CHAPTER 2Second-Order Linear ODEsMajor Changes

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(1)c02.qxd. 6/18/11. 2:55 PM. Page 26. CHAPTER 2. Second-Order Linear ODEs. Major Changes Among linear ODEs those of second order are by far the most important ones from the viewpoint of applications, and from a theoretical standpoint they illustrate the theory of linear ODEs of any order (except for the role of the Wronskian). For these reasons we consider linear ODEs of third and higher order in a relatively short separate chapter, Chap. 3. Section 2.2 combines all three cases of the roots of the characteristic equation governing homogeneous linear ODEs with constant coefficients. (In some of the previous editions the complex case was discussed in a separate section, which seems of no great advantage to the student.) Section 2.3 is a short introduction to differential operators. Modeling begins in Sec. 2.4 with the mass–spring system, which is now derived more simply than before and in a better logical order. After a discussion of the Euler–Cauchy equation and its application to electric fields between concentric spheres in Sec. 2.6, we discuss in Sec. 2.7 the existence and uniqueness of the solution of IVPs involving the homogeneous linear ODE of second order. This is the end of discussing homogeneous ODEs. It is followed in Sec. 2.7 by the method of undetermined coefficients for nonhomogeneous ODEs, which is basic in applications since it is simpler than the general method (variation of parameters, Sec. 2.10) and covers many, if not most of the standard engineering applications. Modeling of forced mechanical oscillations is discussed in Sec. 2.8, and electric RLC-circuits in Sec. 2.9. Note that we have placed the RL-circuit, governed by a firstorder ODE into Sec. 1.5, which the student may perhaps wish to review. This was a request by various users of the book, as a stepping stone that may lessen difficulties and simplify the derivation of the model from physics. SECTION 2.1. Homogeneous Linear ODEs of Second-Order, page 46 Purpose. To extend the basic concepts from first-order to second-order ODEs and to present the basic properties of linear ODEs. Comment on the Standard Form (1) The form (1), with 1 as the coefficient of y s , is practical, because if one starts from f (x)y s g(x)y r h(x)y r苲(x), one usually considers the equation in an interval I in which f (x) is nowhere zero, so that in I one can divide by f (x) and obtain an equation of the form (1). Points at which f (x) 0 require a special study, which we present in Chap. 5. Main Content, Important Concepts Linear and nonlinear ODEs Homogeneous linear ODEs (to be discussed in Secs. 2.1–2.6) Superposition principle for homogeneous ODEs General solution, basis, linear independence Initial value problem (2), (4), particular solution Reduction to first order (text and Probs. 3–10) 26.

(2) c02.qxd. 6/18/11. 2:55 PM. Page 27. Instructor’s Manual. 27. Comment on the Three ODEs after (2) These are for illustration, not for solution, but should a student ask, answers are that the first will be solved by methods in Secs. 2.7 and 2.10, the second is a Bessel equation (Sec. 5.5) and the third has the solutions 1c1x c2 with any c1 and c2. Comment on Footnote 1 In 1760, Lagrange gave the first methodical treatment of the calculus of variations. The book mentioned in the footnote includes all major contributions of others in the field and made him the founder of analytical mechanics. Examples in the Text. The examples show the following. Example 1 shows the superposition of solutions of the homogeneous linear ODE. Examples 2 and 3 are counter-examples to the superposition for a nonhomogeneous linear ODE and a nonlinear ODE. Example 4 is an initial value problem, suggesting the concepts of a general solution, a particular solution, and a basis. Examples 5 and 6 give further illustrations of those concepts. Example 7 shows the reduction of order of y s p(x)y r q(x)y 0 using a known solution y1, followed by the derivation of a general formula for a second solution y2 y1. 冮y. 1 2 1. e ⴚ兰p dx dx.. Hence solving the ODE for finding a second solution is reduced to two integrations, and the student should understand that this is a simpler task. Comment on Terminology p and q are called the coefficients of (1) and (2). The function r on the right is not called a coefficient, to avoid the misunderstanding that r must be constant when we talk about an ODE with constant coefficients.. SOLUTIONS TO PROBLEM SET 2.1, page 53 2. y s . dy r dy dy r dz z dx dy dx dy. 3. y ex c1x c2 4. z y r , 2xz r 3z. Separation of variables and integration gives dz 3 dx, z 2x. ln ƒ z ƒ 32 ln ƒ x ƒ 苲 c,. Integrating once more, we have y. 冮 z dx c x 1. 5>2. c2.. z cx 3>2..

(3) c02.qxd. 6/18/11. 28. 2:55 PM. Page 28. Instructor’s Manual. 6. The formula in the text was derived under the assumption that the ODE is in standard form; in the present case, ys . 2 y r y 0. x. Hence p 2>x, so that eⴚ兰p dx x ⴚ2. It follows from (9) in the text that U. x2. # 1 2. 2. cos x. x. 1 cos2 x. .. The integral of U is tan x; we need no constants of integration because we merely want to obtain a particular solution. The answer is y2 y1 tan x . sin x . x. 7. x cos y c1y c2 8. z r 1 z 2, dz>(1 z 2) dx, arctan z x c1, z tan (x c1), y ln ƒ cos (x c1) ƒ c2 This is an obvious use of problems from Chap. 1 in setting up problems for this section. The only difficulty may be an unpleasant additional integration. 9. y2 1冫x2 dy dz 1 10. z , z a1 y b z2 0 , divide by z, separate variables, and integrate: dx dy dz 1 z a1 y b dy,. ln ƒ z ƒ y ln ƒ y ƒ 苲 c.. Take exponentials, separate again, and integrate: dy c z eⴚy, y dx. 冮 ye. yey dy c dx,. y. dy cx c2.. Evaluation of the integral gives the answer (y 1)ey c1x c2. 12. z r (1 z2) 1>2, (1 z2)ⴚ1>2 dz dx, arcsinh z x c1. From this we have z sinh (x c1), y cosh (x c1) c2. From the boundary conditions y(1) 0, y(1) 0 we get cosh (1 c1) c2 0 cosh (1 c1) c2. Hence c1 0 and then c2 cosh 1. The answer is (see the figure) y cosh x cosh 1. y –1. –0.5. 0.5. –0.54. Section 2.1. Problem 12. 1. x.

(4) c02.qxd. 6/18/11. 2:55 PM. Page 29. Instructor’s Manual. 14. y s 1>y r , y s y r 1,. 29. dz 2 z 1. Integration with respect to y gives dy z3 y c. 3. Hence z 3 3y 苲 c and thus dy z (3y 苲 c )1>3. dt By separation of variables, dy dt. (3y 苲 c )1>3 By integration, 1 2 (3y. 苲苲 c )2>3 t 苲 c.. Hence (3y 苲 c )2>3 2t c2 and thus 3y 苲 c 1 (2t c2)3>2. The answer is y 13 (2t c2)3>2 c1. 15. y 2 cos 3x 13 sin 3x 16. y (2 x)eⴚx 18. y x x ln x SECTION 2.2. Homogeneous Linear ODEs with Constant Coefficients, page 53 Purpose. To show that homogeneous linear ODEs with constant coefficients can be solved by algebra, namely, by solving the quadratic characteristics equation (3). The roots may be: (Case I). Real distinct roots. (Case II). A real double root (“Critical case”). (Case III). Complex conjugate roots. In Case III the roots are conjugate because the coefficients of the ODE, and thus of (3), are real, a fact the student should remember. To help poorer students, we have shifted the derivation of the real form of the solutions in Case III to the end of the section, but the verification of these real solutions is done immediately when they are introduced. This will also help to a better understanding..

(5) c02.qxd. 6/18/11. 2:55 PM. Page 30. 30. Instructor’s Manual. The student should become aware of the fact that Case III includes both undamped (harmonic) oscillations (if c 0) and damped oscillations. Also it should be emphasized that in the transition from the complex to the real form of the solutions we use the superposition principle. Furthermore, one should emphasize the general importance of the Euler formula (11), which we shall use on various occasions. Examples in the Text. The examples show the following. Examples 1 and 2 concern Case I, the case of distinct real roots. In this case, as well as in the other two cases, an initial value problem requires the solution of a system of two linear equations in two unknowns, whose values are determined by the two initial conditions. A typical solution in Case I is shown in Fig. 30. Examples 3 and 4 concern Case II, the case of a real double root, which is the limiting case between Cases I and III. Figure 31 shows a typical solution, having a real root at x 1.5, which is the solution of 3 2x 0, where 3 2x is a factor in the solution of the IVP in Example 4. Example 5 concerns Case III, in which one obtains solutions (9), representing oscillations. These may be damped as in Fig. 32, or of increasing maximum amplitude if a 0, or of constant maximum amplitude if a 0, as in Example 6, giving a harmonic oscillations. Comment on How to Avoid Working in Complex The average engineering student will profit from working a little with complex numbers. However, if one has reasons for avoiding complex numbers here, one may apply the method of eliminating the first derivative from the equation, that is, substituting y = uv and determining v so that the equation for u does not contain u r . For v this gives 2v r av 0.. A solution is. v e ax>2.. With this v, the equation for u takes the form u s (b 14 a 2)u 0 and can be solved by remembering from calculus that cos vx and sin vx reproduce under two differentiations, multiplied by v2. This gives (9), where v 2b 14 a 2. Of course, the present approach can be used to handle all three cases. In particular, u s 0 in Case II gives u c1 c2x at once. SOLUTIONS TO PROBLEM SET 2.2, page 59 1. y c1e x/2 c2ex/2 2. y c1 cos 6x c2 sin 6x 3. 4. 6. 7. 8.. 1. 2. 1. 2. y(x) c1e1/2 4 26 x c2e1/2 4 26 x y eⴚ2x (c1 cos px c2 sin px) y (c1 c2 x)e1.6x 5 y C1 C2e 4 x y e ⴚx>2 (c1 cos (13x) c2 sin (13x)).

(6) c02.qxd. 6/18/11. 2:55 PM. Page 31. Instructor’s Manual. 9. 10. 12. 14. 16. 17. 18. 19. 20. 21. 22.. 31. y c1e4 c2e 2x y eⴚ1.2x (c1 cos (1.4px) c2 sin (1.4px)) y c1e3x c2e5x 2 y (c1 c2x)eⴚk x y s 1.7y r 11.18y 0 y s 222y r 2y 0. y s 4p2y 0 y s 2y r 3y 0 y s 6.2y r 14.02y 0 y 12 sin 3x 15 cos 3x A general solution is x. y(x) eⴚ2x(c1 cos px c2 sin px). The first initial condition gives y(12) e ⴚ1(0 c2) 1, hence c2 e. The derivative is y r (x) eⴚ2x(2c1 cos px 2c2 sin px c1p sin px c2p cos px). From this and the second initial condition we obtain y r (12) eⴚ1(2c2 c1p) eⴚ1(2e c1p) 2 c1eⴚ1p 2. Hence c1 0. This gives the answer y eⴚ2xⴙ1 sin px. Notice that it depends on the initial condition whether both solutions of a basis appear in the particular solution or just one; this is worthwhile pointing out to the students. 23. y 35 e4x 75 ex 3 4 3x x 24. y 13 16 e e 16 e. 26. A general solution is y c1ekx c2eⴚkx. From this and the first initial condition we have c1 c2 1. The derivative is y r c1kekx c2keⴚkx. From this and the second initial condition we obtain c1 c2 1>k..

(7) c02.qxd. 6/18/11. 2:55 PM. Page 32. 32. Instructor’s Manual. The solution of this system is c1 . k1 , 2k. c2 . k1 . 2k. Hence the answer (the particular solution of the IVP) is y [(k 1)ekx (k 1)eⴚkx ]>(2k). 9 3 2 28. y 13 10 e 5 e x. x. 30. A general solution is y (c1 c2x)e5x>3. This is a case of a double root of the characteristic equation. The first initial condition yields y(0) c1 3.3. By differentiation, y r [c2 53 (c1 c2x)]e5x>3. From this and the second initial condition we obtain y r (0) c2 53 c1 c2 5.5 10. Hence c2 4.5, so that the solution of the IVP is y (3.3 4.5x)e5x>3. 32. Independent if a 0 34. Dependent since ln (x 3) 3 ln x 36. If one of the functions is identically zero, the set is linearly dependent because c1 f1(x) c2 # 0 0 holds with any c2 0 (and c1 0). The intervals given in the problems are just a reminder that linear independence or independence always refers to some interval. In the present case we could choose as the interval the real axis, or the positive half-axis if a logarithm is involved. 38. Team Project. (a) We obtain (l l1)(l l2) l2 (l1 l2)l l1l2 l2 al b 0. Comparison of coefficients gives a (l1 l2), b l1l2. (b) y s ay r 0. (i) y c1eⴚax c2e0x c1eⴚax c2. (ii) z r az 0, where z y r , z ceⴚax and the second term comes in by integration: y. 冮z dx 苲c e 1. ⴚax. 苲 c 2..

(8) c02.qxd. 6/18/11. 2:55 PM. Page 33. Instructor’s Manual. 33. (d) e(km)x and ekx satisfy y s (2k m)y r k(k m)y 0, by the coefficient formulas in part (a). By the superposition principle, another solution is e(km)x ekx . m We now let m : 0. This becomes 0>0, and by l’Hôpital’s rule (differentiation of numerator and denominator separately with respect to m, not x!) we obtain xekx>l xekx. The ODE becomes y s 2ky r k 2y 0. The characteristic equation is l2 2kl k 2 (l k) 2 0 and has a double root. Since a 2k, we get k a>2, as expected. SECTION 2.3. Differential Operators. Optional, page 60 Purpose. To take a short look at the operational calculus of second-order differential operators with constant coefficients. This parallels and confirms our discussion of ODEs with constant coefficients. A discussion of the case of variable coefficients would exceed the level and the area of interest of the book, sidetrack the attention of the student, and give no substantial additional insights that might be helpful to our further work. SOLUTIONS TO PROBLEM SET 2.3, page 61 1. 4 cosh 2x 4 sinh 2x, 3ex, sin x 2 cos x 2. The first function gives 6x 3 9x 2 9x 9x 2 3x 3. The second function gives 9e3x 9e3x 0. The third function gives 4 sin 4x 4 cos 4x 3 cos 4x 3 sin 4x 7 cos 4x sin 4x. 3. 4ex, 4ex 4 xex, 16ex 4. For the first function, (D 6I)(6 6 cos 6x 36x 6 sin 6x) 36 sin x 36 36 cos 6x 36 36 cos 6x 216x 36 sin 6x 72 72 cos 6x 216x..

(9) c02.qxd. 6/18/11. 2:55 PM. Page 34. 34. Instructor’s Manual. For the second function, (D 6I )(eⴚ6x 6xeⴚ6x 6xeⴚ6x) 6eⴚ6x 6eⴚ6x 0. 5. 6. 7. 8. 10. 11. 12. 14.. 10e4x, 7e4x 10xe4x, 4e2x (D 2.8I ) (D 1.2I ), y c1eⴚ2.8x c2eⴚ1.2x y c1e 1/3x c2e1/3x (D 13iI )(D 13iI ), y c1 cos 13x c2 sin 13x (D 2.4I )2, y (c1 c2x)eⴚ2.4x y c1 e3/2x c2 e9/2x [D (1.5 0.5i)I][D (1.5 0.5i)I], y e ⴚ1.5x(c1 cos 12 x c2 sin 12 x) y is a solution, as follows from the superposition principle in Sec. 2.1 since the ODE is homogeneous and linear. In applying l’Hopital’s rule, regard y as a function of m , the variable that approaches the limit, whereas l is fixed. Differentiation of the numerator with respect to m gives xe␮x 0 and differentiation of the denominator gives 1. The limit of this is xelx.. SECTION 2.4. Modeling of Free Oscillations of a Mass—Spring System, page 62 Purpose. To present a main application of second-order constant-coefficient ODEs my s cy r ky 0 resulting as models of motions of a mass m on an elastic spring of modulus k (0) under linear damping c (0) applying Newton’s second law and Hooke’s law. These are free motions (no driving force). Forced motions follow in Sec. 2.8. This system should be regarded as a basic building block of more complicated systems, a prototype of a vibrating system that shows the essential features of more sophisticated systems as they occur in various forms and for various purposes in engineering. The quantitative agreement between experiments of the physical system and its mathematical model is surprising. Indeed, the student should not miss performing experiments if there is an opportunity, as I had as a student of Prof. Blaess, the inventor of a (now obscure) graphical method for solving ODEs. Main Content, Important Concepts Restoring force ky, damping force cy r , force of inertia my s. No damping, harmonic oscillations (4), natural frequency v0>(2p) Overdamping, critical damping, nonoscillatory motions (7), (8) Underdamping, damped oscillations (10). In the text, the derivation of the model has been simplified by clarifying the role of the force F0, which has no effect on the motion. The model, like many others, is obtained from Newton’s second law. We discuss the undamped case c 0 and the damped case c 0 separately because the types of motion are basically different, as follows. The undamped case c 0 gives a harmonic motion (4) for an infinite time interval (practically: for a long time). The damped case c 0 gives a damped motion, which is either oscillatory or, if c is large enough, is a nonoscillatory approach to zero..

(10) c02.qxd. 6/18/11. 2:55 PM. Page 35. Instructor’s Manual. 35. Hence it is interesting that the formal distinction of Cases I–III mechanically corresponds to quite different types of motion. No damping (c 0) means no loss of the energy corresponding to the initial displacement and initial velocity. Make sure that the student understands the physics behind (4*), that shows the phase shift. Examples in the Text. The examples illustrate the following. Example 1 discusses the undamped case c 0. Example 2 compares the three cases, the three types of motion, graphically shown in Fig. 40, namely, Case I giving a rapid approach to zero, Case II looking almost the same, also showing a rapid and monotone approach to zero, and, finally, Case III a damped oscillation, of a frequency smaller than that of the harmonic oscillation when c 0. Problem Set 2.4. Problems 1–6 enhance the physical understanding and insight into the basic properties of the undamped model. Team Project 10 shows that the model in the text is in fact the prototype of various physical systems governed by the same mathematical formulas. Problems 11–19 play a role for the damped case similar to that of Probs. 1–9 for the undamped model. CAS Project 20 shows the “continuity” in the transitions between Cases I–III, that is also illustrated in Fig. 47. SOLUTIONS TO PROBLEM SET 2.4, page 69 2. W 20 and s0 2 gives k W>s0 10 by Hooke’s law. Thus f. 2k>m 2k>(W>g) 210>(20>980) v0 3.52 [Hz]. 2p 2p 2p 2p. From this we get the period 1>f 0.284 [sec]. 4. No because the frequency depends only on k>m, not on initial conditions. 6. By Hooke’s law, F1 k1 6 stretches spring S1 by 6, and F2 k2 8 stretches spring S2 by 8. Hence the unknown k of the combination of the springs stretches S1 by k>k1 k>6 and S2 by k>k2 k>8. And k is such that the sum of these stretches equals 1, because k is the force that corresponds to the stretch 1 of the combination. Thus k k 1, k1 k2. 1 1 1 . k1 k2 k. Answer: k 24/7 3.43. 8. my s p # 0.252yg, where p # 0.252y is the volume of water displaced when the buoy is depressed y meters from its equilibrium position, and g 9800 nt is the weight of water per cubic meter. Thus y s v20 y 0, where v20 p # 0.252g>m and the period is 2p>v0 2; hence m p # 0.252g>v20 0.252g> p 194.96 W mg 194.96 # 9.80 1910.66 [nt]. ~gy, where m 1 kg, ay p 0.0152 2y meter3 is the volume of the 9. my s a water that causes the restoring force agy with g 9800 nt (weight/meter3). y s v20 y 0, v20 ag/m ag 0.000707g. Frequency v0/2p 0.6 [sec1]..

(11) c02.qxd. 6/18/11. 36. 2:55 PM. Page 36. Instructor’s Manual. 10. Team Project. (a) By Prob. 7 the frequency is g 9.80 1 1 0.498, 2p B L 2p B 1 so it takes about 2 sec to complete 1 cycle. Answer: It ticks about 30 times per minute. (b) W ks0 8. Now s0 1 because the system has its equilibrium position 1 cm below the horizontal line. Also, m W>g, so that v. W>s0 k 2g 2980 31.3, B m B W>g. and we get the general solution y A cos 31.3t B sin 31.3t. The initial conditions give y(0) A 0 and y r (0) 31.3B 10. Hence B 0.319 and the answer is y 0.319 sin 31.3t [cm]. (c) u(t) 0.5235 cos 3.7t 0.0943 sin 3.7t [rad] 12. y 0 gives c1 c2eⴚ2bt, which has at most one solution because the exponential function is monotone. 14. Case (II) of (5) with c 24mk 24 # 500 # 4500 3000 [kg/sec], where 500 kg is the mass per wheel. 16. 2p>v* since Eq. (10) and y r 0 give tan (v*t d) a>v*; tan is periodic with period p>v*. 17. The positive solutions of sin t cos 2t, that is arctan(2), 1.11 (max.), 4.23 (min.) etc. 18. If an extremum is at t 0, the next one is at t1 t0 p>v*, by Prob. 16. Since the cosine and sine in (10) have period 2p>v*, the amplitude ratio is exp(at)>exp(at1) exp(a(t0 t1)) exp(ap>v*).. The natural logarithm is ap>v*, and maxima alternate with minima. Hence ¢ 2pa>v* follows. For the ODE, ¢ 2p # 1> 213 4 23 p. 19. 0.0462 ln(2) ⴢ 2 ⴢ 1.5>(15 ⴢ 3) [kg>sec]. 20. CAS Project. (a) The three cases appear, along with their typical solution curves, regardless of the numeric values of k>m, y(0), etc. (b) The first step is to see that Case II corresponds to c 2. Then we can choose other values of c by experimentation. In Fig. 47 the values of c (omitted on purpose; the student should choose!) are 0 and 0.1 for the oscillating curves, 1, 1.5, 2, 3 for the others (from below to above). (c) This addresses a general issue arising in various problems involving heating, cooling, mixing, electrical vibrations, and the like. One is generally surprised how quickly certain states are reached whereas the theoretical time is infinite..

(12) c02.qxd. 6/18/11. 2:55 PM. Page 37. Instructor’s Manual. 37. (d) General solution y(t) eⴚct>2(A cos v*t B sin v*t), where v* 12 24 c2. The first initial condition y(0) 1 gives A 1. For the second initial condition we need the derivative (we can set A = 1) c c y r (t) eⴚct>2 a cos v*t B sin v*t v* sin v*t v*B cos v*tb . 2 2 From this we obtain y r (0) c>2 v*B 0, B c>(2v*) c> 24 c2. Hence the particular solution (with c still arbitrary, 0 c 2 ) is y(t) eⴚct>2 acos v*t . c 24 c2. sin v*tb.. It derivative is, since the cosine terms drop out, y r (t) eⴚct>2(sin v*t) a . 2 24 c2. c2. 1 24 c2 b 2 224 c 2. eⴚct>2 sin v*t.. The tangent of the y-curve is horizontal when y r 0, for the first positive time when v*t p, thus t t2 p>v* 2p> 24 c2. Now the y-curve oscillates between eⴚct>2, and (11) is satisfied if eⴚct>2 does not exceed 0.001. Thus ct 2 ln 1000, and t t2 gives the best c satisfying (11). Hence c. 2 ln 1000 t2. ,. c2 . (ln 1000)2. p2. (4 – c2) .. The solution of this is c 1.821, approximately. For this c we get by substitution v* 0.4141, t2 7.587 , and the particular solution y(t) eⴚ0.9103t(cos 0.4139t 2.199 sin 0.4139t) The graph shows a positive maximum near 15, a negative minimum near 23, a positive maximum near 30, and another negative minimum at 38. (e) The main difference is that Case II gives y (1 t)eⴚt which is negative for t 1 . The experiments with the curves are as before in this project. SECTION 2.5. Euler—Cauchy Equations, page 71 Purpose. Algebraic solution of the Euler–Cauchy equation, which appears in certain applications (see our Example 4) and which we shall need again in Sec. 5.4 as the simplest equation to which the Frobenius method applies. We have three cases; this is similar to the situation for constant-coefficient equations, to which the Euler–Cauchy equation can.

(13) c02.qxd. 6/18/11. 2:55 PM. Page 38. 38. Instructor’s Manual. be transformed (Team Project 20(d)); however, this fact is of theoretical rather than of practical interest. Comment on Footnote 4 Euler worked in St. Petersburg 1727–1741 and 1766–1783 and in Berlin 1741–1766. He investigated Euler’s constant (See. 5.6) first in 1734, used Euler’s formula (Secs. 2.2, 13.5, 13.6) beginning in 1740, introduced integrating factors (Sec. 1.4) in 1764, and studied conformal mappings (Chap. 17) starting in 1770. His main influence on the development of mathematics and mathematical physics resulted from his textbooks, in particular from his famous Introductio in analysin infinitorum (1748), in which he also introduced many of the modern notations (for trigonometric functions, etc.). Euler was the central figure of the mathematical activity of the 18th century. His Collected Works are still incomplete, although some seventy volumes have already been published. Cauchy worked in Paris, except during 1830–1838, when he was in Turin and Prague. In his two fundamental works, Cours d’Analyse (1821) and Résumé des leçons données à l’École royale polytechnique (vol. 1, 1823), he introduced more rigorous methods in calculus, based on an exactly defined limit concept; this also includes his convergence principle (Sec. 15.1). Cauchy also was the first to give existence proofs in ODEs. He initiated complex analysis; we discuss his main contributions to this field in Secs. 13.4, 14.2–14.4, and 15.2, His famous integral theorem (Sec. 14.2) was published in 1825 and his paper on complex power series and their radius of convergence (Sec. 15.2), in 1831. Examples in the Text. The examples illustrate the following. Examples 1–3 and Fig. 48 illustrate Cases I–III, respectively. In particular, Example 3 shows the derivation of real solutions from complex ones. Example 4 shows the occurrence of an Euler–Cauchy equation in connection with the electric potential field between concentric spheres kept at different constant potentials. Here, the student may wish to find a solution formula for arbitrary r1, r2 and potentials v1, v2. SOLUTIONS TO PROBLEM SET 2.5, page 73 y c1>x c2x 2 c1 c2>x 3 y c1x 0.5 c2x ⴚ0.2 c1x 6 c2>x y (c1 c2 ln x)x 2 y x(c1 cos (2 ln x) c2 sin (2 ln x)) y 1.2x 2 0.8x 3 1 2 13. y 3>2 1x x 14. y c1 cos (3 ln x) c2 sin (3 ln x) is a general solution, and from the initial conditions we obtain the answer 2. 4. 6. 7. 8. 10. 12.. y 56 sin (3 ln x) because y(1) c1 cos 0 c2 sin 0 c1 0 and 3 3 y r (x) c1(sin (3 ln x) # x c2 cos (3 ln x) # x c1 # 0 c2 # 3 2.5, so that c2 56..

(14) c02.qxd. 6/20/11. 11:21 AM. Page 39. Instructor’s Manual. 39. 3 5 15. y x x ln x 2 4 16. A general solution is y(x) (c1 c2 ln x)x2 , so that y(1) c1 p. The derivative is c2 y r (x) x x2 (c1 c2 ln x) # 2x, so that y r (1) c2 2c1 c2 2p 2p, hence c2 4p. This gives the answer y (p 4p ln x)x2 . 18. The auxiliary equation is 9 (m (m 1) 13 m 19) 9 (m2 23 m 19) 9 (m 13)2 0 has the double root 31. Hence a general solution is y(x) (c1 c2 ln x) x1>3 . ˛. By the first initial condition, y(1) c1 1. Differentiation gives c2 y r (x) x x1>3 (c1 c2 ln x) # 13 xⴚ2>3. The second initial condition thus gives y r (1) c2 c1 #. 1 3. 0.. Hence c2 13. This yields the answer (the solution of the initial value problem) y (1 13 ln x) x1>3 . 3 x2 5 10x 20. Team Project. (a) The student should realize that the present steps are the same as in the general derivation of the method in Sec. 2.1. An advantage of such specific derivations may be that the student gets a somewhat better understanding of the method and feels more comfortable with it. Of course, once a general formula is available, there is no objection to applying it to specific cases, but often a direct derivation may be simpler. In that respect the present situation resembles, for instance, that of the integral solution formula for first-order linear ODEs in Sec. 1.5. (b) The Euler–Cauchy equation to start from is 19. y . 1. 3. x2y s (1 2m s)xy r m(m s)y 0.

(15) c02.qxd. 6/18/11. 40. 2:55 PM. Page 40. Instructor’s Manual. where m (1 a)>2, the exponent of the one solution we first have in the critical case. For s : 0 the ODE becomes x2y s (1 2m)xy r m2y 0 . Here 1 2m 1 (1 a) a, and m2 (1 a) 2>4 , so that this is the Euler–Cauchy equation in the critical case. Now the ODE is homogeneous and linear; hence another solution is Y (xmⴙs xm)>s . L’Hôpital’s rule, applied to Y as a function of s (not x, because the limit process is with respect to s, not x), gives (xmⴙs ln ƒ x ƒ )>1 : xm ln ƒ x ƒ. as s : 0 .. This is the expected result. (c) This is less work than perhaps expected, an exercise in the technique of differentiation (also necessary in other cases). We have y xm ln x, and with (ln x) r 1>x we get y r mxmⴚ1 ln ƒ x ƒ xmⴚ1 y s m(m 1)xmⴚ2 ln ƒ x ƒ mxmⴚ2 (m 1)xmⴚ2 . Since xm x(1a)>2 is a solution, in the substitution into the ODE the ln-terms drop out. Two terms from y s and one from y r remain and give x2(mxmⴚ2 (m 1)xmⴚ2) axm xm(2m 1 a) 0 because 2m 1 a. # # (d) t ln x, dt>dx 1>x, y r y t r y >x, where the dot denotes the derivative with respect to t. By another differentiation, y s (y >x) r y >x2 y >(x2).. #. ##. #. Substitution of y r and y s into (1) gives the constant-coefficient ODE. ##. #. #. ##. #. y y ay by y (a 1)y by 0. The corresponding characteristic equation has the roots l 12 (1 a) 214(1 a)2 b. With these l, solutions are elt (et)l (eln ƒ x ƒ)l xl. (e) telt (ln ƒ x ƒ )el ln ƒ x ƒ (ln ƒ x ƒ )(eln ƒ x ƒ)l xl ln ƒ x ƒ..

(16) c02.qxd. 6/18/11. 2:55 PM. Page 41. Instructor’s Manual. 41. SECTION 2.6. Existence and Uniqueness of Solutions. Wronskian, page 74 Purpose. To explain the theory of existence of solutions of ODEs with variable coefficients in standard form (that is, with y s as the first term, not, say, f (x)y s ) y s p(x)y r q(x)y 0. (1). and of their uniqueness if initial conditions y(x0) K0,. (2). y r (x0) K1. are imposed. Of course, no such theory was needed in the last sections on ODEs for which we were able to write all solutions explicitly. Main Content. The theorems show the following. Theorem 1 shows that the continuity of the coefficients suffices for the existence and uniqueness of a solution of the initial values problem (1), (2). Theorem 2 gives a criterion for linear dependence and independence involving the Wronskian. Simple basic applications are shown in Examples 1 and 2. Theorem 3 on the existence of a general solution follows from Theorems 1 and 2 by the trick of using two special initial value problems; this idea is worth remembering. For Theorem 4 see below. Comment on Wronskian For n 2, where linear independence and dependence can be seen immediately, the Wronskian serves primarily as a tool in our proofs; the practical value of the independence criterion will appear for higher n in Chap. 3. Comment on General Solution Theorem 4 shows that linear ODEs (actually, of any order) have no singular solutions. This also justifies the term “general solution,” on which we commented earlier. We did not pay much attention to singular solutions, which sometimes occur in geometry as envelopes of one-parameter families of straight lines or curves. Altogether, this provides a general theory that is useful in practice. SOLUTIONS TO PROBLEM SET 2.6 page 79 2. W 2. e4x 4e4x. eⴚ1.5x 1 2 e2.5x 2 1.5eⴚ1.5x 4. 1 2 5.5e2.5x 1.5. 3. 2e 3x 1 4. x 6. We use the abbreviations c cos vx, s sin vx. Then W 2. 7.. 1 a 2. eⴚxc eⴚx(c v s). eⴚxs c 2 e 2x 2 eⴚx(s v c) c v s. s 2 veⴚ2x. s v c.

(17) c02.qxd. 6/18/11. 42. 2:55 PM. Page 42. Instructor’s Manual. 8. Equation (6*) saves much work and avoids sources of errors. We obtain y2 r a y b (tan (ln x)) r 1. 1 # 1, cos2 (ln x) x. Multiplication by y 12 x2k cos2 (ln x) gives W x2k1 . 10. x 2y s (m 1 m 2 1)xy r m 1m 2y 0. For the Wronskian we obtain from (6*) W a. xm1 r 2m2 b x (m1 m2)xm1ⴙm2ⴚ1. xm2. From the initial conditions we obtain the particular solution y 2x m1 4x m2. 11. W 0.5e 5x, y s 5y r 6.5y 0 12. x 2y s 3xy r 4y 0. The Wronskian is W. 冟. x2 2x. 冟. x 2 ln x x 3. 2x ln x x. The solution of the initial value problem is y (4 2 ln x)x2 . 13. W 3e3x, y s 3y r 0 14. y1 eⴚkx cos px, y2 eⴚkx sin px . By (6*), W (y2>y1) r y 12 (tan px) r eⴚ2kx cos2 px peⴚ2kx.. The characteristics equation is (l k)2 p2 0. This gives the ODE y s 2ky r 1k2 p22y 0. From the initial conditions we obtain the particular solution y eⴚkx(cos px sin px). 16. Team Project. (a) c1ex c2eⴚx c*1 cosh x c*2 sinh x. Expressing cosh and sinh in terms of exponential functions [see (17) in App. 3.1], we have 1 * 2 (c1. c*2 )e x 12(c*1 c*2 )e ⴚx;. hence c1 12(c*1 c*2 ), c2 12(c*1 c*2 ). The student should become aware that for second-order ODEs there are several possibilities for choosing a basis and making up a general solution. For this reason we say “a general solution,” whereas for first-order ODEs we said “the general solution.”.

(18) c02.qxd. 6/18/11. 2:55 PM. Page 43. Instructor’s Manual. 43. (b) If two solutions are 0 at the same point x 0, their Wronskian is 0 at x 0, so that these solutions are linearly dependent by Theorem 2. (c) The derivatives would be 0 at that point. Hence so would be their Wronskian W, implying linear dependence. (d) y2>y1 is constant in the case of linear dependence; hence the derivative of this quotient is 0, whereas in the case of linear independence this is not the case. This makes it likely that such a formula should exist. (e) The first two derivatives of y1 and y2 are continuous at x 0 (the only x at which something could happen). Hence these functions qualify as solutions of a secondorder ODE. y1 and y2 are linearly dependent for x 0 as well as for x 0 because, in each of these two intervals, one of the functions is identically 0. On 1 x 1 they are linearly independent because c1y1 c2y2 0 gives c1 0 when x 0, and c2 0 when x 0. The Wronskian is W y1 y 2r y2 y 1r e. 0 # 3x2 x3 # 0 f 0 x3 # 0 0 # 3x2. if e. x 0 . x 0. The Euler–Cauchy equation satisfied by these functions has the auxiliary equation (m 3)m m(m 1) 2m 0. Hence the ODE is xy s 2y r 0. Indeed, xy s1 2y1r x # 6x 2 # 3x2 0 if x 0, and 0 0 for x 0. Similarly for y2. Now comes the point. In the present case the standard form, as we use it in all our present theorems, is ys . 2 yr 0 x. and shows that p(x) is not continuous at 0, as required in Theorem 2. Thus there is no contradiction. This illustrates why the continuity assumption for the two coefficients is quite important. (f) According to the hint given in the enunciation, the first step is to write the ODE (1) for y1 and then again for y2. That is, y1s py 1r qy1 0 y s2 py r2 qy2 0 where p and q are variable. The hint then suggests eliminating q from these two ODEs. Multiply the first equation by y2, the second by y1, and add: (y1 y s2 y s1 y2) p(y1 y2r y1r y2) W r pW 0 where the expression for W r results from the fact that y1r y 2r appears twice and drops out. Now solve this by separating variables or as a homogeneous linear ODE..

(19) c02.qxd. 6/18/11. 44. 2:55 PM. Page 44. Instructor’s Manual. In Prob. 6 we have p 2, hence W ceⴚ2x by integration from 0 to x, where c y1(0)y2r (0) y2(0)y1r (0) 1 # v 0 # (1) v. SECTION 2.7. Nonhomogeneous ODEs, page 79 Purpose. We show that for getting a general solution y of a nonhomogeneous linear ODE we must find a general solution yh of the corresponding homogeneous ODE and then—this is our new task—any particular solution yp of the nonhomogeneous ODE, y yh yp. Main Content, Important Concepts General solution, particular solution Continuity of p, q, r suffices for existence and uniqueness. A general solution exists and includes all solutions. Comment on Methods for Particular Solutions The method of undetermined coefficients is simpler than that of variation of parameters (Sec. 2.10), as is mentioned in the text, and it is sufficient for many applications, of which Secs. 2.8 and 2.9 show standard examples. Comment on General Solution Theorem 2 shows that the situation with respect to general solutions is practically the same for homogeneous and nonhomogeneous linear ODEs. Comment on Table 2.1 It is clear that the table could be extended by the inclusion of products of polynomials times cosine or sine and other cases of limited practical value. Also, a 0 in the last pair of lines gives the previous two lines, which are listed separately because of their practical importance. General Comments on Text. Determination of Constants. For a good understanding, it is important to realize that a general solution of a nonhomogeneous linear ODE contains two kinds of constants, namely, (I) the constants in Table 2.1, which depend on the right side of the ODE, but not on the initial conditions, and must be determined first, (II) the two arbitrary constants in a general solution yh of the homogeneous ODE (and thus in a general solution y yh yp of the given ODE) and must be determined after the constants in 1I2 have been determined, and are to be determined by using the initial conditions. Examples 1–3 illustrate this important fact, which, for weaker students, sometimes causes difficulties in understanding. Examples in the Text. Examples 1–3 are very similar, illustrating Rules (a), (b), (c). In particular, the student should realize that Example 3 on the sum rule is not more difficult than the other two examples. The only additional idea in the case, say, r r1 r 2 on the right, is to split yp into a sum, yp yp1 yp2, whose coefficients, according to Table 2.1, can be determined for yp1 and yp2 separately. Hence we have to determine two sets of constants from two systems of algebraic equations, each of which is not larger than it would be had we only yp1 or only yp2 on the right side of the given ODE..

(20) c02.qxd. 6/18/11. 2:55 PM. Page 45. Instructor’s Manual. 45. Problem Set 2.7 This problem set begins with the determination of general solutions of nonhomogeneous linear ODEs, Probs. 1–6 with a single term on the right, Probs. 7–9 with a sum of two terms each, and Prob. 10 showing a simple extension of the method of undetermined coefficients beyond functions r shown in Table 2.1. Problems 11–18 concern IVPs for nonhomogeneous linear ODEs with one term on the right (Probs. 11–14 and 17) and with two terms on the right (Probs. 15, 16, and 18). CAS Project 19 should make the student aware that, depending on the initial conditions and on the kind of the homogeneous ODE, the solution y yh yp may approach yp as: x : , or may contain an increasing ƒ yh ƒ , or may be of the form y yp with yh absent. Team Project 20 is an invitation to explore more general functions on the right, and to what Euler–Cauchy equation the present method can be extended.. SOLUTIONS TO PROBLEM SET 2.7, page 84 1. y c1e 3x c2e 2x e x 2. y c1eⴚ3.2x c2eⴚ1.8x 0.00999 cos x 0.0105 sin x. This is a typical solution of a forced oscillation problem in the overdamped case. The general solution of the homogeneous ODE dies out, practically after some short time (theoretically never), and the transient solution goes over into a harmonic oscillation whose frequency is equal to that of the driving force (or electromotive force). Note that the input (the driving force) is a cosine, whereas the output (the response) is a cosine and sine; this means a phase shift. It is due to the presence of a y r -term, mechanically a linear damping force, as we shall see in the next section. 8 cos p x 4 p2 6. A general solution of the homogeneous ODE is 4. y c1e2x c2eⴚ2x . yh eⴚx>2(c1 cos px c2 sin px). We see that the function on the right side of the ODE is a solution of the homogeneous ODE. Hence we have to apply the Modification Rule, starting from yp xeⴚx>2(K cos px M sin px). Substitution gives K 1>(2p); M 0. Hence the answer is y yh yp eⴚx>2 ac1 cos px c2 sin px . x cos pxb. 2p. Note that the output involves cosine, whereas the input involves sine; and, although we have a y r -term, the output is a single term. Compare this with Probs. 2 and 4, which differ from the present situation. 7. y c1ex c2e3x 14 (1 2x)ex 32 x 2.

(21) c02.qxd. 6/18/11. 2:55 PM. Page 46. 46. Instructor’s Manual 1 8. y A cos 3x B sin 3x 18 cos x 18 x sin 3x. An important point is that the Modification Rule applies to the second term on the right. Hence the best way seems to split yp additively, yp yp1 yp2, where. yp1 K 1 cos x M 1 sin x,. yp2 K 2x cos 3x M 2x sin 3x.. In Prob. 9 the situation is similar. 10. 2x sin x is not listed in the table because it is of minor practical importance. However, by looking at its derivatives, we see that yp Kx cos x Mx sin x N cos x P sin x should be general enough. Indeed, by substitution and collecting cosine and sine terms separately we obtain (1) (2). (2K 2Mx 2P 2M) cos x 0 (2Kx 2M 2N 2K) sin x 2x sin x.. In (1) we must have 2Mx 0; hence M 0 and then P K. In (2) we must have 2Kx 2x; hence K 1, so that P 1 and from (2), finally, 2N 2K 0, hence N 1. Answer: y (c1 c2x)eⴚx (1 – x) cos x sin x. 11. 12. 14. 16.. y 1 2 cos 2x 2x2 The Modification Rule is needed. The answer is y 1.8 cos 2x sin 2x 3x cos 2x. y e 3x 74 xeⴚ3x 18 eⴚx sin 2x yh c1e2x c2, yp C1xe2x C2eⴚ2x by the Modification Rule for a simple root. y e2x 32 3xe2x 12 eⴚ2x.. x 7 1 x 11 15 x 3 3 e sin x e 5 cos x 4e 4 3 5 2 5 18. The Basic Rule and the Sum Rule are needed. We obtain. 17. y . yh eⴚx(A cos 3x B sin 3x) y eⴚx cos 3x 0.4 cos x 1.8 sin x 6 cos 3x sin 3x. 20. Team Project. (b) Perhaps the simplest way is to take a specific ODE, e.g., x 2y s 6xy r 6y r(x) and then experiment by taking various r(x) to find the form of choice functions. The simplest case is a single power of x. However, almost all the functions that work as r(x) in the case of an ODE with constant coefficients can also be used here..

(22) c02.qxd. 6/18/11. 2:55 PM. Page 47. Instructor’s Manual. 47. SECTION 2.8. Modeling: Forced Oscillations. Resonance, page 85 Purpose. To extend Sec. 2.4 from free to forced vibrations by adding an input (a driving force, here assumed to be sinusoidal). Mathematically, we go from a homogeneous to a nonhomogeneous ODE which we solve by undetermined coefficients. New Features Undamped Forced Oscillations. Resonance (Fig. 55) Resonance appears if the physical system is (theoretically) undamped (in practice, if it has small enough damping that the damping effect can be neglected), and if the input frequency is exactly equal to the natural frequency of the system. Then a solution (11). y At sin v0t. has a factor t which makes it increase to (Fig. 55). The approach to resonance as v : v0 ( 1k>m2 is also characterized by the resonance factor in Fig. 54. Undamped Forced Oscillations. Beats (Fig. 56) Beats occur if the input frequently is approximately equal to the natural frequency of the physical system. Then (12). y K (cos vt cos v0t). with K F0>[m(v02 v2)]. Damped Forced Oscillations. For these the transient solution approaches the steady-state solution as t : , practically after some time which may often be rather short. If c 0, there is no more true resonance, but the maximum amplitude (16) may still be large, as Fig. 57 illustrates. Also, there is a phase lag h, discontinuous and equal to 0 or p when c 0, and continuous and monotone when c 0 (Fig. 58). Problem Set 2.8 Problems 3–7 concern damped systems. Hence a general solution of such a physical system is that of a homogeneous linear ODE and approaches 0 as t : , so that solving these problems amounts to determining a particular solution of the corresponding ODE. Problems 8–15 amount to finding a general solution of the nonhomogeneous ODE. Problems 16–20 are IVPs for nonhomogeneous linear ODEs. Problem 17 is of the kind that will occur in connection with partial sums of Fourier series in Chap. 11. Problem 18 is a typical example illustrating the rapidity of approach to the steady-state solution. Beats are considered in Probs. 21, 22, and 25 with v 0.9, whereas for v farther away from v0 1 (corresponding to the natural frequency of the physical system) the form of vibrations cannot be guessed immediately (see Fig. 60). Practical resonance is considered experimentally in Team Experiment 23. Continuity Conditions Nonhomogeneous linear ODEs with a driving force acting for some finite interval of time only will require the idea of continuity conditions of y and y r at the instant of time when the driving force becomes identically zero. This makes such problems more involved, as Prob. 24 illustrates, and motivates the application of an “operational method,” such as the Laplace transform (Chap. 6)..

(23) c02.qxd. 6/18/11. 2:55 PM. Page 48. 48. Instructor’s Manual. SOLUTIONS TO PROBLEM SET 2.8, page 91 2. Problems 2, 8, 18. Note that the damping and restoring terms must have positive coefficients, and that Prob. 12 shows resonance; hence it is not a candidate. 9 3. y c1et c2e3t 36 65 cos 2t 130 sin 2t 4. yp cos 4t 0.6 sin 4t 5. y c1 cos 32 t c2 sin 32 t 59 cos 32 t 56 t sin 32 t 1 1 1 6. yp 10 cos t 90 cos 3t 15 sin t 45 sin 3t 3t 3 3t 1 3t t 8. y c1 et cos 3t c e sin sin 2 2 2 20 2 20 cos 2 10. y A cos 4t B sin 4t 7t sin 4t 1 11. y c1 cos 3t c2 sin 3t 18 (1 3t) cos 3t 16 t sin 3t 12. y eⴚt(A cos 2t B sin 2t) 2 sin t. Note that, whereas a single term on the right side of the ODE will usually produce two terms in the solution (the response), the present problem shows that sometimes the opposite will also occur. sin vt 13. y c1 cos 2t c2 sin 2t 2 v 4 ⴚt 14. y A cos t B sin t e (cos t 2 sin t). Note that this does not give resonance, but, on the contrary, yp : 0, which is understandable because the driving force on the right approaches 0 as t approaches . 4 16. y 11 60 sin 4t cos 4t 15 sin t 18. y eⴚ4t cos t 26.8 sin 0.5t 6.4 cos 0.5t. At t 1.2 the exponential term has decreased to less than 1% of its original value. This marks the end of the transition from a practical point of view. t 1.8 is the time when that term has become less than 1>10 of a percent in absolute value. 19. y 25 e2t sin t 15 e 2t cos t 15 et (cos t 2 sin t) 20. A general solution is y A cos 15 t B sin 15 t (cos pt sin pt)>(p2 5). Using the initial conditions, we obtain the answer y. 1 p acos 15 t sin 15 t cos pt sin ptb. p 5 15 2. 49 22. y 197 2 cos 5t 100 cos 10 t. 24. If 0

(24) t

(25) p, then a particular solution yp K 0 K 1t K 2t 2 gives y sp 2k 2 and y sp yp K0 2K2 K1t K2t2 1 . 1. p2. t2;. thus, K2 . 1. p. 2. ,. K1 0,. K0 1 2K2 1 . 2. p2. ..

(26) c02.qxd. 6/18/11. 2:55 PM. Page 49. Instructor’s Manual. 49. Hence a general solution is y A cos t B sin t 1 . 2. p2. 1. . p2. t2.. From this and the first initial conditions, y(0) A 1 . 2. p2. 0,. A a1 . 2. p2. b.. The derivative is y r A sin t B cos t . 2. p2. t. and gives y r (0) B 0. Hence the solution is (I). y(t) (1 2> p2)(1 cos t) t2> p2. if 0

(27) t

(28) p,. and if t p , then y y2 A2 cos t B2 sin t. (II). with A2 and B2 to be determined from the continuity conditions y(p) y2(p),. y r (p) y 2r (p).. So we need from (I) and (II) y(p) 2(1 2> p2) 1 1 4> p2 y2(p) A2 and y r (t) (1 2> p2) sin t 2t> p2 and from this and (II), y r (p) 2> p B cos p B2. This gives the solution y (1 4> p2) cos t (2> p) sin t. if t p .. Answer: y e. (1 2> p2)(1 cos t) t2> p2. (1 4> p ) cos t (2> p) sin t 2. if 0

(29) t

(30) p if t p. .. The function in the second line gives a harmonic oscillation because we disregarded damping..

(31) c02.qxd. 6/18/11. 2:55 PM. Page 50. 50. Instructor’s Manual. SECTION 2.9. Modeling: Electric Circuits, page 93 Purpose. To discuss the current in an RLC-circuit with sinusoidal input E 0 sin vt, as follows. Modeling the RLC-circuit Solving the model (1) for the current I(t) Discussion of a typical IVP Discussion of a electrical–mechanical analogy Modeling the RLC-circuit. The student should first review the special case of an RL-circuit in Example 2 of Sec. 1.5, which is modeled by a first-order ODE, using Kirchhoff’s KVL. The present addition of a capacitor is very simple in terms of setting up the model, resulting in a second-order ODE. Proceed stepwise in this way: Write the voltage drop across the capacitor in the form (1>C)Q (rather than CQ) is a standard convention to obtain generally more convenient numbers. Solve the model (1) by the method of undetermined coefficients (see Sec. 2.7). ATTENTION! The right side in (1) is E 0v cos vt, because of differentiation. In solving, two quantities of practical importance are introduced, namely, the reactance S vL 1>(vC). (3). and the impedance (also called the apparent resistance) 2R2 C2 . (Its complex analog, the complex impedance Z R iS, is mentioned in the answer to Prob. 20.) Example 1 shows a typical IVP with Ih rapidly going to 0, as illustrated in Fig. 62, so that the transient current rapidly approaches a harmonic steady-state current. Table 2.2 shows a strictly quantitative electrical–mechanical analogy, which is used in transducers, as explained in the text. SOLUTIONS TO PROBLEM SET 2.9, page 98 2. This is another special case of a circuit that leads to an ODE of first order, RI r I>C E r vE0 cos vt . Integration by parts gives the solution I(t) eⴚt>(RC). c. v E0 t>(RC) e cos vt dt c d R. ceⴚt>(RC) ceⴚt>(RC) . 冮. vE0C 1 (vRC)2. (cos vt vRC sin vt). vE0C 21 (vRC)2. sin (vt d),.

(32) c02.qxd. 6/18/11. 2:55 PM. Page 51. Instructor’s Manual. 51. where tan d 1>(vRC). The first term decreases steadily as t increases, and the last term represents the steady-state current, which is sinusoidal. The graph of I(t) is similar to that in Fig. 62. 40t 3. LI r RI E, I (E/R) ce RT/L 11 2 ce 4. The integral that occurs can be evaluated by integration by parts, as is shown (with other notations) in standard calculus texts. From (4) in Sec. 1.5 we obtain I eⴚRt>L. c. E0 Rt>L e sin vt dt c d L. ceⴚRt>L ce ⴚRt>L . 冮. E0 R v2L2 2. (R sin vt vL cos vt). E0 2R v L 2. 2 2. sin (vt d),. 3 6. LI s I>C 6t 2, I(0) I r (0) 0, I(t) 800 . 3 800. d arctan cos 425t . vL. .. R. 3 2 20 t. 8. E r 1000 cos 2t, 0.5I s 4I r 10I 1000 cos 2t, so that the steady-state solution is I 62.5 (cos 2t sin 2t) A. 10. The ODE is I s 2I r 20I 157 # 3. The steady-state solution is I 33 cos 3t 18 sin 3t. Note that if you let C decrease, the sine term in the solution will become smaller and smaller, compared with the cosine term. 22 11. I c1e15t c2e5t 11 3 cos 5t 3 sin 5t 12. The ODE is 0.1I s 0.2I r 0.5I 220 # 314 cos 314t. Its characteristic equation is 0.1[(l 1) 2 4] 0. Hence a general solution of the homogenous ODE is eⴚt(A cos 2t B sin 2t). The transient solution (rounded to 4 decimals) is I eⴚt(A cos 2t B sin 2t ) 7.0064 cos 314t 0.0446 sin 314t A. 15 13. I c1e 4t cos 2 221t c2e 4t sin 2 221t 45 37 cos 50t 74 sin 50t.

(33) c02.qxd. 6/18/11. 2:55 PM. Page 52. 52. Instructor’s Manual. 14. Write l1 a b and l2 a b, as in the text before Example 1. Here a R>(2L) 0 , and b can be real or imaginary. If b is real, then b

(34) R>(2L) because R2 4L>C

(35) R2 . Hence l1 0 (and l2 0, of course) . If b is imaginary, then Ih(t) represents a damped oscillation, which certainly goes to zero as t : . 16. The ODE is 0.2I s 8I r 80I 1000 cos 10t. A general solution is I (c 1 c2t)eⴚ20t 6 cos 10t 8 sin 10t. The initial conditions are I(0) 0, Q(0) 0, which because of (1 r ), that is, LI r (0) RI102 . Q(0) E(0) 0, C. leads to I r (0) 0. This gives I(0) c1 6 0, I r (0) 20c1 c2 80 0,. c1 6 c2 200.. Hence the answer is I (6 200t)eⴚ20t 6 cos 10t 8 sin 10t. 17. I 0.670e 9t sin 13t 1.07e 9t cos13t 0.927 sint 1.07 cost 18. E r 880 sin 4t, LI s RI r C1 I E r , I(0) 0, I r (0) E(0)>L 220, Also, E r 880 sin 4t, The ODE is I s 14I r 40I 880 sin 4t The answer is 165 2750 10t 55 4t 385 e cos 4t sin 4t e 87 3 29 29 20. 苲 I p Keivt, 苲 I pr ivKeivt, 苲 I ps v2Keivt . Substitution gives I. av2L ivR . 1 b Keivt E0veivt . C. Divide this by veivt on both sides and solve the resulting equation algebraically for K, obtaining (A). K. E0 E0 S iR 1 b iR avL vC.

(36) c02.qxd. 6/18/11. 2:55 PM. Page 53. Instructor’s Manual. 53. where S is the reactance given by (3). To make the denominator real, multiply the numerator and the denominator of the last expression by S iR. This gives K. E0(S iR) S2 R2. .. The real part of Keivt is (Re K)(Re eivt) (Im K)(Im eivt) . E0S S R 2. . 2. cos vt . E0 S R2 2. E0R S R2 2. sin vt. (S cos vt R sin vt),. in agreement with (2) and (4). We mention that (A) can be written K. E0 iZ. where Z R iS R i avL . 1 b vC. is called the complex impedance. Note that its absolute value ƒ Z ƒ 2R2 S2 is the impedance, as defined in the text. SECTION 2.10. Solution by Variation of Parameters, page 99 Purpose. To discuss the general method for particular solutions, which applies in any case but may often lead to difficulties in integration (which we, by and large, have avoided in our problems, as the subsequent solutions show). Comments The ODE must be in standard form, with 1 as the coefficient of y s —students tend to forget that. Here we do need the Wronskian, in contrast with Sec. 2.6 where we could get away without it. SOLUTIONS TO PROBLEM SET 2.10, page 102 1. y c1 cos 2x c2 sin 2x 18 cos 2x 14 x sin 2x 2. y1 cos 3x, y2 sin 3x, W 3, r csc 3x. Hence in (2), 3x x dx 冮 W dx 13 冮 sin sin 3x 3 y2r. 3x 1 dx ln ƒ sin 3x ƒ . 冮 W dx 13 冮 cos sin 3x 9 y1r.

(37) c02.qxd. 6/18/11. 2:55 PM. Page 54. 54. Instructor’s Manual. Answer: y A cos 3x B sin 3x . x 1 cos 3x (sin 3x) ln ƒ sin 3x ƒ 3 9. 3. c1 x1 c2 x313 x2 4. y1 e2x cos x,. y2 e2x sin x, W e4x . Hence in (2),. 冮W. dx . 冮W. dx . y2r y1r. 冮冮. (e2x sin x)e2x>sin x. dx x. (e2x cos x)e2x>sin x. dx ln ƒ sin x ƒ .. e4x. e4x. Answer: y c A cos x B sin x x cos x (sin x) ln ƒ sin x ƒ d e2x. 6. y1 eⴚ3x, y2 xeⴚ3x, W e6x. Hence in (2),. 冮W. dx . 冮W. dx . y2r y1r. 冮冮. (xeⴚ3x)16eⴚ3x>(x2 1) ⴚ6x. e. (eⴚ3x)16eⴚ3x>(x2 1) ⴚ6x. e. dx . dx . 冮 x 16x 1 dx 8 ln (x. 2. 2. 1). 冮 x 16 1 dx 16 arctan x. 2. Answer: y (c1 c2x)eⴚ3x 8[ln (x2 1) 2x arctan x] eⴚ3x. 7. y c1 xex c2ex 34 (3 4x 2x2 )ex 8. y c1 cos 2x c2 sin 2x 18 cosh 2x 10. y1 eⴚx cos x, y2 eⴚx sin x, W eⴚ2x . Hence in (2),. 冮W. y2r. 冮W. y1r. dx dx . 冮冮. (eⴚx sin x) 4eⴚx>cos3 x ⴚ2x. e. ⴚx. (e. cos x) 4eⴚx>cos3 x eⴚ2x. dx . 2 cos2 x. dx 4 tan x.. This gives the particular solution eⴚx c (cos x). 2 2 4 sin2 x 4(sin x) tan x d eⴚx a b 2 cos x cos x eⴚx[2(cos 2x)>cos x]..

(38) c02.qxd. 6/18/11. 2:55 PM. Page 55. Instructor’s Manual. 55. Answer: y eⴚx[A cos x B sin x 2(cos 2x)>cos x]. 12. c1ex c2ex 12 ((x ln(sinh x))e2x x ln((sinh x)))ex 14. TEAM PROJECT. (a) y1 eⴚ3x, y2 eⴚx, W 2eⴚ4x, r 65 cos 2x. From (2), yp eⴚ3x . ⴚx. 冮e. 65 cos 2x eⴚ3x65 cos 2x dx eⴚx dx ⴚ4x 2e 2eⴚ4x. 冮. 冮. 冮. 65 aeⴚ3x e3x cos 2x dx eⴚx ex cos 2x dxb 2. ⴚ3x 1 3x ⴚx 1 x 65 2 (e 13 e (3 cos 2x 2 sin 2x) e 5 e (cos 2x 2 sin 2x)). cos 2x 8 sin 2x. Answer: y c1eⴚ3x c2eⴚx cos 2x 8 sin 2x. This was much more work than that for undetermined coefficients. (b) We can treat x2 on the right by undetermined coefficients, obtaining the contribution x2 4x 6 to the solution. We could treat it by the other method, but we would have to evaluate additional integrals of an exponential function times a power of x. We treat the other part, 35x3>2ex , by the method of this section, calling the resulting function yp1. We need y1 ex, y2 xex, W e2x. From this and (2), x. x. 冮 xee 35x e dx xe 冮 ee 35x e 35 ae 冮 x dx xe 冮 x dxb 4e x. yp1 ex. 3>2 x. x. 2x. x. 3>2 x. 2x. 5>2. x. 3>2. dx. x 7>2. .. Complete answer: y (c1 c2x)ex 4exx7>2 x2 4x 6. (c) If the right side is a power of x, say, r r0xk , then substitution of yp Cxk gives x2y s axy r by (k(k 1) ak b)Cxk r0 xk. This can be solved for C. To explore further possibilities, one may work “backwards”; that is, assume a solution, substitute it on the left, and see what from one gets as a right side..

(39) c02.qxd. 6/18/11. 2:55 PM. Page 56. 56. Instructor’s Manual. SOLUTIONS TO CHAPTER 2 REVIEW QUESTIONS AND PROBLEMS, page 102 7. 8. 9. 10.. 9. 5. y c1eⴚ4 x c2eⴚ4 x y c1eⴚ4x c2e3x y c1e2x sin3x c2e2x cos 3x y eⴚ0.1x(A cos 0.4x B sin 0.4x) 2. 2. 11. y c1e3 x c2xe3 x 12. y (c1 c2x)eⴚ2px c1 14. y 4 c2 x4 x 16. y eⴚx(A cos x B sin x) eⴚx cos 2x 18. y c2ec1x 19. y 76 sin3x 92 cos3x 32 ex 20. y 12ex 12 e3x cos x 2 sin x 1 4 ln x 22. y 5 x x5 24. I c1eⴚ1999.87t c2eⴚ0.125008t A 26. E r 220 ⴢ 314 cos 314t, I eⴚ50t(A cos 150t B sin 150t) 0.847001 sin 314t 1.985219 cos 314t A 28. my s ky cos x 2 sin x, y(0) 0, y r (0) 0; y 13 sin 3x 12 cos3x 1 2 cos x sin x, v0 3 3. 3. 30. my s cy r ky F0 sin vt; y c1e 2 t cos 32 23t c2e 2 t sin 23 23t 5 9. sin 32 23t 59 22 cos 32 23t; C* 0.9623.

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