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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 t... | 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 Rma... | 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 displac... | 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... | 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 sect... | 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 ... | 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 flow... | 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 ... | 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... | 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... | 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 t... | 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 differe... | 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... | 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 th... | 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... | 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 ... | 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 ... | 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 conditio... | 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... | 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 mo... | 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 so... | 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. I... | 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)!, −∞... | 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 t... | 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 derivati... | 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 ... | 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... | 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 conv... | 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′ ... | 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... | 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) ∞ ... | 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=... | 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 ... | 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 + 3... | 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 = ∞.) ... | 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)... | 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... | 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 · ... | 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 fo... | 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... | 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.... | 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 se... | 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 ... | 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 (... | 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 +... | 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... | 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 pol... | 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... | 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)... | 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... | 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=... | 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... | 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... | 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, ... | 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 differ... | 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)... | 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 ... | 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... | 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 indici... | 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 . Subst... | 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... | 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) ... | 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 Fr... | 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 + ... | 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 ... | 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 assumpti... | 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 ... | 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),... | 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 ... | 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, y... | 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... | 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 follo... | 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... | 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 ... | 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... | 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... | 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... | 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 + ... | 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) y... | 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, ... | 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)... | 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 term... | 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|... | 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) = ... | 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 −... | 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... | 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)... | 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. 25x2... | 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 formul... | 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) ... | 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 ... | 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 +... | 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... | 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 +... | 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... | 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 ... | 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) ... | 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 i... | 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 T... | Elementary Differential Equations with Boundary Value Problems_Page_396_Chunk2800 |
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