) which solves the euation. To appl the method we substitute the series and its derivatives into the differential euation to determine the coefficients c 0, c 1, c 2, Á. The techniue for finding the coefficients is similar to that used in the method of undetermined coefficients presented in Section 17.2. In our first eample we demonstrate the method in the setting of a simple euation whose general solution we alrea know. This is to help ou become more comfortable with solutions epressed in series form. EXAMPLE 1 Solve the euation + = 0 b the power-series method. Solution We assume the series solution takes the form of = a c n n and calculate the derivatives = a nc n n - 1 n = 1 and Substitution of these forms into the second-order euation gives us a n(n - 1)c n n - 2 + a c n n = 0. n = 2 n = 0 n = 0 = a Net, we euate the coefficients of each power of to zero as summarized in the following table. n = 0 n = 2 n(n - 1)c n n - 2. Power of Coefficient Euation 0 2(1)c 2 + c 0 = 0 or c 2 = - 1 2 c 0 1 2 3 4 o n - 2 3(2)c 3 + c 1 4(3)c 4 + c 2 5(4)c 5 + c 3 6(5)c 6 + c 4 n(n - 1)c n + c n - 2 = 0 = 0 = 0 = 0 o = 0 or c 3 = - 1 3 # c 2 1 or c 4 = - 1 4 # c 3 2 or c 5 = - 1 5 # c 4 3 or c 6 = - 1 6 # c 5 4 o 1 or c n = - n(n - 1) c n - 2

17.5 Power-Series Solutions 17-27 From the table we notice that the coefficients with even indices ( n = 2k, k = 1, 2, 3, Á ) are related to each other and the coefficients with odd indices ( n = 2k + 1) are also interrelated. We treat each group in turn. Even indices: Here n = 2k, so the power is 2k - 2. From the last line of the table, we have or From this recursive relation we find Odd indices: Here n = 2k + 1, so the power is. Substituting this into the last line of the table ields or = (-1)k (2k)! c 0. 2k(2k - 1)c 2k + c 2k - 2 = 0 1 c 2k = - 2k(2k - 1) c 2k - 2. 1 c 2k = c- 2k(2k - 1) dc- 1 (2k - 2)(2k - 3) d Á c- 1 4(3) dc- 1 2 dc 0 2k - 1 (2k + 1)(2k)c 2k + 1 + c 2k - 1 = 0 Thus, 1 c 2k + 1 = - (2k + 1)(2k) c 2k - 1. 1 c 2k + 1 = c- (2k + 1)(2k) dc- 1 (2k - 1)(2k - 2) d Á c- 1 5(4) dc- 1 3(2) dc 1 = (-1) k (2k + 1)! c 1. Writing the power series b grouping its even and odd powers together and substituting for the coefficients ields = a n = 0 = a k = 0 (-1) k (-1) k = c 0 a. (2k)! 2k + c 1 a (2k + 1)! 2k + 1 k = 0 c n n c 2k 2k + a k = 0 c 2k + 1 2k + 1 From Table 9.1 in Section 9.10, we see that the first series on the right-hand side of the last euation represents the cosine function and the second series represents the sine. Thus, the general solution to + = 0 is = c 0 cos + c 1 sin. k = 0

17-28 Chapter 17: Second-Order Differential Euations EXAMPLE 2 Find the general solution to + + = 0. Solution We assume the series solution form = a c n n n = 0 and calculate the derivatives = a nc n n - 1 and = a n(n - 1)c n n - 2. n = 1 n = 2 Substitution of these forms into the second-order euation ields a n(n - 1)c n n - 2 + a nc n n + a c n n = 0. n = 2 n = 1 n = 0 We euate the coefficients of each power of to zero as summarized in the following table. Power of Coefficient Euation 1 2(1)c 2 3(2)c 3 + + c 0 = 0 c 1 + c 1 = 0 or or c 2 = - 1 c 3 = - 1 2 c 0 3 c 1 2 3 4 4(3)c 4 + 2c 2 + c 2 = 0 5(4)c 5 + 3c 3 + c 3 = 0 6(5)c 6 + 4c 4 + c 4 = 0 or or or c 4 = - 1 c 5 = - 1 4 c 2 c 6 = - 1 5 c 3 6 c 4 o o o n (n + 2)(n + 1)c or c n + 2 = - 1 n + 2 + (n + 1)c n = 0 n + 2 c n 0 From the table notice that the coefficients with even indices are interrelated and the coefficients with odd indices are also interrelated. Even indices: Here n = 2k - 2, so the power is 2k - 2. From the last line in the table, we have From this recurrence relation we obtain c 2k = a- 1 2k ba- 1 2k - 2 b Á a- 1 6 ba- 1 4 ba- 1 2 bc 0 Odd indices: Here n = 2k - 1, so the power is 2k - 1. From the last line in the table, we have From this recurrence relation we obtain = (-1) k (2)(4)(6) Á (2k) c 0. 1 c 2k + 1 = a- 2k + 1 ba- 1 2k - 1 b Á a- 1 5 ba- 1 3 bc 1 = c 2k = - 1 2k c 2k - 2. 1 c 2k + 1 = - 2k + 1 c 2k - 1. (-1) k (3)(5) Á (2k + 1) c 1.

17.5 Power-Series Solutions 17-29 Writing the power series b grouping its even and odd powers and substituting for the coefficients ields = a k = 0 = c 0 a k = 0 c 2k 2k + a k = 0 c 2k + 1 2k + 1 (-1) k (2)(4) Á (2k) 2k + c 1 a k = 0 (-1) k (3)(5) Á (2k + 1) 2k + 1. EXAMPLE 3 Find the general solution to (1-2 ) -6-4 = 0, 6 1. Solution Notice that the leading coefficient is zero when = ;1. Thus, we assume the solution interval I: -1 6 6 1. Substitution of the series form and its derivatives gives us a n(n - 1)c n n - 2 - a n(n - 1)c n n - 6 a nc n n - 4 a c n n = 0. n = 2 (1-2 ) a n = 2 n(n - 1)c n n - 2-6 a n = 2 = a n = 0 c n n Net, we euate the coefficients of each power of to zero as summarized in the following table. n = 1 n = 1 nc n n - 4 a n = 0 n = 0 c n n = 0, Power of Coefficient Euation 1 2(1)c 2 3(2)c 3-4c 0 = 0-6(1)c 1-4c 1 = 0 or or c 2 = 4 c 3 = 5 2 c 0 3 c 1 2 3 o 4(3)c 4-2(1)c 2-6(2)c 2-4c 2 = 0 5(4)c 5-3(2)c 3-6(3)c 3-4c 3 = 0 o or or c 4 = 6 c 5 = 7 4 c 2 5 c 3 o n (n + 2)(n + 1)c n + 2 - [n(n - 1) + 6n + 4]c n = 0 (n + 2)(n + 1)c n + 2 - (n + 4)(n + 1)c n = 0 or c n + 2 = n + 4 n + 2 c n 0 Again we notice that the coefficients with even indices are interrelated and those with odd indices are interrelated. Even indices: Here n = 2k - 2, so the power is 2k. From the right-hand column and last line of the table, we get c 2k = 2k + 2 c 2k 2k - 2 = a 2k + 2 2k = (k + 1)c 0. 2k ba 2k - 2 ba2k - 2 2k - 4 b Á 6 4 a4 2 bc 0

17-30 Chapter 17: Second-Order Differential Euations Odd indices: Here n = 2k - 1, so the power is 2k + 1. The right-hand column and last line of the table gives us The general solution is c 2k + 1 = 2k + 3 2k + 1 c 2k - 1 = a 2k + 3 2k + 1 ba2k + 1 2k - 1 ba2k - 1 2k - 3 b Á 7 5 a5 3 bc 1 = 2k + 3 3 = a n = 0 = a k = 0 = c 0 a (k + 1) 2k + c 1 a k = 0 c n n c 1. c 2k 2k + a k = 0 c 2k + 1 2k + 1 k = 0 2k + 3 3 2k + 1. EXAMPLE 4 Find the general solution to -2 + = 0. Solution Assuming that = a c n n, substitution into the differential euation gives us a n(n - 1)c n n - 2-2 a nc n n + a c n n = 0. n = 2 We net determine the coefficients, listing them in the following table. n = 0 n = 1 n = 0 Power of Coefficient Euation 0 1 2 3 4 o n 2(1)c 2 + c 0 = 0 3(2)c 3-2c 1 + c 1 = 0 4(3)c 4-4c 2 + c 2 = 0 5(4)c 5-6c 3 + c 3 = 0 6(5)c 6-8c 4 + c 4 = 0 o (n + 2)(n + 1)c n + 2 - (2n - 1)c n = 0 or or or or or or c n + 2 = c 2 = - 1 2 c 0 c 3 = 1 3 # 2 c 1 c 4 = 3 4 # 3 c 2 c 5 = 5 5 # 4 c 3 c 6 = 7 6 # 5 c 4 o 2n - 1 (n + 2)(n + 1) c n

17.5 Power-Series Solutions 17-31 From the recursive relation c n + 2 = 2n - 1 (n + 2)(n + 1) c n, we write out the first few terms of each series for the general solution: = c 0 a1-1 2 2-3 4! 4-21 6! 6 - Á b + c 1 a + 1 3! 3 + 5 5! 5 + 45 7! 7 + Á b. EXERCISES 17.5 In Eercises 1 18, use power series to find the general solution of the differential euation. 1. +2 =0 2. +2 + = 0 3. +4 = 0 4. -3 +2 = 0 5. 2-2 +2 = 0 6. - + = 0 7. (1 + ) - = 0 8. (1-2 ) -4 +6 = 0 9. ( 2-1) +2-2 = 0 10. + - 2 = 0 11. ( 2-1) -6 = 0 12. -( + 2) +2 = 0 13. ( 2-1) +4 +2 = 0 14. -2 +4 = 0 15. -2 +3 = 0 16. (1-2 ) - +4 = 0 17. - +3 = 0 18. 2-4 +6 = 0