SPS Mathematical Methods. Date for Quiz No. 3: 2nd April 2015.

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1 SPS Mathematical Methods Assignment No. 3 Deadline: 9th April 2015, before 4:45 p.m. INSTRUCTIONS: Answer the following questions. Check your answer for odd number questions at the back of the textbook. Date for Quiz No. 3: 2nd April Exercises 7, 11, 13, 15, Exercises 17, 19, 23, 25, 27, Exercises 13, 15, 19, 33, Exercises 5, 11, 13, 15, Exercises 9, 11, 13, 19, Exercises 5, 7, 9, 11, 13, 19, Exercises 21, 23, 25, 27, Exercises 1, 3, 5 See next page for exercises Equation of the type dn y dx n Equation with y missing. Equation with x missing. = f(y). Equation which one solution is known. 1

2 Equation of the type dn y dx n Solve the following DEs 1. y 3 y = a. 2. e 2y y = sin 3 y y = cos y = f(y). Equation with y missing. Solve the following DEs 1. (1 + x 2 ) y + x y + ax = x d4 y dx 4 cot x d3 y dx 3 = (1 + x 2 ) y (y ) 2 = 0. Equation with x missing. Solve the following DEs 1. y d2 y ( ) dy 2 dx 2 dx + 2 dy = 0. dx 2. y d2 y + dy + ( dy 3 dx 2 dx dx) = y d2 y dx 2 = ( ) dy 2 [ dx 1 dy dx cos y + y dy dx sin y] Equation which one solution is known. Solve the following DEs 1. x y y + (1 x)y = x 2 e x, given that y = e x is a solution. 2. (1 x 2 ) y + x y y = x(1 x 2 ) 3/2, given y = x is a solution. 3. x 2 y + x y y = 0, given y = x + 1 x is a solution. 2

3 Section 4.2 Homogeneous Linear Equations: The General Solution 165 We observe that r ϭ 1 is a root of the above equation, and dividing the polynomial on the left-hand side of (15) by r Ϫ 1 leads to the factorization Ar Ϫ 1B Ar 2 ϩ 4r ϩ 3B ϭ Ar Ϫ 1B Ar ϩ 1B Ar ϩ 3B ϭ 0. Hence, the roots of the auxiliary equation are 1, Ϫ1, and Ϫ3, and so three solutions of (14) are e t, e Ϫt, and e Ϫ3t. The linear independence of these three exponential functions is proved in Problem 40. A general solution to (14) is then (16) yatb ϭ c 1 e t ϩ c 2 e Ϫt ϩ c 3 e Ϫ3t. So far we have seen only exponential solutions to the linear second-order constant coefficient equation. You may wonder where the vibratory solutions that govern mass spring oscillators are. In the next section, it will be seen that they arise when the solutions to the auxiliary equation are complex. 4.2 EXERCISES In Problems 1 12, find a general solution to the given differential equation. 1. y ϩ 6y ϩ 9y ϭ y ϩ 7y Ϫ 4y ϭ 0 3. y Ϫ y Ϫ 2y ϭ 0 4. y ϩ 5y ϩ 6y ϭ 0 5. y Ϫ 5y ϩ 6y ϭ 0 6. y ϩ 8y ϩ 16y ϭ y ϩ y Ϫ 2y ϭ 0 8. z ϩ z Ϫ z ϭ y Ϫ 4y ϩ y ϭ y Ϫ y Ϫ 11y ϭ In Problems 13 20, solve the given initial value problem. 13. y ϩ 2y Ϫ 8y ϭ 0 ; y(0) ϭ 3, y A0B ϭ Ϫ w ϩ 20w ϩ 25w ϭ 0 3y ϩ 11y Ϫ 7y ϭ 0 y ϩ y ϭ 0 ; y A0B ϭ 2, y A0B ϭ 1 y Ϫ 4y Ϫ 5y ϭ 0 ; y AϪ1B ϭ 3, y AϪ1B ϭ 9 y Ϫ 4y ϩ 3y ϭ 0 ; y A0B ϭ 1, y A0B ϭ 1/3 z Ϫ 2z Ϫ 2z ϭ 0 ; z A0B ϭ 0, z A0B ϭ 3 y Ϫ 6y ϩ 9y ϭ 0 ; y A0B ϭ 2, y A0B ϭ 25/3 y ϩ 2y ϩ y ϭ 0 ; y A0B ϭ 1, y A0B ϭ Ϫ3 y Ϫ 4y ϩ 4y ϭ 0 ; y A1B ϭ 1, y A1B ϭ First-Order Constant-Coefficient Equations. (a) Substituting y ϭ e rt, find the auxiliary equation for the first-order linear equation ay ϩ by ϭ 0, where a and b are constants with a 0. (b) Use the result of part (a) to find the general solution. In Problems 22 25, use the method described in Problem 21 to find a general solution to the given equation y Ϫ 7y ϭ y ϩ 4y ϭ z ϩ 11z ϭ w Ϫ 13w ϭ Boundary Value Problems. When the values of a solution to a differential equation are specified at two different points, these conditions are called boundary conditions. (In contrast, initial conditions specify the values of a function and its derivative at the same point.) The purpose of this exercise is to show that for boundary value problems there is no existence uniqueness theorem that is analogous to Theorem 1. Given that every solution to (17) y ϩ y ϭ 0 is of the form y(t) ϭ c 1 cos t ϩ c 2 sin t, where and are arbitrary constants, show that c 1 c 2 (a) There is a unique solution to (17) that satisfies the boundary conditions ya0b ϭ 2 and yap/2b ϭ 0. (b) There is no solution to (17) that satisfies y(0) ϭ 2 and yapb ϭ 0. (c) There are infinitely many solutions to (17) that satisfy ya0b ϭ 2 and yapb ϭ Ϫ2. In Problems 27 32, use Definition 1 to determine whether the functions y 1 and y 2 are linearly dependent on the interval A0, 1B. 27. y 1 AtB ϭ cos t sin t, y 2 AtB ϭ sin 2t 28. y 1 AtB ϭ e 3t, y 2 (tb ϭ e Ϫ4t

4 Section 4.3 Auxiliary Equations with Complex Roots 173 y υ 0 = 6 υ 0 = 0 1 υ 0 = 6 t υ 0 = 18 υ 0 = 12 Figure 4.8 Solution graphs for Example 5 and the plots in Figure 4.8, confirm our prediction that all (nonequilibrium) solutions diverge except for the one with y 0 ϭ Ϫ6. What is the physical significance of this isolated bounded solution? Evidently, if the mass is given an initial inwardly directed velocity of Ϫ6, it has barely enough energy to overcome the effect of the spring banishing it to ϩq but not enough energy to cross the equilibrium point (and get pushed to Ϫq). So it asymptotically approaches the (extremely delicate) equilibrium position y ϭ 0. In Section 4.8, we will see that taking further liberties with the mass spring interpretation enables us to predict qualitative features of more complicated equations. Throughout this section we have assumed that the coefficients a, b, and c in the differential equation were real numbers. If we now allow them to be complex constants, then the roots r 1, r 2 of the auxiliary equation (1) are, in general, also complex but not necessarily conjugates of each other. When r 1 r 2, a general solution to equation (2) still has the form yatb ϭ c 1 e r 1t ϩ c 2 e r 2t, but c 1 and c 2 are now arbitrary complex-valued constants, and we have to resort to the clumsy calculations of Example 1. We also remark that a complex differential equation can be regarded as a system of two real differential equations since we can always work separately with its real and imaginary parts. Systems are discussed in Chapters 5 and EXERCISES In Problems 1 8, the auxiliary equation for the given differential equation has complex roots. Find a general solution. 1. y ϩ y ϭ 0 2. y ϩ 9y ϭ 0 3. y Ϫ 10y ϩ 26y ϭ 0 4. z Ϫ 6z ϩ 10z ϭ 0 5. y Ϫ 4y ϩ 7y ϭ 0 6. w ϩ 4w ϩ 6w ϭ y ϩ 4y ϩ 6y ϭ y Ϫ 4y ϩ 26y ϭ 0 In Problems 9 20, find a general solution. 9. y ϩ 4y ϩ 8y ϭ y Ϫ 8y ϩ 7y ϭ z ϩ 10z ϩ 25z ϭ u ϩ 7u ϭ y ϩ 2y ϩ 5y ϭ y Ϫ 2y ϩ 26y ϭ y ϩ 10y ϩ 41y ϭ y Ϫ 3y Ϫ 11y ϭ y Ϫ y ϩ 7y ϭ y ϩ13y Ϫ 7y ϭ y ϩ y ϩ 3y Ϫ 5y ϭ y Ϫ y ϩ 2y ϭ 0

5 174 Chapter 4 Linear Second-Order Equations In Problems 21 27, solve the given initial value problem. 21. y ϩ 2y ϩ 2y ϭ 0 ; 22. y ϩ 2y ϩ 17y ϭ 0 ; 23. w Ϫ 4w ϩ 2w ϭ 0 ; 24. y ϩ 9y ϭ 0 ; 25. y Ϫ 2y ϩ 2y ϭ 0 ; 26. y Ϫ 2y ϩ y ϭ 0 ; 27. y Ϫ 4y ϩ 7y Ϫ 6y ϭ 0 ; ya0b ϭ 2, ya0b ϭ1, wa0b ϭ0, ya0b ϭ 1, yapb ϭe p, ya0b ϭ1, ya0b ϭ 1, y A0B ϭ 1 y A0B ϭϫ1 w A0B ϭ1 y A0B ϭ 1 y ApB ϭ0 y A0B ϭϫ2 y A0B ϭ 0, y A0B ϭ To see the effect of changing the parameter b in the initial value problem solve the problem for b ϭ 5, 4, and 2 and sketch the solutions. 29. Find a general solution to the following higher-order equations. (a) (b) (c) 30. Using the representation for in (6), verify the differentiation formula (7). 31. Using the mass spring analogy, predict the behavior as t S ϩq of the solution to the given initial value problem. Then confirm your prediction by actually solving the problem. (a) (b) (c) (d) (e) y ϩ by ϩ 4y ϭ 0 ; y A0B ϭ 1, y A0B ϭ 0, y Ϫ y ϩ y ϩ 3y ϭ 0 y ϩ 2y ϩ 5y Ϫ 26y ϭ 0 y iv ϩ 13y ϩ 36y ϭ 0 e AaϩibBt y ϩ 16y ϭ 0 ; y A0B ϭ 2, y A0B ϭ 0 y ϩ 100y ϩ y ϭ 0 ; y A0B ϭ 1, y A0B ϭ 0 y Ϫ 6y ϩ 8y ϭ 0 ; y A0B ϭ 1, y A0B ϭ 0 y ϩ 2y Ϫ 3y ϭ 0 ; y A0B ϭ Ϫ2, y A0B ϭ 0 y Ϫ y Ϫ 6y ϭ 0 ; y A0B ϭ 1, y A0B ϭ Vibrating Spring without Damping. A vibrating spring without damping can be modeled by the initial value problem (11) in Example 3 by taking b ϭ 0. (a) If m ϭ 10 kg, k ϭ 250 kg/sec 2, ya0b ϭ 0.3 m, and y A0B ϭ Ϫ0.1 m/sec, find the equation of motion for this undamped vibrating spring. (b) When the equation of motion is of the form displayed in (9), the motion is said to be oscillatory with frequency b/2p. Find the frequency of oscillation for the spring system of part (a). 33. Vibrating Spring with Damping. Using the model for a vibrating spring with damping discussed in Example 3: (a) Find the equation of motion for the vibrating spring with damping if m ϭ 10 kg, b ϭ 60 kg/sec, k ϭ 250 kg/sec 2, ya0b ϭ 0.3 m, and y A0B ϭ Ϫ0.1 m/sec. (b) Find the frequency of oscillation for the spring system of part (a). [Hint: See the definition of frequency given in Problem 32(b).] (c) Compare the results of Problems 32 and 33 and determine what effect the damping has on the frequency of oscillation. What other effects does it have on the solution? 34. RLC Series Circuit. In the study of an electrical circuit consisting of a resistor, capacitor, inductor, and an electromotive force (see Figure 4.9), we are led to an initial value problem of the form (20) L di dt RI q EAtB ; C qa0b q 0, IA0B I 0, where L is the inductance in henrys, R is the resistance in ohms, C is the capacitance in farads, E AtB is the electromotive force in volts, q AtB is the charge in coulombs on the capacitor at time t, and I ϭ dq/dt is the current in amperes. Find the current at time t if the charge on the capacitor is initially zero, the initial current is zero, L ϭ 10 H, R ϭ 20, C ϭ A6260B Ϫ1 F, and E AtB ϭ 100 V. [Hint: Differentiate both sides of the differential equation in (20) to obtain a homogeneous linear second-order equation for IAtB. Then use (20) to determine di/dt at t ϭ 0.] E ( t ) I ( t ) R L q ( t ) Figure 4.9 RLC series circuit 35. Swinging Door. The motion of a swinging door with an adjustment screw that controls the amount of friction on the hinges is governed by the initial value problem Iu ϩ bu ϩ ku ϭ 0 ; ua0b ϭ u 0, u A0B ϭ y 0, where u is the angle that the door is open, I is the moment of inertia of the door about its hinges, b Ͼ 0 is a damping constant that varies with the amount of friction on the door, k Ͼ 0 is the spring constant associated with the swinging door, is the initial angle that the u 0 C

6 Section 4.4 Nonhomogeneous Equations: The Method of Undetermined Coefficients 175 door is opened, and y 0 is the initial angular velocity imparted to the door (see Figure 4.10). If I and k are fixed, determine for which values of b the door will not continually swing back and forth when closing. Figure 4.10 Top view of swinging door 36. Although the real general solution form (9) is convenient, it is also possible to use the form (21) d 1 e AaϩibB t ϩ d 2 e AaϪibB t to solve initial value problems, as illustrated in Example 1. The coefficients d 1 and d 2 are complex constants. (a) Use the form (21) to solve Problem 21. Verify that your form is equivalent to the one derived using (9). (b) Show that, in general, d 1 and d 2 in (21) must be complex conjugates in order that the solution be real. 37. The auxiliary equations for the following differential equations have repeated complex roots. Adapt the repeated root procedure of Section 4.2 to find their general solutions: (a) (b) y iv ϩ 2y ϩ y ϭ 0. y iv ϩ 4y ϩ 12y ϩ 16y ϩ 16y ϭ 0. [Hint: The auxiliary equation is Ar 2 ϩ 2r ϩ 4B 2 ϭ 0.] 38. Prove the sum of angle formula for the sine function by following these steps. Fix x. (a) Let f(t):ϭ sin (x ϩ t). Show that f Љ(t) ϩ f(t) ϭ 0, f(0) ϭ sin x, and f Ј(0) ϭ cos x. (b) Use the auxiliary equation technique to solve the initial value problem yљ ϩ y ϭ 0, y(0) ϭ sin x, and yј(0) ϭ cos x. (c) By uniqueness, the solution in part (b) is the same as f(t) from part (a). Write this equality; this should be the standard sum of angle formula for sin (x ϩ t). 4.4 NONHOMOGENEOUS EQUATIONS: THE METHOD OF UNDETERMINED COEFFICIENTS In this section we employ judicious guessing to derive a simple procedure for finding a solution to a nonhomogeneous linear equation with constant coefficients (1) ay ϩ by ϩ cy ϭ f(t), when the nonhomogeneity f(t) is a single term of a special type. Our experience in Section 4.3 indicates that (1) will have an infinite number of solutions. For the moment we are content to find one, particular, solution. To motivate the procedure, let s first look at a few instructive examples. Example 1 Solution Find a particular solution to (2) y ϩ 3y ϩ 2y ϭ 3t. We need to find a function y(t) such that the combination y ϩ 3y ϩ 2y is a linear function of t namely, 3t. Now what kind of function y ends up as a linear function after having its zeroth, first, and second derivatives combined? One immediate answer is: another linear function. So we might try y 1 AtB ϭ At and attempt to match up y 1 ϩ 3y 1 ϩ 2y 1 with 3t. Perhaps you can see that this won t work: y 1 ϭ At, y 1 ϭ A and y 1 ϭ 0 gives us y 1 ϩ 3y 1 ϩ 2y 1 ϭ 3A ϩ 2At,

7 182 Chapter 4 Linear Second-Order Equations 4.4 EXERCISES In Problems 1 8, decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation y ϩ 2y Ϫ y ϭ t Ϫ1 e t 5y Ϫ 3y ϩ 2y ϭ t 3 cos 4t 2y AxB Ϫ 6y AxB ϩ yaxb ϭ A sinxb /e 4x x ϩ 5x Ϫ 3x ϭ 3 t 2v AxB Ϫ 3v AxB ϭ 4x sin 2 x ϩ 4x cos 2 x y AuB ϩ 3y AuB Ϫ yaub ϭ sec u ty Ϫ y ϩ 2y ϭ sin 3t 8z AxB Ϫ 2zAxB ϭ 3x 100 e 4x cos 25x In Problems 9 26, find a particular solution to the differential equation. 9. y ϩ 2y Ϫ y ϭ y ϩ 3y ϭ Ϫ9 11. y AxB ϩ yaxb ϭ 2 x 12. 2x ϩ x ϭ 3t y Ϫ y ϩ 9y ϭ 3 sin 3t 14. 2z ϩ z ϭ 9e 2t 15. d 2 y dx 2 Ϫ 5 dy ϩ 6y ϭ xex dx 16. u AtB Ϫ u AtB ϭ t sin t 17. y Ϫ 2y ϩ y ϭ 8e t 18. y ϩ 4y ϭ 8 sin 2t 19. 4y ϩ 11y Ϫ 3y ϭ Ϫ2te Ϫ3t 20. y ϩ 4y ϭ 16t sin 2t 21. x AtB Ϫ 4x AtB ϩ 4xAtB ϭ te 2t x AtB Ϫ 2x AtB ϩ xatb ϭ 24t 2 e t y AuB Ϫ 7y AuB ϭ u 2 y AxB ϩ yaxb ϭ 4x cos x y ϩ 2y ϩ 4y ϭ 111e 2t cos 3t y ϩ 2y ϩ 2y ϭ 4te Ϫt cos t In Problems 27 32, determine the form of a particular solution for the differential equation. (Do not evaluate coefficients.) 27. y ϩ 9y ϭ 4t 3 sin 3t 28. y ϩ 3y Ϫ 7y ϭ t 4 e t 28. y ϩ 3y Ϫ 7y ϭ t 4 e t 29. y Ϫ 6y ϩ 9y ϭ 5t 6 e 3t 30. y Ϫ 2y ϩ y ϭ 7e t cos t 31. y ϩ 2y ϩ 2y ϭ 8t 3 e Ϫt sin t 32. y Ϫ y Ϫ 12y ϭ 2t 6 e Ϫ3t In Problems 33 36, use the method of undetermined coefficients to find a particular solution to the given higher-order equation. 33. y Ϫ y ϩ y ϭ sin t 34. 2y ϩ 3y ϩ y Ϫ 4y ϭ e Ϫt 35. y ϩ y Ϫ 2y ϭ te t 36. y (4) Ϫ 3y Ϫ 8y ϭ sin t 4.5 THE SUPERPOSITION PRINCIPLE AND UNDETERMINED COEFFICIENTS REVISITED The next theorem describes the superposition principle, a very simple observation which nonetheless endows the solution set for our equations with a powerful structure. It extends the applicability of the method of undetermined coefficients and enables us to solve initial value problems for nonhomogeneous differential equations. Superposition Principle Theorem 3. Let be a solution to the differential equation and y 2 y 1 ay ϩ by ϩ cy ϭ f 1AtB, be a solution to ay ϩ by ϩ cy ϭ f 2 AtB. Then for any constants k 1 and k 2, the function k 1 y 1 ϩ k 2 y 2 is a solution to the differential equation ay ϩ by ϩ cy ϭ k 1 f 1 AtB ϩ k 2 f 2 AtB.

8 Section 4.7 Variable-Coefficient Equations EXERCISES In Problems 1 8, find a general solution to the differential equation using the method of variation of parameters. 1. y ϩ y ϭ sec t 2. y ϩ 4y ϭ tan 2t 3. y ϩ 2y ϩ y ϭ e Ϫt 4. y Ϫ 2y ϩ y ϭ t Ϫ1 e t 5. y ϩ 9y ϭ sec 2 A3tB 6. y AuB ϩ 16yAuB ϭ sec 4u In Problems 9 and 10, find a particular solution first by undetermined coefficients, and then by variation of parameters. Which method was quicker? In Problems 11 18, find a general solution to the differential equation y ϩ 4y ϩ 4y ϭ e Ϫ2t ln t y ϩ 4y ϭ csc 2 A2tB y Ϫ y ϭ 2t ϩ 4 2x AtB Ϫ 2x AtB Ϫ 4xAtB ϭ 2e 2t y ϩ y ϭ tan 2 t y ϩ y ϭ tan t ϩ e 3t Ϫ 1 y ϩ 4y ϭ sec 4 A2tB y AuB ϩ yaub ϭ sec 3 u 15. y ϩ y ϭ 3 sec t Ϫ t 2 ϩ y ϩ 5y ϩ 6y ϭ 18t y ϩ 2y ϭ tan 2t Ϫ 1 2 et 18. y Ϫ 6y ϩ 9y ϭ t Ϫ3 e 3t 19. Express the solution to the initial value problem y Ϫ y ϭ 1, ya1b ϭ 0, y A1B ϭ Ϫ2, t using definite integrals. Using numerical integration (Appendix C) to approximate the integrals, find an approximation for y(2) to two decimal places. 20. Use the method of variation of parameters to show that t yatb ϭ c 1 cos t ϩ c 2 sin t ϩ Ύ f AsB sin At Ϫ sb ds is a general solution to the differential equation y ϩ y ϭ f AtB, where f AtB is a continuous function on AϪq, qb. [Hint: Use the trigonometric identity sin At Ϫ sb ϭ sin t cos s Ϫ sin s cos t.] 21. Suppose y satisfies the equation yљ ϩ 10yЈ ϩ 25y ϭ e t 3 subject to y(0) ϭ 1 and yј(0) ϭ Ϫ5. Estimate y(0.2) to within Ϯ by numerically approximating the integrals in the variation of parameters formula VARIABLE-COEFFICIENT EQUATIONS The techniques of Sections 4.2 and 4.3 have explicitly demonstrated that solutions to a linear homogeneous constant-coefficient differential equation, (1) ayљ ϩ byј ϩ cy ϭ 0, are defined and satisfy the equation over the whole interval (Ϫq, ϩq). After all, such solutions are combinations of exponentials, sinusoids, and polynomials. The variation of parameters formula of Section 4.6 extended this to nonhomogeneous constant-coefficient problems, (2) ayљ ϩ byј ϩ cy ϭ ƒ(t), yielding solutions valid over all intervals where ƒ(t) is continuous (ensuring that the integrals in (10) of Section 4.6 containing ƒ(t) exist and are differentiable). We could hardly hope for more; indeed, it is debatable what meaning the differential equation (2) would have at a point where f(t) is undefined, or discontinuous.

9 200 Chapter 4 Linear Second-Order Equations Taking integration constants to be zero yields ln v ϭ 2 A or v ϭ tan 2 lnˇ tan tb t, and v ϭ tan t Ϫ t. Therefore, a second solution to (19) is y 2 ϭ Atan t Ϫ tb cos t ϭ sin t Ϫ tcos t. We conclude that a general solution is c 1 cos t ϩ c 2 Asin t Ϫ t cos tb. In this section we have seen that the theory for variable-coefficient equations differs only slightly from the constant-coefficient case (in that solution domains are restricted to intervals), but explicit solutions can be hard to come by. In the next section, we will supplement our exposition by describing some nonrigorous procedures that sometimes can be used to predict qualitative features of the solutions. 4.7 EXERCISES In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions ya1b ϭ Y 0, y A1B ϭ Y 1, where Y 0 and Y 1 are real constants In Problems 5 through 8, determine whether Theorem 5 applies. If it does, then discuss what conclusions can be drawn. If it does not, explain why In Problems 9 through 14, find a general solution to the given Cauchy Euler equation for t tat Ϫ 3By ϩ 2ty Ϫ y ϭ t 2 A1 ϩ t 2 By ϩ ty Ϫ y ϭ tan t t 2 y ϩ y ϭ cos t e t y Ϫ y t Ϫ 3 ϩ y ϭ ln t t 2 z ϩ tz ϩ z ϭ cos t ; z A0B ϭ 1, z A0B ϭ 0 y ϩ yy ϭ t 2 Ϫ 1 ; ya0b ϭ 1, y A0B ϭ Ϫ1 y ϩ ty Ϫ t 2 y ϭ 0 ; ya0b ϭ 0, ya1b ϭ 0 A1 Ϫ tby ϩ ty Ϫ 2y ϭ sin t ; ya0b ϭ 1, y A0B ϭ 1 t 2 d 2 y dy ϩ 2t 2 dt dt Ϫ 6y ϭ 0 t 2 y AtB ϩ 7ty AtB Ϫ 7yAtB ϭ 0 d 2 w dt ϩ 6 2 t dw dt ϩ 4 t 2 w ϭ 0 t 2 d 2 z dz ϩ 5t 2 dt dt ϩ 4z ϭ 0 9t 2 y AtB ϩ 15ty AtB ϩ yatb ϭ t 2 y AtB Ϫ 3ty AtB ϩ 4yAtB ϭ 0 In Problems 15 through 18, find a general solution for t In Problems 19 and 20, solve the given initial value problem for the Cauchy Euler equation In Problems 21 and 22, devise a modification of the method for Cauchy Euler equations to find a general solution to the given equation y AtB Ϫ 1 t y AtB ϩ 5 t 2 y(t) ϭ 0 t 2 y AtB Ϫ 3ty AtB ϩ 6yAtB ϭ 0 t 2 y AtB ϩ 9ty AtB ϩ 17yAtB ϭ 0 t 2 y AtB ϩ 3ty AtB ϩ 5yAtB ϭ 0 t 2 y AtB Ϫ 4ty AtB ϩ 4yAtB ϭ 0 ; ya1b ϭ Ϫ2, y A1B ϭ Ϫ11 t 2 y AtB ϩ 7ty AtB ϩ 5yAtB ϭ 0 ; ya1b ϭ Ϫ1, y A1B ϭ 13 At Ϫ 2B 2 y AtB Ϫ 7At Ϫ 2By AtB ϩ 7yAtB ϭ 0, t 7 2 At ϩ 1B 2 y AtB ϩ 10At ϩ 1By AtB ϩ 14yAtB ϭ 0, t 7 Ϫ1 23. To justify the solution formulas (8) and (9), perform the following analysis. (a) Show that if the substitution t ϭ e x is made in the function yatb and x is regarded as the new independent variable in YAxB J yae x B, the chain rule implies the following relationships: t dy dt ϭ dy dx, t 2 d 2 y dt 2 ϭ d 2 Y dx 2 Ϫ dy dx.

10 Section 6.2 Homogeneous Linear Equations with Constant Coefficients 331 Example 3 Solution Example 4 Solution Find a general solution to (29) y A4B Ϫ y A3B Ϫ 3y ϩ 5y Ϫ 2y ϭ 0. The auxiliary equation is r 4 Ϫ r 3 Ϫ 3r 2 ϩ 5r Ϫ 2 ϭ Ar Ϫ 1B 3 Ar ϩ 2B ϭ 0, which has roots r 1 ϭ 1, r 2 ϭ 1, r 3 ϭ 1, r 4 ϭ Ϫ2. Because the root at 1 has multiplicity 3, a general solution is (30) yaxb ϭ C 1 e x ϩ C 2 xe x ϩ C 3 x 2 e x ϩ C 4 e Ϫ2x. Find a general solution to (31) y A4B Ϫ 8y A3B ϩ 26y Ϫ 40y ϩ 25y ϭ 0, whose auxiliary equation can be factored as (32) r 4 Ϫ 8r 3 ϩ 26r 2 Ϫ 40r ϩ 25 ϭ Ar 2 Ϫ 4r ϩ 5B 2 ϭ 0. The auxiliary equation (32) has repeated complex roots: r 1 ϭ 2 ϩ i, r 2 ϭ 2 ϩ i, r 3 ϭ 2 Ϫ i, and r 4 ϭ 2 Ϫ i. Hence a general solution is yaxb ϭ C 1 e 2x cos x ϩ C 2 xe 2x cos x ϩ C 3 e 2x sin x ϩ C 4 xe 2x sin x. 6.2 EXERCISES In Problems 1 14, find a general solution for the differential equation with x as the independent variable. 1. y ϩ 2y Ϫ 8y ϭ 0 2. y Ϫ 3y Ϫ y ϩ 3y ϭ z ϩ 7z Ϫ z Ϫ 2z ϭ 0 4. y ϩ 2y Ϫ 19y Ϫ 20y ϭ 0 5. y ϩ 3y ϩ 28y ϩ 26y ϭ 0 6. y Ϫ y ϩ 2y ϭ y Ϫ y Ϫ 10y Ϫ 7y ϭ 0 8. y ϩ 5y Ϫ 13y ϩ 7y ϭ 0 9. u Ϫ 9u ϩ 27u Ϫ 27u ϭ y ϩ 3y Ϫ 4y Ϫ 6y ϭ y A4B ϩ 4y ϩ 6y ϩ 4y ϩ y ϭ y ϩ 5y ϩ 3y Ϫ 9y ϭ y A4B ϩ 4y ϩ 4y ϭ y A4B ϩ 2y ϩ 10y ϩ 18y ϩ 9y ϭ 0 [Hint: yaxb ϭ sin 3x is a solution.] In Problems 15 18, find a general solution to the given homogeneous equation. 15. AD Ϫ 1B 2 AD ϩ 3B AD 2 ϩ 2D ϩ 5B 2 3 y 4 ϭ AD ϩ 1B 2 AD Ϫ 6B 3 AD ϩ 5B AD 2 ϩ 1B # AD 2 ϩ 4B 3 y 4 ϭ In Problems 19 21, solve the given initial value problem. 19. y Ϫ y Ϫ 4y ϩ 4y ϭ 0 ; ya0b ϭ Ϫ4, y A0B ϭ Ϫ1, y A0B ϭ Ϫ y ϩ 7y ϩ 14y ϩ 8y ϭ 0 ; ya0b ϭ 1, y A0B ϭ Ϫ3, y A0B ϭ y Ϫ 4y ϩ 7y Ϫ 6y ϭ 0 ; ya0b ϭ 1, y A0B ϭ 0, y A0B ϭ 0 In Problems 22 and 23, find a general solution for the given linear system using the elimination method of Section AD ϩ 4B AD Ϫ 3B AD ϩ 2B 3 AD 2 ϩ 4D ϩ 5B 2 # D5 3 y 4 ϭ 0 AD Ϫ 1B 3 AD Ϫ 2B AD 2 ϩ D ϩ 1B # AD 2 ϩ 6D ϩ 10B 3 3 y 4 ϭ 0 d 2 x/dt 2 Ϫ x ϩ 5y ϭ 0, 2x ϩ d 2 y/dt 2 ϩ 2y ϭ 0 d 3 x/dt 3 Ϫ x ϩ dy/dt ϩ y ϭ 0, dx/dt Ϫ x ϩ y ϭ 0

11 Section 6.3 Undetermined Coefficients and the Annihilator Method EXERCISES In Problems 1 4, use the method of undetermined coefficients to determine the form of a particular solution for the given equation In Problems 5 10, find a general solution to the given equation In Problems 11 20, find a differential operator that annihilates the given function. 11. x 4 Ϫ x 2 ϩ x 2 Ϫ 6x ϩ e Ϫ7x 14. e 5x 15. e 2x Ϫ 6e x 16. x 2 Ϫ e x 17. x 2 e Ϫx sin 2x 18. xe 3x cos 5x 19. xe Ϫ2 x ϩ xe Ϫ5x sin 3x 20. x 2 e x Ϫ x sin 4x ϩ x 3 In Problems 21 30, use the annihilator method to determine the form of a particular solution for the given equation y Ϫ 2y Ϫ 5y ϩ 6y ϭ e x ϩ x 2 y ϩ y Ϫ 5y ϩ 3y ϭ e Ϫx ϩ sin x y ϩ 3y Ϫ 4y ϭ e Ϫ2x y ϩ y Ϫ 2y ϭ xe x ϩ 1 y Ϫ 2y Ϫ 5y ϩ 6y ϭ e x ϩ x 2 y ϩ y Ϫ 5y ϩ 3y ϭ e Ϫx ϩ sin x y ϩ 3y Ϫ 4y ϭ e Ϫ2x y ϩ y Ϫ 2y ϭ xe x ϩ 1 y Ϫ 3y ϩ 3y Ϫ y ϭ e x y ϩ 4y ϩ y Ϫ 26y ϭ e Ϫ3x sin 2x ϩ x u Ϫ 5u ϩ 6u ϭ cos 2x ϩ 1 y ϩ 6y ϩ 8y ϭ e 3x Ϫ sin x y Ϫ 5y ϩ 6y ϭ e 3x Ϫ x 2 u Ϫ u ϭ xe x y Ϫ 6y ϩ 9y ϭ sin 2x ϩ x y ϩ 2y ϩ y ϭ x 2 Ϫ x ϩ 1 y ϩ 2y ϩ 2y ϭ e Ϫx cos x ϩ x 2 y Ϫ 6y ϩ 10y ϭ e 3x Ϫ x z Ϫ 2z ϩ z ϭ x Ϫ e x y ϩ 2y Ϫ y Ϫ 2y ϭ e x Ϫ 1 In Problems 31 33, solve the given initial value problem. 31. y ϩ 2y Ϫ 9y Ϫ 18y ϭ Ϫ18x 2 Ϫ 18x ϩ 22 ; ya0b ϭ Ϫ2, y A0B ϭ Ϫ8, y A0B ϭ Ϫ Use the annihilator method to show that if a 0 0 in equation (4) and f AxB has the form (17) then is the form of a particular solution to equation (4). 35. Use the annihilator method to show that if a 0 ϭ 0 and a 1 0 in (4) and f AxB has the form given in (17), then equation (4) has a particular solution of the form 36. Use the annihilator method to show that if f AxB in (4) has the form f AxB ϭ Be a x, then equation (4) has a particular solution of the form y p AxB ϭ x s Be a x, where s is chosen to be the smallest nonnegative integer such that x s e a x is not a solution to the corresponding homogeneous equation. 37. Use the annihilator method to show that if f AxB in (4) has the form then equation (4) has a particular solution of the form (18) where s is chosen to be the smallest nonnegative integer such that x s cos bx and x s sin bx are not solutions to the corresponding homogeneous equation. In Problems 38 and 39, use the elimination method of Section 5.2 to find a general solution to the given system y Ϫ 2y ϩ 5y ϭ Ϫ24e 3x ; ya0b ϭ 4, y A0B ϭ Ϫ1, y A0B ϭ Ϫ5 y Ϫ 2y Ϫ 3y ϩ 10y ϭ 34xe Ϫ2x Ϫ 16e Ϫ2x Ϫ 10x 2 ϩ 6x ϩ 34 ; ya0b ϭ 3, y A0B ϭ 0, y A0B ϭ 0 f AxB ϭ b m x m ϩ b mϫ1 x mϫ1 ϩ p ϩ b 1 x ϩ b 0, y p AxB ϭ B m x m ϩ B mϫ1 x mϫ1 ϩ p ϩ B 1 x ϩ B 0 y p AxB ϭ xeb m x m ϩ B mϫ1 x mϫ1 ϩ p ϩ B 1 x ϩ B 0 F. f AxB ϭ a cos bx ϩ b sin bx, y p AxB x s EA cos Bx B sin BxF, x Ϫ d 2 y/dt 2 ϭ t ϩ 1, dx/dt ϩ dy/dt Ϫ 2y ϭ e t d 2 x/dt 2 Ϫ x ϩ y ϭ 0, x ϩ d 2 y/dt 2 Ϫ y ϭ e 3t

12 Section 6.4 Method of Variation of Parameters EXERCISES In Problems 1 6, use the method of variation of parameters to determine a particular solution to the given equation y Ϫ 3y ϩ 4y ϭ e 2x y Ϫ 2y ϩ y ϭ x z ϩ 3z Ϫ 4z ϭ e 2x y Ϫ 3y ϩ 3y Ϫ y ϭ e x y ϩ y ϭ tan x, 0 6 x 6 p/2 y ϩ y ϭ sec u tan u, 0 6 u 6 p/2 7. Find a general solution to the Cauchy Euler equation x 3 y Ϫ 3x 2 y ϩ 6xy Ϫ 6y ϭ x Ϫ1, x 7 0, given that Ex, x 2, x 3 F is a fundamental solution set for the corresponding homogeneous equation. 8. Find a general solution to the Cauchy Euler equation x 3 y Ϫ 2x 2 y ϩ 3xy Ϫ 3y ϭ x 2, x 7 0, given that Ex, x ln x, x 3 F is a fundamental solution set for the corresponding homogeneous equation. 9. Given that Ee x, e Ϫx, e 2x F is a fundamental solution set for the homogeneous equation corresponding to the equation y Ϫ 2y Ϫ y ϩ 2y ϭ gaxb, determine a formula involving integrals for a particular solution. 10. Given that Ex, x Ϫ1, x 4 F is a fundamental solution set for the homogeneous equation corresponding to the equation x 3 y Ϫ x 2 y Ϫ 4xy ϩ 4y ϭ gaxb, x 7 0, determine a formula involving integrals for a particular solution. 11. Find a general solution to the Cauchy Euler equation x 3 y Ϫ 3xy ϩ 3y ϭ x 4 cos x, x Derive the system (7) in the special case when n ϭ 3. [Hint: To determine the last equation, require that L 3 y p 4 ϭ g and use the fact that y 1, y 2, and y 3 satisfy the corresponding homogeneous equation.] 13. Show that W k AxB ϭ AϪ1B AnϪkB W 3 y 1,..., y kϫ1, y kϩ1,..., y n 4 AxB. 14. Deflection of a Beam Under Axial Force. A uniform beam under a load and subject to a constant axial force is governed by the differential equation y A4B AxB Ϫ k 2 y AxB ϭ q AxB, 0 6 x 6 L, where yaxb is the deflection of the beam, L is the length of the beam, k 2 is proportional to the axial force, and q AxB is proportional to the load (see Figure 6.2). (a) Show that a general solution can be written in the form yaxb ϭ C 1 ϩ C 2 x ϩ C 3 e kx ϩ C 4 e Ϫkx (b) Show that the general solution in part (a) can be rewritten in the form yaxb ϭ c 1 ϩ c 2 x ϩ c 3 e kx ϩ c 4 e Ϫkx where ϩ 1 2Ύ k q AxB x dx Ϫ x k2ύ q AxB dx ϩ e kx 2k 3Ύq AxBe Ϫkx dx Ϫ e Ϫkx ϩ Ύ x qasbgas, xb ds, G As, xb J s Ϫ x k 2 Ϫ 0 sin h 3 kas Ϫ xb 4 k 3. 2k 3 Ύq AxBe kx dx. (c) Let q AxB ϵ 1. First compute the general solution using the formula in part (a) and then using the formula in part (b). Compare these two general solutions with the general solution yaxb ϭ B 1 ϩ B 2 x ϩ B 3 e kx ϩ B 4 e Ϫkx Ϫ 1 2k 2 x2, which one would obtain using the method of undetermined coefficients. (d) What are some advantages of the formula in part (b)? x Load L y ( x ) Axial force Figure 6.2 Deformation of a beam under axial force and load