Second-Order Linear Differential Equations. linearly dependent.

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1 SECTION 5. Second-Order Homogeneous Linear Equations 09 SECTION 5. Second-Order Homogeneous Linear Equations Second-Order Linear Differential Equations Higher-Order Linear Differential Equations Applications Second-Order Linear Differential Equations In this section and the following section, we discuss methods for solving higher-order 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 f 0, the equation is homogeneous; otherwise, it is nonhomogeneous. NOTE Notice that this use of the term homogeneous differs from that in Section 5.7. We discuss homogeneous equations in this section, and leave the nonhomogeneous case for 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 C 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.

2 0 CHAPTER 5 Differential Equations 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 5.4 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 We prove this theorem in onl one direction. If and are solutions, ou can 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 5.4 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 Thus, b substitution, e m is a solution if and onl if a b 0 e m. 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.

3 SECTION 5. Second-Order Homogeneous Linear Equations EXPLORATION 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 to the differential equation. Using our results, develop a general solution to differential equations with characteristic equations that have distinct real roots. 9 0 (a) (b) EXAMPLE Characteristic Equation with Distinct Real Roots Solve the differential equation Solution 4 0. In this case, the characteristic equation is m 4 0 Characteristic equation so m ±. Thus, 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 5.4 to conclude that the general solution is C e C e. General solution The characteristic equation in Eample has two distinct real roots. 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 roots m a a 4b and m a a 4b which fall into one of three cases.. Two distinct real roots, m m. Two equal real roots, m m. Two comple conjugate roots, 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 5.5 The solutions of a b 0 Solutions of 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 Roots If m m are distinct real roots of the characteristic equation, then the general solution is C e m C e m.. Equal Real Roots If m m are equal real roots of the characteristic equation, then the general solution is C e m C e m C C e m.. Comple Roots If m i and m i are comple roots of the characteristic equation, then the general solution is C e a cos C e a sin.

4 CHAPTER 5 Differential Equations EXAMPLE Characteristic Equation with Comple Roots Find the general solution of the differential equation Solution 6 0. The characteristic equation m 6m 0 has two comple roots, as follows. 6 ± 6 48 m 6 ± ± ± i, Thus, and and the general solution is C e cos C e sin. NOTE In Eample, note that although the characteristic equation has two comple roots, the solution of the differential equation is real. EXAMPLE 4 Characteristic Equation with Repeated Roots Solve the differential equation subject to the initial conditions 0 and 0. Solution The characteristic equation m 4m 4 m 0 has two equal roots given b m. Thus, the general solution is C e C e. General solution Now, because when 0, we have C C 0 C. Furthermore, because when 0, we 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 SECTION 5. Second-Order Homogeneous Linear 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 roots of the characteristic equation. Then, based on these n roots, ou form a linearl independent collection of n solutions. The major difference is that with equations of third or higher order, roots 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 roots, 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 root 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 roots 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 4 CHAPTER 5 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 5.9 Applications 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 5.9. Assume that the mass of the spring is negligible compared with m. If the object is pulled down 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 Suppose a 4-pound weight stretches a spring 8 inches from its natural length. The weight is pulled down 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. Hence, the resulting differential equation for this undamped motion is d dt Because the characteristic equation m 48 0 has comple roots 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 SECTION 5. Second-Order Homogeneous Linear Equations 5 Suppose the object in Figure 5.0 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 5.0 E X E R C I S E S F O R S E C T I O N 5. 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 6 CHAPTER 5 Differential Equations 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) (c) 4 v (b) (d) If the characteristic equation of the differential equation a b 0 has comple roots 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 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 e a e a e b, a b 57. e a sin b 58. e a cos b, b 0 e a Euler s differential equation is of the form a b 0, > If the characteristic equation of the differential equation a b has two equal real roots given b m r, show that C e r C e r is a solution. 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.