Additional Topics in Differential Equations

Transcription

1 6 Additional Topics in Differential Equations 6. Eact First-Order Equations 6. Second-Order Homogeneous Linear Equations 6.3 Second-Order Nonhomogeneous Linear Equations 6.4 Series Solutions of Differential Equations Electrical Circuits (Eercises 33 and 34, p. 47) Parachute Jump (Section Project, p. 48) Undamped or Damped Motion? (Eercise 47, p. 40) Motion of a Spring (Eample 8, p. 38) Cost (Eercise 45, p. 3) Clockwise from top left, nsj-images/istockphoto; Danshutter/Shutterstock.com; ICHIRO/Photodisc/Gett Images; Mircea BEZERGHEANU/Shutterstock.com; Mont Rakusen/Cultura/Gett Images 5

2 6 Chapter 6 Additional Topics in Differential Equations 6. Eact First-Order Equations Solve an eact differential equation. Use an integrating factor to make a differential equation eact. Eact Differential Equations In Chapter 6, ou studied applications of differential equations to growth and deca problems. You also learned more about the basic ideas of differential equations and studied the solution technique known as separation of variables. In this chapter, ou will learn more about solving differential equations and using them in real-life applications. This section introduces ou to a method for solving the first-order differential equation M, d N, d 0 for the special case in which this equation represents the eact differential of a function z f,. Definition of an Eact Differential Equation The equation M, d N, d 0 is an eact differential equation when there eists a function f of two variables and having continuous partial derivatives such that f, M, and f, N,. The general solution of the equation is f, C. From Section 3.3, ou know that if f has continuous second partials, then M f f N. This suggests the following test for eactness. THEOREM 6. Test for Eactness Let M and N have continuous partial derivatives on an open disk R. The differential equation M, d N, d 0 is eact if and onl if M N. Ever differential equation of the form M d N d 0 is eact. In other words, a separable differential equation is actuall a special tpe of an eact equation. Eactness is a fragile condition in the sense that seemingl minor alterations in an eact equation can destro its eactness. This is demonstrated in the net eample.

3 6. Eact First-Order Equations 7 Testing for Eactness Determine whether each differential equation is eact. a. d d 0 b. cos d sin d 0 Solution a. This differential equation is eact because M and Notice that the equation d d 0 is not eact, even though it is obtained b dividing each side of the first equation b. b. This differential equation is eact because M cos sin and N. N sin sin. Notice that the equation cos d sin d 0 is not eact, even though it differs from the first equation onl b a single sign. Note that the test for eactness of M, d N, d 0 is the same as the test for determining whether F, M, i N, j is the gradient of a potential function (Theorem 5.). This means that a general solution f, C to an eact differential equation can be found b the method used to find a potential function for a conservative vector field C = 000 C = 00 Solving an Eact Differential Equation See LarsonCalculus.com for an interactive version of this tpe of eample. Solve the differential equation 3 d d 0. Solution This differential equation is eact because M 3 and N. The general solution, f, C, is f, M, d 3 d 3 g. In Section 5., ou determined g b integrating N, with respect to and reconciling the two epressions for f,. An alternative method is to partiall differentiate this version of f, with respect to and compare the result with N,. In other words, N, f, 3 g g. g 8 C = 0 C = Figure 6. So, g, and it follows that g C. Therefore, f, 3 C and the general solution is 3 C. Figure 6. shows the solution curves that correspond to C, 0, 00, and 000.

4 8 Chapter 6 Additional Topics in Differential Equations Solving an Eact Differential Equation Find the particular solution of cos sin d d 0 that satisfies the initial condition when. Solution The differential equation is eact because M N cos sin. Because N, is simpler than M,, it is better to begin b integrating N,. f, N, d d g M, TECHNOLOGY A graphing utilit can be used to graph a particular solution that satisfies the initial condition of a differential equation. In Eample 3, the differential equation and initial conditions are satisfied when cos 0, which implies that the particular solution can be written as 0 or ± cos. On a graphing utilit screen, the solution would be represented b Figure 6. together with the -ais. 4 4 Figure f, g g cos sin So, g cos sin and g cos sin d cos C which implies that f, cos C, and the general solution is cos C. General solution Appling the given initial condition produces cos C which implies that C 0. So, the particular solution is cos 0. The graph of the particular solution is shown in Figure 6.3. Notice that the graph consists of two parts: the ovals are given b cos 0, and the -ais is given b 0. g cos sin 3π π Figure π 4 ( π, ) π π 3π In Eample 3, note that for z f, cos, the total differential of z is given b dz f, d f, d cos sin d d M, d N, d. In other words, M d N d 0 is called an eact differential equation because M d N d is eactl the differential of f,.

5 Integrating Factors When the differential equation M, d N, d 0 is not eact, it ma be possible to make it eact b multipling b an appropriate factor u,, which is called an integrating factor for the differential equation. a. When the differential equation d d 0 Multipling b an Integrating Factor Not an eact equation is multiplied b the integrating factor u,, the resulting equation d d 0 Eact equation is eact the left side is the total differential of. b. When the equation d d 0 Not an eact equation is multiplied b the integrating factor u,, the resulting equation d d 0 Eact equation is eact the left side is the total differential of. 6. Eact First-Order Equations 9 Finding an integrating factor can be difficult. There are two classes of differential equations, however, whose integrating factors can be found routinel namel, those that possess integrating factors that are functions of either alone or alone. The net theorem, which is presented without proof, outlines a procedure for finding these two special categories of integrating factors. REMARK When either h or k is constant, Theorem 6. still applies. As an aid to remembering these formulas, note that the subtracted partial derivative identifies both the denominator and the variable for the integrating factor. THEOREM 6. Integrating Factors Consider the differential equation M, d N, d 0.. If N, M, N, h is a function of alone, then e. If M, N, M, k is a function of alone, then e h d k d is an integrating factor. is an integrating factor. Eploration In Chapter 6, ou solved the first-order linear differential equation d P Q d b using the integrating factor u e P d. Show that ou can obtain this integrating factor b using the methods of this section.

6 30 Chapter 6 Additional Topics in Differential Equations Solve the differential equation d d 0. Finding an Integrating Factor Solution This equation is not eact because M, and N, 0. However, because M, N, N, 0 it follows that e h d e d e is an integrating factor. Multipling the differential equation b e produces the eact differential equation e e d e d 0 whose solution is obtained as follows. h f, N, d e d e g M, f, e g e e g e Therefore, g e and g e e C, which implies that f, e e e C. The general solution is e e e C, or Ce. General solution The net eample shows how a differential equation can help in sketching a force field given b F, M, i N, j. Force field: F(, ) = i j + + Famil of curves tangent to F: = + Ce Figure 6.4 Sketch the force field F, An Application to Force Fields i j b finding and sketching the famil of curves tangent to F. Solution At the point, in the plane, the vector F, has a slope of d d which, in differential form, is d d 0. d d From Eample 5, ou know that the general solution of this differential equation is Ce. Figure 6.4 shows several representative curves from this famil. Note that the force vector at, is tangent to the curve passing through,.

7 6. Eact First-Order Equations 3 6. Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Testing for Eactness In Eercises 4, determine whether the differential equation is eact. Eplain our reasoning.. d 3 d 0. d d 0 3. sin d cos d 0 4. e d e d 0 Solving an Eact Differential Equation In Eercises 5 4, determine whether the differential equation is eact. If it is, find the general solution d 3 d 0 6. e d e d d 6 0 d 0 8. cos d cos d d d 0 0. e d e d 0. d d 0. e d d 0 3. d d 0 4. e cos d tan d 0 Graphical and Analtic Analsis In Eercises 5 and 6, (a) sketch an approimate solution of the differential equation satisfing the initial condition on the slope field, (b) find the particular solution that satisfies the initial condition, and (c) use a graphing utilit to graph the particular solution. Compare the graph with the sketch in part (a). Differential Equation 5. tan 5 d sec d 0 6. d d Figure for 5 Figure for 6 4 Initial Condition, 4 4 4, 3 4 Finding a Particular Solution In Eercises 7, find the particular solution that satisfies the initial condition. Differential Equation Initial Condition d ln d 0 d d e 3 sin 3 d cos 3 d 0 0. d d 0. 9 d d 0. 4 d 6 d Finding an Integrating Factor In Eercises 3 3, find the integrating factor that is a function of or alone and use it to find the general solution of the differential equation. 3. d 6 d d d d d d d 0 7. d tan d 0 8. d 3 d 0 9. d d d d 0 3. d sin d d 3 3 d 0 Using an Integrating Factor In Eercises 33 36, use the integrating factor to find the general solution of the differential equation. Integrating Factor Differential Equation 33. u, 4 d d u, 3 5 d 3 3 d u, 3 5 d 4 3 d u, 3 d d Integrating Factor Show that each epression is an integrating factor for the differential equation d d 0. (a) (b) (c) (d) 38. Integrating Factor Show that the differential equation a b d b a d 0 is eact onl when a b. For a b, show that m n is an integrating factor, where a m b a b, a b n a b.

8 3 Chapter 6 Additional Topics in Differential Equations Tangent Curves In Eercises 39 4, use a graphing utilit to graph the famil of curves tangent to the force field. 39. F, i j 40. F, i j 4. F, 4 i j 4. F, i j Finding an Equation of a Curve In Eercises 43 and 44, find an equation of the curve with the specified slope passing through the given point. Slope Point d d 3 d d, 0, 45. Cost In a manufacturing process where C represents the cost of producing units, the elasticit of cost is defined as E Find the cost function when the elasticit function is 0 E 0 where C and 00. marginal cost C average cost C d d. 46. HOW DO YOU SEE IT? The graph of the cost function in Eercise 45 is shown below. Use the figure to estimate the limit of the cost function as approaches 00 from the right. Euler s Method In Eercises 47 and 48, (a) use Euler s Method and a graphing utilit to graph the particular solution of the initial value problem over the indicated interval with the specified value of h and initial condition, (b) find the eact solution of the differential equation analticall, and (c) use a graphing utilit to graph the particular solution and compare the result with the graph in part (a). Differential Equation 47., , 5 0. Interval 49. Euler s Method Repeat Eercise 47 for h and discuss how the accurac of the result changes. 50. Euler s Method Repeat Eercise 48 for h 0.5 and discuss how the accurac of the result changes. WRITING ABOUT CONCEPTS True or False? In Eercises 53 56, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 53. The differential equation d d 0 is eact. 54. If M d N d 0 is eact, then M d N d 0 is also eact. 55. If M d N d 0 is eact, then f M d g N d 0 is also eact. 56. The differential equation f d g d 0 is eact. Eact Differential Equation In Eercises 57 and 58, find all values of k such that the differential equation is eact. 57. k 3 d 3 d e d ke d 0 h Initial Condition 0 5. Testing for Eactness Eplain how to determine whether a differential equation is eact. 5. Finding an Integrating Factor Outline the procedure for finding an integrating factor for the differential equation M, d N, d Eact Differential Equation Find all nonzero functions f and g such that g sin d f d 0 is eact. 60. Eact Differential Equation Find all nonzero functions g such that g e d d 0 is eact. Mircea BEZERGHEANU/Shutterstock.com

9 6. Second-Order Homogeneous Linear Equations Second-Order Homogeneous Linear Equations Solve a second-order linear differential equation. Solve a higher-order linear differential equation. Use a second-order linear differential equation to solve an applied problem. Second-Order Linear Differential Equations In this section and the net section, ou will learn methods for solving higher-order linear differential equations. REMARK Notice that this use of the term homogeneous differs from that in Section 6.3. Definition of Linear Differential Equation of Order n Let g, g,..., g n and f be functions of with a common (interval) domain. An equation of the form n g n g n... g n g n f is a linear differential equation of order n. If f 0, then the equation is homogeneous; otherwise, it is nonhomogeneous. Homogeneous equations are discussed in this section, and the nonhomogeneous case is discussed in the net section. The functions,,..., n are linearl independent when the onl solution of the equation C C... C n n 0 is the trivial one C C... C n 0. Otherwise, this set of functions is linearl dependent. Linearl Independent and Dependent Functions a. The functions sin and are linearl independent because the onl values of and for which C sin C 0 for all are C 0 and C 0. b. It can be shown that two functions form a linearl dependent set if and onl if one is a constant multiple of the other. For eample, and 3 are linearl dependent because C C 3 0 has the nonzero solutions C 3 and C. C C

10 34 Chapter 6 Additional Topics in Differential Equations The net theorem points out the importance of linear independence in constructing the general solution of a second-order linear homogeneous differential equation with constant coefficients. THEOREM 6.3 Linear Combinations of Solutions If and are linearl independent solutions of the differential equation then the general solution is a b 0, C C General solution where C and C are constants. Proof This theorem is proved in onl one direction. Letting and be solutions of a b 0 ou obtain the following sstem of equations. a b 0 a b 0 Multipling the first equation b C, multipling the second b C, and adding the resulting equations together produces C C a C C b C C 0 which means that C C is a solution, as desired. The proof that all solutions are of this form is best left to a full course on differential equations. See LarsonCalculus.com for Bruce Edwards s video of this proof. Theorem 6.3 states that when ou can find two linearl independent solutions, ou can obtain the general solution b forming a linear combination of the two solutions. To find two linearl independent solutions, note that the nature of the equation suggests that it ma have solutions of the form If so, then and So, b substitution, e m is a solution if and onl if a b 0 em. me m m e m. a b 0 m e m ame m be m 0 e m m am b 0. Because e m is never 0, e m is a solution if and onl if m am b 0. Characteristic equation This is the characteristic equation of the differential equation a b 0. Note that the characteristic equation can be determined from its differential equation simpl b replacing with m, with m, and with.

11 6. Second-Order Homogeneous Linear Equations 35 Eploration For each differential equation below, find the characteristic equation. Solve the characteristic equation for m, and use the values of m to find a general solution of the differential equation. Using our results, develop a general solution of differential equations with characteristic equations that have distinct real roots. (a) (b) Solve the differential equation 4 0. Solution m 4 0. Characteristic Equation: Distinct Real Zeros In this case, the characteristic equation is Characteristic equation So, m ±. Therefore, and e m e e m e are particular solutions of the differential equation. Furthermore, because these two solutions are linearl independent, ou can appl Theorem 6.3 to conclude that the general solution is C e C e. General solution The characteristic equation in Eample has two distinct real zeros. From algebra, ou know that this is onl one of three possibilities for quadratic equations. In general, the quadratic equation m am b 0 has zeros m a a 4b and which fall into one of three cases.. Two distinct real zeros, m m. Two equal real zeros, m m 3. Two comple conjugate zeros, m i and m i In terms of the differential equation a b 0 m a a 4b these three cases correspond to three different tpes of general solutions. FOR FURTHER INFORMATION For more information on Theorem 6.4, see the article A Note on a Differential Equation b Russell Euler in the 989 winter issue of the Missouri Journal of Mathematical Sciences. THEOREM 6.4 Solutions of The solutions of a b 0 a b 0 fall into one of the following three cases, depending on the solutions of the characteristic equation, m am b 0.. Distinct Real Zeros If m m are distinct real zeros of the characteristic equation, then the general solution is C e m C e m.. Equal Real Zeros If m m are equal real zeros of the characteristic equation, then the general solution is C e m C e m C C e m. 3. Comple Zeros If m i and m i are comple zeros of the characteristic equation, then the general solution is C e cos C e sin.

12 36 Chapter 6 Additional Topics in Differential Equations Characteristic Equation: Comple Zeros f 3 g g f f + g 3 4 The basic solutions in Eample 3, f e 3 cos 3 and g e 3 sin 3, are shown in the graph along with other members of the famil of solutions. Notice that as, all of these solutions approach 0. Figure 6.5 Find the general solution of the differential equation 6 0. Solution The characteristic equation m 6m 0 has two comple zeros, as follows. m 3 ± 3 3 ± 3i 3 6 ± ± 6 ± 3 3, So, and and the general solution is C e 3 cos 3 C e 3 sin 3. The graphs of the basic solutions f e 3 cos 3 and g e 3 sin 3 along with other members of the famil of solutions, are shown in Figure 6.5. In Eample 3, note that although the characteristic equation has two comple zeros, the solution of the differential equation is real. Characteristic Equation: Repeated Zeros Solve the differential equation subject to the initial conditions 0 and 0. Solution The characteristic equation m 4m 4 0 m 0 has two equal zeros given b m. So, the general solution is C e C e. General solution Now, because when 0, ou have C C 0 C. Furthermore, because when 0, ou have C e C e e C 0 5 C. Therefore, the solution is e 5e. Particular solution Tr checking this solution in the original differential equation.

13 Higher-Order Linear Differential Equations For higher-order homogeneous linear differential equations, ou can find the general solution in much the same wa as ou do for second-order equations. That is, ou begin b determining the n zeros of the characteristic equation. Then, based on these n zeros, ou form a linearl independent collection of n solutions. The major difference is that with equations of third or higher order, zeros of the characteristic equation ma occur more than twice. When this happens, the linearl independent solutions are formed b multipling b increasing powers of, as demonstrated in Eamples 6 and 7. Solving a Third-Order Equation Find the general solution of Second-Order Homogeneous Linear Equations 37 Solution The characteristic equation and its zeros are m 3 m 0 m m m 0 m 0,,. Because the characteristic equation has three distinct zeros, the general solution is C C e C 3 e. General solution Solving a Third-Order Equation Find the general solution of Solution The characteristic equation and its zeros are m 3 3m 3m 0 m 3 0 m. Because the zero m occurs three times, the general solution is C e C e C 3 e. General solution Solving a Fourth-Order Equation See LarsonCalculus.com for an interactive version of this tpe of eample. Find the general solution of 4 0. Solution The characteristic equation and its zeros are m 4 m 0 m 0 m ±i. Because each of the zeros m i 0 i and m i 0 i occurs twice, the general solution is C cos C sin C 3 cos C 4 sin. General solution

14 38 Chapter 6 Additional Topics in Differential Equations m l = natural length = displacement A rigid object of mass m attached to the end of the spring causes a displacement of. Figure 6.6 Application One of the man applications of linear differential equations is describing the motion of an oscillating spring. According to Hooke s Law, a spring that is stretched (or compressed) units from its natural length l tends to restore itself to its natural length b a force F that is proportional to. That is, F k, where k is the spring constant and indicates the stiffness of the spring. A rigid object of mass m is attached to the end of a spring and causes a displacement, as shown in Figure 6.6. Assume that the mass of the spring is negligible compared with m. When the object is pulled downward and released, the resulting oscillations are a product of two opposing forces the spring force F k and the weight mg of the object. Under such conditions, ou can use a differential equation to find the position of the object as a function of time t. According to Newton s Second Law of Motion, the force acting on the weight is F ma, where a d dt is the acceleration. Assuming that the motion is undamped that is, there are no other eternal forces acting on the object it follows that m d dt k, and ou have d dt m k 0. Undamped motion of a spring Undamped Motion of a Spring A common tpe of spring is a coil spring, also called a helical spring because the shape of the spring is a heli. A tension coil spring resists being stretched (see photo above and Eample 8). A compression coil spring resists being compressed, such as the spring in a car suspension. A 4-pound weight stretches a spring 8 inches from its natural length. The weight is pulled downward an additional 6 inches and released with an initial upward velocit of 8 feet per second. Find a formula for the position of the weight as a function of time t. Solution B Hooke s Law, 4 k 3, so k 6. Moreover, because the weight w is given b mg, it follows that m w g So, the resulting differential equation for this undamped motion is d dt The characteristic equation m 48 0 has comple zeros m 0 ± 4 3i, so the general solution is C e 0 cos 4 3 t C e 0 sin 4 3 t C cos 4 3 t C sin 4 3 t. When t 0 seconds, 6 inches foot. Using this initial condition, ou have C C 0 To determine C, note that feet per second when t 0 seconds. t 4 3 C sin 4 3 t 4 3 C cos 4 3 t C 3 C 3 Consequentl, the position at time t is given b ICHIRO/Photodisc/Gett Images 8 C. 3 cos 4 3 t sin 4 3 t

15 6. Second-Order Homogeneous Linear Equations 39 A damped vibration could be caused b friction and movement through a liquid. Figure 6.7 The object in Figure 6.7 undergoes an additional damping or frictional force that is proportional to its velocit. A case in point would be the damping force resulting from friction and movement through a fluid. Considering this damping force p d dt the differential equation for the oscillation is m d d k p dt dt or, in standard linear form, d dt p m d dt k 0. m Damping force Damped motion of a spring 6. Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Verifing a Solution In Eercises 4, verif the solution of the differential equation. Then use a graphing utilit to graph the particular solutions for several different values of C and C. What do ou observe? Solution Differential Equation. C C e C e C e 3. C cos C sin 4. C e cos 3 C e sin 3 Finding a General Solution In Eercises 5 30, find the general solution of the linear differential equation Finding a Particular Solution Consider the differential equation and the solution C cos 0 C sin 0. Find the particular solution satisfing each initial condition (a) 0, 0 0 (b) 0 0, 0 (c) 0, Finding a Particular Solution Determine C and such that C sin 3 t is a particular solution of the differential equation where 0 5. Finding a Particular Solution: Initial Conditions In Eercises 33 38, find the particular solution of the linear differential equation that satisfies the initial conditions , , , 0 Finding a Particular Solution: Boundar Conditions In Eercises 39 44, find the particular solution of the linear differential equation that satisfies the boundar conditions, if possible , , , , 0 4, WRITING ABOUT CONCEPTS , , , , , Characteristic Equation The solutions of the differential equation fall into what three cases? What is the relationship of these solutions to the characteristic equation of the differential equation? 46. Linearl Independent Functions Two functions are said to be linearl independent provided what? a b 0

16 40 Chapter 6 Additional Topics in Differential Equations Vibrating Spring In Eercises 49 5, match the differential equation with the graph of a particular solution. [The graphs are labeled (a), (b), (c), and (d).] The correct match can be made b comparing the frequenc of the oscillations or the rate at which the oscillations are being damped with the appropriate coefficient in the differential equation. (a) 47. Undamped or Damped Motion? Several shock absorbers are shown at the right. Do ou think the motion of the spring in a shock absorber is undamped or damped? 48. HOW DO YOU SEE IT? Give a geometric argument to eplain wh the graph cannot be a solution of the differential equation. (It is not necessar to solve the differential equation.) (a) (b) (b) Vibrating Spring In Eercises 53 58, describe the motion of a 3-pound weight suspended on a spring. Assume that the weight stretches the spring foot from its natural position. 53. The weight is pulled foot below the equilibrium position and released. 54. The weight is raised 3 foot above the equilibrium position and released. 55. The weight is raised 3 foot above the equilibrium position and started off with a downward velocit of foot per second. 56. The weight is pulled foot below the equilibrium position and started off with an upward velocit of foot per second. 57. The weight is pulled foot below the equilibrium position and released. The motion takes place in a medium that furnishes a damping force of magnitude 8 speed at all times. 58. The weight is pulled foot below the equilibrium position and released. The motion takes place in a medium that furnishes a damping force of magnitude at all times. 59. Real Zeros The characteristic equation of the differential equation has two equal real zeros given b m r. Show that C e r C e r is a solution. 60. Comple Zeros The characteristic equation of the differential equation a b 0 a b 0 has comple zeros given b m i and m i. Show that C e cos C e sin is a solution. True or False? In Eercises 6 64, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 6. C e 3 C e 3 is the general solution of C C sin C 3 C 4 cos is the general solution of is a solution of a n n a n n... a a 0 0 if and onl if a a It is possible to choose a and b such that e is a solution of a b v 6 (c) (d) Wronskian The Wronskian of two differentiable functions f and g, denoted b W f, g, is defined as the function given b the determinant W f, g f f g g. The functions f and g are linearl independent when there eists at least one value of for which W f, g 0. In Eercises 65 68, use the Wronskian to verif the linear independence of the two functions e a e a e b, a b e a 67. e a sin b 68. e a cos b, b 0 Mont Rakusen/Cultura/Gett Images

17 6.3 Second-Order Nonhomogeneous Linear Equations Second-Order Nonhomogeneous Linear Equations Recognize the general solution of a second-order nonhomogeneous linear differential equation. Use the method of undetermined coefficients to solve a second-order nonhomogeneous linear differential equation. Use the method of variation of parameters to solve a second-order nonhomogeneous linear differential equation. SOPHIE GERMAIN (776 83) Man of the earl contributors to calculus were interested in forming mathematical models for vibrating strings and membranes, oscillating springs, and elasticit. One of these was the French mathematician Sophie Germain, who in 86 was awarded a prize b the French Academ for a paper entitled Memoir on the Vibrations of Elastic Plates. See LarsonCalculus.com to read more of this biograph. Nonhomogeneous Equations In the preceding section, damped oscillations of a spring were represented b the homogeneous second-order linear equation d dt p m d dt k 0. m Free motion This tpe of oscillation is called free because it is determined solel b the spring and gravit and is free of the action of other eternal forces. If such a sstem is also subject to an eternal periodic force, such as a sin bt, caused b vibrations at the opposite end of the spring, then the motion is called forced, and it is characterized b the nonhomogeneous equation d dt p m d dt k a sin bt. m Forced motion In this section, ou will stud two methods for finding the general solution of a nonhomogeneous linear differential equation. In both methods, the first step is to find the general solution of the corresponding homogeneous equation. h General solution of homogeneous equation Having done this, ou tr to find a particular solution of the nonhomogeneous equation. p Particular solution of nonhomogeneous equation B combining these two results, ou can conclude that the general solution of the nonhomogeneous equation is h p as stated in the net theorem. THEOREM 6.5 Solution of Nonhomogeneous Linear Equation Let a b F be a second-order nonhomogeneous linear differential equation. If p is a particular solution of this equation and h is the general solution of the corresponding homogeneous equation, then h p is the general solution of the nonhomogeneous equation. The Granger Collection, NYC

18 4 Chapter 6 Additional Topics in Differential Equations Method of Undetermined Coefficients You alread know how to find the solution of a linear homogeneous differential equation. The remainder of this section looks at was to find the particular solution p. When F in a b F consists of sums or products of n, e m, cos, or sin, ou can find a particular solution p b the method of undetermined coefficients. The object of this method is to guess that the solution is a generalized form of F. Here are some eamples. p. For F 3, choose p A B C.. For F 4e, choose p Ae Be. 3. For F sin, choose p A B C sin D cos. Then, b substitution, determine the coefficients for the generalized solution. Method of Undetermined Coefficients Find the general solution of the equation 3 sin. Solution To find h, solve the characteristic equation. m m 3 0 m m 3 0 m or m 3 So, h C e C e 3. Net, let be a generalized form of sin. p A cos B sin p A sin B cos p A cos B sin Substitution into the original differential equation ields p 3 sin A cos B sin A sin B cos 3A cos 3B sin sin 4A B cos A 4B sin sin. B equating coefficients of like terms, ou obtain 4A B 0 and A 4B with solutions h A 5 Therefore, and B 5. p 5 cos 5 sin and the general solution is h p C e C e 3 5 cos sin. 5

19 6.3 Second-Order Nonhomogeneous Linear Equations 43 In Eample, the form of the homogeneous solution h C e C e 3 has no overlap with the function F in the equation a b F. However, suppose the given differential equation in Eample were of the form 3 e. Now it would make no sense to guess that the particular solution was Ae because this solution would ield 0. In such cases, ou should alter our guess b multipling b the lowest power of that removes the duplication. For this particular problem, ou would guess p Ae. Method of Undetermined Coefficients Find the general solution of e. Solution The characteristic equation m m 0 has solutions m 0 and m. So, h C C e. Because F e, our first choice for would be A B Ce p. However, because h alread contains a constant term C, ou should multipl the polnomial part b and use p A B Ce p A B Ce p B Ce. Substitution into the differential equation produces e B Ce A B Ce e B A 4B Ce e. Equating coefficients of like terms ields the sstem B A 0, 4B, C with solutions A B 4 and C. Therefore, p 4 4 e and the general solution is h p C C e 4 4 e.

20 44 Chapter 6 Additional Topics in Differential Equations In Eample, the polnomial part of the initial guess A B Ce for overlapped b a constant term with h C C e and it was necessar to multipl the polnomial part b a power of that removed the overlap. The net eample further illustrates some choices for p that eliminate overlap with h. Remember that in all cases, the first guess for p should match the tpes of functions occurring in F. p Choosing the Form of the Particular Solution Determine a suitable choice for p for each differential equation, given its general solution of the homogeneous equation. a b F h a. C C b. c. C e C e Solution a. Because F, the normal choice for would be A B C p. However, because alread contains a linear term, ou should multipl b h C C to obtain 0 4 sin 3 C e cos 3 C e sin e p A B 3 C 4. b. Because F 4 sin 3 and each term in contains a factor of e h, ou can simpl let p A cos 3 B sin 3. c. Because F e, the normal choice for would be Ae p. However, because alread contains an e h C e C e term, ou should multipl b to get p A e. Solving a Third-Order Equation See LarsonCalculus.com for an interactive version of this tpe of eample. Find the general solution of 3 3. Solution From Eample 6 in Section 6., ou know that the homogeneous solution is h C e C e C 3 e. Because F, let p A B and obtain p B and p 0. So, b substitution into the general solution, ou have B A B 3B A B. So, B and A 3, which implies that p 3. Therefore, the general solution is h p C e C e C 3 e 3.

21 6.3 Second-Order Nonhomogeneous Linear Equations 45 Variation of Parameters The method of undetermined coefficients works well when F is made up of polnomials or functions whose successive derivatives have a cclical pattern. For functions such as and tan, which do not have such characteristics, it is better to use a more general method called variation of parameters. In this method, ou assume that p has the same form as h, ecept that the constants in are replaced b variables. h Variation of Parameters To find the general solution of the equation steps.. Find h C C.. Replace the constants b variables to form p u u. 3. Solve the following sstem for u and u. u u 0 u u F a b F, 4. Integrate to find u and u. The general solution is h p. use these Solve the differential equation e Solution Variation of Parameters The characteristic equation m m 0 has one repeated solution, m. So, the homogeneous solution is h C C C e C e. Replacing and b and produces C C, u > 0. p u u u e u e. The resulting sstem of equations is u e u e 0 u u e u e e e. m 0 Subtracting the second equation from the first produces u. Then, b substitution in the first equation, ou have u. Finall, integration ields u d and u d ln ln. From this result, it follows that a particular solution is p e ln e and the general solution is C e C e e e ln.

22 46 Chapter 6 Additional Topics in Differential Equations Eploration Notice in Eample 5 that the constants of integration were not introduced when finding and u. Show that for u u a the general solution and u ln a h p C e C e e e ln ields the same result as the solution obtained in the eample. Variation of Parameters Solve the differential equation Solution Because the characteristic equation m 0 has solutions m ±i, the homogeneous solution is Replacing C and C b u and u produces p u cos u sin. The resulting sstem of equations is Multipling the first equation b sin and the second b cos produces Adding these two equations produces u sin, which implies that Integration ields u cos sec d and u so that h C cos C sin. u cos u sin 0 u sin u cos tan. u sin cos u sin 0 u sin cos u cos sin. u sin cos cos cos cos sec. sin d cos p sin cos cos ln sec tan sin cos cos ln sec tan and the general solution is tan. sin ln sec tan h p C cos C sin cos ln sec tan.

23 6.3 Second-Order Nonhomogeneous Linear Equations Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Verifing a Solution In Eercises 4, verif the solution of the differential equation. Solution Differential Equation. e cos 0e. sin 3. 3 sin cos ln sec tan 4. 5 ln sin cos sin Finding a Particular Solution In Eercises 5 0, find a particular solution of the differential equation e 3 3 e 5 sin 4 5 e cos Method of Undetermined Coefficients In Eercises 8, solve the differential equation b the method of undetermined coefficients. 3 3 e 9 5e e e 9 sin 3 3 e Method of Undetermined Coefficients In Eercises 9 4, solve the differential equation b the method of undetermined coefficients that satisfies the initial condition(s) , 0 0 sin.. 0 0, 0 3 0, 0 4 e e 4 sin Method of Variation of Parameters In Eercises 5 30, solve the differential equation b the method of variation of parameters. sec csc e ln nsj-images/istockphoto 4 4 0, cos 5 cos tan csc cot sec tan 4 4 e 4 4 e WRITING ABOUT CONCEPTS 3. Choosing p Using the method of undetermined coefficients, determine a suitable choice for p for each differential equation. Eplain our reasoning. (You do not need to solve the differential equations.) (a) (b) 3. Variation of Parameters Describe the steps for solving a differential equation b the method of variation of parameters. Electrical Circuits Vibrating Spring In Eercises 35 38, find the particular solution of the differential equation w g t b t k t w g F t for the oscillating motion of an object on the end of a spring. Use a graphing utilit to graph the solution. In the equation, is the displacement from equilibrium (positive direction is downward), measured in feet, and t is time in seconds (see figure). The constant w is the weight of the object, g is the acceleration due to gravit, b is the magnitude of l = natural the resistance to the motion, k length is the spring constant from Hooke s Law, and F t is the = displacement m acceleration imposed on the Spring displacement sstem sin 4t 0 4, e 4 In Eercises 33 and 34, use the electrical circuit differential equation d q dt R L dq dt LC q L E t where R is the resistance (in ohms), C is the capacitance (in farads), L is the inductance (in henrs), E t is the electromotive force (in volts), and q is the charge on the capacitor (in coulombs). Find the charge q as a function of time for the electrical circuit described. Assume that q 0 0 and q R 0, C 0.0, L, E t sin 5t 34. R 0, C 0.0, L, E t 0 sin 5t

24 48 Chapter 6 Additional Topics in Differential Equations sin 8t 0 4, sin 8t 0 4, , Vibrating Spring Rewrite h in the solution to Eercise 35 b using the identit a cos t b sin t a b sin t where arctan a b. 40. HOW DO YOU SEE IT? The figure shows the particular solution of the differential equation 4 3 b 5 0 that satisfies the initial conditions 0 and 0 4 for values of the resistance component b in the interval 0,. (Note that when b, the problem is identical to that of Eercise 38.) According to the figure, is the motion damped or undamped when b 0? when b > 0? (You do not need to solve the differential equation.) t 4. Vibrating Spring Refer to the differential equation and the initial conditions given in Eercise 40. (a) When there is no resistance to the motion b 0, describe the motion. (b) For b > 0, what is the ultimate effect of the retarding force? (c) Is there a real number M such that there will be no oscillations of the spring for b > M? Eplain our answer. 4. Solving a Differential Equation Solve the differential equation given that and are solutions of the corresponding homogeneous equation. (a) 4 ln, ln (b) 4 sin ln sin ln, cos ln Danshutter/Shutterstock.com b = 0 Generated b Maple b = b = b True or False? In Eercises 43 and 44, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 43. p e cos e is a particular solution of the differential equation 3 cos e. 44. p 8e is a particular solution of the differential equation 6 e. PUTNAM EXAM CHALLENGE 45. For all real, the real-valued function f satisfies e. (a) If f > 0 for all real, must > 0 for all real? Eplain. (b) If > 0 for all real, must f > 0 for all real? Eplain. This problem was composed b the Committee on the Putnam Prize Competition. The Mathematical Association of America. All rights reserved. f Parachute Jump The fall of a parachutist is described b the second-order linear differential equation w d d k g dt dt w where w is the weight of the parachutist, is the height at time t, g is the acceleration due to gravit, and k is the drag factor of the parachute. (a) The parachute is opened at 000 feet, so At that time, the velocit is 0 00 feet per second. For a 60-pound parachutist, using k 8, the differential equation is Using the initial conditions, verif that the solution of the differential equation is e.6t 0t. (b) Consider a 9-pound parachutist who has a parachute with a drag factor of k 9. Using the initial conditions given in part (a), write and solve a differential equation that describes the fall of the parachutist. f

25 6.4 Series Solutions of Differential Equations Series Solutions of Differential Equations Use a power series to solve a differential equation. Use a Talor series to find the series solution of a differential equation. Power Series Solution of a Differential Equation Power series can be used to solve certain tpes of differential equations. This section begins with the general power series solution method. Recall from Chapter 9 that a power series represents a function f on an interval of convergence, and that ou can successivel differentiate the power series to obtain a series for f, f, and so on. These properties are used in the power series solution method demonstrated in the first two eamples. Power Series Solution Use a power series to solve the differential equation Solution Assume that a n n is a solution. Then, n Substituting for and, ou obtain the following series form of the differential equation. (Note that, from the third step to the fourth, the inde of summation is changed to ensure that n occurs in both sums.) Now, b equating coefficients of like terms, ou obtain the recursion formula which implies that na n n. n a n a n 0 na n n a n n 0 n na n n a n n n n a n n a n n 0. Eploration In Eample, the differential equation could be solved easil without using a series. Determine which method should be used to solve the differential equation 0 and show that the result is the same as that obtained in the eample. a n This formula generates the following results.... a 0 a 0 a a 0... Using these values as the coefficients for the solution series, ou have n a 0 n! n a 0 n n! a 0 e. a n n, a a 0 n 0. a 3 3 a 0 3! a 4 4 a 0 4! a 5 5 a 0 5!

26 50 Chapter 6 Additional Topics in Differential Equations In Eample, the differential equation could be solved easil without using a series. The differential equation in Eample cannot be solved b an of the methods discussed in previous sections. Power Series Solution Use a power series to solve the differential equation 0. Solution Assume that a n n is a solution. Then ou have Substituting for,, and in the given differential equation, ou obtain the following series. To obtain equal powers of, adjust the summation indices b replacing n b n in the left-hand sum, to obtain B equating coefficients, ou have from which ou obtain the recursion formula and the coefficients of the solution series are as follows. a k n a a 0 a 4 a 4 a 0 4 a 6 a 4 6 a 0 So, ou can represent the general solution as the sum of two series one for the even-powered terms with coefficients in terms of a 0, and one for the odd-powered terms with coefficients in terms of a. a a a 0 k a k k a 0 k k! k 0 na n n, n a n n n a n a n n, k k k k! 4 6 n n a n n n a n n. n n a n n a n a n n n a n n na n n a n n 0 n n n a n n n a n n n k 0 na n n, n 0, k k a 3 a 3 a k k a 5 a 3 5 a 3 5 a 7 a 5 7 a k a k The solution has two arbitrar constants, a 0 and a, as ou would epect in the general solution of a second-order differential equation. n n n a n n

27 Approimation b Talor Series A second tpe of series solution method involves a differential equation with initial conditions and makes use of Talor series, as given in Section 9.0. Approimation b Talor Series See LarsonCalculus.com for an interactive version of this tpe of eample. Use a Talor series to find the first si terms of the series solution of for the initial condition when 0. Then, use this polnomial to approimate values of for 0. Solution Recall from Section 9.0 that, for c 0, 0 0 0! Because 0 and ou obtain the following., So, can be approimated b the first si terms of the series solution shown below ! 6.4 Series Solutions of Differential Equations 5 0 3! ! 4 3! 3 4 4! ! ! ! 5 Using this polnomial, ou can approimate values for in the interval 0, as shown in the table below In addition to approimating values of a function, ou can also use a series solution to sketch a graph. In Figure 6.8, the series solutions of using the first two, four, and si terms are shown, along with an approimation found using a computer algebra sstem. The approimations are nearl the same for values of close to 0. As approaches, however, there is a noticeable difference among the approimations. For a series solution that is more accurate near, repeat Eample 3 using c Figure terms 4 terms terms

28 5 Chapter 6 Additional Topics in Differential Equations 6.4 Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Equivalent Solution Techniques In Eercises 6, verif that the power series solution of the differential equation is equivalent to the solution found using previousl learned solution techniques Power Series Solution In Eercises 7 0, use power series to solve the differential equation and find the interval of convergence of the series k 0 k 0 k Finding Terms of a Power Series Solution In Eercises and, find the first three terms of each of the power series representing independent solutions of the differential equation. 8. HOW DO YOU SEE IT? Consider the differential equation 9 0 with initial conditions 0 and 0 6. The figure shows the graph of the solution of the differential equation and the third-degree and fifth-degree polnomial approimations of the solution. Identif each Approimation b Talor Series In Eercises 3 and 4, use a Talor series to find the first n terms of the series solution of the differential equation under the specified initial conditions. Use this polnomial to approimate for the given value of and compare the result with the approimation given b Euler s Method for h , 0, n 5, 0, 0, n 4, WRITING ABOUT CONCEPTS 5. Power Series Solution Method Describe how to use power series to solve a differential equation. 6. Recursion Formula What is a recursion formula? Give an eample. 7. Investigation Consider the differential equation 0 with the initial conditions 0 0 and 0. (See Eercise 9.) (a) Find the series solution satisfing the initial conditions. (b) Use a graphing utilit to graph the third-degree and fifthdegree series approimations of the solution. Identif the approimations. (c) Identif the smmetr of the solution. Approimation b Talor Series In Eercises 9, use a Talor series to find the first n terms of the series solution of the differential equation under the specified initial conditions. Use this polnomial to approimate for the given value of Verifing that a Series Converges In Eercises 3 6, verif that the series converges to the given function on the indicated interval. (Hint: Use the given differential equation.) 3. n n! e,, Differential equation: , 0, 0 3, n 6, 4 0, 0, 0, n 8, cos 0, e sin 0, Differential equation: 0 n n cos,, n! 0 n n arctan,, n 0 3, 0, n 4, 3 0, 0, n 4, 5 Differential equation: 0 6. n! n arcsin,, n n! n Differential equation: 0 7. Air s Equation Find the first si terms in the series solution of Air s equation, 0.

29 Review Eercises 53 Review Eercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered eercises. Testing for Eactness In Eercises and, determine whether the differential equation is eact. Eplain our reasoning.. 3 d d 0. 5 d 5 d 0 Solving an Eact Differential Equation In Eercises 3 8, determine whether the differential equation is eact. If it is, find the general solution d 8 5 d d 6 d d 3 d d 3 5 d 0 7. d d 0 8. sin d sin d 0 Graphical and Analtic Analsis In Eercises 9 and 0, (a) sketch an approimate solution of the differential equation satisfing the initial condition on the slope field, (b) find the particular solution that satisfies the initial condition, and (c) use a graphing utilit to graph the particular solution. Compare the graph with the sketch in part (a). 9. d d 0, 4 Finding an Integrating Factor In Eercises 3 6, find the integrating factor that is a function of or alone and use it to find the general solution of the differential equation d d 0 4. d d 0 5. d 3 e d 0 6. cos d sin cos d 0 Verifing a Solution In Eercises 7 and 8, verif the solution of the differential equation. Then use a graphing utilit to graph the particular solutions for several different values of and C. What do ou observe? C Solution 7. C e C e 8. C cos C sin Differential Equation Finding a Particular Solution: Initial Conditions In Eercises 9, find the particular solution of the differential equation that satisfies the initial conditions. Use a graphing utilit to graph the solution. Differential Equation Initial Conditions 0 0, 0 3 0, 0 7 0, 0 0 0, Finding a Particular Solution: Boundar Conditions In Eercises 3 and 4, find the particular solution of the differential equation that satisfies the boundar conditions. Use a graphing utilit to graph the solution d d 0, Differential Equation Boundar Conditions 4, 0 0, Finding a Particular Solution In Eercises and, find the particular solution that satisfies the initial condition.. 3 d 3 d 0, 0. 3 d 3 3 d 0, 5. Think About It Is the differential equation 5 sin homogeneous? Wh or wh not? 6. Solutions of a Differential Equation Find all values of k for which the differential equation k k 0 has a general solution of the indicated form. (a) C e m 3 C e m (b) C e m C e m (c) C e cos C e