Second-Order Linear Differential Equations C 2
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1 C8 APPENDIX C Additional Topics in Differential Equations APPENDIX C. Second-Order Homogeneous Linear Equations Second-Order Linear Differential Equations Higher-Order Linear Differential Equations Application Second-Order Linear Differential Equations In this section and the following section, ou will learn methods for solving higherorder linear differential equations. 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 called a linear differential equation of order n. If homogeneous; otherwise, it is nonhomogeneous. f 0, the equation is Homogeneous equations are discussed in this section, and the nonhomogeneous case is discussed in the net section. The functions,,..., n are linearl independent if the onl solution of the equation C C... C n n 0 is the trivial one, C C... C n 0. linearl dependent. Otherwise, this set of functions is EXAMPLE 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 are linearl dependent because C C 0 has the nonzero solutions C and C. C C
2 APPENDIX C. Second-Order Homogeneous Linear Equations C9 The following 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 C. Linear Combinations of Solutions If and are linearl independent solutions of the differential equation then the general solution is a b 0, C C where C and C are constants. Proof This theorem is proved in onl one direction. If and are solutions, 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. Theorem C. states that if 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 e m. me m m e m. m e m ame m be m 0 e m m am b 0. Because is never 0, e m is a solution if and onl if e m a b 0 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.
3 C0 APPENDIX C Additional Topics in Differential Equations EXAMPLE Characteristic Equation with Distinct Real Zeros Solve the differential equation Solution 4 0. In this case, the characteristic equation is m 4 0 Characteristic equation so, m ±. Therefore, and e m e e m e are particular solutions of the given differential equation. Furthermore, because these two solutions are linearl independent, ou can appl Theorem C. 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 m a a 4b which fall into one of three cases.. Two distinct real zeros, m m. Two equal real zeros, m m. Two comple conjugate zeros, m i and m i In terms of the differential equation these three cases correspond to three different tpes of general solutions. a b 0, THEOREM C.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.. Comple Zeros If m i and m i are comple zeros of the characteristic equation, then the general solution is C e a cos C e a sin.
4 APPENDIX C. Second-Order Homogeneous Linear Equations C EXAMPLE Characteristic Equation with Comple Zeros Find the general solution of the differential equation Solution 6 0. The characteristic equation m 6m 0 has two comple zeros, as follows. m 6 ± ± 6 ± ± ± i, So, and and the general solution is C e cos C e sin. NOTE In Eample, note that although the characteristic equation has two comple zeros, the solution of the differential equation is real. EXAMPLE 4 Characteristic Equation with Repeated Zeros Solve the differential equation subject to the initial conditions 0 and 0. Solution The characteristic equation m 4m 4 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.
5 C APPENDIX C Additional Topics in Differential Equations 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. EXAMPLE 5 Solving a Third-Order Equation Find the general solution of 0. Solution The characteristic equation is m m 0 m m m 0 m 0,,. Because the characteristic equation has three distinct zeros, the general solution is C C e C e. General solution EXAMPLE 6 Solving a Third-Order Equation Find the general solution of 0. Solution The characteristic equation is m m m 0 m 0 m. Because the zero m occurs three times, the general solution is C e C e C e. General solution EXAMPLE 7 Solving a Fourth-Order Equation Find the general solution of 4 0. Solution The characteristic equation is as follows. 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 cos C 4 sin. General solution
6 APPENDIX C. Second-Order Homogeneous Linear Equations C m l = natural length = displacement A rigid object of mass m attached to the end of the spring causes a displacement of. Figure C.4 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 given spring. Suppose a rigid object of mass m is attached to the end of a spring and causes a displacement, as shown in Figure C.4. Assume that the mass of the spring is negligible compared with m. If 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 EXAMPLE 8 Undamped Motion of a Spring A four-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, so k 6. Moreover, because the weight w is given b mg, it follows that m w g 4 8. So, the resulting differential equation for this undamped motion is d dt Because the characteristic equation m 48 0 has comple zeros m 0 ± 4 i, the general solution is C e 0 cos 4 t C e 0 sin 4 t C cos 4 t C sin 4 t. Using the initial conditions, ou have C C C 0 t 4 C sin 4 t 4 C cos 4 t Consequentl, the position at time t is given b C cos 4 t sin 4 t. C
7 C4 APPENDIX C Additional Topics in Differential Equations Suppose the object in Figure C.5 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 dt k p dt or, in standard linear form, d dt p m d dt k 0. m Damped motion of a spring A damped vibration could be caused b friction and movement through a liquid. Figure C.5 EXERCISES FOR APPENDIX C. In Eercises 4, verif the solution of the differential equation. Solution In Eercises 5 0, find the general solution of the linear differential equation Differential Equation C C e C e C e C cos C sin e sin Consider the differential equation and the solution C cos 0 C sin 0. Find the particular solution satisfing each of the following initial conditions. (a) 0, 0 0 (b) 0 0, 0 (c) 0, 0. Determine C and such that C sin t is a particular solution of the differential equation where 0 5. In Eercises 6, find the particular solution of the linear differential equation , , 0 Think About It In Eercises 7 and 8, 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 , 0 0, 0 0 0,
8 APPENDIX C. Second-Order Homogeneous Linear Equations C5 Vibrating Spring In Eercises 9 44, describe the motion of a -pound weight suspended on a spring. Assume that the weight stretches the spring foot from its natural position. 9. The weight is pulled foot below the equilibrium position and released. 40. The weight is raised foot above the equilibrium position and released. 4. The weight is raised foot above the equilibrium position and started off with a downward velocit of foot per second. 4. The weight is pulled foot below the equilibrium position and started off with an upward velocit of foot per second. 4. 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. 44. 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. Vibrating Spring In Eercises 45 48, 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) v (b) If the characteristic equation of the differential equation 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 5 54, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 5. C e C e is the general solution of C C sin C 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 0. 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 g g. f 6 The functions f and g are linearl independent if there eists at least one value of for which W f, g 0. In Eercises 55 58, use the Wronskian to verif the linear independence of the two functions. (c) (d) 55. e a 56. e b, a b 57. e a sin b 58. e a cos b, b 0 e a e a If the characteristic equation of the differential equation a b has two equal real zeros given b m r, show that C e r C e r is a solution. 59. Euler s differential equation is of the form a b 0, > 0 where a and b are constants. (a) Show that this equation can be transformed into a secondorder linear equation with constant coefficients b using the substitution e t. (b) Solve Solve A 0 where A is constant, subject to the conditions 0 0 and 0.
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