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286 Chapter 6 Applications of Linear Second Order Equations is of the form y = yc + yp, where yc has one of the three forms yc = e−ct/2m(c1 cos ω1t + c2 sinω1t), yc = e−ct/2m(c1 + c2t), yc = c1er1t + c2er2t (r1, r2 < 0). In all three cases limt→∞yc(t) = 0 for any choice of c1 and c2. For this reason we say that yc is the transient component of the solution y. The behavior of y for large t is determined by yp, which we call the steady state component of y. Thus, for large t the motion is like simple harmonic motion at the frequency of the external force. The amplitude of yp in (6.2.17) is R = F0 p m2(ω2 0 −ω2)2 + c2ω2 , (6.2.18) which is finite for all ω; that is, the presence of damping precludes the phenomenon of resonance that we encountered in studying undamped vibrations under a periodic forcing function. We’ll now find the value ωmax of ω for which R is maximized. This is the value of ω for which the function ρ(ω) = m2(ω2 0 −ω2)2 + c2ω2 in the denominator of (6.2.18) attains its minimum value. By rewriting this as ρ(ω) = m2(ω4 0 + ω4) + (c2 −2m2ω2 0)ω2, (6.2.19) you can see that ρ is a strictly increasing function of ω2 if c ≥ q 2m2ω2 0 = √ 2mk. (Recall that ω2 0 = k/m). Therefore ωmax = 0 if this inequality holds. From (6.2.15), you can see that φ = 0 if ω = 0. In this case, (6.2.14) reduces to yp = F0 p m2ω4 0 = F0 k , which is consistent with Hooke’s law: if the mass is subjected to a constant force F0, its displacement should approach a constant yp such that kyp = F0. Now suppose c < √ 2mk. Then, from (6.2.19), ρ′(ω) = 2ω(2m2ω2 + c2 −2m2ω2 0), and ωmax is the value of ω for which the expression in parentheses equals zero; that is, ωmax = r ω2 0 − c2 2m2 = s k m 1 − c2 2km . (To see that ρ(ωmax) is the minimum value of ρ(ω), note that ρ′(ω) < 0 if ω < ωmax and ρ′(ω) > 0 if ω > ωmax.) Substituting ω = ωmax in (6.2.18) and simplifying shows that the maximum amplitude Rmax is Rmax = 2mF0 c √ 4mk −c2 if c < √ 2mk. We summarize our results as follows. | Elementary Differential Equations with Boundary Value Problems_Page_296_Chunk2701 |
Section 6.2 Spring Problems II 287 Theorem 6.2.1 Suppose we consider the amplitude R of the steady state component of the solution of my′′ + cy′ + ky = F0 cos ωt as a function of ω. (a) If c ≥ √ 2mk, the maximum amplitude is Rmax = F0/k and it’s attained when ω = ωmax = 0. (b) If c < √ 2mk, the maximum amplitude is Rmax = 2mF0 c √ 4mk −c2 , (6.2.20) and it’s attained when ω = ωmax = s k m 1 − c2 2km . (6.2.21) Note that Rmax and ωmax are continuous functions of c, for c ≥0, since (6.2.20) and (6.2.21) reduce to Rmax = F0/k and ωmax = 0 if c = √ 2km. | Elementary Differential Equations with Boundary Value Problems_Page_297_Chunk2702 |
288 Chapter 6 Applications of Linear Second Order Equations 6.2 Exercises 1. A 64 lb object stretches a spring 4 ft in equilibrium. It is attached to a dashpot with damping constant c = 8 lb-sec/ft. The object is initially displaced 18 inches above equilibrium and given a downward velocity of 4 ft/sec. Find its displacement and time–varying amplitude for t > 0. 2. C/G A 16 lb weight is attached to a spring with natural length 5 ft. With the weight attached, the spring measures 8.2 ft. The weight is initially displaced 3 ft below equilibrium and given an upward velocity of 2 ft/sec. Find and graph its displacement for t > 0 if the medium resists the motion with a force of one lb for each ft/sec of velocity. Also, find its time–varying amplitude. 3. C/G An 8 lb weight stretches a spring 1.5 inches. It is attached to a dashpot with damping constant c=8 lb-sec/ft. The weight is initially displaced 3 inches above equilibrium and given an upward velocity of 6 ft/sec. Find and graph its displacement for t > 0. 4. A 96 lb weight stretches a spring 3.2 ft in equilibrium. It is attached to a dashpot with damping constant c=18 lb-sec/ft. The weight is initially displaced 15 inches below equilibrium and given a downward velocity of 12 ft/sec. Find its displacement for t > 0. 5. A 16 lb weight stretches a spring 6 inches in equilibrium. It is attached to a damping mechanism with constant c. Find all values of c such that the free vibration of the weight has infinitely many oscillations. 6. An 8 lb weight stretches a spring .32 ft. The weight is initiallydisplaced 6 inches above equilibrium and given an upward velocity of 4 ft/sec. Find its displacement for t > 0 if the medium exerts a damping force of 1.5 lb for each ft/sec of velocity. 7. A 32 lb weight stretches a spring 2 ft in equilibrium. It is attached to a dashpot with constant c = 8 lb-sec/ft. The weight is initially displaced 8 inches below equilibrium and released from rest. Find its displacement for t > 0. 8. A mass of 20 gm stretches a spring 5 cm. The spring is attached to a dashpot with damping constant 400 dyne sec/cm. Determine the displacement for t > 0 if the mass is initially displaced 9 cm above equilibrium and released from rest. 9. A 64 lb weight is suspended from a spring with constant k = 25 lb/ft. It is initially displaced 18 inches above equilibrium and released from rest. Find its displacement for t > 0 if the medium resists the motion with 6 lb of force for each ft/sec of velocity. 10. A 32 lb weight stretches a spring 1 ft in equilibrium. The weight is initially displaced 6 inches above equilibrium and given a downward velocity of 3 ft/sec. Find its displacement for t > 0 if the medium resists the motion with a force equal to 3 times the speed in ft/sec. 11. An 8 lb weight stretches a spring 2 inches. It is attached to a dashpot with damping constant c=4 lb-sec/ft. The weight is initially displaced 3 inches above equilibrium and given a downward velocity of 4 ft/sec. Find its displacement for t > 0. 12. C/G A 2 lb weight stretches a spring .32 ft. The weight is initially displaced 4 inches below equilibriumand given an upward velocity of 5 ft/sec. The medium provides damping with constant c = 1/8 lb-sec/ft. Find and graph the displacement for t > 0. 13. An 8 lb weight stretches a spring 8 inches in equilibrium. It is attached to a dashpot with damping constant c = .5 lb-sec/ft and subjected to an external force F (t) = 4 cos 2t lb. Determine the steady state component of the displacement for t > 0. | Elementary Differential Equations with Boundary Value Problems_Page_298_Chunk2703 |
Section 6.2 Spring Problems II 289 14. A 32 lb weight stretches a spring 1 ft in equilibrium. It is attached to a dashpot with constant c = 12 lb-sec/ft. The weight is initially displaced 8 inches above equilibrium and released from rest. Find its displacement for t > 0. 15. A mass of one kg stretches a spring 49 cm in equilibrium. A dashpot attached to the spring supplies a damping force of 4 N for each m/sec of speed. The mass is initially displaced 10 cm above equilibrium and given a downward velocity of 1 m/sec. Find its displacement for t > 0. 16. A mass of 100 grams stretches a spring 98 cm in equilibrium. A dashpot attached to the spring supplies a damping force of 600 dynes for each cm/sec of speed. The mass is initially displaced 10 cm above equilibrium and given a downward velocity of 1 m/sec. Find its displacement for t > 0. 17. A 192 lb weight is suspended from a spring with constant k = 6 lb/ft and subjected to an external force F (t) = 8 cos 3t lb. Find the steady state component of the displacement for t > 0 if the medium resists the motion with a force equal to 8 times the speed in ft/sec. 18. A 2 gm mass is attached to a spring with constant 20 dyne/cm. Find the steady state component of the displacement if the mass is subjected to an external force F (t) = 3 cos 4t −5 sin 4t dynes and a dashpot supplies 4 dynes of damping for each cm/sec of velocity. 19. C/G A 96 lb weight is attached to a spring with constant 12 lb/ft. Find and graph the steady state component of the displacement if the mass is subjected to an external force F (t) = 18 cos t−9 sin t lb and a dashpot supplies 24 lb of damping for each ft/sec of velocity. 20. A mass of one kg stretches a spring 49 cm in equilibrium. It is attached to a dashpot that supplies a damping force of 4 N for each m/sec of speed. Find the steady state component of its displacement if it’s subjected to an external force F (t) = 8 sin 2t −6 cos 2t N. 21. A mass m is suspended from a spring with constant k and subjected to an external force F (t) = α cos ω0t+β sin ω0t, where ω0 is the natural frequency of the spring–mass system without damp- ing. Find the steady state component of the displacement if a dashpot with constant c supplies damping. 22. Show that if c1 and c2 are not both zero then y = er1t(c1 + c2t) can’t equal zero for more than one value of t. 23. Show that if c1 and c2 are not both zero then y = c1er1t + c2er2t can’t equal zero for more than one value of t. 24. Find the solution of the initial value problem my′′ + cy′ + ky = 0, y(0) = y0, y′(0) = v0, given that the motion is underdamped, so the general solution of the equation is y = e−ct/2m(c1 cos ω1t + c2 sin ω1t). 25. Find the solution of the initial value problem my′′ + cy′ + ky = 0, y(0) = y0, y′(0) = v0, given that the motion is overdamped, so the general solution of the equation is y = c1er1t + c2er2t (r1, r2 < 0). | Elementary Differential Equations with Boundary Value Problems_Page_299_Chunk2704 |
290 Chapter 6 Applications of Linear Second Order Equations 26. Find the solution of the initial value problem my′′ + cy′ + ky = 0, y(0) = y0, y′(0) = v0, given that the motion is critically damped, so that the general solution of the equation is of the form y = er1t(c1 + c2t) (r1 < 0). 6.3 THE RLC CIRCUIT In this section we consider the RLC circuit, shown schematically in Figure 6.3.1. As we’ll see, the RLC circuit is an electrical analog of a spring-mass system with damping. Nothing happens while the switch is open (dashed line). When the switch is closed (solid line) we say that the circuit is closed. Differences in electrical potential in a closed circuit cause current to flow in the circuit. The battery or generator in Figure 6.3.1 creates a difference in electrical potential E = E(t) between its two terminals, which we’ve marked arbitrarily as positive and negative. (We could just as well interchange the markings.) We’ll say that E(t) > 0 if the potential at the positive terminal is greater than the potential at the negative terminal, E(t) < 0 if the potential at the positive terminal is less than the potential at the negative terminal, and E(t) = 0 if the potential is the same at the two terminals. We call E the impressed voltage. Induction Coil (Inductance L) + −− Capacitor (Capacitance C) + −− Resistor (Resistance R) + −− Battery or Generator (Impressed Voltage E=E(t)) + −− Switch Figure 6.3.1 An RLC circuit At any time t, the same current flows in all points of the circuit. We denote current by I = I(t). We say that I(t) > 0 if the direction of flow is around the circuit from the positive terminal of the battery or generator back to the negative terminal, as indicated by the arrows in Figure 6.3.1 I(t) < 0 if the flow is in the opposite direction, and I(t) = 0 if no current flows at time t. | Elementary Differential Equations with Boundary Value Problems_Page_300_Chunk2705 |
Section 6.3 The RLC Circuit 291 Differences in potential occur at the resistor, induction coil, and capacitor in Figure 6.3.1. Note that the two sides of each of these components are also identified as positive and negative. The voltagedrop across each component is defined to be the potential on the positive side of the component minus the potential on the negative side. This terminology is somewhat misleading, since “drop” suggests a decrease even though changes in potential are signed quantities and therefore may be increases. Nevertheless, we’ll go along with tradition and call them voltage drops. The voltage drop across the resistor in Figure 6.3.1 is given by VR = IR, (6.3.1) where I is current and R is a positive constant, the resistance of the resistor. The voltage drop across the induction coil is given by VI = LdI dt = LI′, (6.3.2) where L is a positive constant, the inductance of the coil. A capacitor stores electrical charge Q = Q(t), which is related to the current in the circuit by the equation Q(t) = Q0 + Z t 0 I(τ) dτ, (6.3.3) where Q0 is the charge on the capacitor at t = 0. The voltage drop across a capacitor is given by VC = Q C , (6.3.4) where C is a positive constant, the capacitance of the capacitor. Table 6.3.8 names the units for the quantities that we’ve discussed. The units are defined so that 1 volt = 1 ampere · 1 ohm = 1 henry · 1 ampere/second = 1 coulomb/ farad and 1 ampere = 1 coulomb/second. Table 6.3.8. Electrical Units Symbol Name Unit E Impressed Voltage volt I Current ampere Q Charge coulomb R Resistance ohm L Inductance henry C Capacitance farad According to Kirchoff’s law, the sum of the voltage drops in a closed RLC circuit equals the impressed voltage. Therefore, from (6.3.1), (6.3.2), and (6.3.4), LI′ + RI + 1 C Q = E(t). (6.3.5) | Elementary Differential Equations with Boundary Value Problems_Page_301_Chunk2706 |
292 Chapter 6 Applications of Linear Second Order Equations This equation contains two unknowns, the current I in the circuit and the charge Q on the capacitor. However, (6.3.3) implies that Q′ = I, so (6.3.5) can be converted into the second order equation LQ′′ + RQ′ + 1 C Q = E(t) (6.3.6) in Q. To find the current flowing in an RLC circuit, we solve (6.3.6) for Q and then differentiate the solution to obtain I. In Sections 6.1 and 6.2 we encountered the equation my′′ + cy′ + ky = F (t) (6.3.7) in connection with spring-mass systems. Except for notation this equation is the same as (6.3.6). The correspondence between electrical and mechanical quantities connected with (6.3.6) and (6.3.7) is shown in Table 6.3.9. Table 6.3.9. Electrical and Mechanical Units Electrical Mechanical charge Q displacement y curent I velocity y′ impressed voltage E(t) external force F (t) inductance L Mass m resistance R damping c 1/capacitance 1/C cpring constant k The equivalence between (6.3.6) and (6.3.7) is an example of how mathematics unifies fundamental similarities in diverse physical phenomena. Since we’ve already studied the properties of solutions of (6.3.7) in Sections 6.1 and 6.2, we can obtain results concerning solutions of (6.3.6) by simpling changing notation, according to Table 6.3.8. Free Oscillations We say that an RLC circuit is in free oscillation if E(t) = 0 for t > 0, so that (6.3.6) becomes LQ′′ + RQ′ + 1 C Q = 0. (6.3.8) The characteristic equation of (6.3.8) is Lr2 + Rr + 1 C = 0, with roots r1 = −R − p R2 −4L/C 2L and r2 = −R + p R2 −4L/C 2L . (6.3.9) There are three cases to consider, all analogous to the cases considered in Section 6.2 for free vibrations of a damped spring-mass system. CASE 1. The oscillation is underdamped if R < p 4L/C. In this case, r1 and r2 in (6.3.9) are complex conjugates, which we write as r1 = −R 2L + iω1 and r2 = −R 2L −iω1, | Elementary Differential Equations with Boundary Value Problems_Page_302_Chunk2707 |
Section 6.3 The RLC Circuit 293 where ω1 = p 4L/C −R2 2L . The general solution of (6.3.8) is Q = e−Rt/2L(c1 cos ω1t + c2 sin ω1t), which we can write as Q = Ae−Rt/2L cos(ω1t −φ), (6.3.10) where A = q c2 1 + c2 2, A cos φ = c1, and A sin φ = c2. In the idealized case where R = 0, the solution (6.3.10) reduces to Q = A cos t √ LC −φ , which is analogous to the simple harmonic motion of an undamped spring-mass system in free vibration. Actual RLC circuits are usually underdamped, so the case we’ve just considered is the most important. However, for completeness we’ll consider the other two possibilities. CASE 2. The oscillation is overdamped if R > p 4L/C. In this case, the zeros r1 and r2 of the characteristic polynomial are real, with r1 < r2 < 0 (see (6.3.9)), and the general solution of (6.3.8) is Q = c1er1t + c2er2t. (6.3.11) CASE 3. The oscillation is critically damped if R = p 4L/C. In this case, r1 = r2 = −R/2L and the general solution of (6.3.8) is Q = e−Rt/2L(c1 + c2t). (6.3.12) If R ̸= 0, the exponentials in (6.3.10), (6.3.11), and (6.3.12) are negative, so the solution of any homogeneous initial value problem LQ′′ + RQ′ + 1 C Q = 0, Q(0) = Q0, Q′(0) = I0, approaches zero exponentially as t →∞. Thus, all such solutions are transient, in the sense defined Section 6.2 in the discussion of forced vibrations of a spring-mass system with damping. Example 6.3.1 At t = 0 a current of 2 amperes flows in an RLC circuit with resistance R = 40 ohms, inductance L = .2 henrys, and capacitance C = 10−5 farads. Find the current flowing in the circuit at t > 0 if the initial charge on the capacitor is 1 coulomb. Assume that E(t) = 0 for t > 0. Solution The equation for the charge Q is 1 5Q′′ + 40Q′ + 10000Q = 0, or Q′′ + 200Q′ + 50000Q = 0. (6.3.13) Therefore we must solve the initial value problem Q′′ + 200Q′ + 50000Q = 0, Q(0) = 1, Q′(0) = 2. (6.3.14) | Elementary Differential Equations with Boundary Value Problems_Page_303_Chunk2708 |
294 Chapter 6 Applications of Linear Second Order Equations The desired current is the derivative of the solution of this initial value problem. The characteristic equation of (6.3.13) is r2 + 200r + 50000 = 0, which has complex zeros r = −100 ± 200i. Therefore the general solution of (6.3.13) is Q = e−100t(c1 cos 200t + c2 sin 200t). (6.3.15) Differentiating this and collecting like terms yields Q′ = −e−100t [(100c1 −200c2) cos 200t + (100c2 + 200c1) sin 200t]. (6.3.16) To find the solution of the initial value problem (6.3.14), we set t = 0 in (6.3.15) and (6.3.16) to obtain c1 = Q(0) = 1 and −100c1 + 200c2 = Q′(0) = 2; therefore, c1 = 1 and c2 = 51/100, so Q = e−100t cos 200t + 51 100 sin 200t is the solution of (6.3.14). Differentiating this yields I = e−100t(2 cos 200t −251 sin200t). Forced Oscillations With Damping An initial value problem for (6.3.6) has the form LQ′′ + RQ′ + 1 C Q = E(t), Q(0) = Q0, Q′(0) = I0, (6.3.17) where Q0 is the initial charge on the capacitor and I0 is the initial current in the circuit. We’ve already seen that if E ≡0 then all solutions of (6.3.17) are transient. If E ̸≡0, we know that the solution of (6.3.17) has the form Q = Qc + Qp, where Qc satisfies the complementary equation, and approaches zero exponentially as t →∞for any initial conditions , while Qp depends only on E and is independent of the initial conditions. As in the case of forced oscillations of a spring-mass system with damping, we call Qp the steady state charge on the capacitor of the RLC circuit. Since I = Q′ = Q′ c + Q′ p and Q′ c also tends to zero exponentially as t →∞, we say that Ic = Q′ c is the transient current and Ip = Q′ p is the steady state current. In most applications we’re interested only in the steady state charge and current. Example 6.3.2 Find the amplitude-phase form of the steady state current in the RLC circuit in Fig- ure 6.3.1 if the impressed voltage, provided by an alternating current generator, is E(t) = E0 cos ωt. Solution We’ll first find the steady state charge on the capacitor as a particular solution of LQ′′ + RQ′ + 1 C Q = E0 cos ωt. To do, this we’ll simply reinterpret a result obtained in Section 6.2, where we found that the steady state solution of my′′ + cy′ + ky = F0 cos ωt | Elementary Differential Equations with Boundary Value Problems_Page_304_Chunk2709 |
Section 6.3 The RLC Circuit 295 is yp = F0 p (k −mω2)2 + c2ω2 cos(ωt −φ), where cos φ = k −mω2 p (k −mω2)2 + c2ω2 and sin φ = cω p (k −mω2)2 + c2ω2 . (See Equations (6.2.14) and (6.2.15).) By making the appropriate changes in the symbols (according to Table 2) yields the steady state charge Qp = E0 p (1/C −Lω2)2 + R2ω2 cos(ωt −φ), where cos φ = 1/C −Lω2 p (1/C −Lω2)2 + R2ω2 and sin φ = Rω p (1/C −Lω2)2 + R2ω2 . Therefore the steady state current in the circuit is Ip = Q′ p = − ωE0 p (1/C −Lω2)2 + R2ω2 sin(ωt −φ). 6.3 Exercises In Exercises 1-5 find the current in the RLC circuit, assuming that E(t) = 0 for t > 0. 1. R = 3 ohms; L = .1 henrys; C = .01 farads; Q0 = 0 coulombs; I0 = 2 amperes. 2. R = 2 ohms; L = .05 henrys; C = .01 farads’; Q0 = 2 coulombs; I0 = −2 amperes. 3. R = 2 ohms; L = .1 henrys; C = .01 farads; Q0 = 2 coulombs; I0 = 0 amperes. 4. R = 6 ohms; L = .1 henrys; C = .004 farads’; Q0 = 3 coulombs; I0 = −10 amperes. 5. R = 4 ohms; L = .05 henrys; C = .008 farads; Q0 = −1 coulombs; I0 = 2 amperes. In Exercises 6-10 find the steady state current in the circuit described by the equation. 6. 1 10Q′′ + 3Q′ + 100Q = 5 cos 10t −5 sin10t 7. 1 20Q′′ + 2Q′ + 100Q = 10 cos 25t −5 sin 25t 8. 1 10Q′′ + 2Q′ + 100Q = 3 cos 50t −6 sin50t 9. 1 10Q′′ + 6Q′ + 250Q = 10 cos 100t + 30 sin100t 10. 1 20Q′′ + 4Q′ + 125Q = 15 cos 30t −30 sin 30t 11. Show that if E(t) = U cos ωt+V sin ωt where U and V are constants then the steady state current in the RLC circuit shown in Figure 6.3.1 is Ip = ω2RE(t) + (1/C −Lω2)E′(t) ∆ , where ∆= (1/C −Lω2)2 + R2ω2. | Elementary Differential Equations with Boundary Value Problems_Page_305_Chunk2710 |
296 Chapter 6 Applications of Linear Second Order Equations 12. Find the amplitude of the steady state current Ip in the RLC circuit shown in Figure 6.3.1 if E(t) = U cos ωt+V sinωt, where U and V are constants. Then find the value ω0 of ω maximizes the amplitude, and find the maximum amplitude. In Exercises 13-17 plot the amplitude of the steady state current against ω. Estimate the value of ω that maximizes the amplitude of the steady state current, and estimate this maximum amplitude. HINT: You can confirm your results by doing Exercise 12. 13. L 1 10Q′′ + 3Q′ + 100Q = U cos ωt + V sin ωt 14. L 1 20Q′′ + 2Q′ + 100Q = U cos ωt + V sin ωt 15. L 1 10Q′′ + 2Q′ + 100Q = U cos ωt + V sin ωt 16. L 1 10Q′′ + 6Q′ + 250Q = U cos ωt + V sin ωt 17. L 1 20Q′′ + 4Q′ + 125Q = U cos ωt + V sin ωt 6.4 MOTION UNDER A CENTRAL FORCE We’ll now study the motion of a object moving under the influence of a central force; that is, a force whose magnitude at any point P other than the origin depends only on the distance from P to the origin, and whose direction at P is parallel to the line connecting P and the origin, as indicated in Figure 6.4.1 for the case where the direction of the force at every point is toward the origin. Gravitational forces are central forces; for example, as mentioned in Section 4.3, if we assume that Earth is a perfect sphere with constant mass density then Newton’s law of gravitation asserts that the force exerted on an object by Earth’s gravitational field is proportional to the mass of the object and inversely proportional to the square of its distance from the center of Earth, which we take to be the origin. If the initial position and velocity vectors of an object moving under a central force are parallel, then the subsequent motion is along the line from the origin to the initial position. Here we’ll assume that the initial position and velocity vectors are not parallel; in this case the subsequent motion is in the plane determined by them. For convenience we take this to be the xy-plane. We’ll consider the problem of determining the curve traversed by the object. We call this curve the orbit. We can represent a central force in terms of polar coordinates x = r cos θ, y = r sin θ as F(r, θ) = f(r)(cos θ i + sin θ j). We assume that f is continuous for all r > 0. The magnitude of F at (x, y) = (r cos θ, r sin θ) is |f(r)|, so it depends only on the distance r from the point to the origin the direction of F is from the point to the origin if f(r) < 0, or from the origin to the point if f(r) > 0. We’ll show that the orbit of an object with mass m moving under this force is given by r(θ) = 1 u(θ), | Elementary Differential Equations with Boundary Value Problems_Page_306_Chunk2711 |
Section 6.4 Motion Under a Central Force 297 x y Figure 6.4.1 where u is solution of the differential equation d2u dθ2 + u = − 1 mh2u2 f(1/u), (6.4.1) and h is a constant defined below. Newton’s second law of motion (F = ma) says that the polar coordinates r = r(t) and θ = θ(t) of the particle satisfy the vector differential equation m(r cos θ i + r sin θ j)′′ = f(r)(cos θ i + sin θ j). (6.4.2) To deal with this equation we introduce the unit vectors e1 = cos θ i + sin θ j and e2 = −sin θ i + cos θ j. Note that e1 points in the direction of increasing r and e2 points in the direction of increasing θ (Fig- ure 6.4.2); moreover, de1 dθ = e2, de2 dθ = −e1, (6.4.3) and e1 · e2 = cos θ(−sin θ) + sin θ cos θ = 0, so e1 and e2 are perpendicular. Recalling that the single prime (′) stands for differentiation with respect to t, we see from (6.4.3) and the chain rule that e′ 1 = θ′e2 and e′ 2 = −θ′e1. (6.4.4) | Elementary Differential Equations with Boundary Value Problems_Page_307_Chunk2712 |
298 Chapter 6 Applications of Linear Second Order Equations e1 e2 x y Figure 6.4.2 Now we can write (6.4.2) as m(re1)′′ = f(r)e1. (6.4.5) But (re1)′ = r′e1 + re′ 1 = r′e1 + rθ′e2 (from (6.4.4)), and (re1)′′ = (r′e1 + rθ′e2)′ = r′′e1 + r′e′ 1 + (rθ′′ + r′θ′)e2 + rθ′e′ 2 = r′′e1 + r′θ′e2 + (rθ′′ + r′θ′)e2 −r(θ′)2e1 (from (6.4.4)) = | Elementary Differential Equations with Boundary Value Problems_Page_308_Chunk2713 |
Section 6.4 Motion Under a Central Force 299 so r2θ′ = h, (6.4.7) where h is a constant that we can write in terms of the initial conditions as h = r2(0)θ′(0). Since the initial position and velocity vectors are r(0)e1(0) and r′(0)e1(0) + r(0)θ′(0)e2(0), our assumption that these two vectors are not parallel implies that θ′(0) ̸= 0, so h ̸= 0. Now let u = 1/r. Then u2 = θ′/h (from (6.4.7)) and r′ = −u′ u2 = −h u′ θ′ , which implies that r′ = −hdu dθ , (6.4.8) since u′ θ′ = du dt dθ dt = du dθ . Differentiating (6.4.8) with respect to t yields r′′ = −h d dt du dθ = −hd2u dθ2 θ′, which implies that r′′ = −h2u2 d2u dθ2 since θ′ = hu2. Substituting from these equalities into (6.4.6) and recalling that r = 1/u yields −m h2u2 d2u dθ2 + 1 uh2u4 = f(1/u), and dividing through by −mh2u2 yields (6.4.1). Eqn. (6.4.7) has the following geometrical interpretation, which is known as Kepler’s Second Law. Theorem 6.4.1 The position vector of an object moving under a central force sweeps out equal areas in equal times; more precisely, if θ(t1) ≤θ(t2) then the (signed) area of the sector {(x, y) = (r cos θ, r sinθ) : 0 ≤r ≤r(θ), θ(t1) ≤θ(t2)} (Figure 6.4.3) is given by A = h(t2 −t1) 2 , where h = r2θ′, which we have shown to be constant. | Elementary Differential Equations with Boundary Value Problems_Page_309_Chunk2714 |
300 Chapter 6 Applications of Linear Second Order Equations θ = θ ( t2 ) θ = θ ( t1 ) x y Figure 6.4.3 Proof Recall from calculus that the area of the shaded sector in Figure 6.4.3 is A = 1 2 Z θ(t2) θ(t1) r2(θ) dθ, where r = r(θ) is the polar representation of the orbit. Making the change of variable θ = θ(t) yields A = 1 2 Z t2 t1 r2(θ(t))θ′(t) dt. (6.4.9) But (6.4.7) and (6.4.9) imply that A = 1 2 Z t2 t1 h dt = h(t2 −t1) 2 , which completes the proof. Motion Under an Inverse Square Law Force In the special case where f(r) = −mk/r2 = −mku2, so F can be interpreted as a gravitational force, (6.4.1) becomes d2u dθ2 + u = k h2 . (6.4.10) The general solution of the complementary equation d2u dθ2 + u = 0 | Elementary Differential Equations with Boundary Value Problems_Page_310_Chunk2715 |
Section 6.4 Motion Under a Central Force 301 can be written in amplitude–phase form as u = A cos(θ −φ), where A ≥0 and φ is a phase angle. Since up = k/h2 is a particular solution of (6.4.10), the general solution of (6.4.10) is u = A cos(θ −φ) + k h2 ; hence, the orbit is given by r = A cos(θ −φ) + k h2 −1 , which we rewrite as r = ρ 1 + e cos(θ −φ), (6.4.11) where ρ = h2 k and e = Aρ. A curve satisfying (6.4.11) is a conic section with a focus at the origin (Exercise 1). The nonnegative constant e is the eccentricity of the orbit, which is an ellipse if e < 1 ellipse (a circle if e = 0), a parabola if e = 1, or a hyperbola if e > 1. A P x y Figure 6.4.4 If the orbit is an ellipse, then the minimum and maximum values of r are rmin = ρ 1 + e (the perihelion distance, attained when θ = φ) rmax = ρ 1 −e (the aphelion distance, attained when θ = φ + π). Figure 6.4.4 shows a typical elliptic orbit. The point P on the orbit where r = rmin is the perigee and the point A where r = rmax is the apogee. | Elementary Differential Equations with Boundary Value Problems_Page_311_Chunk2716 |
302 Chapter 6 Applications of Linear Second Order Equations For example, Earth’s orbit around the Sun is approximately an ellipse with e ≈.017, rmin ≈91 × 106 miles, and rmax ≈95 × 106 miles. Halley’s comet has a very elongated approximately elliptical orbit around the sun, with e ≈.967, rmin ≈55 × 106 miles, and rmax ≈33 × 108 miles. Some comets (the nonrecurring type) have parabolic or hyperbolic orbits. 6.4 Exercises 1. Find the equation of the curve r = ρ 1 + e cos(θ −φ) (A) in terms of (X, Y ) = (r cos(θ −φ), r sin(θ −φ)), which are rectangular coordinates with respect to the axes shown in Figure 6.4.5. Use your results to verify that (A) is the equation of an ellipse if 0 < e < 1, a parabola if e = 1, or a hyperbola if e > 1. If e < 1, leave your answer in the form (X −X0)2 a2 + (Y −Y0)2 b2 = 1, and show that the area of the ellipse is A = πρ2 (1 −e2)3/2 . Then use Theorem 6.4.1 to show that the time required for the object to traverse the entire orbit is T = 2πρ2 h(1 −e2)3/2 . (This is Kepler’s third law; T is called the period of the orbit.) x y X Y φ Figure 6.4.5 | Elementary Differential Equations with Boundary Value Problems_Page_312_Chunk2717 |
Section 6.4 Motion Under a Central Force 303 2. Suppose an object with mass m moves in the xy-plane under the central force F(r, θ) = −mk r2 (cos θ i + sin θ j), where k is a positive constant. As we shown, the orbit of the object is given by r = ρ 1 + e cos(θ −φ). Determine ρ, e, and φ in terms of the initial conditions r(0) = r0, r′(0) = r′ 0, and θ(0) = θ0, θ′(0) = θ′ 0. Assume that the initial position and velocity vectors are not collinear. 3. Suppose we wish to put a satellite with mass m into an elliptical orbit around Earth. Assume that the only force acting on the object is Earth’s gravity, given by F(r, θ) = −mg R2 r2 (cos θ i + sin θ j), where R is Earth’s radius, g is the acceleration due to gravity at Earth’s surface, and r and θ are polar coordinates in the plane of the orbit, with the origin at Earth’s center. (a) Find the eccentricity required to make the aphelion and perihelion distances equal to Rγ1 and Rγ2, respectively, where 1 < γ1 < γ2. (b) Find the initial conditions r(0) = r0, r′(0) = r′ 0, and θ(0) = θ0, θ′(0) = θ′ 0 required to make the initial point the perigee, and the motion along the orbit in the direction of increasing θ. HINT: Use the results of Exercise 2. 4. An object with mass m moves in a spiral orbit r = cθ2 under a central force F(r, θ) = f(r)(cos θ i + sin θ j). Find f. 5. An object with mass m moves in the orbit r = r0eγθ under a central force F(r, θ) = f(r)(cos θ i + sin θ j). Find f. 6. Suppose an object with mass m moves under the central force F(r, θ) = −mk r3 (cos θ i + sin θ j), with r(0) = r0, r′(0) = r′ 0, and θ(0) = θ0, θ′(0) = θ′ 0, where h = r2 0θ′ 0 ̸= 0. (a) Set up a second order initial value problem for u = 1/r as a function of θ. (b) Determine r = r(θ) if (i) h2 < k; (ii) h2 = k; (iii) h2 > k. | Elementary Differential Equations with Boundary Value Problems_Page_313_Chunk2718 |
CHAPTER 7 Series Solutions of Linear Second Equations IN THIS CHAPTER we study a class of second order differential equations that occur in many applica- tions, but can’t be solved in closed form in terms of elementary functions. Here are some examples: (1) Bessel’s equation x2y′′ + xy′ + (x2 −ν2)y = 0, which occurs in problems displaying cylindrical symmetry, such as diffraction of light through a circular aperture, propagation of electromagnetic radiation through a coaxial cable, and vibrations of a circular drum head. (2) Airy’s equation, y′′ −xy = 0, which occurs in astronomy and quantum physics. (3) Legendre’s equation (1 −x2)y′′ −2xy′ + α(α + 1)y = 0, which occurs in problems displaying spherical symmetry, particularly in electromagnetism. These equations and others considered in this chapter can be written in the form P0(x)y′′ + P1(x)y′ + P2(x)y = 0, (A) where P0, P1, and P2 are polynomials with no common factor. For most equations that occur in appli- cations, these polynomials are of degree two or less. We’ll impose this restriction, although the methods that we’ll develop can be extended to the case where the coefficient functions are polynomials of arbitrary degree, or even power series that converge in some circle around the origin in the complex plane. Since (A) does not in general have closed form solutions, we seek series representations for solutions. We’ll see that if P0(0) ̸= 0 then solutions of (A) can be written as power series y = ∞ X n=0 anxn that converge in an open interval centered at x = 0. 305 | Elementary Differential Equations with Boundary Value Problems_Page_315_Chunk2719 |
306 Chapter 6 Applications of Linear Second Order Equations SECTION 7.1 reviews the properties of power series. SECTIONS 7.2 AND 7.3 are devoted to finding power series solutionsof (A) in the case where P0(0) ̸= 0. The situation is more complicated if P0(0) = 0; however, if P1 and P2 satisfy assumptions that apply to most equations of interest, then we’re able to use a modified series method to obtain solutions of (A). SECTION 7.4 introduces the appropriate assumptions on P1 and P2 in the case where P0(0) = 0, and deals with Euler’s equation ax2y′′ + bxy′ + cy = 0, where a, b, and c are constants. This is the simplest equation that satisfies these assumptions. SECTIONS 7.5 –7.7 deal with three distinct cases satisfying the assumptions introduced in Section 7.4. In all three cases, (A) has at least one solution of the form y1 = xr ∞ X n=0 anxn, where r need not be an integer. The problem is that there are three possibilities– each requiring a different approach – for the form of a second solution y2 such that {y1, y2} is a fundamental pair of solutionsof (A). | Elementary Differential Equations with Boundary Value Problems_Page_316_Chunk2720 |
Section 7.1 Review of Power Series 307 7.1 REVIEW OF POWER SERIES Many applications give rise to differential equations with solutions that can’t be expressed in terms of elementary functions such as polynomials, rational functions, exponential and logarithmic functions, and trigonometric functions. The solutions of some of the most important of these equations can be expressed in terms of power series. We’ll study such equations in this chapter. In this section we review relevant properties of power series. We’ll omit proofs, which can be found in any standard calculus text. Definition 7.1.1 An infinite series of the form ∞ X n=0 an(x −x0)n, (7.1.1) where x0 and a0, a1, ..., an, ...are constants, is called a power series in x −x0. We say that the power series (7.1.1) converges for a given x if the limit lim N→∞ N X n=0 an(x −x0)n exists; otherwise, we say that the power series diverges for the given x. A power series in x −x0 must converge if x = x0, since the positive powers of x −x0 are all zero in this case. This may be the only value of x for which the power series converges. However, the next theorem shows that if the power series converges for some x ̸= x0 then the set of all values of x for which it converges forms an interval. Theorem 7.1.2 For any power series ∞ X n=0 an(x −x0)n, exactly one of the these statements is true: (i) The power series converges only for x = x0. (ii) The power series converges for all values of x. (iii) There’s a positive number R such that the power series converges if |x −x0| < R and diverges if |x −x0| > R. In case (iii) we say that R is the radius of convergence of the power series. For convenience, we include the other two cases in this definition by defining R = 0 in case (i) and R = ∞in case (ii). We define the open interval of convergence of P∞ n=0 an(x −x0)n to be (x0 −R, x0 + R) if 0 < R < ∞, or (−∞, ∞) if R = ∞. If R is finite, no general statement can be made concerning convergence at the endpoints x = x0 ± R of the open interval of convergence; the series may converge at one or both points, or diverge at both. Recall from calculus that a series of constants P∞ n=0 αn is said to converge absolutely if the series of absolute values P∞ n=0 |αn| converges. It can be shown that a power series P∞ n=0 an(x −x0)n with a positive radius of convergence R converges absolutely in its open interval of convergence; that is, the series ∞ X n=0 |an||x −x0|n | Elementary Differential Equations with Boundary Value Problems_Page_317_Chunk2721 |
308 Chapter 7 Series Solutions of Linear Second Equations of absolute values converges if |x −x0| < R. However, if R < ∞, the series may fail to converge absolutely at an endpoint x0 ± R, even if it converges there. The next theorem provides a useful method for determining the radius of convergence of a power series. It’s derived in calculus by applying the ratio test to the corresponding series of absolute values. For related theorems see Exercises 2 and 4. Theorem 7.1.3 Suppose there’s an integer N such that an ̸= 0 if n ≥N and lim n→∞ an+1 an = L, where 0 ≤L ≤∞. Then the radius of convergence of P∞ n=0 an(x −x0)n is R = 1/L, which should be interpreted to mean that R = 0 if L = ∞, or R = ∞if L = 0. Example 7.1.1 Find the radius of convergence of the series: (a) ∞ X n=0 n!xn (b) ∞ X n=10 (−1)n xn n! (c) ∞ X n=0 2nn2(x −1)n. SOLUTION(a) Here an = n!, so lim n→∞ an+1 an = lim n→∞ (n + 1)! n! = lim n→∞(n + 1) = ∞. Hence, R = 0. SOLUTION(b) Here an = (1)n/n! for n ≥N = 10, so lim n→∞ an+1 an = lim n→∞ n! (n + 1)! = lim n→∞ 1 n + 1 = 0. Hence, R = ∞. SOLUTION(c) Here an = 2nn2, so lim n→∞ an+1 an = lim n→∞ 2n+1(n + 1)2 2nn2 = 2 lim n→∞ 1 + 1 n 2 = 2. Hence, R = 1/2. Taylor Series If a function f has derivatives of all orders at a point x = x0, then the Taylor series of f about x0 is defined by ∞ X n=0 f(n)(x0) n! (x −x0)n. In the special case where x0 = 0, this series is also called the Maclaurin series of f. | Elementary Differential Equations with Boundary Value Problems_Page_318_Chunk2722 |
Section 7.1 Review of Power Series 309 Taylor series for most of the common elementary functions converge to the functions on their open intervals of convergence. For example, you are probably familiar with the following Maclaurin series: ex = ∞ X n=0 xn n! , −∞< x < ∞, (7.1.2) sin x = ∞ X n=0 (−1)n x2n+1 (2n + 1)!, −∞< x < ∞, (7.1.3) cos x = ∞ X n=0 (−1)n x2n (2n)!, −∞< x < ∞, (7.1.4) 1 1 −x = ∞ X n=0 xn, −1 < x < 1. (7.1.5) Differentiation of Power Series A power series with a positive radius of convergence defines a function f(x) = ∞ X n=0 an(x −x0)n on its open interval of convergence. We say that the series represents f on the open interval of conver- gence. A function f represented by a power series may be a familiar elementary function as in (7.1.2)– (7.1.5); however, it often happens that f isn’t a familiar function, so the series actually defines f. The next theorem shows that a function represented by a power series has derivatives of all orders on the open interval of convergence of the power series, and provides power series representations of the derivatives. Theorem 7.1.4 A power series f(x) = ∞ X n=0 an(x −x0)n with positive radius of convergence R has derivatives of all orders in its open interval of convergence, and successive derivatives can be obtained by repeatedly differentiating term by term; that is, f′(x) = ∞ X n=1 nan(x −x0)n−1, (7.1.6) f′′(x) = ∞ X n=2 n(n −1)an(x −x0)n−2, (7.1.7) ... f(k)(x) = ∞ X n=k n(n −1) · · ·(n −k + 1)an(x −x0)n−k. (7.1.8) Moreover, all of these series have the same radius of convergence R. | Elementary Differential Equations with Boundary Value Problems_Page_319_Chunk2723 |
310 Chapter 7 Series Solutions of Linear Second Equations Example 7.1.2 Let f(x) = sin x. From (7.1.3), f(x) = ∞ X n=0 (−1)n x2n+1 (2n + 1)!. From (7.1.6), f′(x) = ∞ X n=0 (−1)n d dx x2n+1 (2n + 1)! = ∞ X n=0 (−1)n x2n (2n)!, which is the series (7.1.4) for cos x. Uniqueness of Power Series The next theorem shows that if f is defined by a power series in x −x0 with a positive radius of conver- gence, then the power series is the Taylor series of f about x0. Theorem 7.1.5 If the power series f(x) = ∞ X n=0 an(x −x0)n has a positive radius of convergence, then an = f(n)(x0) n! ; (7.1.9) that is, P∞ n=0 an(x −x0)n is the Taylor series of f about x0. This result can be obtained by setting x = x0 in (7.1.8), which yields f(k)(x0) = k(k −1) · · · 1 · ak = k!ak. This implies that ak = f(k)(x0) k! . Except for notation, this is the same as (7.1.9). The next theorem lists two important properties of power series that follow from Theorem 7.1.5. Theorem 7.1.6 (a) If ∞ X n=0 an(x −x0)n = ∞ X n=0 bn(x −x0)n for all x in an open interval that contains x0, then an = bn for n = 0, 1, 2, .... (b) If ∞ X n=0 an(x −x0)n = 0 for all x in an open interval that contains x0, then an = 0 for n = 0, 1, 2, .... | Elementary Differential Equations with Boundary Value Problems_Page_320_Chunk2724 |
Section 7.1 Review of Power Series 311 To obtain (a) we observe that the two series represent the same function f on the open interval; hence, Theorem 7.1.5 implies that an = bn = f(n)(x0) n! , n = 0, 1, 2, . . .. (b) can be obtained from (a) by taking bn = 0 for n = 0, 1, 2, .... Taylor Polynomials If f has N derivatives at a point x0, we say that TN(x) = N X n=0 f(n)(x0) n! (x −x0)n is the N-th Taylor polynomial of f about x0. This definition and Theorem 7.1.5 imply that if f(x) = ∞ X n=0 an(x −x0)n, where the power series has a positive radius of convergence, then the Taylor polynomials of f about x0 are given by TN(x) = N X n=0 an(x −x0)n. In numerical applications, we use the Taylor polynomials to approximate f on subintervals of the open interval of convergence of the power series. For example, (7.1.2) implies that the Taylor polynomial TN of f(x) = ex is TN(x) = N X n=0 xn n! . The solid curve in Figure 7.1.1 is the graph of y = ex on the interval [0, 5]. The dotted curves in Figure 7.1.1 are the graphs of the Taylor polynomials T1, ..., T6 of y = ex about x0 = 0. From this figure, we conclude that the accuracy of the approximation of y = ex by its Taylor polynomial TN improves as N increases. Shifting the Summation Index In Definition 7.1.1 of a power series in x −x0, the n-th term is a constant multiple of (x −x0)n. This isn’t true in (7.1.6), (7.1.7), and (7.1.8), where the general terms are constant multiples of (x −x0)n−1, (x −x0)n−2, and (x −x0)n−k, respectively. However, these series can all be rewritten so that their n-th terms are constant multiples of (x −x0)n. For example, letting n = k + 1 in the series in (7.1.6) yields f′(x) = ∞ X k=0 (k + 1)ak+1(x −x0)k, (7.1.10) where we start the new summation index k from zero so that the first term in (7.1.10) (obtained by setting k = 0) is the same as the first term in (7.1.6) (obtained by setting n = 1). However, the sum of a series is independent of the symbol used to denote the summation index, just as the value of a definite integral is independent of the symbol used to denote the variable of integration. Therefore we can replace k by n in (7.1.10) to obtain f′(x) = ∞ X n=0 (n + 1)an+1(x −x0)n, (7.1.11) | Elementary Differential Equations with Boundary Value Problems_Page_321_Chunk2725 |
312 Chapter 7 Series Solutions of Linear Second Equations x y 1 2 3 4 5 N = 1 N = 2 N = 3 N = 4 N = 5 N = 6 Figure 7.1.1 Approximation of y = ex by Taylor polynomials about x = 0 where the general term is a constant multiple of (x −x0)n. It isn’t really necessary to introduce the intermediate summation index k. We can obtain (7.1.11) directly from (7.1.6) by replacing n by n + 1 in the general term of (7.1.6) and subtracting 1 from the lower limit of (7.1.6). More generally, we use the following procedure for shifting indices. Shifting the Summation Index in a Power Series For any integer k, the power series ∞ X n=n0 bn(x −x0)n−k can be rewritten as ∞ X n=n0−k bn+k(x −x0)n; that is, replacing n by n + k in the general term and subtracting k from the lower limit of summation leaves the series unchanged. Example 7.1.3 Rewrite the following power series from (7.1.7) and (7.1.8) so that the general term in | Elementary Differential Equations with Boundary Value Problems_Page_322_Chunk2726 |
Section 7.1 Review of Power Series 313 each is a constant multiple of (x −x0)n: (a) ∞ X n=2 n(n −1)an(x −x0)n−2 (b) ∞ X n=k n(n −1) · · ·(n −k + 1)an(x −x0)n−k. SOLUTION(a) Replacing n by n + 2 in the general term and subtracting 2 from the lower limit of summation yields ∞ X n=2 n(n −1)an(x −x0)n−2 = ∞ X n=0 (n + 2)(n + 1)an+2(x −x0)n. SOLUTION(b) Replacing n by n + k in the general term and subtracting k from the lower limit of summation yields ∞ X n=k n(n −1) · · ·(n −k + 1)an(x −x0)n−k = ∞ X n=0 (n + k)(n + k −1) · · ·(n + 1)an+k(x −x0)n. Example 7.1.4 Given that f(x) = ∞ X n=0 anxn, write the function xf′′ as a power series in which the general term is a constant multiple of xn. Solution From Theorem 7.1.4 with x0 = 0, f′′(x) = ∞ X n=2 n(n −1)anxn−2. Therefore xf′′(x) = ∞ X n=2 n(n −1)anxn−1. Replacing n by n + 1 in the general term and subtracting 1 from the lower limit of summation yields xf′′(x) = ∞ X n=1 (n + 1)nan+1xn. We can also write this as xf′′(x) = ∞ X n=0 (n + 1)nan+1xn, since the first term in this last series is zero. (We’ll see later that sometimes it’s useful to include zero terms at the beginning of a series.) Linear Combinations of Power Series If a power series is multiplied by a constant, then the constant can be placed inside the summation; that is, c ∞ X n=0 an(x −x0)n = ∞ X n=0 can(x −x0)n. | Elementary Differential Equations with Boundary Value Problems_Page_323_Chunk2727 |
314 Chapter 7 Series Solutions of Linear Second Equations Two power series f(x) = ∞ X n=0 an(x −x0)n and g(x) = ∞ X n=0 bn(x −x0)n with positive radii of convergence can be added term by term at points common to their open intervals of convergence; thus, if the first series converges for |x −x0| < R1 and the second converges for |x −x0| < R2, then f(x) + g(x) = ∞ X n=0 (an + bn)(x −x0)n for |x −x0| < R, where R is the smaller of R1 and R2. More generally, linear combinations of power series can be formed term by term; for example, c1f(x) + c2f(x) = ∞ X n=0 (c1an + c2bn)(x −x0)n. Example 7.1.5 Find the Maclaurin series for cosh x as a linear combination of the Maclaurin series for ex and e−x. Solution By definition, cosh x = 1 2ex + 1 2e−x. Since ex = ∞ X n=0 xn n! and e−x = ∞ X n=0 (−1)n xn n! , it follows that cosh x = ∞ X n=0 1 2[1 + (−1)n]xn n! . (7.1.12) Since 1 2[1 + (−1)n] = 1 if n = 2m, an even integer, 0 if n = 2m + 1, an odd integer, we can rewrite (7.1.12) more simply as cosh x = ∞ X m=0 x2m (2m)!. This result is valid on (−∞, ∞), since this is the open interval of convergence of the Maclaurin series for ex and e−x. Example 7.1.6 Suppose y = ∞ X n=0 anxn on an open interval I that contains the origin. (a) Express (2 −x)y′′ + 2y as a power series in x on I. | Elementary Differential Equations with Boundary Value Problems_Page_324_Chunk2728 |
Section 7.1 Review of Power Series 315 (b) Use the result of (a) to find necessary and sufficient conditions on the coefficients {an} for y to be a solution of the homogeneous equation (2 −x)y′′ + 2y = 0 (7.1.13) on I. SOLUTION(a) From (7.1.7) with x0 = 0, y′′ = ∞ X n=2 n(n −1)anxn−2. Therefore (2 −x)y′′ + 2y = 2y′′ −xy′ + 2y = ∞ X n=2 2n(n −1)anxn−2 − ∞ X n=2 n(n −1)anxn−1 + ∞ X n=0 2anxn. (7.1.14) To combine the three series we shift indices in the first two to make their general terms constant multiples of xn; thus, ∞ X n=2 2n(n −1)anxn−2 = ∞ X n=0 2(n + 2)(n + 1)an+2xn (7.1.15) and ∞ X n=2 n(n −1)anxn−1 = ∞ X n=1 (n + 1)nan+1xn = ∞ X n=0 (n + 1)nan+1xn, (7.1.16) where we added a zero term in the last series so that when we substitute from (7.1.15) and (7.1.16) into (7.1.14) all three series will start with n = 0; thus, (2 −x)y′′ + 2y = ∞ X n=0 [2(n + 2)(n + 1)an+2 −(n + 1)nan+1 + 2an]xn. (7.1.17) SOLUTION(b) From (7.1.17) we see that y satisfies (7.1.13) on I if 2(n + 2)(n + 1)an+2 −(n + 1)nan+1 + 2an = 0, n = 0, 1, 2, . . .. (7.1.18) Conversely, Theorem 7.1.6 (b) implies that if y = P∞ n=0 anxn satisfies (7.1.13) on I, then (7.1.18) holds. Example 7.1.7 Suppose y = ∞ X n=0 an(x −1)n on an open interval I that contains x0 = 1. Express the function (1 + x)y′′ + 2(x −1)2y′ + 3y (7.1.19) as a power series in x −1 on I. | Elementary Differential Equations with Boundary Value Problems_Page_325_Chunk2729 |
316 Chapter 7 Series Solutions of Linear Second Equations Solution Since we want a power series in x −1, we rewrite the coefficient of y′′ in (7.1.19) as 1 + x = 2 + (x −1), so (7.1.19) becomes 2y′′ + (x −1)y′′ + 2(x −1)2y′ + 3y. From (7.1.6) and (7.1.7) with x0 = 1, y′ = ∞ X n=1 nan(x −1)n−1 and y′′ = ∞ X n=2 n(n −1)an(x −1)n−2. Therefore 2y′′ = ∞ X n=2 2n(n −1)an(x −1)n−2, (x −1)y′′ = ∞ X n=2 n(n −1)an(x −1)n−1, 2(x −1)2y′ = ∞ X n=1 2nan(x −1)n+1, 3y = ∞ X n=0 3an(x −1)n. Before adding these four series we shift indices in the first three so that their general terms become constant multiples of (x −1)n. This yields 2y′′ = ∞ X n=0 2(n + 2)(n + 1)an+2(x −1)n, (7.1.20) (x −1)y′′ = ∞ X n=0 (n + 1)nan+1(x −1)n, (7.1.21) 2(x −1)2y′ = ∞ X n=1 2(n −1)an−1(x −1)n, (7.1.22) 3y = ∞ X n=0 3an(x −1)n, (7.1.23) where we added initial zero terms to the series in (7.1.21) and (7.1.22). Adding (7.1.20)–(7.1.23) yields (1 + x)y′′ + 2(x −1)2y′ + 3y = 2y′′ + (x −1)y′′ + 2(x −1)2y′ + 3y = ∞ X n=0 bn(x −1)n, where b0 = 4a2 + 3a0, (7.1.24) bn = 2(n + 2)(n + 1)an+2 + (n + 1)nan+1 + 2(n −1)an−1 + 3an, n ≥1. (7.1.25) The formula (7.1.24) for b0 can’t be obtained by setting n = 0 in (7.1.25), since the summation in (7.1.22) begins with n = 1, while those in (7.1.20), (7.1.21), and (7.1.23) begin with n = 0. | Elementary Differential Equations with Boundary Value Problems_Page_326_Chunk2730 |
Section 7.1 Review of Power Series 317 7.1 Exercises 1. For each power series use Theorem 7.1.3 to find the radius of convergence R. If R > 0, find the open interval of convergence. (a) ∞ X n=0 (−1)n 2nn (x −1)n (b) ∞ X n=0 2nn(x −2)n (c) ∞ X n=0 n! 9n xn (d) ∞ X n=0 n(n + 1) 16n (x −2)n (e) ∞ X n=0 (−1)n 7n n! xn (f) ∞ X n=0 3n 4n+1(n + 1)2 (x + 7)n 2. Suppose there’s an integer M such that bm ̸= 0 for m ≥M, and lim m→∞ bm+1 bm = L, where 0 ≤L ≤∞. Show that the radius of convergence of ∞ X m=0 bm(x −x0)2m is R = 1/ √ L, which is interpreted to mean that R = 0 if L = ∞or R = ∞if L = 0. HINT: Apply Theorem 7.1.3 to the series P∞ m=0 bmzm and then let z = (x −x0)2. 3. For each power series, use the result of Exercise 2 to find the radius of convergence R. If R > 0, find the open interval of convergence. (a) ∞ X m=0 (−1)m(3m + 1)(x −1)2m+1 (b) ∞ X m=0 (−1)m m(2m + 1) 2m (x + 2)2m (c) ∞ X m=0 m! (2m)!(x −1)2m (d) ∞ X m=0 (−1)m m! 9m (x + 8)2m (e) ∞ X m=0 (−1)m (2m −1) 3m x2m+1 (f) ∞ X m=0 (x −1)2m 4. Suppose there’s an integer M such that bm ̸= 0 for m ≥M, and lim m→∞ bm+1 bm = L, where 0 ≤L ≤∞. Let k be a positive integer. Show that the radius of convergence of ∞ X m=0 bm(x −x0)km is R = 1/ k√ L, which is interpreted to mean that R = 0 if L = ∞or R = ∞if L = 0. HINT: Apply Theorem 7.1.3 to the series P∞ m=0 bmzm and then let z = (x −x0)k. 5. For each power series use the result of Exercise 4 to find the radius of convergence R. If R > 0, find the open interval of convergence. | Elementary Differential Equations with Boundary Value Problems_Page_327_Chunk2731 |
318 Chapter 7 Series Solutions of Linear Second Equations (a) ∞ X m=0 (−1)m (27)m (x −3)3m+2 (b) ∞ X m=0 x7m+6 m (c) ∞ X m=0 9m(m + 1) (m + 2) (x −3)4m+2 (d) ∞ X m=0 (−1)m 2m m! x4m+3 (e) ∞ X m=0 m! (26)m (x + 1)4m+3 (f) ∞ X m=0 (−1)m 8mm(m + 1)(x −1)3m+1 6. L Graph y = sin x and the Taylor polynomial T2M+1(x) = M X n=0 (−1)nx2n+1 (2n + 1)! on the interval (−2π, 2π) for M = 1, 2, 3, ..., until you find a value of M for which there’s no perceptible difference between the two graphs. 7. L Graph y = cos x and the Taylor polynomial T2M(x) = M X n=0 (−1)nx2n (2n)! on the interval (−2π, 2π) for M = 1, 2, 3, ..., until you find a value of M for which there’s no perceptible difference between the two graphs. 8. L Graph y = 1/(1 −x) and the Taylor polynomial TN(x) = N X n=0 xn on the interval [0, .95] for N = 1, 2, 3, ..., until you find a value of N for which there’s no perceptible difference between the two graphs. Choose the scale on the y-axis so that 0 ≤y ≤20. 9. L Graph y = cosh x and the Taylor polynomial T2M(x) = M X n=0 x2n (2n)! on the interval (−5, 5) for M = 1, 2, 3, ..., until you find a value of M for which there’s no perceptible difference between the two graphs. Choose the scale on the y-axis so that 0 ≤y ≤75. 10. L Graph y = sinh x and the Taylor polynomial T2M+1(x) = M X n=0 x2n+1 (2n + 1)! on the interval (−5, 5) for M = 0, 1, 2, ..., until you find a value of M for which there’s no per- ceptible difference between the two graphs. Choose the scale on the y-axis so that −75 ≤y ≤75. In Exercises 11–15 find a power series solution y(x) = P∞ n=0 anxn. 11. (2 + x)y′′ + xy′ + 3y 12. (1 + 3x2)y′′ + 3x2y′ −2y 13. (1 + 2x2)y′′ + (2 −3x)y′ + 4y 14. (1 + x2)y′′ + (2 −x)y′ + 3y | Elementary Differential Equations with Boundary Value Problems_Page_328_Chunk2732 |
Section 7.1 Review of Power Series 319 15. (1 + 3x2)y′′ −2xy′ + 4y 16. Suppose y(x) = P∞ n=0 an(x + 1)n on an open interval that contains x0 = −1. Find a power series in x + 1 for xy′′ + (4 + 2x)y′ + (2 + x)y. 17. Suppose y(x) = P∞ n=0 an(x −2)n on an open interval that contains x0 = 2. Find a power series in x −2 for x2y′′ + 2xy′ −3xy. 18. L Do the following experiment for various choices of real numbers a0 and a1. (a) Use differential equations software to solve the initial value problem (2 −x)y′′ + 2y = 0, y(0) = a0, y′(0) = a1, numerically on (−1.95, 1.95). Choose the most accurate method your software package provides. (See Section 10.1 for a brief discussion of one such method.) (b) For N = 2, 3, 4, ..., compute a2, ..., aN from Eqn.(7.1.18) and graph TN(x) = N X n=0 anxn and the solution obtained in (a) on the same axes. Continue increasing N until it’s obvious that there’s no point in continuing. (This sounds vague, but you’ll know when to stop.) 19. L Follow the directions of Exercise 18 for the initial value problem (1 + x)y′′ + 2(x −1)2y′ + 3y = 0, y(1) = a0, y′(1) = a1, on the interval (0, 2). Use Eqns. (7.1.24) and (7.1.25) to compute {an}. 20. Suppose the series P∞ n=0 anxn converges on an open interval (−R, R), let r be an arbitrary real number, and define y(x) = xr ∞ X n=0 anxn = ∞ X n=0 anxn+r on (0, R). Use Theorem 7.1.4 and the rule for differentiating the product of two functions to show that y′(x) = ∞ X n=0 (n + r)anxn+r−1, y′′(x) = ∞ X n=0 (n + r)(n + r −1)anxn+r−2, ... y(k)(x) = ∞ X n=0 (n + r)(n + r −1) · · ·(n + r −k)anxn+r−k on (0, R) | Elementary Differential Equations with Boundary Value Problems_Page_329_Chunk2733 |
320 Chapter 7 Series Solutions of Linear Second Order Equations In Exercises 21–26 let y be as defined in Exercise 20, and write the given expression in the form xr P∞ n=0 bnxn. 21. x2(1 −x)y′′ + x(4 + x)y′ + (2 −x)y 22. x2(1 + x)y′′ + x(1 + 2x)y′ −(4 + 6x)y 23. x2(1 + x)y′′ −x(1 −6x −x2)y′ + (1 + 6x + x2)y 24. x2(1 + 3x)y′′ + x(2 + 12x + x2)y′ + 2x(3 + x)y 25. x2(1 + 2x2)y′′ + x(4 + 2x2)y′ + 2(1 −x2)y 26. x2(2 + x2)y′′ + 2x(5 + x2)y′ + 2(3 −x2)y 7.2 SERIES SOLUTIONS NEAR AN ORDINARY POINT I Many physical applications give rise to second order homogeneous linear differential equations of the form P0(x)y′′ + P1(x)y′ + P2(x)y = 0, (7.2.1) where P0, P1, and P2 are polynomials. Usually the solutions of these equations can’t be expressed in terms of familiar elementary functions. Therefore we’ll consider the problem of representing solutions of (7.2.1) with series. We assume throughout that P0, P1 and P2 have no common factors. Then we say that x0 is an ordinary point of (7.2.1) if P0(x0) ̸= 0, or a singular point if P0(x0) = 0. For Legendre’s equation, (1 −x2)y′′ −2xy′ + α(α + 1)y = 0, (7.2.2) x0 = 1 and x0 = −1 are singular points and all other points are ordinary points. For Bessel’s equation, x2y′′ + xy′ + (x2 −ν2)y = 0, x0 = 0 is a singular point and all other points are ordinary points. If P0 is a nonzero constant as in Airy’s equation, y′′ −xy = 0, (7.2.3) then every point is an ordinary point. Since polynomials are continuous everywhere, P1/P0 and P2/P0 are continuous at any point x0 that isn’t a zero of P0. Therefore, if x0 is an ordinary point of (7.2.1) and a0 and a1 are arbitrary real numbers, then the initial value problem P0(x)y′′ + P1(x)y′ + P2(x)y = 0, y(x0) = a0, y′(x0) = a1 (7.2.4) has a unique solution on the largest open interval that contains x0 and does not contain any zeros of P0. To see this, we rewrite the differential equation in (7.2.4) as y′′ + P1(x) P0(x)y′ + P2(x) P0(x)y = 0 and apply Theorem 5.1.1 with p = P1/P0 and q = P2/P0. In this section and the next we consider the problem of representing solutions of (7.2.1) by power series that converge for values of x near an ordinary point x0. We state the next theorem without proof. | Elementary Differential Equations with Boundary Value Problems_Page_330_Chunk2734 |
Section 7.2 Series Solutions Near an Ordinary Point I 321 Theorem 7.2.1 Suppose P0, P1, and P2 are polynomials with no common factor and P0 isn’t identically zero. Let x0 be a point such that P0(x0) ̸= 0, and let ρ be the distance from x0 to the nearest zero of P0 in the complex plane. (If P0 is constant, then ρ = ∞.) Then every solution of P0(x)y′′ + P1(x)y′ + P2(x)y = 0 (7.2.5) can be represented by a power series y = ∞ X n=0 an(x −x0)n (7.2.6) that converges at least on the open interval (x0−ρ, x0+ρ). ( If P0 is nonconstant, so that ρ is necessarily finite, then the open interval of convergence of (7.2.6) may be larger than (x0−ρ, x0+ρ). If P0 is constant then ρ = ∞and (x0 −ρ, x0 + ρ) = (−∞, ∞).) We call (7.2.6) a power series solution in x −x0 of (7.2.5). We’ll now develop a method for finding power series solutions of (7.2.5). For this purpose we write (7.2.5) as Ly = 0, where Ly = P0y′′ + P1y′ + P2y. (7.2.7) Theorem 7.2.1 implies that every solution of Ly = 0 on (x0 −ρ, x0 + ρ) can be written as y = ∞ X n=0 an(x −x0)n. Setting x = x0 in this series and in the series y′ = ∞ X n=1 nan(x −x0)n−1 shows that y(x0) = a0 and y′(x0) = a1. Since every initial value problem (7.2.4) has a unique solution, this means that a0 and a1 can be chosen arbitrarily, and a2, a3, ...are uniquely determined by them. To find a2, a3, ..., we write P0, P1, and P2 in powers of x −x0, substitute y = ∞ X n=0 an(x −x0)n, y′ = ∞ X n=1 nan(x −x0)n−1, y′′ = ∞ X n=2 n(n −1)an(x −x0)n−2 into (7.2.7), and collect the coefficients of like powers of x −x0. This yields Ly = ∞ X n=0 bn(x −x0)n, (7.2.8) where {b0, b1, . . ., bn, . . .} are expressed in terms of {a0, a1, . . ., an, . . .} and the coefficients of P0, P1, and P2, written in powers of x −x0. Since (7.2.8) and (a) of Theorem 7.1.6 imply that Ly = 0 if and only if bn = 0 for n ≥0, all power series solutions in x −x0 of Ly = 0 can be obtained by choosing a0 and a1 arbitrarily and computing a2, a3, ..., successively so that bn = 0 for n ≥0. For simplicity, we call the power series obtained this way the power series in x −x0 for the general solution of Ly = 0, without explicitly identifying the open interval of convergence of the series. | Elementary Differential Equations with Boundary Value Problems_Page_331_Chunk2735 |
322 Chapter 7 Series Solutions of Linear Second Order Equations Example 7.2.1 Let x0 be an arbitrary real number. Find the power series in x−x0 for the general solution of y′′ + y = 0. (7.2.9) Solution Here Ly = y′′ + y. If y = ∞ X n=0 an(x −x0)n, then y′′ = ∞ X n=2 n(n −1)an(x −x0)n−2, so Ly = ∞ X n=2 n(n −1)an(x −x0)n−2 + ∞ X n=0 an(x −x0)n. To collect coefficients of like powers of x −x0, we shift the summation index in the first sum. This yields Ly = ∞ X n=0 (n + 2)(n + 1)an+2(x −x0)n + ∞ X n=0 an(x −x0)n = ∞ X n=0 bn(x −x0)n, with bn = (n + 2)(n + 1)an+2 + an. Therefore Ly = 0 if and only if an+2 = −an (n + 2)(n + 1), n ≥0, (7.2.10) where a0 and a1 are arbitrary. Since the indices on the left and right sides of (7.2.10) differ by two, we write (7.2.10) separately for n even (n = 2m) and n odd (n = 2m + 1). This yields a2m+2 = −a2m (2m + 2)(2m + 1), m ≥0, (7.2.11) and a2m+3 = −a2m+1 (2m + 3)(2m + 2), m ≥0. (7.2.12) Computing the coefficients of the even powers of x −x0 from (7.2.11) yields a2 = −a0 2 · 1 a4 = −a2 4 · 3 = −1 4 · 3 −a0 2 · 1 = a0 4 · 3 · 2 · 1, a6 = −a4 6 · 5 = −1 6 · 5 a0 4 · 3 · 2 · 1 = − a0 6 · 5 · 4 · 3 · 2 · 1, and, in general, a2m = (−1)m a0 (2m)! , m ≥0. (7.2.13) | Elementary Differential Equations with Boundary Value Problems_Page_332_Chunk2736 |
Section 7.2 Series Solutions Near an Ordinary Point I 323 Computing the coefficients of the odd powers of x −x0 from (7.2.12) yields a3 = −a1 3 · 2 a5 = −a3 5 · 4 = −1 5 · 4 −a1 3 · 2 = a1 5 · 4 · 3 · 2, a7 = −a5 7 · 6 = −1 7 · 6 a1 5 · 4 · 3 · 2 = − a1 7 · 6 · 5 · 4 · 3 · 2, and, in general, a2m+1 = (−1)ma1 (2m + 1)! m ≥0. (7.2.14) Thus, the general solution of (7.2.9) can be written as y = ∞ X m=0 a2m(x −x0)2m + ∞ X m=0 a2m+1(x −x0)2m+1, or, from (7.2.13) and (7.2.14), as y = a0 ∞ X m=0 (−1)m (x −x0)2m (2m)! + a1 ∞ X m=0 (−1)m (x −x0)2m+1 (2m + 1)! . (7.2.15) If we recall from calculus that ∞ X m=0 (−1)m (x −x0)2m (2m)! = cos(x −x0) and ∞ X m=0 (−1)m (x −x0)2m+1 (2m + 1)! = sin(x −x0), then (7.2.15) becomes y = a0 cos(x −x0) + a1 sin(x −x0), which should look familiar. Equations like (7.2.10), (7.2.11), and (7.2.12), which define a given coefficient in the sequence {an} in terms of one or more coefficients with lesser indices are called recurrence relations. When we use a recurrence relation to compute terms of a sequence we’re computing recursively. In the remainder of this section we consider the problem of finding power series solutions in x −x0 for equations of the form | Elementary Differential Equations with Boundary Value Problems_Page_333_Chunk2737 |
324 Chapter 7 Series Solutions of Linear Second Order Equations if α ̸= 0, or on (−∞, ∞) if α = 0. We’ll see that the coefficients in these power series can be obtained by methods similar to the one used in Example 7.2.1. To simplify finding the coefficients, we introduce some notation for products: s Y j=r bj = brbr+1 · · ·bs if s ≥r. Thus, 7 Y j=2 bj = b2b3b4b5b6b7, 4 Y j=0 (2j + 1) = (1)(3)(5)(7)(9) = 945, and 2 Y j=2 j2 = 22 = 4. We define s Y j=r bj = 1 if s < r, no matter what the form of bj. Example 7.2.2 Find the power series in x for the general solution of (1 + 2x2)y′′ + 6xy′ + 2y = 0. (7.2.17) Solution Here Ly = (1 + 2x2)y′′ + 6xy′ + 2y. If y = ∞ X n=0 anxn then y′ = ∞ X n=1 nanxn−1 and y′′ = ∞ X n=2 n(n −1)anxn−2, so Ly = (1 + 2x2) ∞ X n=2 n(n −1)anxn−2 + 6x ∞ X n=1 nanxn−1 + 2 ∞ X n=0 anxn = ∞ X n=2 n(n −1)anxn−2 + ∞ X n=0 [2n(n −1) + 6n + 2] anxn = ∞ X n=2 n(n −1)anxn−2 + 2 ∞ X n=0 (n + 1)2anxn. | Elementary Differential Equations with Boundary Value Problems_Page_334_Chunk2738 |
Section 7.2 Series Solutions Near an Ordinary Point I 325 To collect coefficients of xn, we shift the summation index in the first sum. This yields Ly = ∞ X n=0 (n + 2)(n + 1)an+2xn + 2 ∞ X n=0 (n + 1)2anxn = ∞ X n=0 bnxn, with bn = (n + 2)(n + 1)an+2 + 2(n + 1)2an, n ≥0. To obtain solutions of (7.2.17), we set bn = 0 for n ≥0. This is equivalent to the recurrence relation an+2 = −2n + 1 n + 2an, n ≥0. (7.2.18) Since the indices on the left and right differ by two, we write (7.2.18) separately for n = 2m and n = 2m + 1, as in Example 7.2.1. This yields a2m+2 = −22m + 1 2m + 2a2m = −2m + 1 m + 1 a2m, m ≥0, (7.2.19) and a2m+3 = −22m + 2 2m + 3a2m+1 = −4 m + 1 2m + 3a2m+1, m ≥0. (7.2.20) Computing the coefficients of even powers of x from (7.2.19) yields a2 = −1 1a0, a4 = −3 2a2 = −3 2 −1 1 a0 = 1 · 3 1 · 2a0, a6 = −5 3a4 = −5 3 1 · 3 1 · 2 a0 = −1 · 3 · 5 1 · 2 · 3a0, a8 = −7 4a6 = −7 4 −1 · 3 · 5 1 · 2 · 3 a0 = 1 · 3 · 5 · 7 1 · 2 · 3 · 4a0. In general, a2m = (−1)m Qm j=1(2j −1) m! a0, m ≥0. (7.2.21) (Note that (7.2.21) is correct for m = 0 because we defined Q0 j=1 bj = 1 for any bj.) Computing the coefficients of odd powers of x from (7.2.20) yields a3 = −4 1 3a1, a5 = −4 2 5a3 = −4 2 5 −41 3 a1 = 42 1 · 2 3 · 5a1, a7 = −4 3 7a5 = −4 3 7 42 1 · 2 3 · 5 a1 = −43 1 · 2 · 3 3 · 5 · 7a1, a9 = −4 4 9a7 = −4 4 9 43 1 · 2 · 3 3 · 5 · 7 a1 = 44 1 · 2 · 3 · 4 3 · 5 · 7 · 9a1. | Elementary Differential Equations with Boundary Value Problems_Page_335_Chunk2739 |
326 Chapter 7 Series Solutions of Linear Second Order Equations In general, a2m+1 = (−1)m4mm! Qm j=1(2j + 1)a1, m ≥0. (7.2.22) From (7.2.21) and (7.2.22), y = a0 ∞ X m=0 (−1)m Qm j=1(2j −1) m! x2m + a1 ∞ X m=0 (−1)m 4mm! Qm j=1(2j + 1)x2m+1. is the power series in x for the general solution of (7.2.17). Since P0(x) = 1+2x2 has no real zeros, Theo- rem 5.1.1 implies that every solutionof (7.2.17) is defined on (−∞, ∞). However, since P0(±i/ √ 2) = 0, Theorem 7.2.1 implies only that the power series converges in (−1/ √ 2, 1/ √ 2) for any choice of a0 and a1. The results in Examples 7.2.1 and 7.2.2 are consequences of the following general theorem. Theorem 7.2.2 The coefficients {an} in any solution y = P∞ n=0 an(x −x0)n of | Elementary Differential Equations with Boundary Value Problems_Page_336_Chunk2740 |
Section 7.2 Series Solutions Near an Ordinary Point I 327 from (7.2.25). To collect coefficients of powers of x −x0, we shift the summation index in the first sum. This yields Ly = ∞ X n=0 [(n + 2)(n + 1)an+2 + p(n)an] (x −x0)n. Thus, Ly = 0 if and only if (n + 2)(n + 1)an+2 + p(n)an = 0, n ≥0, which is equivalent to (7.2.24). Writing (7.2.24) separately for the cases where n = 2m and n = 2m + 1 yields (7.2.26) and (7.2.27). Example 7.2.3 Find the power series in x −1 for the general solution of (2 + 4x −2x2)y′′ −12(x −1)y′ −12y = 0. (7.2.28) Solution We must first write the coefficient P0(x) = 2 + 4x −x2 in powers of x −1. To do this, we write x = (x −1) + 1 in P0(x) and then expand the terms, collecting powers of x −1; thus, 2 + 4x −2x2 = 2 + 4[(x −1) + 1] −2[(x −1) + 1]2 = 4 −2(x −1)2. Therefore we can rewrite (7.2.28) as | Elementary Differential Equations with Boundary Value Problems_Page_337_Chunk2741 |
328 Chapter 7 Series Solutions of Linear Second Order Equations which implies that the power series in x −1 for the general solution of (7.2.28) is y = a0 ∞ X m=0 2m + 1 2m (x −1)2m + a1 ∞ X m=0 m + 1 2m (x −1)2m+1. In the examples considered so far we were able to obtain closed formulas for coefficients in the power series solutions. In some cases this is impossible, and we must settle for computing a finite number of terms in the series. The next example illustrates this with an initial value problem. Example 7.2.4 Compute a0, a1, ..., a7 in the series solution y = P∞ n=0 anxn of the initial value prob- lem (1 + 2x2)y′′ + 10xy′ + 8y = 0, y(0) = 2, y′(0) = −3. (7.2.29) Solution Since α = 2, β = 10, and γ = 8 in (7.2.29), p(n) = 2n(n −1) + 10n + 8 = 2(n + 2)2. Therefore an+2 = −2 (n + 2)2 (n + 2)(n + 1)an = −2n + 2 n + 1an, n ≥0. Writing this equation separately for n = 2m and n = 2m + 1 yields a2m+2 = −2(2m + 2) 2m + 1 a2m = −4 m + 1 2m + 1a2m, m ≥0 (7.2.30) and a2m+3 = −22m + 3 2m + 2a2m+1 = −2m + 3 m + 1 a2m+1, m ≥0. (7.2.31) Starting with a0 = y(0) = 2, we compute a2, a4, and a6 from (7.2.30): a2 = −4 1 12 = −8, a4 = −4 2 3(−8) = 64 3 , a6 = −4 3 5 64 3 = −256 5 . Starting with a1 = y′(0) = −3, we compute a3, a5 and a7 from (7.2.31): a3 = −3 1(−3) = 9, a5 = −5 29 = −45 2 , a7 = −7 3 −45 2 = 105 2 . Therefore the solution of (7.2.29) is y = 2 −3x −8x2 + 9x3 + 64 3 x4 −45 2 x5 −256 5 x6 + 105 2 x7 + · · · . | Elementary Differential Equations with Boundary Value Problems_Page_338_Chunk2742 |
Section 7.2 Series Solutions Near an Ordinary Point I 329 USING TECHNOLOGY Computing coefficients recursively as in Example 7.2.4 is tedious. We recommend that you do this kind of computation by writing a short program to implement the appropriate recurrence relation on a calculator or computer. You may wish to do this in verifying examples and doing exercises (identified by the symbol C ) in this chapter that call for numerical computation of the coefficients in series solutions. We obtained the answers to these exercises by using software that can produce answers in the form of rational numbers. However, it’s perfectly acceptable - and more practical - to get your answers in decimal form. You can always check them by converting our fractions to decimals. If you’re interested in actually using series to compute numerical approximations to solutions of a differential equation, then whether or not there’s a simple closed form for the coefficents is essentially irrelevant. For computational purposes it’s usually more efficient to start with the given coefficients a0 = y(x0) and a1 = y′(x0), compute a2, ..., aN recursively, and then compute approximate values of the solution from the Taylor polynomial TN(x) = N X n=0 an(x −x0)n. The trick is to decide how to choose N so the approximation y(x) ≈TN(x) is sufficiently accurate on the subinterval of the interval of convergence that you’re interested in. In the computational exercises in this and the next two sections, you will often be asked to obtain the solution of a given problem by numerical integration with software of your choice (see Section 10.1 for a brief discussion of one such method), and to compare the solution obtained in this way with the approximations obtained with TN for various values of N. This is a typical textbook kind of exercise, designed to give you insight into how the accuracy of the approximation y(x) ≈TN(x) behaves as a function of N and the interval that you’re working on. In real life, you would choose one or the other of the two methods (numerical integration or series solution). If you choose the method of series solution, then a practical procedure for determining a suitable value of N is to continue increasing N until the maximum of |TN −TN−1| on the interval of interest is within the margin of error that you’re willing to accept. In doing computational problems that call for numerical solution of differential equations you should choose the most accurate numerical integration procedure your software supports, and experiment with the step size until you’re confident that the numerical results are sufficiently accurate for the problem at hand. 7.2 Exercises In Exercises 1 –8 find the power series in x for the general solution. 1. (1 + x2)y′′ + 6xy′ + 6y = 0 2. (1 + x2)y′′ + 2xy′ −2y = 0 3. (1 + x2)y′′ −8xy′ + 20y = 0 4. (1 −x2)y′′ −8xy′ −12y = 0 5. (1 + 2x2)y′′ + 7xy′ + 2y = 0 6. (1 + x2)y′′ + 2xy′ + 1 4y = 0 7. (1 −x2)y′′ −5xy′ −4y = 0 8. (1 + x2)y′′ −10xy′ + 28y = 0 | Elementary Differential Equations with Boundary Value Problems_Page_339_Chunk2743 |
330 Chapter 7 Series Solutions of Linear Second Order Equations 9. L (a) Find the power series in x for the general solution of y′′ + xy′ + 2y = 0. (b) For several choices of a0 and a1, use differential equations software to solve the initial value problem y′′ + xy′ + 2y = 0, y(0) = a0, y′(0) = a1, (A) numerically on (−5, 5). (c) For fixed r in {1, 2, 3, 4, 5} graph TN(x) = N X n=0 anxn and the solution obtained in (a) on (−r, r). Continueincreasing N until there’s no perceptible difference between the two graphs. 10. L Follow the directions of Exercise 9 for the differential equation y′′ + 2xy′ + 3y = 0. In Exercises 11 –13 find a0, ..., aN for N at least 7 in the power series solution y = P∞ n=0 anxn of the initial value problem. 11. C (1 + x2)y′′ + xy′ + y = 0, y(0) = 2, y′(0) = −1 12. C (1 + 2x2)y′′ −9xy′ −6y = 0, y(0) = 1, y′(0) = −1 13. C (1 + 8x2)y′′ + 2y = 0, y(0) = 2, y′(0) = −1 14. L Do the next experiment for various choices of real numbers a0, a1, and r, with 0 < r < 1/ √ 2. (a) Use differential equations software to solve the initial value problem (1 −2x2)y′′ −xy′ + 3y = 0, y(0) = a0, y′(0) = a1, (A) numerically on (−r, r). (b) For N = 2, 3, 4, ..., compute a2, ..., aN in the power series solution y = P∞ n=0 anxn of (A), and graph TN(x) = N X n=0 anxn and the solution obtained in (a) on (−r, r). Continueincreasing N until there’s no perceptible difference between the two graphs. 15. L Do (a) and (b) for several values of r in (0, 1): (a) Use differential equations software to solve the initial value problem (1 + x2)y′′ + 10xy′ + 14y = 0, y(0) = 5, y′(0) = 1, (A) numerically on (−r, r). (b) For N = 2, 3, 4, ..., compute a2, ..., aN in the power series solution y = P∞ n=0 anxn of (A) , and graph TN(x) = N X n=0 anxn and the solution obtained in (a) on (−r, r). Continue increasing N until there’s no percep- tible difference between the two graphs. What happens to the required N as r →1? (c) Try (a) and (b) with r = 1.2. Explain your results. | Elementary Differential Equations with Boundary Value Problems_Page_340_Chunk2744 |
Section 7.2 Series Solutions Near an Ordinary Point I 331 In Exercises 16 –20 find the power series in x −x0 for the general solution. 16. y′′ −y = 0; x0 = 3 17. y′′−(x−3)y′ −y = 0; x0 = 3 18. (1 −4x + 2x2)y′′ + 10(x −1)y′ + 6y = 0; x0 = 1 19. (11 −8x + 2x2)y′′ −16(x −2)y′ + 36y = 0; x0 = 2 20. (5 + 6x + 3x2)y′′ + 9(x + 1)y′ + 3y = 0; x0 = −1 In Exercises 21 –26 find a0, ..., aN for N at least 7 in the power series y = P∞ n=0 an(x −x0)n for the solution of the initial value problem. Take x0 to be the point where the initial conditions are imposed. 21. C (x2 −4)y′′ −xy′ −3y = 0, y(0) = −1, y′(0) = 2 22. C y′′ + (x −3)y′ + 3y = 0, y(3) = −2, y′(3) = 3 23. C (5 −6x + 3x2)y′′ + (x −1)y′ + 12y = 0, y(1) = −1, y′(1) = 1 24. C (4x2 −24x + 37)y′′ + y = 0, y(3) = 4, y′(3) = −6 25. C (x2 −8x + 14)y′′ −8(x −4)y′ + 20y = 0, y(4) = 3, y′(4) = −4 26. C (2x2 + 4x + 5)y′′ −20(x + 1)y′ + 60y = 0, y(−1) = 3, y′(−1) = −3 27. (a) Find a power series in x for the general solution of (1 + x2)y′′ + 4xy′ + 2y = 0. (A) (b) Use (a) and the formula 1 1 −r = 1 + r + r2 + · · · + rn + · · · (−1 < r < 1) for the sum of a geometric series to find a closed form expression for the general solution of (A) on (−1, 1). (c) Show that the expression obtained in (b) is actually the general solution of of (A) on (−∞, ∞). 28. Use Theorem 7.2.2 to show that the power series in x for the general solution of (1 + αx2)y′′ + βxy′ + γy = 0 is y = a0 ∞ X m=0 (−1)m m−1 Y j=0 p(2j) x2m (2m)! + a1 ∞ X m=0 (−1)m m−1 Y j=0 p(2j + 1) x2m+1 (2m + 1)!. 29. Use Exercise 28 to show that all solutions of (1 + αx2)y′′ + βxy′ + γy = 0 are polynomials if and only if αn(n −1) + βn + γ = α(n −2r)(n −2s −1), where r and s are nonnegative integers. | Elementary Differential Equations with Boundary Value Problems_Page_341_Chunk2745 |
332 Chapter 7 Series Solutions of Linear Second Order Equations 30. (a) Use Exercise 28 to show that the power series in x for the general solution of (1 −x2)y′′ −2bxy′ + α(α + 2b −1)y = 0 is y = a0y1 + a1y2, where y1 = ∞ X m=0 m−1 Y j=0 (2j −α)(2j + α + 2b −1) x2m (2m)! and y2 = ∞ X m=0 m−1 Y j=0 (2j + 1 −α)(2j + α + 2b) x2m+1 (2m + 1)!. (b) Suppose 2b isn’t a negative odd integer and k is a nonnegative integer. Show that y1 is a polynomial of degree 2k such that y1(−x) = y1(x) if α = 2k, while y2 is a polynomial of degree 2k+1 such that y2(−x) = −y2(−x) if α = 2k+1. Conclude that if n is a nonnegative integer, then there’s a polynomial Pn of degree n such that Pn(−x) = (−1)nPn(x) and (1 −x2)P ′′ n −2bxP ′ n + n(n + 2b −1)Pn = 0. (A) (c) Show that (A) implies that [(1 −x2)bP ′ n]′ = −n(n + 2b −1)(1 −x2)b−1Pn, and use this to show that if m and n are nonnegative integers, then [(1 −x2)bP ′ n]′Pm −[(1 −x2)bP ′ m]′Pn = [m(m + 2b −1) −n(n + 2b −1)] (1 −x2)b−1PmPn. (B) (d) Now suppose b > 0. Use (B) and integration by parts to show that if m ̸= n, then Z 1 −1 (1 −x2)b−1Pm(x)Pn(x) dx = 0. (We say that Pm and Pn are orthogonal on (−1, 1) with respect to the weighting function (1 −x2)b−1.) 31. (a) Use Exercise 28 to show that the power series in x for the general solution of Hermite’s equation y′′ −2xy′ + 2αy = 0 is y = a0y1 + a1y1, where y1 = ∞ X m=0 m−1 Y j=0 (2j −α) 2mx2m (2m)! and y2 = ∞ X m=0 m−1 Y j=0 (2j + 1 −α) 2mx2m+1 (2m + 1)!. | Elementary Differential Equations with Boundary Value Problems_Page_342_Chunk2746 |
Section 7.2 Series Solutions Near an Ordinary Point I 333 (b) Suppose k is a nonnegative integer. Show that y1 is a polynomial of degree 2k such that y1(−x) = y1(x) if α = 2k, while y2 is a polynomial of degree 2k + 1 such that y2(−x) = −y2(−x) if α = 2k+1. Conclude that if n is a nonnegative integer then there’s a polynomial Pn of degree n such that Pn(−x) = (−1)nPn(x) and P ′′ n −2xP ′ n + 2nPn = 0. (A) (c) Show that (A) implies that [e−x2P ′ n]′ = −2ne−x2Pn, and use this to show that if m and n are nonnegative integers, then [e−x2P ′ n]′Pm −[e−x2P ′ m]′Pn = 2(m −n)e−x2PmPn. (B) (d) Use (B) and integration by parts to show that if m ̸= n, then Z ∞ −∞ e−x2Pm(x)Pn(x) dx = 0. (We say that Pm and Pn are orthogonal on (−∞, ∞) with respect to the weighting function e−x2.) 32. Consider the equation | Elementary Differential Equations with Boundary Value Problems_Page_343_Chunk2747 |
334 Chapter 7 Series Solutions of Linear Second Order Equations In Exercises 33 –37 use the method of Exercise 32 to find the power series in x for the general solution. 33. y′′ −xy = 0 34. (1 −2x3)y′′ −10x2y′ −8xy = 0 35. (1 + x3)y′′ + 7x2y′ + 9xy = 0 36. (1 −2x3)y′′ + 6x2y′ + 24xy = 0 37. (1 −x3)y′′ + 15x2y′ −63xy = 0 38. Consider the equation | Elementary Differential Equations with Boundary Value Problems_Page_344_Chunk2748 |
Section 7.3 Series Solutions Near an Ordinary Point II 335 7.3 SERIES SOLUTIONS NEAR AN ORDINARY POINT II In this section we continue to find series solutions y = ∞ X n=0 an(x −x0)n of initial value problems P0(x)y′′ + P1(x)y′ + P2(x)y = 0, y(x0) = a0, y′(x0) = a1, (7.3.1) where P0, P1, and P2 are polynomials and P0(x0) ̸= 0, so x0 is an ordinary point of (7.3.1). However, here we consider cases where the differential equation in (7.3.1) is not of the form | Elementary Differential Equations with Boundary Value Problems_Page_345_Chunk2749 |
336 Chapter 7 Series Solutions of Linear Second Order Equations where bn = (n + 2)(n + 1)an+2 + (n + 1)2an+1 + (n + 2)(2n + 1)an. Therefore y = P∞ n=0 anxn is a solution of Ly = 0 if and only if an+2 = −n + 1 n + 2 an+1 −2n + 1 n + 1 an, n ≥0. (7.3.3) From the initial conditions in (7.3.2), a0 = y(0) = −1 and a1 = y′(0) = −2. Setting n = 0 in (7.3.3) yields a2 = −1 2a1 −a0 = −1 2 (−2) −(−1) = 2. Setting n = 1 in (7.3.3) yields a3 = −2 3a2 −3 2a1 = −2 3(2) −3 2(−2) = 5 3. We leave it to you to compute a4, a5, a6, a7 from (7.3.3) and show that y = −1 −2x + 2x2 + 5 3x3 −55 12x4 + 3 4x5 + 61 8 x6 −443 56 x7 + · · · . We also leave it to you (Exercise 13) to verify numerically that the Taylor polynomialsTN(x) = PN n=0 anxn converge to the solution of (7.3.2) on (−1/ √ 2, 1/ √ 2). Example 7.3.2 Find the coefficients a0, ..., a5 in the series solution y = ∞ X n=0 an(x + 1)n of the initial value problem (3 + x)y′′ + (1 + 2x)y′ −(2 −x)y = 0, y(−1) = 2, y′(−1) = −3. (7.3.4) Solution Since the desired series is in powers of x + 1 we rewrite the differential equation in (7.3.4) as Ly = 0, with Ly = (2 + (x + 1)) y′′ −(1 −2(x + 1)) y′ −(3 −(x + 1)) y. Since y = ∞ X n=0 an(x + 1)n, y′ = ∞ X n=1 nan(x + 1)n−1 and y′′ = ∞ X n=2 n(n −1)an(x + 1)n−2, Ly = 2 ∞ X n=2 n(n −1)an(x + 1)n−2 + ∞ X n=2 n(n −1)an(x + 1)n−1 − ∞ X n=1 nan(x + 1)n−1 + 2 ∞ X n=1 nan(x + 1)n −3 ∞ X n=0 an(x + 1)n + ∞ X n=0 an(x + 1)n+1. | Elementary Differential Equations with Boundary Value Problems_Page_346_Chunk2750 |
Section 7.3 Series Solutions Near an Ordinary Point II 337 Shifting indices so that the general term in each series is a constant multiple of (x + 1)n yields Ly = 2 ∞ X n=0 (n + 2)(n + 1)an+2(x + 1)n + ∞ X n=0 (n + 1)nan+1(x + 1)n − ∞ X n=0 (n + 1)an+1(x + 1)n + ∞ X n=0 (2n −3)an(x + 1)n + ∞ X n=1 an−1(x + 1)n = ∞ X n=0 bn(x + 1)n, where b0 = 4a2 −a1 −3a0 and bn = 2(n + 2)(n + 1)an+2 + (n2 −1)an+1 + (2n −3)an + an−1, n ≥1. Therefore y = P∞ n=0 an(x + 1)n is a solution of Ly = 0 if and only if a2 = 1 4(a1 + 3a0) (7.3.5) and an+2 = − 1 2(n + 2)(n + 1) (n2 −1)an+1 + (2n −3)an + an−1 , n ≥1. (7.3.6) From the initial conditions in (7.3.4), a0 = y(−1) = 2 and a1 = y′(−1) = −3. We leave it to you to compute a2, ..., a5 with (7.3.5) and (7.3.6) and show that the solution of (7.3.4) is y = −2 −3(x + 1) + 3 4(x + 1)2 −5 12(x + 1)3 + 7 48(x + 1)4 −1 60(x + 1)5 + · · · . We also leave it to you (Exercise 14) to verify numerically that the Taylor polynomialsTN(x) = PN n=0 anxn converge to the solution of (7.3.4) on the interval of convergence of the power series solution. Example 7.3.3 Find the coefficients a0, ..., a5 in the series solution y = P∞ n=0 anxn of the initial value problem y′′ + 3xy′ + (4 + 2x2)y = 0, y(0) = 2, y′(0) = −3. (7.3.7) Solution Here Ly = y′′ + 3xy′ + (4 + 2x2)y. Since y = ∞ X n=0 anxn, y′ = ∞ X n=1 nanxn−1, and y′′ = ∞ X n=2 n(n −1)anxn−2, Ly = ∞ X n=2 n(n −1)anxn−2 + 3 ∞ X n=1 nanxn + 4 ∞ X n=0 anxn + 2 ∞ X n=0 anxn+2. Shifting indices so that the general term in each series is a constant multiple of xn yields Ly = ∞ X n=0 (n + 2)(n + 1)an+2xn + ∞ X n=0 (3n + 4)anxn + 2 ∞ X n=2 an−2xn = ∞ X n=0 bnxn | Elementary Differential Equations with Boundary Value Problems_Page_347_Chunk2751 |
338 Chapter 7 Series Solutions of Linear Second Order Equations where b0 = 2a2 + 4a0, b1 = 6a3 + 7a1, and bn = (n + 2)(n + 1)an+2 + (3n + 4)an + 2an−2, n ≥2. Therefore y = P∞ n=0 anxn is a solution of Ly = 0 if and only if a2 = −2a0, a3 = −7 6a1, (7.3.8) and an+2 = − 1 (n + 2)(n + 1) [(3n + 4)an + 2an−2] , n ≥2. (7.3.9) From the initial conditions in (7.3.7), a0 = y(0) = 2 and a1 = y′(0) = −3. We leave it to you to compute a2, ..., a5 with (7.3.8) and (7.3.9) and show that the solution of (7.3.7) is y = 2 −3x −4x2 + 7 2x3 + 3x4 −79 40x5 + · · · . We also leave it to you (Exercise 15) to verify numerically that the Taylor polynomialsTN(x) = PN n=0 anxn converge to the solution of (7.3.9) on the interval of convergence of the power series solution. 7.3 Exercises In Exercises 1–12 find the coefficients a0,..., aN for N at least 7 in the series solution y = P∞ n=0 anxn of the initial value problem. 1. C (1 + 3x)y′′ + xy′ + 2y = 0, y(0) = 2, y′(0) = −3 2. C (1 + x + 2x2)y′′ + (2 + 8x)y′ + 4y = 0, y(0) = −1, y′(0) = 2 3. C (1 −2x2)y′′ + (2 −6x)y′ −2y = 0, y(0) = 1, y′(0) = 0 4. C (1 + x + 3x2)y′′ + (2 + 15x)y′ + 12y = 0, y(0) = 0, y′(0) = 1 5. C (2 + x)y′′ + (1 + x)y′ + 3y = 0, y(0) = 4, y′(0) = 3 6. C (3 + 3x + x2)y′′ + (6 + 4x)y′ + 2y = 0, y(0) = 7, y′(0) = 3 7. C (4 + x)y′′ + (2 + x)y′ + 2y = 0, y(0) = 2, y′(0) = 5 8. C (2 −3x + 2x2)y′′ −(4 −6x)y′ + 2y = 0, y(1) = 1, y′(1) = −1 9. C (3x + 2x2)y′′ + 10(1 + x)y′ + 8y = 0, y(−1) = 1, y′(−1) = −1 10. C (1 −x + x2)y′′ −(1 −4x)y′ + 2y = 0, y(1) = 2, y′(1) = −1 11. C (2 + x)y′′ + (2 + x)y′ + y = 0, y(−1) = −2, y′(−1) = 3 12. C x2y′′ −(6 −7x)y′ + 8y = 0, y(1) = 1, y′(1) = −2 13. L Do the following experiment for various choices of real numbers a0, a1, and r, with 0 < r < 1/ √ 2. | Elementary Differential Equations with Boundary Value Problems_Page_348_Chunk2752 |
Section 7.3 Series Solutions Near an Ordinary Point II 339 (a) Use differential equations software to solve the initial value problem (1 + x + 2x2)y′′ + (1 + 7x)y′ + 2y = 0, y(0) = a0, y′(0) = a1, (A) numerically on (−r, r). (See Example 7.3.1.) (b) For N = 2, 3, 4, ..., compute a2, ..., aN in the power series solution y = P∞ n=0 anxn of (A), and graph TN(x) = N X n=0 anxn and the solution obtained in (a) on (−r, r). Continueincreasing N until there’s no perceptible difference between the two graphs. 14. L Do the followingexperiment for various choices of real numbers a0, a1, and r, with 0 < r < 2. (a) Use differential equations software to solve the initial value problem (3 + x)y′′ + (1 + 2x)y′ −(2 −x)y = 0, y(−1) = a0, y′(−1) = a1, (A) numerically on (−1 −r, −1 + r). (See Example 7.3.2. Why this interval?) (b) For N = 2, 3, 4, ..., compute a2, . . ., aN in the power series solution y = ∞ X n=0 an(x + 1)n of (A), and graph TN(x) = N X n=0 an(x + 1)n and the solution obtained in (a) on (−1 −r, −1 + r). Continue increasing N until there’s no perceptible difference between the two graphs. 15. L Do the following experiment for several choices of a0, a1, and r, with r > 0. (a) Use differential equations software to solve the initial value problem y′′ + 3xy′ + (4 + 2x2)y = 0, y(0) = a0, y′(0) = a1, (A) numerically on (−r, r). (See Example 7.3.3.) (b) Find the coefficients a0, a1, ..., aN in the power series solution y = P∞ n=0 anxn of (A), and graph TN(x) = N X n=0 anxn and the solution obtained in (a) on (−r, r). Continueincreasing N until there’s no perceptible difference between the two graphs. | Elementary Differential Equations with Boundary Value Problems_Page_349_Chunk2753 |
340 Chapter 7 Series Solutions of Linear Second Order Equations 16. L Do the following experiment for several choices of a0 and a1. (a) Use differential equations software to solve the initial value problem (1 −x)y′′ −(2 −x)y′ + y = 0, y(0) = a0, y′(0) = a1, (A) numerically on (−r, r). (b) Find the coefficients a0, a1, ..., aN in the power series solution y = PN n=0 anxn of (A), and graph TN(x) = N X n=0 anxn and the solution obtained in (a) on (−r, r). Continueincreasing N until there’s no perceptible difference between the two graphs. What happens as you let r →1? 17. L Follow the directions of Exercise 16 for the initial value problem (1 + x)y′′ + 3y′ + 32y = 0, y(0) = a0, y′(0) = a1. 18. L Follow the directions of Exercise 16 for the initial value problem (1 + x2)y′′ + y′ + 2y = 0, y(0) = a0, y′(0) = a1. In Exercises 19–28 find the coefficients a0, ..., aN for N at least 7 in the series solution y = ∞ X n=0 an(x −x0)n of the initial value problem. Take x0 to be the point where the initial conditions are imposed. 19. C (2 + 4x)y′′ −4y′ −(6 + 4x)y = 0, y(0) = 2, y′(0) = −7 20. C (1 + 2x)y′′ −(1 −2x)y′ −(3 −2x)y = 0, y(1) = 1, y′(1) = −2 21. C (5 + 2x)y′′ −y′ + (5 + x)y = 0, y(−2) = 2, y′(−2) = −1 22. C (4 + x)y′′ −(4 + 2x)y′ + (6 + x)y = 0, y(−3) = 2, y′(−3) = −2 23. C (2 + 3x)y′′ −xy′ + 2xy = 0, y(0) = −1, y′(0) = 2 24. C (3 + 2x)y′′ + 3y′ −xy = 0, y(−1) = 2, y′(−1) = −3 25. C (3 + 2x)y′′ −3y′ −(2 + x)y = 0, y(−2) = −2, y′(−2) = 3 26. C (10 −2x)y′′ + (1 + x)y = 0, y(2) = 2, y′(2) = −4 27. C (7 + x)y′′ + (8 + 2x)y′ + (5 + x)y = 0, y(−4) = 1, y′(−4) = 2 28. C (6 + 4x)y′′ + (1 + 2x)y = 0, y(−1) = −1, y′(−1) = 2 29. Show that the coefficients in the power series in x for the general solution of (1 + αx + βx2)y′′ + (γ + δx)y′ + ϵy = 0 satisfy the recurrrence relation an+2 = −γ + αn n + 2 an+1 −βn(n −1) + δn + ϵ (n + 2)(n + 1) an. | Elementary Differential Equations with Boundary Value Problems_Page_350_Chunk2754 |
Section 7.3 Series Solutions Near an Ordinary Point II 341 30. (a) Let α and β be constants, with β ̸= 0. Show that y = P∞ n=0 anxn is a solution of (1 + αx + βx2)y′′ + (2α + 4βx)y′ + 2βy = 0 (A) if and only if an+2 + αan+1 + βan = 0, n ≥0. (B) An equation of this form is called a second order homogeneous linear difference equation. The polynomial p(r) = r2 + αr + β is called the characteristic polynomial of (B). If r1 and r2 are the zeros of p, then 1/r1 and 1/r2 are the zeros of P0(x) = 1 + αx + βx2. (b) Suppose p(r) = (r −r1)(r −r2) where r1 and r2 are real and distinct, and let ρ be the smaller of the two numbers {1/|r1|, 1/|r2|}. Show that if c1 and c2 are constants then the sequence an = c1rn 1 + c2rn 2 , n ≥0 satisfies (B). Conclude from this that any function of the form y = ∞ X n=0 (c1rn 1 + c2rn 2 )xn is a solution of (A) on (−ρ, ρ). (c) Use (b) and the formula for the sum of a geometric series to show that the functions y1 = 1 1 −r1x and y2 = 1 1 −r2x form a fundamental set of solutions of (A) on (−ρ, ρ). (d) Show that {y1, y2} is a fundamental set of solutions of (A) on any interval that does’nt contain either 1/r1 or 1/r2. (e) Suppose p(r) = (r −r1)2, and let ρ = 1/|r1|. Show that if c1 and c2 are constants then the sequence an = (c1 + c2n)rn 1 , n ≥0 satisfies (B). Conclude from this that any function of the form y = ∞ X n=0 (c1 + c2n)rn 1 xn is a solution of (A) on (−ρ, ρ). (f) Use (e) and the formula for the sum of a geometric series to show that the functions y1 = 1 1 −r1x and y2 = x (1 −r1x)2 form a fundamental set of solutions of (A) on (−ρ, ρ). (g) Show that {y1, y2} is a fundamental set of solutions of (A) on any interval that does not contain 1/r1. | Elementary Differential Equations with Boundary Value Problems_Page_351_Chunk2755 |
342 Chapter 7 Series Solutions of Linear Second Order Equations 31. Use the results of Exercise 30 to find the general solution of the given equation on any interval on which polynomial multiplying y′′ has no zeros. (a) (1 + 3x + 2x2)y′′ + (6 + 8x)y′ + 4y = 0 (b) (1 −5x + 6x2)y′′ −(10 −24x)y′ + 12y = 0 (c) (1 −4x + 4x2)y′′ −(8 −16x)y′ + 8y = 0 (d) (4 + 4x + x2)y′′ + (8 + 4x)y′ + 2y = 0 (e) (4 + 8x + 3x2)y′′ + (16 + 12x)y′ + 6y = 0 In Exercises 32–38 find the coefficients a0, ..., aN for N at least 7 in the series solution y = P∞ n=0 anxn of the initial value problem. 32. C y′′ + 2xy′ + (3 + 2x2)y = 0, y(0) = 1, y′(0) = −2 33. C y′′ −3xy′ + (5 + 2x2)y = 0, y(0) = 1, y′(0) = −2 34. C y′′ + 5xy′ −(3 −x2)y = 0, y(0) = 6, y′(0) = −2 35. C y′′ −2xy′ −(2 + 3x2)y = 0, y(0) = 2, y′(0) = −5 36. C y′′ −3xy′ + (2 + 4x2)y = 0, y(0) = 3, y′(0) = 6 37. C 2y′′ + 5xy′ + (4 + 2x2)y = 0, y(0) = 3, y′(0) = −2 38. C 3y′′ + 2xy′ + (4 −x2)y = 0, y(0) = −2, y′(0) = 3 39. Find power series in x for the solutions y1 and y2 of y′′ + 4xy′ + (2 + 4x2)y = 0 such that y1(0) = 1, y′ 1(0) = 0, y2(0) = 0, y′ 2(0) = 1, and identify y1 and y2 in terms of familiar elementary functions. In Exercises 40–49 find the coefficients a0, ..., aN for N at least 7 in the series solution y = ∞ X n=0 an(x −x0)n of the initial value problem. Take x0 to be the point where the initial conditions are imposed. 40. C (1 + x)y′′ + x2y′ + (1 + 2x)y = 0, y(0) −2, y′(0) = 3 41. C y′′ + (1 + 2x + x2)y′ + 2y = 0, y(0) = 2, y′(0) = 3 42. C (1 + x2)y′′ + (2 + x2)y′ + xy = 0, y(0) = −3, y′(0) = 5 43. C (1 + x)y′′ + (1 −3x + 2x2)y′ −(x −4)y = 0, y(1) = −2, y′(1) = 3 44. C y′′ + (13 + 12x + 3x2)y′ + (5 + 2x), y(−2) = 2, y′(−2) = −3 45. C (1 + 2x + 3x2)y′′ + (2 −x2)y′ + (1 + x)y = 0, y(0) = 1, y′(0) = −2 46. C (3 + 4x + x2)y′′ −(5 + 4x −x2)y′ −(2 + x)y = 0, y(−2) = 2, y′(−2) = −1 47. C (1 + 2x + x2)y′′ + (1 −x)y = 0, y(0) = 2, y′(0) = −1 48. C (x −2x2)y′′ + (1 + 3x −x2)y′ + (2 + x)y = 0, y(1) = 1, y′(1) = 0 49. C (16 −11x + 2x2)y′′ + (10 −6x + x2)y′ −(2 −x)y, y(3) = 1, y′(3) = −2 | Elementary Differential Equations with Boundary Value Problems_Page_352_Chunk2756 |
Section 7.4 Regular Singular Points: Euler Equations 343 7.4 REGULAR SINGULAR POINTS EULER EQUATIONS This section sets the stage for Sections 1.5, 1.6, and 1.7. If you’re not interested in those sections, but wish to learn about Euler equations, omit the introductory paragraphs and start reading at Definition 7.4.2. In the next three sections we’ll continue to study equations of the form P0(x)y′′ + P1(x)y′ + P2(x)y = 0 (7.4.1) where P0, P1, and P2 are polynomials, but the emphasis will be different from that of Sections 7.2 and 7.3, where we obtained solutions of (7.4.1) near an ordinary point x0 in the form of power series in x −x0. If x0 is a singular point of (7.4.1) (that is, if P (x0) = 0), the solutions can’t in general be represented by power series in x −x0. Nevertheless, it’s often necessary in physical applications to study the behavior of solutions of (7.4.1) near a singular point. Although this can be difficult in the absence of some sort of assumption on the nature of the singular point, equations that satisfy the requirements of the next definition can be solved by series methods discussed in the next three sections. Fortunately, many equations arising in applications satisfy these requirements. Definition 7.4.1 Let P0, P1, and P2 be polynomials with no common factor and suppose P0(x0) = 0. Then x0 is a regular singular point of the equation P0(x)y′′ + P1(x)y′ + P2(x)y = 0 (7.4.2) if (7.4.2) can be written as (x −x0)2A(x)y′′ + (x −x0)B(x)y′ + C(x)y = 0 (7.4.3) where A, B, and C are polynomials and A(x0) ̸= 0; otherwise, x0 is an irregular singular point of (7.4.2). Example 7.4.1 Bessel’s equation, x2y′′ + xy′ + (x2 −ν2)y = 0, (7.4.4) has the singular point x0 = 0. Since this equation is of the form (7.4.3) with x0 = 0, A(x) = 1, B(x) = 1, and C(x) = x2 −ν2, it follows that x0 = 0 is a regular singular point of (7.4.4). Example 7.4.2 Legendre’s equation, (1 −x2)y′′ −2xy′ + α(α + 1)y = 0, (7.4.5) has the singular points x0 = ±1. Mutiplying through by 1 −x yields (x −1)2(x + 1)y′′ + 2x(x −1)y′ −α(α + 1)(x −1)y = 0, which is of the form (7.4.3) with x0 = 1, A(x) = x + 1, B(x) = 2x, and C(x) = −α(α + 1)(x −1). Therefore x0 = 1 is a regular singular point of (7.4.5). We leave it to you to show that x0 = −1 is also a regular singular point of (7.4.5). | Elementary Differential Equations with Boundary Value Problems_Page_353_Chunk2757 |
344 Chapter 7 Series Solutions of Linear Second Order Equations Example 7.4.3 The equation x3y′′ + xy′ + y = 0 has an irregular singular point at x0 = 0. (Verify.) For convenience we restrict our attention to the case where x0 = 0 is a regular singular point of (7.4.2). This isn’t really a restriction, since if x0 ̸= 0 is a regular singular point of (7.4.2) then introducing the new independent variable t = x −x0 and the new unknown Y (t) = y(t + x0) leads to a differential equation with polynomial coefficients that has a regular singular point at t0 = 0. This is illustrated in Exercise 22 for Legendre’s equation, and in Exercise 23 for the general case. Euler Equations The simplest kind of equation with a regular singular point at x0 = 0 is the Euler equation, defined as follows. Definition 7.4.2 An Euler equation is an equation that can be written in the form ax2y′′ + bxy′ + cy = 0, (7.4.6) where a, b, and c are real constants and a ̸= 0. Theorem 5.1.1 implies that (7.4.6) has solutions defined on (0, ∞) and (−∞, 0), since (7.4.6) can be rewritten as ay′′ + b xy′ + c x2 y = 0. For convenience we’ll restrict our attention to the interval (0, ∞). (Exercise 19 deals with solutions of (7.4.6) on (−∞, 0).) The key to finding solutions on (0, ∞) is that if x > 0 then xr is defined as a real-valued function on (0, ∞) for all values of r, and substituting y = xr into (7.4.6) produces ax2(xr)′′ + bx(xr)′ + cxr = ax2r(r −1)xr−2 + bxrxr−1 + cxr = [ar(r −1) + br + c]xr. (7.4.7) The polynomial p(r) = ar(r −1) + br + c is called the indicial polynomial of (7.4.6), and p(r) = 0 is its indicial equation. From (7.4.7) we can see that y = xr is a solution of (7.4.6) on (0, ∞) if and only if p(r) = 0. Therefore, if the indicial equation has distinct real roots r1 and r2 then y1 = xr1 and y2 = xr2 form a fundamental set of solutions of (7.4.6) on (0, ∞), since y2/y1 = xr2−r1 is nonconstant. In this case y = c1xr1 + c2xr2 is the general solution of (7.4.6) on (0, ∞). Example 7.4.4 Find the general solution of x2y′′ −xy′ −8y = 0 (7.4.8) on (0, ∞). Solution The indicial polynomial of (7.4.8) is p(r) = r(r −1) −r −8 = (r −4)(r + 2). Therefore y1 = x4 and y2 = x−2 are solutions of (7.4.8) on (0, ∞), and its general solution on (0, ∞) is y = c1x4 + c2 x2 . | Elementary Differential Equations with Boundary Value Problems_Page_354_Chunk2758 |
Section 7.4 Regular Singular Points: Euler Equations 345 Example 7.4.5 Find the general solution of 6x2y′′ + 5xy′ −y = 0 (7.4.9) on (0, ∞). Solution The indicial polynomial of (7.4.9) is p(r) = 6r(r −1) + 5r −1 = (2r −1)(3r + 1). Therefore the general solution of (7.4.9) on (0, ∞) is y = c1x1/2 + c2x−1/3. If the indicial equation has a repeated root r1, then y1 = xr1 is a solution of ax2y′′ + bxy′ + cy = 0, (7.4.10) on (0, ∞), but (7.4.10) has no other solution of the form y = xr. If the indicial equation has complex conjugate zeros then (7.4.10) has no real–valued solutions of the form y = xr. Fortunately we can use the results of Section 5.2 for constant coefficient equations to solve (7.4.10) in any case. Theorem 7.4.3 Suppose the roots of the indicial equation ar(r −1) + br + c = 0 (7.4.11) are r1 and r2. Then the general solution of the Euler equation ax2y′′ + bxy′ + cy = 0 (7.4.12) on (0, ∞) is y = c1xr1 + c2xr2 if r1 and r2 are distinct real numbers ; y = xr1(c1 + c2 ln x) if r1 = r2 ; y = xλ [c1 cos (ω ln x) + c2 sin (ω ln x)] if r1, r2 = λ ± iω with ω > 0. Proof We first show that y = y(x) satisfies (7.4.12) on (0, ∞) if and only if Y (t) = y(et) satisfies the constant coefficient equation ad2Y dt2 + (b −a)dY dt + cY = 0 (7.4.13) on (−∞, ∞). To do this, it’s convenient to write x = et, or, equivalently, t = lnx; thus, Y (t) = y(x), where x = et. From the chain rule, dY dt = dy dx dx dt and, since dx dt = et = x, it follows that dY dt = x dy dx. (7.4.14) | Elementary Differential Equations with Boundary Value Problems_Page_355_Chunk2759 |
346 Chapter 7 Series Solutions of Linear Second Order Equations Differentiating this with respect to t and using the chain rule again yields d2Y dt2 = d dt dY dt = d dt x dy dx = dx dt dy dx + x d2y dx2 dx dt = x dy dx + x2 d2y dx2 since dx dt = x . From this and (7.4.14), x2 d2y dx2 = d2Y dt2 −dY dt . Substituting this and (7.4.14) into (7.4.12) yields (7.4.13). Since (7.4.11) is the characteristic equation of (7.4.13), Theorem 5.2.1 implies that the general solution of (7.4.13) on (−∞, ∞) is Y (t) = c1er1t + c2er2t if r1 and r2 are distinct real numbers; Y (t) = er1t(c1 + c2t) if r1 = r2; Y (t) = eλt (c1 cos ωt + c2 sin ωt) if r1, r2 = λ ± iω with ω ̸= 0. Since Y (t) = y(et), substituting t = lnx in the last three equations shows that the general solution of (7.4.12) on (0, ∞) has the form stated in the theorem. Example 7.4.6 Find the general solution of x2y′′ −5xy′ + 9y = 0 (7.4.15) on (0, ∞). Solution The indicial polynomial of (7.4.15) is p(r) = r(r −1) −5r + 9 = (r −3)2. Therefore the general solution of (7.4.15) on (0, ∞) is y = x3(c1 + c2 ln x). Example 7.4.7 Find the general solution of x2y′′ + 3xy′ + 2y = 0 (7.4.16) on (0, ∞). Solution The indicial polynomial of (7.4.16) is p(r) = r(r −1) + 3r + 2 = (r + 1)2 + 1. The roots of the indicial equation are r = −1 ± i and the general solution of (7.4.16) on (0, ∞) is y = 1 x [c1 cos(lnx) + c2 sin(ln x)] . | Elementary Differential Equations with Boundary Value Problems_Page_356_Chunk2760 |
Section 7.4 Regular Singular Points: Euler Equations 347 7.4 Exercises In Exercises 1–18 find the general solution of the given Euler equation on (0, ∞). 1. x2y′′ + 7xy′ + 8y = 0 2. x2y′′ −7xy′ + 7y = 0 3. x2y′′ −xy′ + y = 0 4. x2y′′ + 5xy′ + 4y = 0 5. x2y′′ + xy′ + y = 0 6. x2y′′ −3xy′ + 13y = 0 7. x2y′′ + 3xy′ −3y = 0 8. 12x2y′′ −5xy′′ + 6y = 0 9. 4x2y′′ + 8xy′ + y = 0 10. 3x2y′′ −xy′ + y = 0 11. 2x2y′′ −3xy′ + 2y = 0 12. x2y′′ + 3xy′ + 5y = 0 13. 9x2y′′ + 15xy′ + y = 0 14. x2y′′ −xy′ + 10y = 0 15. x2y′′ −6y = 0 16. 2x2y′′ + 3xy′ −y = 0 17. x2y′′ −3xy′ + 4y = 0 18. 2x2y′′ + 10xy′ + 9y = 0 19. (a) Adapt the proof of Theorem 7.4.3 to show that y = y(x) satisfies the Euler equation ax2y′′ + bxy′ + cy = 0 (7.4.1) on (−∞, 0) if and only if Y (t) = y(−et) ad2Y dt2 + (b −a)dY dt + cY = 0. on (−∞, ∞). (b) Use (a) to show that the general solution of (7.4.1) on (−∞, 0) is y = c1|x|r1 + c2|x|r2 if r1 and r2 are distinct real numbers; y = |x|r1(c1 + c2 ln |x|) if r1 = r2; y = |x|λ [c1 cos (ω ln |x|) + c2 sin (ω ln |x|)] if r1, r2 = λ ± iω with ω > 0. 20. Use reduction of order to show that if ar(r −1) + br + c = 0 has a repeated root r1 then y = xr1(c1 + c2 ln x) is the general solution of ax2y′′ + bxy′ + cy = 0 on (0, ∞). | Elementary Differential Equations with Boundary Value Problems_Page_357_Chunk2761 |
348 Chapter 7 Series Solutions of Linear Second Order Equations 21. A nontrivial solution of P0(x)y′′ + P1(x)y′ + P2(x)y = 0 is said to be oscillatory on an interval (a, b) if it has infinitely many zeros on (a, b). Otherwise y is said to be nonoscillatory on (a, b). Show that the equation x2y′′ + ky = 0 (k = constant) has oscillatory solutions on (0, ∞) if and only if k > 1/4. 22. In Example 7.4.2 we saw that x0 = 1 and x0 = −1 are regular singular points of Legendre’s equation (1 −x2)y′′ −2xy′ + α(α + 1)y = 0. (A) (a) Introduce the new variables t = x −1 and Y (t) = y(t + 1), and show that y is a solution of (A) if and only if Y is a solution of t(2 + t)d2Y dt2 + 2(1 + t)dY dt −α(α + 1)Y = 0, which has a regular singular point at t0 = 0. (b) Introduce the new variables t = x + 1 and Y (t) = y(t −1), and show that y is a solution of (A) if and only if Y is a solution of t(2 −t)d2Y dt2 + 2(1 −t)dY dt + α(α + 1)Y = 0, which has a regular singular point at t0 = 0. 23. Let P0, P1, and P2 be polynomials with no common factor, and suppose x0 ̸= 0 is a singular point of P0(x)y′′ + P1(x)y′ + P2(x)y = 0. (A) Let t = x −x0 and Y (t) = y(t + x0). (a) Show that y is a solution of (A) if and only if Y is a solution of R0(t)d2Y dt2 + R1(t)dY dt + R2(t)Y = 0. (B) where Ri(t) = Pi(t + x0), i = 0, 1, 2. (b) Show that R0, R1, and R2 are polynomials in t with no common factors, and R0(0) = 0; thus, t0 = 0 is a singular point of (B). 7.5 THE METHOD OF FROBENIUS I In this section we begin to study series solutions of a homogeneous linear second order differential equa- tion with a regular singular point at x0 = 0, so it can be written as x2A(x)y′′ + xB(x)y′ + C(x)y = 0, (7.5.1) where A, B, C are polynomials and A(0) ̸= 0. | Elementary Differential Equations with Boundary Value Problems_Page_358_Chunk2762 |
Section 7.5 The Method of Frobenius I 349 We’ll see that (7.5.1) always has at least one solution of the form y = xr ∞ X n=0 anxn where a0 ̸= 0 and r is a suitably chosen number. The method we will use to find solutions of this form and other forms that we’ll encounter in the next two sections is called the method of Frobenius, and we’ll call them Frobenius solutions. It can be shown that the power series P∞ n=0 anxn in a Frobenius solution of (7.5.1) converges on some open interval (−ρ, ρ), where 0 < ρ ≤∞. However, since xr may be complex for negative x or undefined if x = 0, we’ll consider solutions defined for positive values of x. Easy modifications of our results yield solutions defined for negative values of x. (Exercise 54). We’ll restrict our attention to the case where A, B, and C are polynomials of degree not greater than two, so (7.5.1) becomes x2(α0 + α1x + α2x2)y′′ + x(β0 + β1x + β2x2)y′ + (γ0 + γ1x + γ2x2)y = 0, (7.5.2) where αi, βi, and γi are real constants and α0 ̸= 0. Most equations that arise in applications can be written this way. Some examples are αx2y′′ + βxy′ + γy = 0 (Euler’s equation), x2y′′ + xy′ + (x2 −ν2)y = 0 (Bessel’s equation), and xy′′ + (1 −x)y′ + λy = 0, (Laguerre’s equation), where we would multiply the last equation through by x to put it in the form (7.5.2). However, the method of Frobenius can be extended to the case where A, B, and C are functions that can be represented by power series in x on some interval that contains zero, and A0(0) ̸= 0 (Exercises 57 and 58). The next two theorems will enable us to develop systematic methods for finding Frobenius solutions of (7.5.2). Theorem 7.5.1 Let Ly = x2(α0 + α1x + α2x2)y′′ + x(β0 + β1x + β2x2)y′ + (γ0 + γ1x + γ2x2)y, and define p0(r) = α0r(r −1) + β0r + γ0, p1(r) = α1r(r −1) + β1r + γ1, p2(r) = α2r(r −1) + β2r + γ2. Suppose the series y = ∞ X n=0 anxn+r (7.5.3) converges on (0, ρ). Then Ly = ∞ X n=0 bnxn+r (7.5.4) | Elementary Differential Equations with Boundary Value Problems_Page_359_Chunk2763 |
350 Chapter 7 Series Solutions of Linear Second Order Equations on (0, ρ), where b0 = p0(r)a0, b1 = p0(r + 1)a1 + p1(r)a0, bn = p0(n + r)an + p1(n + r −1)an−1 + p2(n + r −2)an−2, n ≥2. (7.5.5) Proof We begin by showing that if y is given by (7.5.3) and α, β, and γ are constants, then αx2y′′ + βxy′ + γy = ∞ X n=0 p(n + r)anxn+r, (7.5.6) where p(r) = αr(r −1) + βr + γ. Differentiating (3) twice yields y′ = ∞ X n=0 (n + r)anxn+r−1 (7.5.7) and y′′ = ∞ X n=0 (n + r)(n + r −1)anxn+r−2. (7.5.8) Multiplying (7.5.7) by x and (7.5.8) by x2 yields xy′ = ∞ X n=0 (n + r)anxn+r and x2y′′ = ∞ X n=0 (n + r)(n + r −1)anxn+r. Therefore αx2y′′ + βxy′ + γy = ∞ X n=0 [α(n + r)(n + r −1) + β(n + r) + γ] anxn+r = ∞ X n=0 p(n + r)anxn+r, which proves (7.5.6). Multiplying (7.5.6) by x yields x(αx2y′′ + βxy′ + γy) = ∞ X n=0 p(n + r)anxn+r+1 = ∞ X n=1 p(n + r −1)an−1xn+r. (7.5.9) Multiplying (7.5.6) by x2 yields x2(αx2y′′ + βxy′ + γy) = ∞ X n=0 p(n + r)anxn+r+2 = ∞ X n=2 p(n + r −2)an−2xn+r. (7.5.10) | Elementary Differential Equations with Boundary Value Problems_Page_360_Chunk2764 |
Section 7.5 The Method of Frobenius I 351 To use these results, we rewrite Ly = x2(α0 + α1x + α2x2)y′′ + x(β0 + β1x + β2x2)y′ + (γ0 + γ1x + γ2x2)y as Ly = | Elementary Differential Equations with Boundary Value Problems_Page_361_Chunk2765 |
352 Chapter 7 Series Solutions of Linear Second Order Equations Suppose r is a real number such that p0(n + r) is nonzero for all positive integers n. Define a0(r) = 1, a1(r) = − p1(r) p0(r + 1), an(r) = −p1(n + r −1)an−1(r) + p2(n + r −2)an−2(r) p0(n + r) , n ≥2. (7.5.12) Then the Frobenius series y(x, r) = xr ∞ X n=0 an(r)xn (7.5.13) converges and satisfies Ly(x, r) = p0(r)xr (7.5.14) on the interval (0, ρ), where ρ is the distance from the origin to the nearest zero of A(x) = α0 + α1x + α2x2 in the complex plane. (If A is constant, then ρ = ∞.) If {an(r)} is determined by the recurrence relation (7.5.12) then substituting an = an(r) into (7.5.5) yields b0 = p0(r) and bn = 0 for n ≥1, so (7.5.4) reduces to (7.5.14). We omit the proof that the series (7.5.13) converges on (0, ρ). If αi = βi = γi = 0 for i = 1, 2, then Ly = 0 reduces to the Euler equation α0x2y′′ + β0xy′ + γ0y = 0. Theorem 7.4.3 shows that the solutions of this equation are determined by the zeros of the indicial poly- nomial p0(r) = α0r(r −1) + β0r + γ0. Since (7.5.14) implies that this is also true for the solutions of Ly = 0, we’ll also say that p0 is the indicial polynomial of (7.5.2), and that p0(r) = 0 is the indicial equation of Ly = 0. We’ll consider only cases where the indicial equation has real roots r1 and r2, with r1 ≥r2. Theorem 7.5.3 Let L and {an(r)} be as in Theorem 7.5.2, and suppose the indicial equation p0(r) = 0 of Ly = 0 has real roots r1 and r2, where r1 ≥r2. Then y1(x) = y(x, r1) = xr1 ∞ X n=0 an(r1)xn is a Frobenius solution of Ly = 0. Moreover, if r1 −r2 isn’t an integer then y2(x) = y(x, r2) = xr2 ∞ X n=0 an(r2)xn is also a Frobenius solution of Ly = 0, and {y1, y2} is a fundamental set of solutions. Proof Since r1 and r2 are roots of p0(r) = 0, the indicial polynomial can be factored as p0(r) = α0(r −r1)(r −r2). (7.5.15) Therefore p0(n + r1) = nα0(n + r1 −r2), | Elementary Differential Equations with Boundary Value Problems_Page_362_Chunk2766 |
Section 7.5 The Method of Frobenius I 353 which is nonzero if n > 0, since r1 −r2 ≥0. Therefore the assumptions of Theorem 7.5.2 hold with r = r1, and (7.5.14) implies that Ly1 = p0(r1)xr1 = 0. Now suppose r1 −r2 isn’t an integer. From (7.5.15), p0(n + r2) = nα0(n −r1 + r2) ̸= 0 if n = 1, 2, · · · . Hence, the assumptions of Theorem 7.5.2 hold with r = r2, and (7.5.14) implies that Ly2 = p0(r2)xr2 = 0. We leave the proof that {y1, y2} is a fundamental set of solutions as an exercise (Exercise 52). It isn’t always possible to obtain explicit formulas for the coefficients in Frobenius solutions. However, we can always set up the recurrence relations and use them to compute as many coefficients as we want. The next example illustrates this. Example 7.5.1 Find a fundamental set of Frobenius solutions of 2x2(1 + x + x2)y′′ + x(9 + 11x + 11x2)y′ + (6 + 10x + 7x2)y = 0. (7.5.16) Compute just the first six coefficients a0,..., a5 in each solution. Solution For the given equation, the polynomials defined in Theorem 7.5.2 are p0(r) = 2r(r −1) + 9r + 6 = (2r + 3)(r + 2), p1(r) = 2r(r −1) + 11r + 10 = (2r + 5)(r + 2), p2(r) = 2r(r −1) + 11r + 7 = (2r + 7)(r + 1). The zeros of the indicial polynomial p0 are r1 = −3/2 and r2 = −2, so r1 −r2 = 1/2. Therefore Theorem 7.5.3 implies that y1 = x−3/2 ∞ X n=0 an(−3/2)xn and y2 = x−2 ∞ X n=0 an(−2)xn (7.5.17) form a fundamental set of Frobenius solutions of (7.5.16). To find the coefficients in these series, we use the recurrence relation of Theorem 7.5.2; thus, a0(r) = 1, a1(r) = − p1(r) p0(r + 1) = −(2r + 5)(r + 2) (2r + 5)(r + 3) = −r + 2 r + 3, an(r) = −p1(n + r −1)an−1 + p2(n + r −2)an−2 p0(n + r) = −(n + r + 1)(2n + 2r + 3)an−1(r) + (n + r −1)(2n + 2r + 3)an−2(r) (n + r + 2)(2n + 2r + 3) = −(n + r + 1)an−1(r) + (n + r −1)an−2(r) n + r + 2 , n ≥2. Setting r = −3/2 in these equations yields a0(−3/2) = 1, a1(−3/2) = −1/3, an(−3/2) = −(2n −1)an−1(−3/2) + (2n −5)an−2(−3/2) 2n + 1 , n ≥2, (7.5.18) | Elementary Differential Equations with Boundary Value Problems_Page_363_Chunk2767 |
354 Chapter 7 Series Solutions of Linear Second Order Equations and setting r = −2 yields a0(−2) = 1, a1(−2) = 0, an(−2) = −(n −1)an−1(−2) + (n −3)an−2(−2) n , n ≥2. (7.5.19) Calculating with (7.5.18) and (7.5.19) and substituting the results into (7.5.17) yields the fundamental set of Frobenius solutions y1 = x−3/2 1 −1 3x + 2 5x2 −5 21x3 + 7 135x4 + 76 1155x5 + · · · , y2 = x−2 1 + 1 2x2 −1 3x3 + 1 8x4 + 1 30x5 + · · · . Special Cases With Two Term Recurrence Relations For n ≥2, the recurrence relation (7.5.12) of Theorem 7.5.2 involves the three coefficients an(r), an−1(r), and an−2(r). We’ll now consider some special cases where (7.5.12) reduces to a two term recurrence relation; that is, a relation involving only an(r) and an−1(r) or only an(r) and an−2(r). This simplification often makes it possible to obtain explicit formulas for the coefficents of Frobenius solutions. We first consider equations of the form x2(α0 + α1x)y′′ + x(β0 + β1x)y′ + (γ0 + γ1x)y = 0 with α0 ̸= 0. For this equation, α2 = β2 = γ2 = 0, so p2 ≡0 and the recurrence relations in Theorem 7.5.2 simplify to a0(r) = 1, an(r) = −p1(n + r −1) p0(n + r) an−1(r), n ≥1. (7.5.20) Example 7.5.2 Find a fundamental set of Frobenius solutions of x2(3 + x)y′′ + 5x(1 + x)y′ −(1 −4x)y = 0. (7.5.21) Give explicit formulas for the coefficients in the solutions. Solution For this equation, the polynomials defined in Theorem 7.5.2 are p0(r) = 3r(r −1) + 5r −1 = (3r −1)(r + 1), p1(r) = r(r −1) + 5r + 4 = (r + 2)2, p2(r) = 0. The zeros of the indicial polynomial p0 are r1 = 1/3 and r2 = −1, so r1 −r2 = 4/3. Therefore Theorem 7.5.3 implies that y1 = x1/3 ∞ X n=0 an(1/3)xn and y2 = x−1 ∞ X n=0 an(−1)xn | Elementary Differential Equations with Boundary Value Problems_Page_364_Chunk2768 |
Section 7.5 The Method of Frobenius I 355 form a fundamental set of Frobenius solutions of (7.5.21). To find the coefficients in these series, we use the recurrence relationss (7.5.20); thus, a0(r) = 1, an(r) = −p1(n + r −1) p0(n + r) an−1(r) = − (n + r + 1)2 (3n + 3r −1)(n + r + 1)an−1(r) = −n + r + 1 3n + 3r −1an−1(r), n ≥1. (7.5.22) Setting r = 1/3 in (7.5.22) yields a0(1/3) = 1, an(1/3) = −3n + 4 9n an−1(1/3), n ≥1. By using the product notation introduced in Section 7.2 and proceeding as we did in the examples in that section yields an(1/3) = (−1)n Qn j=1(3j + 4) 9nn! , n ≥0. Therefore y1 = x1/3 ∞ X n=0 (−1)n Qn j=1(3j + 4) 9nn! xn is a Frobenius solution of (7.5.21). Setting r = −1 in (7.5.22) yields a0(−1) = 1, an(−1) = − n 3n −4an−1(−1), n ≥1, so an(−1) = (−1)nn! Qn j=1(3j −4). Therefore y2 = x−1 ∞ X n=0 (−1)nn! Qn j=1(3j −4)xn is a Frobenius solution of (7.5.21), and {y1, y2} is a fundamental set of solutions. We now consider equations of the form x2(α0 + α2x2)y′′ + x(β0 + β2x2)y′ + (γ0 + γ2x2)y = 0 (7.5.23) with α0 ̸= 0. For this equation, α1 = β1 = γ1 = 0, so p1 ≡0 and the recurrence relations in Theorem 7.5.2 simplify to a0(r) = 1, a1(r) = 0, an(r) = −p2(n + r −2) p0(n + r) an−2(r), n ≥2. | Elementary Differential Equations with Boundary Value Problems_Page_365_Chunk2769 |
356 Chapter 7 Series Solutions of Linear Second Order Equations Since a1(r) = 0, the last equation implies that an(r) = 0 if n is odd, so the Frobenius solutions are of the form y(x, r) = xr ∞ X m=0 a2m(r)x2m, where a0(r) = 1, a2m(r) = −p2(2m + r −2) p0(2m + r) a2m−2(r), m ≥1. (7.5.24) Example 7.5.3 Find a fundamental set of Frobenius solutions of x2(2 −x2)y′′ −x(3 + 4x2)y′ + (2 −2x2)y = 0. (7.5.25) Give explicit formulas for the coefficients in the solutions. Solution For this equation, the polynomials defined in Theorem 7.5.2 are p0(r) = 2r(r −1) −3r + 2 = (r −2)(2r −1), p1(r) = 0 p2(r) = −[r(r −1) + 4r + 2] = −(r + 1)(r + 2). The zeros of the indicial polynomial p0 are r1 = 2 and r2 = 1/2, so r1 −r2 = 3/2. Therefore Theorem 7.5.3 implies that y1 = x2 ∞ X m=0 a2m(1/3)x2m and y2 = x1/2 ∞ X m=0 a2m(1/2)x2m form a fundamental set of Frobenius solutions of (7.5.25). To find the coefficients in these series, we use the recurrence relation (7.5.24); thus, a0(r) = 1, a2m(r) = −p2(2m + r −2) p0(2m + r) a2m−2(r) = (2m + r)(2m + r −1) (2m + r −2)(4m + 2r −1)a2m−2(r), m ≥1. (7.5.26) Setting r = 2 in (7.5.26) yields a0(2) = 1, a2m(2) = (m + 1)(2m + 1) m(4m + 3) a2m−2(2), m ≥1, so a2m(2) = (m + 1) m Y j=1 2j + 1 4j + 3. Therefore y1 = x2 ∞ X m=0 (m + 1) m Y j=1 2j + 1 4j + 3 x2m is a Frobenius solution of (7.5.25). | Elementary Differential Equations with Boundary Value Problems_Page_366_Chunk2770 |
Section 7.5 The Method of Frobenius I 357 Setting r = 1/2 in (7.5.26) yields a0(1/2) = 1, a2m(1/2) = (4m −1)(4m + 1) 8m(4m −3) a2m−2(1/2), m ≥1, so a2m(1/2) = 1 8mm! m Y j=1 (4j −1)(4j + 1) 4j −3 . Therefore y2 = x1/2 ∞ X m=0 1 8mm! m Y j=1 (4j −1)(4j + 1) 4j −3 x2m is a Frobenius solution of (7.5.25) and {y1, y2} is a fundamental set of solutions. REMARK: Thus far, we considered only the case where the indicial equation has real roots that don’t differ by an integer, which allows us to apply Theorem 7.5.3. However, for equations of the form (7.5.23), the sequence {a2m(r)} in (7.5.24) is defined for r = r2 if r1 −r2 isn’t an even integer. It can be shown (Exercise 56) that in this case y1 = xr1 ∞ X m=0 a2m(r1)x2m and y2 = xr2 ∞ X m=0 a2m(r2)x2m form a fundamental set Frobenius solutions of (7.5.23). USING TECHNOLOGY As we said at the end of Section 7.2, if you’re interested in actually using series to compute numerical approximations to solutions of a differential equation, then whether or not there’s a simple closed form for the coefficents is essentially irrelevant; recursive computation is usually more efficient. Since it’s also laborious, we encourage you to write short programs to implement recurrence relations on a calculator or computer, even in exercises where this is not specifically required. In practical use of the method of Frobenius when x0 = 0 is a regular singular point, we’re interested in how well the functions yN(x, ri) = xri N X n=0 an(ri)xn, i = 1, 2, approximate solutions to a given equation when ri is a zero of the indicial polynomial. In dealing with the corresponding problem for the case where x0 = 0 is an ordinary point, we used numerical integration to solve the differential equation subject to initial conditions y(0) = a0, y′(0) = a1, and compared the result with values of the Taylor polynomial TN(x) = N X n=0 anxn. We can’t do that here, since in general we can’t prescribe arbitrary initial values for solutions of a dif- ferential equation at a singular point. Therefore, motivated by Theorem 7.5.2 (specifically, (7.5.14)), we suggest the following procedure. | Elementary Differential Equations with Boundary Value Problems_Page_367_Chunk2771 |
358 Chapter 7 Series Solutions of Linear Second Order Equations Verification Procedure Let L and Yn(x; ri) be defined by Ly = x2(α0 + α1x + α2x2)y′′ + x(β0 + β1x + β2x2)y′ + (γ0 + γ1x + γ2x2)y and yN(x; ri) = xri N X n=0 an(ri)xn, where the coefficients {an(ri)}N n=0 are computed as in (7.5.12), Theorem 7.5.2. Compute the error EN(x; ri) = x−riLyN(x; ri)/α0 (7.5.27) for various values of N and various values of x in the interval (0, ρ), with ρ as defined in Theorem 7.5.2. The multiplier x−ri/α0 on the right of (7.5.27) eliminates the effects of small or large values of xri near x = 0, and of multiplication by an arbitrary constant. In some exercises you will be asked to estimate the maximum value of EN(x; ri) on an interval (0, δ] by computing EN(xm; ri) at the M points xm = mδ/M, m = 1, 2, ..., M, and finding the maximum of the absolute values: σN(δ) = max{|EN(xm; ri)|, m = 1, 2, . . ., M}. (7.5.28) (For simplicity, this notation ignores the dependence of the right side of the equation on i and M.) To implement this procedure, you’ll have to write a computer program to calculate {an(ri)} from the applicable recurrence relation, and to evaluate EN(x; ri). The next exercise set contains five exercises specifically identified by L that ask you to implement the verification procedure. These particular exercises were chosen arbitrarily you can just as well formulate such laboratory problems for any of the equations in any of the Exercises 1–10, 14-25, and 28–51 7.5 Exercises This set contains exercises specifically identified by L that ask you to implement the verification pro- cedure. These particular exercises were chosen arbitrarily you can just as well formulate such laboratory problems for any of the equations in Exercises 1–10, 14-25, and 28–51. In Exercises 1–10 find a fundamental set of Frobenius solutions. Compute a0, a1 ..., aN for N at least 7 in each solution. 1. C 2x2(1 + x + x2)y′′ + x(3 + 3x + 5x2)y′ −y = 0 2. C 3x2y′′ + 2x(1 + x −2x2)y′ + (2x −8x2)y = 0 3. C x2(3 + 3x + x2)y′′ + x(5 + 8x + 7x2)y′ −(1 −2x −9x2)y = 0 4. C 4x2y′′ + x(7 + 2x + 4x2)y′ −(1 −4x −7x2)y = 0 5. C 12x2(1 + x)y′′ + x(11 + 35x + 3x2)y′ −(1 −10x −5x2)y = 0 6. C x2(5 + x + 10x2)y′′ + x(4 + 3x + 48x2)y′ + (x + 36x2)y = 0 7. C 8x2y′′ −2x(3 −4x −x2)y′ + (3 + 6x + x2)y = 0 8. C 18x2(1 + x)y′′ + 3x(5 + 11x + x2)y′ −(1 −2x −5x2)y = 0 9. C x(3 + x + x2)y′′ + (4 + x −x2)y′ + xy = 0 | Elementary Differential Equations with Boundary Value Problems_Page_368_Chunk2772 |
Section 7.5 The Method of Frobenius I 359 10. C 10x2(1 + x + 2x2)y′′ + x(13 + 13x + 66x2)y′ −(1 + 4x + 10x2)y = 0 11. L The Frobenius solutions of 2x2(1 + x + x2)y′′ + x(9 + 11x + 11x2)y′ + (6 + 10x + 7x2)y = 0 obtained in Example 7.5.1 are defined on (0, ρ), where ρ is defined in Theorem 7.5.2. Find ρ. Then do the following experiments for each Frobenius solution, with M = 20 and δ = .5ρ, .7ρ, and .9ρ in the verification procedure described at the end of this section. (a) Compute σN(δ) (see Eqn. (7.5.28)) for N = 5, 10, 15,..., 50. (b) Find N such that σN(δ) < 10−5. (c) Find N such that σN(δ) < 10−10. 12. L By Theorem 7.5.2 the Frobenius solutions of the equation in Exercise 4 are defined on (0, ∞). Do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with M = 20 and δ = 1, 2, and 3 in the verification procedure described at the end of this section. 13. L The Frobenius solutions of the equation in Exercise 6 are defined on (0, ρ), where ρ is defined in Theorem 7.5.2. Find ρ and do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with M = 20 and δ = .3ρ, .4ρ, and .5ρ, in the verification procedure described at the end of this section. In Exercises 14–25 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi- cients in each solution. 14. 2x2y′′ + x(3 + 2x)y′ −(1 −x)y = 0 15. x2(3 + x)y′′ + x(5 + 4x)y′ −(1 −2x)y = 0 16. 2x2y′′ + x(5 + x)y′ −(2 −3x)y = 0 17. 3x2y′′ + x(1 + x)y′ −y = 0 18. 2x2y′′ −xy′ + (1 −2x)y = 0 19. 9x2y′′ + 9xy′ −(1 + 3x)y = 0 20. 3x2y′′ + x(1 + x)y′ −(1 + 3x)y = 0 21. 2x2(3 + x)y′′ + x(1 + 5x)y′ + (1 + x)y = 0 22. x2(4 + x)y′′ −x(1 −3x)y′ + y = 0 23. 2x2y′′ + 5xy′ + (1 + x)y = 0 24. x2(3 + 4x)y′′ + x(5 + 18x)y′ −(1 −12x)y = 0 25. 6x2y′′ + x(10 −x)y′ −(2 + x)y = 0 26. L By Theorem 7.5.2 the Frobenius solutionsof the equation in Exercise 17 are defined on (0, ∞). Do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with M = 20 and δ = 3, 6, 9, and 12 in the verification procedure described at the end of this section. 27. L The Frobenius solutionsof the equation in Exercise 22 are defined on (0, ρ), where ρ is defined in Theorem 7.5.2. Find ρ and do experiments (a), (b), and (c) of Exercise 11 for each Frobenius solution, with M = 20 and δ = .25ρ, .5ρ, and .75ρ in the verification procedure described at the end of this section. | Elementary Differential Equations with Boundary Value Problems_Page_369_Chunk2773 |
360 Chapter 7 Series Solutions of Linear Second Order Equations In Exercises 28–32 find a fundamental set of Frobenius solutions. Compute coefficients a0, ..., aN for N at least 7 in each solution. 28. C x2(8 + x)y′′ + x(2 + 3x)y′ + (1 + x)y = 0 29. C x2(3 + 4x)y′′ + x(11 + 4x)y′ −(3 + 4x)y = 0 30. C 2x2(2 + 3x)y′′ + x(4 + 11x)y′ −(1 −x)y = 0 31. C x2(2 + x)y′′ + 5x(1 −x)y′ −(2 −8x)y 32. C x2(6 + x)y′′ + x(11 + 4x)y′ + (1 + 2x)y = 0 In Exercises 33–46 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi- cients in each solution. 33. 8x2y′′ + x(2 + x2)y′ + y = 0 34. 8x2(1 −x2)y′′ + 2x(1 −13x2)y′ + (1 −9x2)y = 0 35. x2(1 + x2)y′′ −2x(2 −x2)y′ + 4y = 0 36. x(3 + x2)y′′ + (2 −x2)y′ −8xy = 0 37. 4x2(1 −x2)y′′ + x(7 −19x2)y′ −(1 + 14x2)y = 0 38. 3x2(2 −x2)y′′ + x(1 −11x2)y′ + (1 −5x2)y = 0 39. 2x2(2 + x2)y′′ −x(12 −7x2)y′ + (7 + 3x2)y = 0 40. 2x2(2 + x2)y′′ + x(4 + 7x2)y′ −(1 −3x2)y = 0 41. 2x2(1 + 2x2)y′′ + 5x(1 + 6x2)y′ −(2 −40x2)y = 0 42. 3x2(1 + x2)y′′ + 5x(1 + x2)y′ −(1 + 5x2)y = 0 43. x(1 + x2)y′′ + (4 + 7x2)y′ + 8xy = 0 44. x2(2 + x2)y′′ + x(3 + x2)y′ −y = 0 45. 2x2(1 + x2)y′′ + x(3 + 8x2)y′ −(3 −4x2)y = 0 46. 9x2y′′ + 3x(3 + x2)y′ −(1 −5x2)y = 0 In Exercises 47–51 find a fundamental set of Frobenius solutions. Compute the coefficients a0, ..., a2M for M at least 7 in each solution. 47. C 6x2y′′ + x(1 + 6x2)y′ + (1 + 9x2)y = 0 48. C x2(8 + x2)y′′ + 7x(2 + x2)y′ −(2 −9x2)y = 0 49. C 9x2(1 + x2)y′′ + 3x(3 + 13x2)y′ −(1 −25x2)y = 0 50. C 4x2(1 + x2)y′′ + 4x(1 + 6x2)y′ −(1 −25x2)y = 0 51. C 8x2(1 + 2x2)y′′ + 2x(5 + 34x2)y′ −(1 −30x2)y = 0 52. Suppose r1 > r2, a0 = b0 = 1, and the Frobenius series y1 = xr1 ∞ X n=0 anxn and y2 = xr2 ∞ X n=0 bnxn both converge on an interval (0, ρ). | Elementary Differential Equations with Boundary Value Problems_Page_370_Chunk2774 |
Section 7.5 The Method of Frobenius I 361 (a) Show that y1 and y2 are linearly independent on (0, ρ). HINT: Show that if c1 and c2 are constants such that c1y1 + c2y2 ≡0 on (0, ρ), then c1xr1−r2 ∞ X n=0 anxn + c2 ∞ X n=0 bnxn = 0, 0 < x < ρ. Then let x →0+ to conclude that c2 = 0. (b) Use the result of (b) to complete the proof of Theorem 7.5.3. 53. The equation x2y′′ + xy′ + (x2 −ν2)y = 0 (7.5.1) is Bessel’s equation of order ν. (Here ν is a parameter, and this use of “order” should not be con- fused with its usual use as in “the order of the equation.”) The solutionsof (7.5.1) are Bessel functions of order ν. (a) Assuming that ν isn’t an integer, find a fundamental set of Frobenius solutions of (7.5.1). (b) If ν = 1/2, the solutions of (7.5.1) reduce to familiar elementary functions. Identify these functions. 54. (a) Verify that d dx (|x|rxn) = (n + r)|x|rxn−1 and d2 dx2 (|x|rxn) = (n + r)(n + r −1)|x|rxn−2 if x ̸= 0. (b) Let Ly = x2(α0 + α1x + α2x2)y′′ + x(β0 + β1x + β2x2)y′ + (γ0 + γ1x + γ2x2)y = 0. Show that if xr P∞ n=0 anxn is a solution of Ly = 0 on (0, ρ) then |x|r P∞ n=0 anxn is a solution on (−ρ, 0) and (0, ρ). 55. (a) Deduce from Eqn. (7.5.20) that an(r) = (−1)n n Y j=1 p1(j + r −1) p0(j + r) . (b) Conclude that if p0(r) = α0(r −r1)(r −r2) where r1 −r2 is not an integer, then y1 = xr1 ∞ X n=0 an(r1)xn and y2 = xr2 ∞ X n=0 an(r2)xn form a fundamental set of Frobenius solutions of x2(α0 + α1x)y′′ + x(β0 + β1x)y′ + (γ0 + γ1x)y = 0. (c) Show that if p0 satisfies the hypotheses of (b) then y1 = xr1 ∞ X n=0 (−1)n n! Qn j=1(j + r1 −r2) γ1 α0 n xn and y2 = xr2 ∞ X n=0 (−1)n n! Qn j=1(j + r2 −r1) γ1 α0 n xn form a fundamental set of Frobenius solutions of α0x2y′′ + β0xy′ + (γ0 + γ1x)y = 0. | Elementary Differential Equations with Boundary Value Problems_Page_371_Chunk2775 |
362 Chapter 7 Series Solutions of Linear Second Order Equations 56. Let Ly = x2(α0 + α2x2)y′′ + x(β0 + β2x2)y′ + (γ0 + γ2x2)y = 0 and define p0(r) = α0r(r −1) + β0r + γ0 and p2(r) = α2r(r −1) + β2r + γ2. (a) Use Theorem 7.5.2 to show that if a0(r) = 1, p0(2m + r)a2m(r) + p2(2m + r −2)a2m−2(r) = 0, m ≥1, (7.5.1) then the Frobenius series y(x, r) = xr P∞ m=0 a2mx2m satisfies Ly(x, r) = p0(r)xr. (b) Deduce from (7.5.1) that if p0(2m + r) is nonzero for every positive integer m then a2m(r) = (−1)m m Y j=1 p2(2j + r −2) p0(2j + r) . (c) Conclude that if p0(r) = α0(r −r1)(r −r2) where r1 −r2 is not an even integer, then y1 = xr1 ∞ X m=0 a2m(r1)x2m and y2 = xr2 ∞ X m=0 a2m(r2)x2m form a fundamental set of Frobenius solutions of Ly = 0. (d) Show that if p0 satisfies the hypotheses of (c) then y1 = xr1 ∞ X m=0 (−1)m 2mm! Qm j=1(2j + r1 −r2) γ2 α0 m x2m and y2 = xr2 ∞ X m=0 (−1)m 2mm! Qm j=1(2j + r2 −r1) γ2 α0 m x2m form a fundamental set of Frobenius solutions of α0x2y′′ + β0xy′ + (γ0 + γ2x2)y = 0. 57. Let Ly = x2q0(x)y′′ + xq1(x)y′ + q2(x)y, where q0(x) = ∞ X j=0 αjxj, q1(x) = ∞ X j=0 βjxj, q2(x) = ∞ X j=0 γjxj, and define pj(r) = αjr(r −1) + βjr + γj, j = 0, 1, . . .. Let y = xr P∞ n=0 anxn. Show that Ly = xr ∞ X n=0 bnxn, where bn = n X j=0 pj(n + r −j)an−j. | Elementary Differential Equations with Boundary Value Problems_Page_372_Chunk2776 |
Section 7.5 The Method of Frobenius I 363 58. (a) Let L be as in Exercise 57. Show that if y(x, r) = xr ∞ X n=0 an(r)xn where a0(r) = 1, an(r) = − 1 p0(n + r) n X j=1 pj(n + r −j)an−j(r), n ≥1, then Ly(x, r) = p0(r)xr. (b) Conclude that if p0(r) = α0(r −r1)(r −r2) where r1 −r2 isn’t an integer then y1 = y(x, r1) and y2 = y(x, r2) are solutions of Ly = 0. 59. Let Ly = x2(α0 + αqxq)y′′ + x(β0 + βqxq)y′ + (γ0 + γqxq)y where q is a positive integer, and define p0(r) = α0r(r −1) + β0r + γ0 and pq(r) = αqr(r −1) + βqr + γq. (a) Show that if y(x, r) = xr ∞ X m=0 aqm(r)xqm where a0(r) = 1, aqm(r) = −pq (q(m −1) + r) p0(qm + r) aq(m−1)(r), m ≥1, (7.5.1) then Ly(x, r) = p0(r)xr. (b) Deduce from (7.5.1) that aqm(r) = (−1)m m Y j=1 pq (q(j −1) + r) p0(qj + r) . (c) Conclude that if p0(r) = α0(r −r1)(r −r2) where r1 −r2 is not an integer multiple of q, then y1 = xr1 ∞ X m=0 aqm(r1)xqm and y2 = xr2 ∞ X m=0 aqm(r2)xqm form a fundamental set of Frobenius solutions of Ly = 0. | Elementary Differential Equations with Boundary Value Problems_Page_373_Chunk2777 |
364 Chapter 7 Series Solutions of Linear Second Order Equations (d) Show that if p0 satisfies the hypotheses of (c) then y1 = xr1 ∞ X m=0 (−1)m qmm! Qm j=1(qj + r1 −r2) γq α0 m xqm and y2 = xr2 ∞ X m=0 (−1)m qmm! Qm j=1(qj + r2 −r1) γq α0 m xqm form a fundamental set of Frobenius solutions of α0x2y′′ + β0xy′ + (γ0 + γqxq)y = 0. 60. (a) Suppose α0, α1, and α2 are real numbers with α0 ̸= 0, and {an}∞ n=0 is defined by α0a1 + α1a0 = 0 and α0an + α1an−1 + α2an−2 = 0, n ≥2. Show that (α0 + α1x + α2x2) ∞ X n=0 anxn = α0a0, and infer that ∞ X n=0 anxn = α0a0 α0 + α1x + α2x2 . (b) With α0, α1, and α2 as in (a), consider the equation x2(α0 + α1x + α2x2)y′′ + x(β0 + β1x + β2x2)y′ + (γ0 + γ1x + γ2x2)y = 0, (7.5.1) and define pj(r) = αjr(r −1) + βjr + γj, j = 0, 1, 2. Suppose p1(r −1) p0(r) = α1 α0 , p2(r −2) p0(r) = α2 α0 , and p0(r) = α0(r −r1)(r −r2), where r1 > r2. Show that y1 = xr1 α0 + α1x + α2x2 and y2 = xr2 α0 + α1x + α2x2 form a fundamental set of Frobenius solutions of (7.5.1) on any interval (0, ρ) on which α0 + α1x + α2x2 has no zeros. In Exercises 61–68 use the method suggested by Exercise 60 to find the general solution on some interval (0, ρ). 61. 2x2(1 + x)y′′ −x(1 −3x)y′ + y = 0 | Elementary Differential Equations with Boundary Value Problems_Page_374_Chunk2778 |
Section 7.6 The Method of Frobenius II 365 62. 6x2(1 + 2x2)y′′ + x(1 + 50x2)y′ + (1 + 30x2)y = 0 63. 28x2(1 −3x)y′′ −7x(5 + 9x)y′ + 7(2 + 9x)y = 0 64. 9x2(5 + x)y′′ + 9x(5 + 3x)y′ −(5 −8x)y = 0 65. 8x2(2 −x2)y′′ + 2x(10 −21x2)y′ −(2 + 35x2)y = 0 66. 4x2(1 + 3x + x2)y′′ −4x(1 −3x −3x2)y′ + 3(1 −x + x2)y = 0 67. 3x2(1 + x)2y′′ −x(1 −10x −11x2)y′ + (1 + 5x2)y = 0 68. 4x2(3 + 2x + x2)y′′ −x(3 −14x −15x2)y′ + (3 + 7x2)y = 0 7.6 THE METHOD OF FROBENIUS II In this section we discuss a method for finding two linearly independent Frobenius solutions of a homo- geneous linear second order equation near a regular singular point in the case where the indicial equation has a repeated real root. As in the preceding section, we consider equations that can be written as x2(α0 + α1x + α2x2)y′′ + x(β0 + β1x + β2x2)y′ + (γ0 + γ1x + γ2x2)y = 0 (7.6.1) where α0 ̸= 0. We assume that the indicial equation p0(r) = 0 has a repeated real root r1. In this case Theorem 7.5.3 implies that (7.6.1) has one solution of the form y1 = xr1 ∞ X n=0 anxn, but does not provide a second solution y2 such that {y1, y2} is a fundamental set of solutions. The following extension of Theorem 7.5.2 provides a way to find a second solution. Theorem 7.6.1 Let Ly = x2(α0 + α1x + α2x2)y′′ + x(β0 + β1x + β2x2)y′ + (γ0 + γ1x + γ2x2)y, (7.6.2) where α0 ̸= 0 and define p0(r) = α0r(r −1) + β0r + γ0, p1(r) = α1r(r −1) + β1r + γ1, p2(r) = α2r(r −1) + β2r + γ2. Suppose r is a real number such that p0(n + r) is nonzero for all positive integers n, and define a0(r) = 1, a1(r) = − p1(r) p0(r + 1), an(r) = −p1(n + r −1)an−1(r) + p2(n + r −2)an−2(r) p0(n + r) , n ≥2. Then the Frobenius series y(x, r) = xr ∞ X n=0 an(r)xn (7.6.3) | Elementary Differential Equations with Boundary Value Problems_Page_375_Chunk2779 |
366 Chapter 7 Series Solutions of Linear Second Order Equations satisfies Ly(x, r) = p0(r)xr. (7.6.4) Moreover, ∂y ∂r (x, r) = y(x, r) ln x + xr ∞ X n=1 a′ n(r)xn, (7.6.5) and L ∂y ∂r (x, r) = p′ 0(r)xr + xrp0(r) lnx. (7.6.6) Proof Theorem 7.5.2 implies (7.6.4). Differentiating formally with respect to r in (7.6.3) yields ∂y ∂r (x, r) = ∂ ∂r (xr) ∞ X n=0 an(r)xn + xr ∞ X n=1 a′ n(r)xn = xr ln x ∞ X n=0 an(r)xn + xr ∞ X n=1 a′ n(r)xn = y(x, r) ln x + xr ∞ X n=1 a′ n(r)xn, which proves (7.6.5). To prove that ∂y(x, r)/∂r satisfies (7.6.6), we view y in (7.6.2) as a function y = y(x, r) of two variables, where the prime indicates partial differentiation with respect to x; thus, y′ = y′(x, r) = ∂y ∂x(x, r) and y′′ = y′′(x, r) = ∂2y ∂x2 (x, r). With this notation we can use (7.6.2) to rewrite (7.6.4) as x2q0(x)∂2y ∂x2 (x, r) + xq1(x)∂y ∂x(x, r) + q2(x)y(x, r) = p0(r)xr, (7.6.7) where q0(x) = α0 + α1x + α2x2, q1(x) = β0 + β1x + β2x2, q2(x) = γ0 + γ1x + γ2x2. Differentiating both sides of (7.6.7) with respect to r yields x2q0(x) ∂3y ∂r∂x2 (x, r) + xq1(x) ∂2y ∂r∂x (x, r) + q2(x)∂y ∂r (x, r) = p′ 0(r)xr + p0(r)xr ln x. By changing the order of differentiation in the first two terms on the left we can rewrite this as x2q0(x) ∂3y ∂x2∂r (x, r) + xq1(x) ∂2y ∂x∂r (x, r) + q2(x)∂y ∂r (x, r) = p′ 0(r)xr + p0(r)xr ln x, | Elementary Differential Equations with Boundary Value Problems_Page_376_Chunk2780 |
Section 7.6 The Method of Frobenius II 367 or x2q0(x) ∂2 ∂x2 ∂y ∂r (x, r) + xq1(x) ∂ ∂r ∂y ∂x(x, r) + q2(x)∂y ∂r (x, r) = p′ 0(r)xr + p0(r)xr ln x, which is equivalent to (7.6.6). Theorem 7.6.2 Let L be as in Theorem 7.6.1 and suppose the indicial equation p0(r) = 0 has a repeated real root r1. Then y1(x) = y(x, r1) = xr1 ∞ X n=0 an(r1)xn and y2(x) = ∂y ∂r (x, r1) = y1(x) ln x + xr1 ∞ X n=1 a′ n(r1)xn (7.6.8) form a fundamental set of solutions of Ly = 0. Proof Since r1 is a repeated root of p0(r) = 0, the indicial polynomial can be factored as p0(r) = α0(r −r1)2, so p0(n + r1) = α0n2, which is nonzero if n > 0. Therefore the assumptions of Theorem 7.6.1 hold with r = r1, and (7.6.4) implies that Ly1 = p0(r1)xr1 = 0. Since p′ 0(r) = 2α(r −r1) it follows that p′ 0(r1) = 0, so (7.6.6) implies that Ly2 = p′ 0(r1)xr1 + xr1p0(r1) lnx = 0. This proves that y1 and y2 are both solutions of Ly = 0. We leave the proof that {y1, y2} is a fundamental set as an exercise (Exercise 53). Example 7.6.1 Find a fundamental set of solutions of x2(1 −2x + x2)y′′ −x(3 + x)y′ + (4 + x)y = 0. (7.6.9) Compute just the terms involving xn+r1, where 0 ≤n ≤4 and r1 is the root of the indicial equation. Solution For the given equation, the polynomials defined in Theorem 7.6.1 are p0(r) = r(r −1) −3r + 4 = (r −2)2, p1(r) = −2r(r −1) −r + 1 = −(r −1)(2r + 1), p2(r) = r(r −1). Since r1 = 2 is a repeated root of the indicial polynomial p0, Theorem 7.6.2 implies that y1 = x2 ∞ X n=0 an(2)xn and y2 = y1 lnx + x2 ∞ X n=1 a′ n(2)xn (7.6.10) | Elementary Differential Equations with Boundary Value Problems_Page_377_Chunk2781 |
368 Chapter 7 Series Solutions of Linear Second Order Equations form a fundamental set of Frobenius solutions of (7.6.9). To find the coefficients in these series, we use the recurrence formulas from Theorem 7.6.1: a0(r) = 1, a1(r) = − p1(r) p0(r + 1) = −(r −1)(2r + 1) (r −1)2 = 2r + 1 r −1 , an(r) = −p1(n + r −1)an−1(r) + p2(n + r −2)an−2(r) p0(n + r) = (n + r −2) [(2n + 2r −1)an−1(r) −(n + r −3)an−2(r)] (n + r −2)2 = (2n + 2r −1) (n + r −2) an−1(r) −(n + r −3) (n + r −2)an−2(r), n ≥2. (7.6.11) Differentiating yields a′ 1(r) = − 3 (r −1)2 , a′ n(r) = 2n + 2r −1 n + r −2 a′ n−1(r) −n + r −3 n + r −2a′ n−2(r) − 3 (n + r −2)2 an−1(r) − 1 (n + r −2)2 an−2(r), n ≥2. (7.6.12) Setting r = 2 in (7.6.11) and (7.6.12) yields a0(2) = 1, a1(2) = 5, an(2) = (2n + 3) n an−1(2) −(n −1) n an−2(2), n ≥2 (7.6.13) and a′ 1(2) = −3, a′ n(2) = 2n + 3 n a′ n−1(2) −n −1 n a′ n−2(2) −3 n2 an−1(2) −1 n2 an−2(2), n ≥2. (7.6.14) Computing recursively with (7.6.13) and (7.6.14) yields a0(2) = 1, a1(2) = 5, a2(2) = 17, a3(2) = 143 3 , a4(2) = 355 3 , and a′ 1(2) = −3, a′ 2(2) = −29 2 , a′ 3(2) = −859 18 , a′ 4(2) = −4693 36 . Substituting these coefficients into (7.6.10) yields y1 = x2 1 + 5x + 17x2 + 143 3 x3 + 355 3 x4 + · · · and y2 = y1 ln x −x3 3 + 29 2 x + 859 18 x2 + 4693 36 x3 + · · · . | Elementary Differential Equations with Boundary Value Problems_Page_378_Chunk2782 |
Section 7.6 The Method of Frobenius II 369 Since the recurrence formula (7.6.11) involves three terms, it’s not possible to obtain a simple explicit formula for the coefficients in the Frobenius solutions of (7.6.9). However, as we saw in the preceding sections, the recurrrence formula for {an(r)} involves only two terms if either α1 = β1 = γ1 = 0 or α2 = β2 = γ2 = 0 in (7.6.1). In this case, it’s often possible to find explicit formulas for the coefficients. The next two examples illustrate this. Example 7.6.2 Find a fundamental set of Frobenius solutions of 2x2(2 + x)y′′ + 5x2y′ + (1 + x)y = 0. (7.6.15) Give explicit formulas for the coefficients in the solutions. Solution For the given equation, the polynomials defined in Theorem 7.6.1 are p0(r) = 4r(r −1) + 1 = (2r −1)2, p1(r) = 2r(r −1) + 5r + 1 = (r + 1)(2r + 1), p2(r) = 0. Since r1 = 1/2 is a repeated zero of the indicial polynomial p0, Theorem 7.6.2 implies that y1 = x1/2 ∞ X n=0 an(1/2)xn (7.6.16) and y2 = y1 lnx + x1/2 ∞ X n=1 a′ n(1/2)xn (7.6.17) form a fundamental set of Frobenius solutions of (7.6.15). Since p2 ≡0, the recurrence formulas in Theorem 7.6.1 reduce to a0(r) = 1, an(r) = −p1(n + r −1) p0(n + r) an−1(r), = −(n + r)(2n + 2r −1) (2n + 2r −1)2 an−1(r), = − n + r 2n + 2r −1an−1(r), n ≥0. We leave it to you to show that an(r) = (−1)n n Y j=1 j + r 2j + 2r −1, n ≥0. (7.6.18) Setting r = 1/2 yields an(1/2) = (−1)n n Y j=1 j + 1/2 2j = (−1)n n Y j=1 2j + 1 4j , = (−1)n Qn j=1(2j + 1) 4nn! , n ≥0. (7.6.19) | Elementary Differential Equations with Boundary Value Problems_Page_379_Chunk2783 |
370 Chapter 7 Series Solutions of Linear Second Order Equations Substituting this into (7.6.16) yields y1 = x1/2 ∞ X n=0 (−1)n Qn j=1(2j + 1) 4nn! xn. To obtain y2 in (7.6.17), we must compute a′ n(1/2) for n = 1, 2,.... We’ll do this by logarithmic differentiation. From (7.6.18), |an(r)| = n Y j=1 |j + r| |2j + 2r −1|, n ≥1. Therefore ln|an(r)| = n X j=1 (ln |j + r| −ln|2j + 2r −1|) . Differentiating with respect to r yields a′ n(r) an(r) = n X j=1 1 j + r − 2 2j + 2r −1 . Therefore a′ n(r) = an(r) n X j=1 1 j + r − 2 2j + 2r −1 . Setting r = 1/2 here and recalling (7.6.19) yields a′ n(1/2) = (−1)n Qn j=1(2j + 1) 4nn! n X j=1 1 j + 1/2 − n X j=1 1 j . (7.6.20) Since 1 j + 1/2 −1 j = j −j −1/2 j(j + 1/2) = − 1 j(2j + 1), (7.6.20) can be rewritten as a′ n(1/2) = − (−1)n Qn j=1(2j + 1) 4nn! n X j=1 1 j(2j + 1). Therefore, from (7.6.17), y2 = y1 ln x −x1/2 ∞ X n=1 (−1)n Qn j=1(2j + 1) 4nn! n X j=1 1 j(2j + 1) xn. Example 7.6.3 Find a fundamental set of Frobenius solutions of x2(2 −x2)y′′ −2x(1 + 2x2)y′ + (2 −2x2)y = 0. (7.6.21) Give explicit formulas for the coefficients in the solutions. | Elementary Differential Equations with Boundary Value Problems_Page_380_Chunk2784 |
Section 7.6 The Method of Frobenius II 371 Solution For (7.6.21), the polynomials defined in Theorem 7.6.1 are p0(r) = 2r(r −1) −2r + 2 = 2(r −1)2, p1(r) = 0, p2(r) = −r(r −1) −4r −2 = −(r + 1)(r + 2). As in Section 7.5, since p1 ≡0, the recurrence formulas of Theorem 7.6.1 imply that an(r) = 0 if n is odd, and a0(r) = 1, a2m(r) = −p2(2m + r −2) p0(2m + r) a2m−2(r) = (2m + r −1)(2m + r) 2(2m + r −1)2 a2m−2(r) = 2m + r 2(2m + r −1)a2m−2(r), m ≥1. Since r1 = 1 is a repeated root of the indicial polynomial p0, Theorem 7.6.2 implies that y1 = x ∞ X m=0 a2m(1)x2m (7.6.22) and y2 = y1 lnx + x ∞ X m=1 a′ 2m(1)x2m (7.6.23) form a fundamental set of Frobenius solutions of (7.6.21). We leave it to you to show that a2m(r) = 1 2m m Y j=1 2j + r 2j + r −1. (7.6.24) Setting r = 1 yields a2m(1) = 1 2m m Y j=1 2j + 1 2j = Qm j=1(2j + 1) 4mm! , (7.6.25) and substituting this into (7.6.22) yields y1 = x ∞ X m=0 Qm j=1(2j + 1) 4mm! x2m. To obtain y2 in (7.6.23), we must compute a′ 2m(1) for m = 1, 2, .... Again we use logarithmic differentiation. From (7.6.24), |a2m(r)| = 1 2m m Y j=1 |2j + r| |2j + r −1|. Taking logarithms yields ln|a2m(r)| = −m ln 2 + m X j=1 (ln|2j + r| −ln|2j + r −1|) . | Elementary Differential Equations with Boundary Value Problems_Page_381_Chunk2785 |
372 Chapter 7 Series Solutions of Linear Second Order Equations Differentiating with respect to r yields a′ 2m(r) a2m(r) = m X j=1 1 2j + r − 1 2j + r −1 . Therefore a′ 2m(r) = a2m(r) m X j=1 1 2j + r − 1 2j + r −1 . Setting r = 1 and recalling (7.6.25) yields a′ 2m(1) = Qm j=1(2j + 1) 4mm! m X j=1 1 2j + 1 −1 2j . (7.6.26) Since 1 2j + 1 −1 2j = − 1 2j(2j + 1), (7.6.26) can be rewritten as a′ 2m(1) = − Qm j=1(2j + 1) 2 · 4mm! m X j=1 1 j(2j + 1). Substituting this into (7.6.23) yields y2 = y1 lnx −x 2 ∞ X m=1 Qm j=1(2j + 1) 4mm! m X j=1 1 j(2j + 1) x2m. If the solution y1 = y(x, r1) of Ly = 0 reduces to a finite sum, then there’s a difficulty in using logarithmic differentiation to obtain the coefficients {a′ n(r1)} in the second solution. The next example illustrates this difficulty and shows how to overcome it. Example 7.6.4 Find a fundamental set of Frobenius solutions of x2y′′ −x(5 −x)y′ + (9 −4x)y = 0. (7.6.27) Give explicit formulas for the coefficients in the solutions. Solution For (7.6.27) the polynomials defined in Theorem 7.6.1 are p0(r) = r(r −1) −5r + 9 = (r −3)2, p1(r) = r −4, p2(r) = 0. Since r1 = 3 is a repeated zero of the indicial polynomial p0, Theorem 7.6.2 implies that y1 = x3 ∞ X n=0 an(3)xn (7.6.28) | Elementary Differential Equations with Boundary Value Problems_Page_382_Chunk2786 |
Section 7.6 The Method of Frobenius II 373 and y2 = y1 ln x + x3 ∞ X n=1 a′ n(3)xn (7.6.29) are linearly independent Frobenius solutions of (7.6.27). To find the coefficients in (7.6.28) we use the recurrence formulas a0(r) = 1, an(r) = −p1(n + r −1) p0(n + r) an−1(r) = −n + r −5 (n + r −3)2 an−1(r), n ≥1. We leave it to you to show that an(r) = (−1)n n Y j=1 j + r −5 (j + r −3)2 . (7.6.30) Setting r = 3 here yields an(3) = (−1)n n Y j=1 j −2 j2 , so a1(3) = 1 and an(3) = 0 if n ≥2. Substituting these coefficients into (7.6.28) yields y1 = x3(1 + x). To obtain y2 in (7.6.29) we must compute a′ n(3) for n = 1, 2, .... Let’s first try logarithmic differenti- ation. From (7.6.30), |an(r)| = n Y j=1 |j + r −5| |j + r −3|2 , n ≥1, so ln|an(r)| = n X j=1 (ln|j + r −5| −2 ln|j + r −3|). Differentiating with respect to r yields a′ n(r) an(r) = n X j=1 1 j + r −5 − 2 j + r −3 . Therefore a′ n(r) = an(r) n X j=1 1 j + r −5 − 2 j + r −3 . (7.6.31) However, we can’t simply set r = 3 here if n ≥2, since the bracketed expression in the sum correspond- ing to j = 2 contains the term 1/(r −3). In fact, since an(3) = 0 for n ≥2, the formula (7.6.31) for a′ n(r) is actually an indeterminate form at r = 3. We overcome this difficulty as follows. From (7.6.30) with n = 1, a1(r) = −r −4 (r −2)2 . | Elementary Differential Equations with Boundary Value Problems_Page_383_Chunk2787 |
374 Chapter 7 Series Solutions of Linear Second Order Equations Therefore a′ 1(r) = r −6 (r −2)3 , so a′ 1(3) = −3. (7.6.32) From (7.6.30) with n ≥2, an(r) = (−1)n(r −4)(r −3) Qn j=3(j + r −5) Qn j=1(j + r −3)2 = (r −3)cn(r), where cn(r) = (−1)n(r −4) Qn j=3(j + r −5) Qn j=1(j + r −3)2 , n ≥2. Therefore a′ n(r) = cn(r) + (r −3)c′ n(r), n ≥2, which implies that a′ n(3) = cn(3) if n ≥3. We leave it to you to verify that a′ n(3) = cn(3) = (−1)n+1 n(n −1)n!, n ≥2. Substituting this and (7.6.32) into (7.6.29) yields y2 = x3(1 + x) ln x −3x4 −x3 ∞ X n=2 (−1)n n(n −1)n!xn. 7.6 Exercises In Exercises 1–11 find a fundamental set of Frobenius solutions. Compute the terms involving xn+r1, where 0 ≤n ≤N (N at least 7) and r1 is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take N > 7. 1. C x2y′′ −x(1 −x)y′ + (1 −x2)y = 0 2. C x2(1 + x + 2x2)y′ + x(3 + 6x + 7x2)y′ + (1 + 6x −3x2)y = 0 3. C x2(1 + 2x + x2)y′′ + x(1 + 3x + 4x2)y′ −x(1 −2x)y = 0 4. C 4x2(1 + x + x2)y′′ + 12x2(1 + x)y′ + (1 + 3x + 3x2)y = 0 5. C x2(1 + x + x2)y′′ −x(1 −4x −2x2)y′ + y = 0 6. C 9x2y′′ + 3x(5 + 3x −2x2)y′ + (1 + 12x −14x2)y = 0 7. C x2y′′ + x(1 + x + x2)y′ + x(2 −x)y = 0 8. C x2(1 + 2x)y′′ + x(5 + 14x + 3x2)y′ + (4 + 18x + 12x2)y = 0 9. C 4x2y′′ + 2x(4 + x + x2)y′ + (1 + 5x + 3x2)y = 0 10. C 16x2y′′ + 4x(6 + x + 2x2)y′ + (1 + 5x + 18x2)y = 0 11. C 9x2(1 + x)y′′ + 3x(5 + 11x −x2)y′ + (1 + 16x −7x2)y = 0 | Elementary Differential Equations with Boundary Value Problems_Page_384_Chunk2788 |
Section 7.6 The Method of Frobenius II 375 In Exercises 12–22 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi- cients. 12. 4x2y′′ + (1 + 4x)y = 0 13. 36x2(1 −2x)y′′ + 24x(1 −9x)y′ + (1 −70x)y = 0 14. x2(1 + x)y′′ −x(3 −x)y′ + 4y = 0 15. x2(1 −2x)y′′ −x(5 −4x)y′ + (9 −4x)y = 0 16. 25x2y′′ + x(15 + x)y′ + (1 + x)y = 0 17. 2x2(2 + x)y′′ + x2y′ + (1 −x)y = 0 18. x2(9 + 4x)y′′ + 3xy′ + (1 + x)y = 0 19. x2y′′ −x(3 −2x)y′ + (4 + 3x)y = 0 20. x2(1 −4x)y′′ + 3x(1 −6x)y′ + (1 −12x)y = 0 21. x2(1 + 2x)y′′ + x(3 + 5x)y′ + (1 −2x)y = 0 22. 2x2(1 + x)y′′ −x(6 −x)y′ + (8 −x)y = 0 In Exercises 23–27 find a fundamental set of Frobenius solutions. Compute the terms involving xn+r1, where 0 ≤n ≤N (N at least 7) and r1 is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take N > 7. 23. C x2(1 + 2x)y′′ + x(5 + 9x)y′ + (4 + 3x)y = 0 24. C x2(1 −2x)y′′ −x(5 + 4x)y′ + (9 + 4x)y = 0 25. C x2(1 + 4x)y′′ −x(1 −4x)y′ + (1 + x)y = 0 26. C x2(1 + x)y′′ + x(1 + 2x)y′ + xy = 0 27. C x2(1 −x)y′′ + x(7 + x)y′ + (9 −x)y = 0 In Exercises 28–38 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi- cients. 28. x2y′′ −x(1 −x2)y′ + (1 + x2)y = 0 29. x2(1 + x2)y′′ −3x(1 −x2)y′ + 4y = 0 30. 4x2y′′ + 2x3y′ + (1 + 3x2)y = 0 31. x2(1 + x2)y′′ −x(1 −2x2)y′ + y = 0 32. 2x2(2 + x2)y′′ + 7x3y′ + (1 + 3x2)y = 0 33. x2(1 + x2)y′′ −x(1 −4x2)y′ + (1 + 2x2)y = 0 34. 4x2(4 + x2)y′′ + 3x(8 + 3x2)y′ + (1 −9x2)y = 0 35. 3x2(3 + x2)y′′ + x(3 + 11x2)y′ + (1 + 5x2)y = 0 36. 4x2(1 + 4x2)y′′ + 32x3y′ + y = 0 37. 9x2y′′ −3x(7 −2x2)y′ + (25 + 2x2)y = 0 38. x2(1 + 2x2)y′′ + x(3 + 7x2)y′ + (1 −3x2)y = 0 | Elementary Differential Equations with Boundary Value Problems_Page_385_Chunk2789 |
376 Chapter 7 Series Solutions of Linear Second Order Equations In Exercises 39–43 find a fundamental set of Frobenius solutions. Compute the terms involving x2m+r1, where 0 ≤m ≤M (M at least 3) and r1 is the root of the indicial equation. Optionally,write a computer program to implement the applicable recurrence formulas and take M > 3. 39. C x2(1 + x2)y′′ + x(3 + 8x2)y′ + (1 + 12x2)y 40. C x2y′′ −x(1 −x2)y′ + (1 + x2)y = 0 41. C x2(1 −2x2)y′′ + x(5 −9x2)y′ + (4 −3x2)y = 0 42. C x2(2 + x2)y′′ + x(14 −x2)y′ + 2(9 + x2)y = 0 43. C x2(1 + x2)y′′ + x(3 + 7x2)y′ + (1 + 8x2)y = 0 In Exercises 44–52 find a fundamental set of Frobenius solutions. Give explicit formulas for the coeffi- cients. 44. x2(1 −2x)y′′ + 3xy′ + (1 + 4x)y = 0 45. x(1 + x)y′′ + (1 −x)y′ + y = 0 46. x2(1 −x)y′′ + x(3 −2x)y′ + (1 + 2x)y = 0 47. 4x2(1 + x)y′′ −4x2y′ + (1 −5x)y = 0 48. x2(1 −x)y′′ −x(3 −5x)y′ + (4 −5x)y = 0 49. x2(1 + x2)y′′ −x(1 + 9x2)y′ + (1 + 25x2)y = 0 50. 9x2y′′ + 3x(1 −x2)y′ + (1 + 7x2)y = 0 51. x(1 + x2)y′′ + (1 −x2)y′ −8xy = 0 52. 4x2y′′ + 2x(4 −x2)y′ + (1 + 7x2)y = 0 53. Under the assumptions of Theorem 7.6.2, suppose the power series ∞ X n=0 an(r1)xn and ∞ X n=1 a′ n(r1)xn converge on (−ρ, ρ). (a) Show that y1 = xr1 ∞ X n=0 an(r1)xn and y2 = y1 ln x + xr1 ∞ X n=1 a′ n(r1)xn are linearly independent on (0, ρ). HINT: Show that if c1 and c2 are constants such that c1y1 + c2y2 ≡0 on (0, ρ), then (c1 + c2 ln x) ∞ X n=0 an(r1)xn + c2 ∞ X n=1 a′ n(r1)xn = 0, 0 < x < ρ. Then let x →0+ to conclude that c2 = 0. (b) Use the result of (a) to complete the proof of Theorem 7.6.2. 54. Let Ly = x2(α0 + α1x)y′′ + x(β0 + β1x)y′ + (γ0 + γ1x)y | Elementary Differential Equations with Boundary Value Problems_Page_386_Chunk2790 |
Section 7.6 The Method of Frobenius II 377 and define p0(r) = α0r(r −1) + β0r + γ0 and p1(r) = α1r(r −1) + β1r + γ1. Theorem 7.6.1 and Exercise 7.5.55(a) imply that if y(x, r) = xr ∞ X n=0 an(r)xn where an(r) = (−1)n n Y j=1 p1(j + r −1) p0(j + r) , then Ly(x, r) = p0(r)xr. Now suppose p0(r) = α0(r −r1)2 and p1(k + r1) ̸= 0 if k is a nonnegative integer. (a) Show that Ly = 0 has the solution y1 = xr1 ∞ X n=0 an(r1)xn, where an(r1) = (−1)n αn 0(n!)2 n Y j=1 p1(j + r1 −1). (b) Show that Ly = 0 has the second solution y2 = y1 lnx + xr1 ∞ X n=1 an(r1)Jnxn, where Jn = n X j=1 p′ 1(j + r1 −1) p1(j + r1 −1) −2 n X j=1 1 j . (c) Conclude from (a) and (b) that if γ1 ̸= 0 then y1 = xr1 ∞ X n=0 (−1)n (n!)2 γ1 α0 n xn and y2 = y1 ln x −2xr1 ∞ X n=1 (−1)n (n!)2 γ1 α0 n n X j=1 1 j xn are solutions of α0x2y′′ + β0xy′ + (γ0 + γ1x)y = 0. (The conclusion is also valid if γ1 = 0. Why?) | Elementary Differential Equations with Boundary Value Problems_Page_387_Chunk2791 |
378 Chapter 7 Series Solutions of Linear Second Order Equations 55. Let Ly = x2(α0 + αqxq)y′′ + x(β0 + βqxq)y′ + (γ0 + γqxq)y where q is a positive integer, and define p0(r) = α0r(r −1) + β0r + γ0 and pq(r) = αqr(r −1) + βqr + γq. Suppose p0(r) = α0(r −r1)2 and pq(r) ̸≡0. (a) Recall from Exercise 7.5.59 that Ly = 0 has the solution y1 = xr1 ∞ X m=0 aqm(r1)xqm, where aqm(r1) = (−1)m (q2α0)m(m!)2 m Y j=1 pq (q(j −1) + r1) . (b) Show that Ly = 0 has the second solution y2 = y1 ln x + xr1 ∞ X m=1 a′ qm(r1)Jmxqm, where Jm = m X j=1 p′ q (q(j −1) + r1) pq (q(j −1) + r1) −2 q m X j=1 1 j . (c) Conclude from (a) and (b) that if γq ̸= 0 then y1 = xr1 ∞ X m=0 (−1)m (m!)2 γq q2α0 m xqm and y2 = y1 lnx −2 q xr1 ∞ X m=1 (−1)m (m!)2 γq q2α0 m m X j=1 1 j xqm are solutions of α0x2y′′ + β0xy′ + (γ0 + γqxq)y = 0. 56. The equation xy′′ + y′ + xy = 0 is Bessel’s equation of order zero. (See Exercise 53.) Find two linearly independent Frobenius solutions of this equation. 57. Suppose the assumptions of Exercise 7.5.53 hold, except that p0(r) = α0(r −r1)2. Show that y1 = xr1 α0 + α1x + α2x2 and y2 = xr1 lnx α0 + α1x + α2x2 are linearly independent Frobenius solutions of x2(α0 + α1x + α2x2)y′′ + x(β0 + β1x + β2x2)y′ + (γ0 + γ1x + γ2x2)y = 0 on any interval (0, ρ) on which α0 + α1x + α2x2 has no zeros. | Elementary Differential Equations with Boundary Value Problems_Page_388_Chunk2792 |
Section 7.7 The Method of Frobenius III 379 In Exercises 58–65 use the method suggested by Exercise 57 to find the general solution on some interval (0, ρ). 58. 4x2(1 + x)y′′ + 8x2y′ + (1 + x)y = 0 59. 9x2(3 + x)y′′ + 3x(3 + 7x)y′ + (3 + 4x)y = 0 60. x2(2 −x2)y′′ −x(2 + 3x2)y′ + (2 −x2)y = 0 61. 16x2(1 + x2)y′′ + 8x(1 + 9x2)y′ + (1 + 49x2)y = 0 62. x2(4 + 3x)y′′ −x(4 −3x)y′ + 4y = 0 63. 4x2(1 + 3x + x2)y′′ + 8x2(3 + 2x)y′ + (1 + 3x + 9x2)y = 0 64. x2(1 −x)2y′′ −x(1 + 2x −3x2)y′ + (1 + x2)y = 0 65. 9x2(1 + x + x2)y′′ + 3x(1 + 7x + 13x2)y′ + (1 + 4x + 25x2)y = 0 66. (a) Let L and y(x, r) be as in Exercises 57 and 58. Extend Theorem 7.6.1 by showing that L ∂y ∂r (x, r) = p′ 0(r)xr + xrp0(r) ln x. (b) Show that if p0(r) = α0(r −r1)2 then y1 = y(x, r1) and y2 = ∂y ∂r (x, r1) are solutions of Ly = 0. 7.7 THE METHOD OF FROBENIUS III In Sections 7.5 and 7.6 we discussed methods for finding Frobenius solutions of a homogeneous linear second order equation near a regular singular point in the case where the indicial equation has a repeated root or distinct real roots that don’t differ by an integer. In this section we consider the case where the indicial equation has distinct real roots that differ by an integer. We’ll limit our discussion to equations that can be written as x2(α0 + α1x)y′′ + x(β0 + β1x)y′ + (γ0 + γ1x)y = 0 (7.7.1) or x2(α0 + α2x2)y′′ + x(β0 + β2x2)y′ + (γ0 + γ2x2)y = 0, where the roots of the indicial equation differ by a positive integer. We begin with a theorem that provides a fundamental set of solutions of equations of the form (7.7.1). Theorem 7.7.1 Let Ly = x2(α0 + α1x)y′′ + x(β0 + β1x)y′ + (γ0 + γ1x)y, where α0 ̸= 0, and define p0(r) = α0r(r −1) + β0r + γ0, p1(r) = α1r(r −1) + β1r + γ1. | Elementary Differential Equations with Boundary Value Problems_Page_389_Chunk2793 |
380 Chapter 7 Series Solutions of Linear Second Order Equations Suppose r is a real number such that p0(n + r) is nonzero for all positive integers n, and define a0(r) = 1, an(r) = −p1(n + r −1) p0(n + r) an−1(r), n ≥1. (7.7.2) Let r1 and r2 be the roots of the indicial equation p0(r) = 0, and suppose r1 = r2 + k, where k is a positive integer. Then y1 = xr1 ∞ X n=0 an(r1)xn is a Frobenius solution of Ly = 0. Moreover, if we define a0(r2) = 1, an(r2) = −p1(n + r2 −1) p0(n + r2) an−1(r2), 1 ≤n ≤k −1, (7.7.3) and C = −p1(r1 −1) kα0 ak−1(r2), (7.7.4) then y2 = xr2 k−1 X n=0 an(r2)xn + C y1 lnx + xr1 ∞ X n=1 a′ n(r1)xn ! (7.7.5) is also a solution of Ly = 0, and {y1, y2} is a fundamental set of solutions. Proof Theorem 7.5.3 implies that Ly1 = 0. We’ll now show that Ly2 = 0. Since L is a linear operator, this is equivalent to showing that L xr2 k−1 X n=0 an(r2)xn ! + CL y1 ln x + xr1 ∞ X n=1 a′ n(r1)xn ! = 0. (7.7.6) To verify this, we’ll show that L xr2 k−1 X n=0 an(r2)xn ! = p1(r1 −1)ak−1(r2)xr1 (7.7.7) and L y1 lnx + xr1 ∞ X n=1 a′ n(r1)xn ! = kα0xr1. (7.7.8) This will imply that Ly2 = 0, since substituting (7.7.7) and (7.7.8) into (7.7.6) and using (7.7.4) yields Ly2 = [p1(r1 −1)ak−1(r2) + Ckα0] xr1 = [p1(r1 −1)ak−1(r2) −p1(r1 −1)ak−1(r2)] xr1 = 0. We’ll prove (7.7.8) first. From Theorem 7.6.1, L y(x, r) ln x + xr ∞ X n=1 a′ n(r)xn ! = p′ 0(r)xr + xrp0(r) ln x. | Elementary Differential Equations with Boundary Value Problems_Page_390_Chunk2794 |
Section 7.7 The Method of Frobenius III 381 Setting r = r1 and recalling that p0(r1) = 0 and y1 = y(x, r1) yields L y1 ln x + xr1 ∞ X n=1 a′ n(r1)xn ! = p′ 0(r1)xr1. (7.7.9) Since r1 and r2 are the roots of the indicial equation, the indicial polynomial can be written as p0(r) = α0(r −r1)(r −r2) = α0 r2 −(r1 + r2)r + r1r2 . Differentiating this yields p′ 0(r) = α0(2r −r1 −r2). Therefore p′ 0(r1) = α0(r1 −r2) = kα0, so (7.7.9) implies (7.7.8). Before proving (7.7.7), we first note an(r2) is well defined by (7.7.3) for 1 ≤n ≤k −1, since p0(n + r2) ̸= 0 for these values of n. However, we can’t define an(r2) for n ≥k with (7.7.3), since p0(k + r2) = p0(r1) = 0. For convenience, we define an(r2) = 0 for n ≥k. Then, from Theorem 7.5.1, L xr2 k−1 X n=0 an(r2)xn ! = L xr2 ∞ X n=0 an(r2)xn ! = xr2 ∞ X n=0 bnxn, (7.7.10) where b0 = p0(r2) = 0 and bn = p0(n + r2)an(r2) + p1(n + r2 −1)an−1(r2), n ≥1. If 1 ≤n ≤k −1, then (7.7.3) implies that bn = 0. If n ≥k + 1, then bn = 0 because an−1(r2) = an(r2) = 0. Therefore (7.7.10) reduces to L xr2 k−1 X n=0 an(r2)xn ! = [p0(k + r2)ak(r2) + p1(k + r2 −1)ak−1(r2)] xk+r2. Since ak(r2) = 0 and k + r2 = r1, this implies (7.7.7). We leave the proof that {y1, y2} is a fundamental set as an exercise (Exercise 41). Example 7.7.1 Find a fundamental set of Frobenius solutions of 2x2(2 + x)y′′ −x(4 −7x)y′ −(5 −3x)y = 0. Give explicit formulas for the coefficients in the solutions. Solution For the given equation, the polynomials defined in Theorem 7.7.1 are p0(r) = 4r(r −1) −4r −5 = (2r + 1)(2r −5), p1(r) = 2r(r −1) + 7r + 3 = (r + 1)(2r + 3). The roots of the indicial equation are r1 = 5/2 and r2 = −1/2, so k = r1 −r2 = 3. Therefore Theorem 7.7.1 implies that y1 = x5/2 ∞ X n=0 an(5/2)xn (7.7.11) and y2 = x−1/2 2 X n=0 an(−1/2) + C y1 lnx + x5/2 ∞ X n=1 a′ n(5/2)xn ! (7.7.12) | Elementary Differential Equations with Boundary Value Problems_Page_391_Chunk2795 |
382 Chapter 7 Series Solutions of Linear Second Order Equations (with C as in (7.7.4)) form a fundamental set of solutions of Ly = 0. The recurrence formula (7.7.2) is a0(r) = 1, an(r) = −p1(n + r −1) p0(n + r) an−1(r) = − (n + r)(2n + 2r + 1) (2n + 2r + 1)(2n + 2r −5)an−1(r), = − n + r 2n + 2r −5an−1(r), n ≥1, (7.7.13) which implies that an(r) = (−1)n n Y j=1 j + r 2j + 2r −5, n ≥0. (7.7.14) Therefore an(5/2) = (−1)n Qn j=1(2j + 5) 4nn! . (7.7.15) Substituting this into (7.7.11) yields y1 = x5/2 ∞ X n=0 (−1)n Qn j=1(2j + 5) 4nn! xn. To compute the coefficients a0(−1/2), a1(−1/2) and a2(−1/2) in y2, we set r = −1/2 in (7.7.13) and apply the resulting recurrence formula for n = 1, 2; thus, a0(−1/2) = 1, an(−1/2) = −2n −1 4(n −3)an−1(−1/2), n = 1, 2. The last formula yields a1(−1/2) = 1/8 and a2(−1/2) = 3/32. Substituting r1 = 5/2, r2 = −1/2, k = 3, and α0 = 4 into (7.7.4) yields C = −15/128. Therefore, from (7.7.12), y2 = x−1/2 1 + 1 8x + 3 32x2 −15 128 y1 lnx + x5/2 ∞ X n=1 a′ n(5/2)xn ! . (7.7.16) We use logarithmic differentiation to obtain obtain a′ n(r). From (7.7.14), |an(r)| = n Y j=1 |j + r| |2j + 2r −5|, n ≥1. Therefore ln|an(r)| = n X j=1 (ln |j + r| −ln|2j + 2r −5|) . Differentiating with respect to r yields a′ n(r) an(r) = n X j=1 1 j + r − 2 2j + 2r −5 . | Elementary Differential Equations with Boundary Value Problems_Page_392_Chunk2796 |
Section 7.7 The Method of Frobenius III 383 Therefore a′ n(r) = an(r) n X j=1 1 j + r − 2 2j + 2r −5 . Setting r = 5/2 here and recalling (7.7.15) yields a′ n(5/2) = (−1)n Qn j=1(2j + 5) 4nn! n X j=1 1 j + 5/2 −1 j . (7.7.17) Since 1 j + 5/2 −1 j = − 5 j(2j + 5), we can rewrite (7.7.17) as a′ n(5/2) = −5 (−1)n Qn j=1(2j + 5) 4nn! n X j=1 1 j(2j + 5) . Substituting this into (7.7.16) yields y2 = x−1/2 1 + 1 8x + 3 32x2 −15 128y1 ln x + 75 128x5/2 ∞ X n=1 (−1)n Qn j=1(2j + 5) 4nn! n X j=1 1 j(2j + 5) xn. If C = 0 in (7.7.4), there’s no need to compute y1 ln x + xr1 ∞ X n=1 a′ n(r1)xn in the formula (7.7.5) for y2. Therefore it’s best to compute C before computing {a′ n(r1)}∞ n=1. This is illustrated in the next example. (See also Exercises 44 and 45.) Example 7.7.2 Find a fundamental set of Frobenius solutions of x2(1 −2x)y′′ + x(8 −9x)y′ + (6 −3x)y = 0. Give explicit formulas for the coefficients in the solutions. Solution For the given equation, the polynomials defined in Theorem 7.7.1 are p0(r) = r(r −1) + 8r + 6 = (r + 1)(r + 6) p1(r) = −2r(r −1) −9r −3 = −(r + 3)(2r + 1). The roots of the indicial equation are r1 = −1 and r2 = −6, so k = r1−r2 = 5. Therefore Theorem 7.7.1 implies that y1 = x−1 ∞ X n=0 an(−1)xn (7.7.18) | Elementary Differential Equations with Boundary Value Problems_Page_393_Chunk2797 |
384 Chapter 7 Series Solutions of Linear Second Order Equations and y2 = x−6 4 X n=0 an(−6) + C y1 lnx + x−1 ∞ X n=1 a′ n(−1)xn ! (7.7.19) (with C as in (7.7.4)) form a fundamental set of solutions of Ly = 0. The recurrence formula (7.7.2) is a0(r) = 1, an(r) = −p1(n + r −1) p0(n + r) an−1(r) = (n + r + 2)(2n + 2r −1) (n + r + 1)(n + r + 6) an−1(r), n ≥1, (7.7.20) which implies that an(r) = n Y j=1 (j + r + 2)(2j + 2r −1) (j + r + 1)(j + r + 6) = n Y j=1 j + r + 2 j + r + 1 n Y j=1 2j + 2r −1 j + r + 6 . (7.7.21) Since n Y j=1 j + r + 2 j + r + 1 = (r + 3)(r + 4) · · ·(n + r + 2) (r + 2)(r + 3) · · ·(n + r + 1) = n + r + 2 r + 2 because of cancellations, (7.7.21) simplifies to an(r) = n + r + 2 r + 2 n Y j=1 2j + 2r −1 j + r + 6 . Therefore an(−1) = (n + 1) n Y j=1 2j −3 j + 5 . Substituting this into (7.7.18) yields y1 = x−1 ∞ X n=0 (n + 1) n Y j=1 2j −3 j + 5 xn. To compute the coefficients a0(−6), . . ., a4(−6) in y2, we set r = −6 in (7.7.20) and apply the resulting recurrence formula for n = 1, 2, 3, 4; thus, a0(−6) = 1, an(−6) = (n −4)(2n −13) n(n −5) an−1(−6), n = 1, 2, 3, 4. The last formula yields a1(−6) = −33 4 , a2(−6) = 99 4 , a3(−6) = −231 8 , a4(−6) = 0. Since a4(−6) = 0, (7.7.4) implies that the constant C in (7.7.19) is zero. Therefore (7.7.19) reduces to y2 = x−6 1 −33 4 x + 99 4 x2 −231 8 x3 . | Elementary Differential Equations with Boundary Value Problems_Page_394_Chunk2798 |
Section 7.7 The Method of Frobenius III 385 We now consider equations of the form x2(α0 + α2x2)y′′ + x(β0 + β2x2)y′ + (γ0 + γ2x2)y = 0, where the roots of the indicial equation are real and differ by an even integer. The case where the roots are real and differ by an odd integer can be handled by the method discussed in 56. The proof of the next theorem is similar to the proof of Theorem 7.7.1 (Exercise 43). Theorem 7.7.2 Let Ly = x2(α0 + α2x2)y′′ + x(β0 + β2x2)y′ + (γ0 + γ2x2)y, where α0 ̸= 0, and define p0(r) = α0r(r −1) + β0r + γ0, p2(r) = α2r(r −1) + β2r + γ2. Suppose r is a real number such that p0(2m + r) is nonzero for all positive integers m, and define a0(r) = 1, a2m(r) = −p2(2m + r −2) p0(2m + r) a2m−2(r), m ≥1. (7.7.22) Let r1 and r2 be the roots of the indicial equation p0(r) = 0, and suppose r1 = r2 + 2k, where k is a positive integer. Then y1 = xr1 ∞ X m=0 a2m(r1)x2m is a Frobenius solution of Ly = 0. Moreover, if we define a0(r2) = 1, a2m(r2) = −p2(2m + r2 −2) p0(2m + r2) a2m−2(r2), 1 ≤m ≤k −1 and C = −p2(r1 −2) 2kα0 a2k−2(r2), (7.7.23) then y2 = xr2 k−1 X m=0 a2m(r2)x2m + C y1 lnx + xr1 ∞ X m=1 a′ 2m(r1)x2m ! (7.7.24) is also a solution of Ly = 0, and {y1, y2} is a fundamental set of solutions. Example 7.7.3 Find a fundamental set of Frobenius solutions of x2(1 + x2)y′′ + x(3 + 10x2)y′ −(15 −14x2)y = 0. Give explicit formulas for the coefficients in the solutions. | Elementary Differential Equations with Boundary Value Problems_Page_395_Chunk2799 |
386 Chapter 7 Series Solutions of Linear Second Order Equations Solution For the given equation, the polynomials defined in Theorem 7.7.2 are p0(r) = r(r −1) + 3r −15 = (r −3)(r + 5) p2(r) = r(r −1) + 10r + 14 = (r + 2)(r + 7). The roots of the indicial equation are r1 = 3 and r2 = −5, so k = (r1 −r2)/2 = 4. Therefore Theorem 7.7.2 implies that y1 = x3 ∞ X m=0 a2m(3)x2m (7.7.25) and y2 = x−5 3 X m=0 a2m(−5)x2m + C y1 lnx + x3 ∞ X m=1 a′ 2m(3)x2m ! (with C as in (7.7.23)) form a fundamental set of solutions of Ly = 0. The recurrence formula (7.7.22) is a0(r) = 1, a2m(r) = −p2(2m + r −2) p0(2m + r) a2m−2(r) = − (2m + r)(2m + r + 5) (2m + r −3)(2m + r + 5)a2m−2(r) = − 2m + r 2m + r −3a2m−2(r), m ≥1, (7.7.26) which implies that a2m(r) = (−1)m m Y j=1 2j + r 2j + r −3, m ≥0. (7.7.27) Therefore a2m(3) = (−1)m Qm j=1(2j + 3) 2mm! . (7.7.28) Substituting this into (7.7.25) yields y1 = x3 ∞ X m=0 (−1)m Qm j=1(2j + 3) 2mm! x2m. To compute the coefficients a2(−5), a4(−5), and a6(−5) in y2, we set r = −5 in (7.7.26) and apply the resulting recurrence formula for m = 1, 2, 3; thus, a2m(−5) = −2m −5 2(m −4)a2m−2(−5), m = 1, 2, 3. This yields a2(−5) = −1 2, a4(−5) = 1 8, a6(−5) = 1 16. Substituting r1 = 3, r2 = −5, k = 4, and α0 = 1 into (7.7.23) yields C = −3/16. Therefore, from (7.7.24), y2 = x−5 1 −1 2x2 + 1 8x4 + 1 16x6 −3 16 y1 lnx + x3 ∞ X m=1 a′ 2m(3)x2m ! . (7.7.29) | Elementary Differential Equations with Boundary Value Problems_Page_396_Chunk2800 |
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