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Section 7.7 The Method of Frobenius III 387 To obtain a′ 2m(r) we use logarithmic differentiation. From (7.7.27), |a2m(r)| = m Y j=1 |2j + r| |2j + r −3|, m ≥1. Therefore ln |a2m(r)| = n X j=1 (ln |2j + r| −ln |2j + r −3|). Differentiating with respect to r yields a′ 2m(r) a2m(r) = m X j=1 1 2j + r − 1 2j + r −3 . ... | Elementary Differential Equations with Boundary Value Problems_Page_397_Chunk2801 |
388 Chapter 7 Series Solutions of Linear Second Order Equations The roots of the indicial equation are r1 = 0 and r2 = −6, so k = (r1 −r2)/2 = 3. Therefore Theorem 7.7.2 implies that y1 = ∞ X m=0 a2m(0)x2m, (7.7.31) and y2 = x−6 2 X m=0 a2m(−6)x2m + C y1 ln x + ∞ X m=1 a′ 2m(0)x2m ! (7.7.32) (with C as in (7.7.23)) for... | Elementary Differential Equations with Boundary Value Problems_Page_398_Chunk2802 |
Section 7.7 The Method of Frobenius III 389 1. x2y′′ −3xy′ + (3 + 4x)y = 0 2. xy′′ + y = 0 3. 4x2(1 + x)y′′ + 4x(1 + 2x)y′ −(1 + 3x)y = 0 4. xy′′ + xy′ + y = 0 5. 2x2(2 + 3x)y′′ + x(4 + 21x)y′ −(1 −9x)y = 0 6. x2y′′ + x(2 + x)y′ −(2 −3x)y = 0 7. 4x2y′′ + 4xy′ −(9 −x)y = 0 8. x2y′′ + 10xy′ + (14 + x)y = 0 9. 4x2(1 + x)y... | Elementary Differential Equations with Boundary Value Problems_Page_399_Chunk2803 |
390 Chapter 7 Series Solutions of Linear Second Order Equations 38. x2(1 + x2)y′′ + x(5 + 2x2)y′ −21y = 0 39. x2(1 + 2x2)y′′ −x(3 + x2)y′ −2x2y = 0 40. 4x2(1 + x2)y′′ + 4x(2 + x2)y′ −(15 + x2)y = 0 41. (a) Under the assumptions of Theorem 7.7.1, show that y1 = xr1 ∞ X n=0 an(r1)xn and y2 = xr2 k−1 X n=0 an(r2)xn + C y1... | Elementary Differential Equations with Boundary Value Problems_Page_400_Chunk2804 |
Section 7.7 The Method of Frobenius III 391 47. Let Ly = α0x2y′′ + β0xy′ + (γ0 + γ2x2)y and define p0(r) = α0r(r −1) + β0r + γ0. Show that if p0(r) = α0(r −r1)(r −r2) where r1 −r2 = 2k, an even positive integer, then Ly = 0 has the solutions y1 = xr1 ∞ X m=0 (−1)m 4mm! Qm j=1(j + k) γ2 α0 m x2m and y2 = xr2 k−1 X m=0... | Elementary Differential Equations with Boundary Value Problems_Page_401_Chunk2805 |
392 Chapter 7 Series Solutions of Linear Second Order Equations where A = k X j=1 pj(r1 −j)ak−j(r2). (c) Show that y1 and y2 = xr2 ∞ X n=0 an(r2)xn −A kα0 y1 ln x + xr1 ∞ X n=1 a′ n(r1)xn ! form a fundamental set of Frobenius solutions of Ly = 0. (d) Show that choosing the arbitrary quantity ak(r2) to be nonzero merely... | Elementary Differential Equations with Boundary Value Problems_Page_402_Chunk2806 |
CHAPTER 8 Laplace Transforms IN THIS CHAPTER we study the method of Laplace transforms, which illustrates one of the basic prob- lem solving techniques in mathematics: transform a difficult problem into an easier one, solve the lat- ter, and then use its solution to obtain a solution of the original problem. The method ... | Elementary Differential Equations with Boundary Value Problems_Page_403_Chunk2807 |
394 Chapter 8 Laplace Transforms 8.1 INTRODUCTION TO THE LAPLACE TRANSFORM Definition of the Laplace Transform To define the Laplace transform, we first recall the definition of an improper integral. If g is integrable over the interval [a, T] for every T > a, then the improper integral of g over [a, ∞) is defined as Z ∞ a ... | Elementary Differential Equations with Boundary Value Problems_Page_404_Chunk2808 |
Section 8.1 Introduction to the Laplace Transform 395 If s = 0 the integrand reduces to the constant 1, and lim T →∞ Z T 0 1 dt = lim T →∞ Z T 0 1 dt = lim T →∞T = ∞. Therefore F (0) is undefined, and F (s) = Z ∞ 0 e−stdt = 1 s, s > 0. This result can be written in operator notation as L(1) = 1 s, s > 0, or as the trans... | Elementary Differential Equations with Boundary Value Problems_Page_405_Chunk2809 |
396 Chapter 8 Laplace Transforms Example 8.1.3 Find the Laplace transform of f(t) = eat, where a is a constant. Solution From (8.1.2) with f(t) = eat, F (s) = Z ∞ 0 e−steat dt. Combining the exponentials yields F (s) = Z ∞ 0 e−(s−a)t dt. However, we know from Example 8.1.1 that Z ∞ 0 e−st dt = 1 s, s > 0. Replacing s b... | Elementary Differential Equations with Boundary Value Problems_Page_406_Chunk2810 |
Section 8.1 Introduction to the Laplace Transform 397 Solving this for G(s) yields G(s) = s s2 + ω2 , s > 0. This and (8.1.8) imply that F (s) = ω s2 + ω2 , s > 0. Tables of Laplace transforms Extensive tables of Laplace transforms have been compiled and are commonly used in applications. The brief table of Laplace tra... | Elementary Differential Equations with Boundary Value Problems_Page_407_Chunk2811 |
398 Chapter 8 Laplace Transforms Solution By definition, cosh bt = ebt + e−bt 2 . Therefore L(cosh bt) = L 1 2ebt + 1 2e−bt = 1 2L(ebt) + 1 2L(e−bt) (linearity property) = 1 2 1 s −b + 1 2 1 s + b, (8.1.9) where the first transform on the right is defined for s > b and the second for s > −b; hence, both are defined for ... | Elementary Differential Equations with Boundary Value Problems_Page_408_Chunk2812 |
Section 8.1 Introduction to the Laplace Transform 399 f(t) ↔F (s) eatf(t) ↔F (s −a) 1 ↔1 s, s > 0 eat ↔ 1 (s −a), s > a t ↔1 s2 , s > 0 teat ↔ 1 (s −a)2 , s > a sin ωt ↔ ω s2 + ω2 , s > 0 eλt sin ωt ↔ ω (s −λ)2 + ω2 , s > λ cos ωt ↔ s s2 + ω2 , s > 0 eλt sin ωt ↔ s −λ (s −λ)2 + ω2 , s > λ Existence of Laplace Transform... | Elementary Differential Equations with Boundary Value Problems_Page_409_Chunk2813 |
400 Chapter 8 Laplace Transforms t0 x y f (t0+) f (t0−) Figure 8.1.1 A jump discontinuity If f(t0+) and f(t0−) are finite and equal, but either f isn’t defined at t0 or it’s defined but f(t0) ̸= f(t0+) = f(t0−), we say that f has a removable discontinuity at t0 (Figure 8.1.2). This terminolgy is appropriate since a functi... | Elementary Differential Equations with Boundary Value Problems_Page_410_Chunk2814 |
Section 8.1 Introduction to the Laplace Transform 401 t0 x f(t0) f(t0−) = f(t0+) Figure 8.1.2 a b x y Figure 8.1.3 A piecewise continuous function on [a, b] exists for every T > 0. However, piecewise continuity alone does not guarantee that the improper integral Z ∞ 0 e−stf(t) dt = lim T →∞ Z T 0 e−stf(t) dt (8.1.13) c... | Elementary Differential Equations with Boundary Value Problems_Page_411_Chunk2815 |
402 Chapter 8 Laplace Transforms Example 8.1.9 It can be shown that if limt→∞e−s0tf(t) exists and is finite then f is of exponential order s0 (Exercise 9). If α is any real number and s0 > 0 then f(t) = tα is of exponential order s0, since lim t→∞e−s0ttα = 0, by L’Hôpital’s rule. If α ≥0, f is also continuous on [0, ∞).... | Elementary Differential Equations with Boundary Value Problems_Page_412_Chunk2816 |
Section 8.1 Introduction to the Laplace Transform 403 8.1 Exercises 1. Find the Laplace transforms of the followingfunctions by evaluating the integral F (s) = R ∞ 0 e−stf(t) dt. (a) t (b) te−t (c) sinh bt (d) e2t −3et (e) t2 2. Use the table of Laplace transforms to find the Laplace transforms of the following function... | Elementary Differential Equations with Boundary Value Problems_Page_413_Chunk2817 |
404 Chapter 8 Laplace Transforms (c) Show that if f is of exponential order s0 and g(t) = f(t + τ) where τ > 0, then g is also of exponential order s0. 10. Recall the next theorem from calculus. THEOREM A. Let g be integrable on [0, T] for every T > 0. Suppose there’s a function w defined on some interval [τ, ∞) (with τ... | Elementary Differential Equations with Boundary Value Problems_Page_414_Chunk2818 |
Section 8.2 The Inverse Laplace Transform 405 (a) Use integration by parts to show that Γ(α + 1) = αΓ(α), α > 0. (b) Show that Γ(n + 1) = n! if n = 1, 2, 3,.... (c) From (b) and the table of Laplace transforms, L(tα) = Γ(α + 1) sα+1 , s > 0, if α is a nonnegative integer. Show that this formula is valid for any α > −1.... | Elementary Differential Equations with Boundary Value Problems_Page_415_Chunk2819 |
406 Chapter 8 Laplace Transforms To solve differential equations with the Laplace transform, we must be able to obtain f from its transform F . There’s a formula for doing this, but we can’t use it because it requires the theory of functions of a complex variable. Fortunately, we can use the table of Laplace transforms... | Elementary Differential Equations with Boundary Value Problems_Page_416_Chunk2820 |
Section 8.2 The Inverse Laplace Transform 407 Example 8.2.3 Find L−1 3s + 8 s2 + 2s + 5 . Solution Completing the square in the denominator yields 3s + 8 s2 + 2s + 5 = 3s + 8 (s + 1)2 + 4. Because of the form of the denominator, we consider the transform pairs e−t cos 2t ↔ s + 1 (s + 1)2 + 4 and e−t sin 2t ↔ 2 (s +... | Elementary Differential Equations with Boundary Value Problems_Page_417_Chunk2821 |
408 Chapter 8 Laplace Transforms Multiplying this by (s −1)(s −2) yields 3s + 2 = (s −2)A + (s −1)B. Setting s = 2 yields B = 8 and setting s = 1 yields A = −5. Therefore F (s) = − 5 s −1 + 8 s −2 and L−1(F ) = −5L−1 1 s −1 + 8L−1 1 s −2 = −5et + 8e2t. Solution (METHOD 2) We don’t really have to multiply (8.2.3... | Elementary Differential Equations with Boundary Value Problems_Page_418_Chunk2822 |
Section 8.2 The Inverse Laplace Transform 409 Solution The partial fraction expansion of (8.2.7) is of the form F (s) = A s + B s −1 + C s −2 + D s + 1. (8.2.8) To find A, we ignore the factor s in the denominator of (8.2.7) and set s = 0 elsewhere. This yields A = 6 + (1)(11) (−1)(−2)(1) = 17 2 . Similarly, the other c... | Elementary Differential Equations with Boundary Value Problems_Page_419_Chunk2823 |
410 Chapter 8 Laplace Transforms The two sides of this equation are polynomials of degree two. From a theorem of algebra, they will be equal for all s if they are equal for any three distinct values of s. We may determine A, B and C by choosing convenient values of s. The left side of (8.2.12) suggests that we take s =... | Elementary Differential Equations with Boundary Value Problems_Page_420_Chunk2824 |
Section 8.2 The Inverse Laplace Transform 411 Solution One form for the partial fraction expansion of F is F (s) = A s + Bs + C (s + 1)2 + 1. (8.2.14) However, we see from the table of Laplace transforms that the inverse transform of the second fraction on the right of (8.2.14) will be a linear combination of the inver... | Elementary Differential Equations with Boundary Value Problems_Page_421_Chunk2825 |
412 Chapter 8 Laplace Transforms Solution The form for the partial fraction expansion is F (s) = A + Bs s2 + 1 + C + Ds s2 + 4 . The coefficients A, B, C and D can be obtained by finding a common denominator and equating the resulting numerator to the numerator in (8.2.17). However, since there’s no first power of s in th... | Elementary Differential Equations with Boundary Value Problems_Page_422_Chunk2826 |
Section 8.2 The Inverse Laplace Transform 413 (a) 2s + 3 (s −7)4 (b) s2 −1 (s −2)6 (c) s + 5 s2 + 6s + 18 (d) 2s + 1 s2 + 9 (e) s s2 + 2s + 1 (f) s + 1 s2 −9 (g) s3 + 2s2 −s −3 (s + 1)4 (h) 2s + 3 (s −1)2 + 4 (i) 1 s − s s2 + 1 (j) 3s + 4 s2 −1 (k) 3 s −1 + 4s + 1 s2 + 9 (l) 3 (s + 2)2 −2s + 6 s2 + 4 3. Use Heaviside’s... | Elementary Differential Equations with Boundary Value Problems_Page_423_Chunk2827 |
414 Chapter 8 Laplace Transforms (a) 2s + 1 (s2 + 1)(s −1)(s −3) (b) s + 2 (s2 + 2s + 2)(s2 −1) (c) 2s −1 (s2 −2s + 2)(s + 1)(s −2) (d) s −6 (s2 −1)(s2 + 4) (e) 2s −3 s(s −2)(s2 −2s + 5) (f) 5s −15 (s2 −4s + 13)(s −2)(s −1) 9. Given that f(t) ↔F (s), find the inverse Laplace transform of F (as −b), where a > 0. 10. (a) ... | Elementary Differential Equations with Boundary Value Problems_Page_424_Chunk2828 |
Section 8.3 Solution of Initial Value Problems 415 which proves (8.3.1). Now suppose T > 0 and f′ is only piecewise continuous on [0, T], with discon- tinuities at t1 < t2 < · · · < tn−1. For convenience, let t0 = 0 and tn = T. Integrating by parts yields Z ti ti−1 e−stf′(t) dt = e−stf(t) ti ti−1 + s Z ti ti−1 e−stf(t)... | Elementary Differential Equations with Boundary Value Problems_Page_425_Chunk2829 |
416 Chapter 8 Laplace Transforms Theorem 8.3.2 Suppose f and f′ are continuous on [0, ∞) and of exponential order s0, and that f′′ is piecewise continuous on [0, ∞). Then f, f′, and f′′ have Laplace transforms for s > s0, L(f′) = sL(f) −f(0), (8.3.4) and L(f′′) = s2L(f) −f′(0) −sf(0). (8.3.5) Proof Theorem 8.3.1 implie... | Elementary Differential Equations with Boundary Value Problems_Page_426_Chunk2830 |
Section 8.3 Solution of Initial Value Problems 417 and Y (s) = 3 + (s −2)(2s −9) (s −2)(s −5)(s −1). Heaviside’s method yields the partial fraction expansion Y (s) = − 1 s −2 + 1 2 1 s −5 + 5 2 1 s −1, and taking the inverse transform of this yields y = −e2t + 1 2e5t + 5 2et as the solution of (8.3.6). It isn’t necessa... | Elementary Differential Equations with Boundary Value Problems_Page_427_Chunk2831 |
418 Chapter 8 Laplace Transforms and F (s) = L(8e−2t) = 8 s + 2, so (8.3.13) becomes (2s + 1)(s + 1)Y (s) = 8 s + 2 + 2(2 −4s) + 3(−4). Solving for Y (s) yields Y (s) = 4 (1 −(s + 2)(s + 1)) (s + 1/2)(s + 1)(s + 2). Heaviside’s method yields the partial fraction expansion Y (s) = 4 3 1 s + 1/2 − 8 s + 1 + 8 3 1 s + 2, ... | Elementary Differential Equations with Boundary Value Problems_Page_428_Chunk2832 |
Section 8.3 Solution of Initial Value Problems 419 so (8.3.13) becomes (s + 1)2 + 1 Y (s) = 1 s + 1 · (1 −3s) + 2(−3). Solving for Y (s) yields Y (s) = 1 −s(5 + 3s) s [(s + 1)2 + 1]. In Example 8.2.8 we found the inverse transform of this function to be y = 1 2 −7 2e−t cos t −5 2e−t sin t (Figure 8.3.2), which is t... | Elementary Differential Equations with Boundary Value Problems_Page_429_Chunk2833 |
420 Chapter 8 Laplace Transforms 21. y′′ + y = t −3 sin 2t, y(0) = 1, y′(0) = −3 22. y′′ + 5y′ + 6y = 2e−t, y(0) = 1, y′(0) = 3 23. y′′ + 2y′ + y = 6 sint −4 cos t, y(0) = −1, y′(0) = 1 24. y′′ −2y′ −3y = 10 cos t, y(0) = 2, y′(0) = 7 25. y′′ + y = 4 sint + 6 cos t, y(0) = −6, y′(0) = 2 26. y′′ + 4y = 8 sin 2t + 9 cos ... | Elementary Differential Equations with Boundary Value Problems_Page_430_Chunk2834 |
Section 8.4 The Unit Step Function 421 Example 8.4.1 Use the table of Laplace transforms to find the Laplace transform of f(t) = ( 2t + 1, 0 ≤t < 2, 3t, t ≥2 (8.4.1) (Figure 8.4.1). Solution Since the formula for f changes at t = 2, we write L(f) = Z ∞ 0 e−stf(t) dt = Z 2 0 e−st(2t + 1) dt + Z ∞ 2 e−st(3t) dt. (8.4.2) T... | Elementary Differential Equations with Boundary Value Problems_Page_431_Chunk2835 |
422 Chapter 8 Laplace Transforms 1 2 3 4 1 2 3 4 5 6 7 8 9 10 11 12 y t Figure 8.4.1 The piecewise continuous function (8.4.1) 1 τ t y Figure 8.4.2 y = u(t −τ) Laplace Transforms of Piecewise Continuous Functions We’ll now develop the method of Example 8.4.1 into a systematic way to find the Laplace transform of a piece... | Elementary Differential Equations with Boundary Value Problems_Page_432_Chunk2836 |
Section 8.4 The Unit Step Function 423 Theorem 8.4.1 Let g be defined on [0, ∞). Suppose τ ≥0 and L (g(t + τ)) exists for s > s0. Then L (u(t −τ)g(t)) exists for s > s0, and L(u(t −τ)g(t)) = e−sτL (g(t + τ)) . Proof By definition, L (u(t −τ)g(t)) = Z ∞ 0 e−stu(t −τ)g(t) dt. From this and the definition of u(t −τ), L (u(t ... | Elementary Differential Equations with Boundary Value Problems_Page_433_Chunk2837 |
424 Chapter 8 Laplace Transforms Therefore L(f) = L(2t + 1) + L (u(t −2)(t −1)) = L(2t + 1) + e−2sL(t + 1) (from Theorem 8.4.1) = 2 s2 + 1 s + e−2s 1 s2 + 1 s , which is the result obtained in Example 8.4.1. Formula (8.4.6) can be extended to more general piecewise continuous functions. For example, we can write f(... | Elementary Differential Equations with Boundary Value Problems_Page_434_Chunk2838 |
Section 8.4 The Unit Step Function 425 1 2 3 4 5 6 2 4 6 8 10 12 14 16 −6 −4 −2 t y Figure 8.4.3 The piecewise contnuous function (8.4.7) Example 8.4.5 Find the Laplace transform of f(t) = sin t, 0 ≤t < π 2 , cos t −3 sint, π 2 ≤t < π, 3 cos t, t ≥π (8.4.10) (Figure 8.4.4). Solution In terms of step f... | Elementary Differential Equations with Boundary Value Problems_Page_435_Chunk2839 |
426 Chapter 8 Laplace Transforms 1 2 3 4 5 6 1 −1 2 −2 3 −3 t y Figure 8.4.4 The piecewise continuous function (8.4.10) The Second Shifting Theorem Replacing g(t) by g(t −τ) in Theorem 8.4.1 yields the next theorem. Theorem 8.4.2 [Second Shifting Theorem] If τ ≥0 and L(g) exists for s > s0 then L (u(t −τ)g(t −τ)) exist... | Elementary Differential Equations with Boundary Value Problems_Page_436_Chunk2840 |
Section 8.4 The Unit Step Function 427 Example 8.4.7 Find the inverse Laplace transform h of H(s) = 1 s2 −e−s 1 s2 + 2 s + e−4s 4 s3 + 1 s , and find distinct formulas for h on appropriate intervals. Solution Let G0(s) = 1 s2 , G1(s) = 1 s2 + 2 s, G2(s) = 4 s3 + 1 s. Then g0(t) = t, g1(t) = t + 2, g2(t) = 2t2 + ... | Elementary Differential Equations with Boundary Value Problems_Page_437_Chunk2841 |
428 Chapter 8 Laplace Transforms Using the trigonometric identities (8.4.8) and (8.4.9), we can rewrite this as h(t) = 2 cos 2t + u(t −π/2) | Elementary Differential Equations with Boundary Value Problems_Page_438_Chunk2842 |
Section 8.4 The Unit Step Function 429 In Exercises 7–18 express the given function f in terms of unit step functions and use Theorem 8.4.1 to find L(f). Where indicated by C/G , graph f. 7. f(t) = ( 0, 0 ≤t < 2, t2 + 3t, t ≥2. 8. f(t) = ( t2 + 2, 0 ≤t < 1, t, t ≥1. 9. f(t) = ( tet, 0 ≤t < 1, et, t ≥1. 10. f(t) = ( e −t... | Elementary Differential Equations with Boundary Value Problems_Page_439_Chunk2843 |
430 Chapter 8 Laplace Transforms 23. H(s) = 5 s −1 s2 + e−3s 6 s + 7 s2 + 3e−6s s3 24. H(s) = e−πs(1 −2s) s2 + 4s + 5 25. C/G H(s) = 1 s − s s2 + 1 + e−π 2 s 3s −1 s2 + 1 26. H(s) = e−2s 3(s −3) (s + 1)(s −2) − s + 1 (s −1)(s −2) 27. H(s) = 1 s + 1 s2 + e−s 3 s + 2 s2 + e−3s 4 s + 3 s2 28. H(s) = ... | Elementary Differential Equations with Boundary Value Problems_Page_440_Chunk2844 |
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 431 (b) Show that (D) can be rewritten as L(f) = ∞ X m=0 Z ∞ tm e−stfm(t) dt − Z ∞ tm+1 e−stfm(t) dt ! . (E) (c) Use (A), the assumed convergence of (B), and the comparison test to show that the series ∞ X m=0 Z ∞ tm e−stfm(t) dt and... | Elementary Differential Equations with Boundary Value Problems_Page_441_Chunk2845 |
432 Chapter 8 Laplace Transforms Theorem 8.5.1 Suppose a, b, and c are constants (a ̸= 0), and f is piecewise continuous on [0, ∞). with jump discontinuities at t1, ..., tn, where 0 < t1 < · · · < tn. Let k0 and k1 be arbitrary real numbers. Then there is a unique function y defined on [0, ∞) with these properties: (a) ... | Elementary Differential Equations with Boundary Value Problems_Page_442_Chunk2846 |
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 433 1 2 3 4 5 6 1 2 −1 −2 t y Figure 8.5.1 Graph of (8.5.4) Solution The initial value problem in Step 1 is y′′ + y = 1, y(0) = 2, y′(0) = −1. We leave it to you to verify that its solution is y0 = 1 + cos t −sin t. Doing Step 2 yiel... | Elementary Differential Equations with Boundary Value Problems_Page_443_Chunk2847 |
434 Chapter 8 Laplace Transforms Example 8.5.2 Use the Laplace transform to solve the initial value problem y′′ + y = f(t), y(0) = 2, y′(0) = −1, (8.5.5) where f(t) = 1, 0 ≤t < π 2 , −1, t ≥π 2 . Solution Here f(t) = 1 −2u t −π 2 , so Theorem 8.4.1 (with g(t) = 1) implies that L(f) = 1 −2e−πs/2 s . Therefore,... | Elementary Differential Equations with Boundary Value Problems_Page_444_Chunk2848 |
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 435 or y = 1 + cos t −sint, 0 ≤t < π 2 , −1 + cos t + sint, t ≥π 2 , which is the result obtained in Example 8.5.1. REMARK: It isn’t obvious that using the Laplace transform to solve (8.5.2) as we did in Example 8.5.2 yields a ... | Elementary Differential Equations with Boundary Value Problems_Page_445_Chunk2849 |
436 Chapter 8 Laplace Transforms Example 8.5.4 Solve the initial value problem y′′ + y = f(t), y(0) = 0, y′(0) = 0, (8.5.12) where f(t) = 0, 0 ≤t < π 4 , cos 2t, π 4 ≤t < π, 0, t ≥π. Solution Here f(t) = u(t −π/4) cos 2t −u(t −π) cos 2t, so L(f) = L (u(t −π/4) cos 2t) −L (u(t −π) cos 2t) = e−πs/4L (co... | Elementary Differential Equations with Boundary Value Problems_Page_446_Chunk2850 |
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 437 1.0 0.5 −0.5 −1.0 1 2 3 4 5 6 t y Figure 8.5.2 Graph of (8.5.16) and h2(t −π) = −1 3 cos(t −π) + 1 3 cos 2(t −π) = 1 3 cos t + 1 3 cos 2t, (8.5.15) can be rewritten as y = −1 3u t −π 4 √ 2(sin t −cos t) + cos 2t + 1 3u(t −π... | Elementary Differential Equations with Boundary Value Problems_Page_447_Chunk2851 |
438 Chapter 8 Laplace Transforms 1. y′′ + y = ( 3, 0 ≤t < π, 0, t ≥π, y(0) = 0, y′(0) = 0 2. y′′ + y = 3, 0 ≤t < 4, ; 2t −5, t > 4, y(0) = 1, y′(0) = 0 3. y′′ −2y′ = ( 4, 0 ≤t < 1, 6, t ≥1, y(0) = −6, y′(0) = 1 4. y′′ −y = ( e2t, 0 ≤t < 2, 1, t ≥2, y(0) = 3, y′(0) = −1 5. y′′ −3y′ + 2y = 0, 0 ≤t < 1, 1,... | Elementary Differential Equations with Boundary Value Problems_Page_448_Chunk2852 |
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 439 18. y′′ −4y′ + 4y = e2t, 0 ≤t < 2, −e2t, t ≥2, y(0) = 0, y′(0) = −1 19. C/G y′′ = t2, 0 ≤t < 1, −t, 1 ≤t < 2, t + 1, t ≥2, y(0) = 1, y′(0) = 0 20. y′′ + 2y′ + 2y = 1, 0 ≤t < 2π, t, 2π ≤t < 3π, −1, t ≥3π, y(0) = 2, y... | Elementary Differential Equations with Boundary Value Problems_Page_449_Chunk2853 |
440 Chapter 8 Laplace Transforms (c) Conclude from (a) that if g is differentiable on (α, β) then g′ can’t have a jump discontinuity on (α, β). 24. (a) Let a, b, and c be constants, with a ̸= 0. Let f be piecewise continuous on an interval (α, β), with a single jump discontinuity at a point t0 in (α, β). Suppose y and ... | Elementary Differential Equations with Boundary Value Problems_Page_450_Chunk2854 |
Section 8.6 Convolution 441 In this section we consider the problem of finding the inverse Laplace transform of a product H(s) = F (s)G(s), where F and G are the Laplace transforms of known functions f and g. To motivate our interest in this problem, consider the initial value problem ay′′ + by′ + cy = f(t), y(0) = 0, y... | Elementary Differential Equations with Boundary Value Problems_Page_451_Chunk2855 |
442 Chapter 8 Laplace Transforms and y = L−1(F G), respectively. Therefore L−1(F G) = Z t 0 f(τ)g(t −τ) dτ (8.6.5) in this case. This motivates the next definition. Definition 8.6.1 The convolution f ∗g of two functions f and g is defined by (f ∗g)(t) = Z t 0 f(τ)g(t −τ) dτ. It can be shown (Exercise 6) that f ∗g = g ∗f; ... | Elementary Differential Equations with Boundary Value Problems_Page_452_Chunk2856 |
Section 8.6 Convolution 443 t = τ t τ Figure 8.6.1 Example 8.6.1 Let f(t) = eat and g(t) = ebt (a ̸= b). Verify that L(f ∗g) = L(f)L(g), as implied by the convolution theorem. Solution We first compute (f ∗g)(t) = Z t 0 eaτeb(t−τ) dτ = ebt Z t 0 e(a−b)τdτ = ebt e(a−b)τ a −b t 0 = ebt e(a−b)t −1 a −b = eat −ebt a −b . ... | Elementary Differential Equations with Boundary Value Problems_Page_453_Chunk2857 |
444 Chapter 8 Laplace Transforms A Formula for the Solution of an Initial Value Problem The convolution theorem provides a formula for the solution of an initial value problem for a linear constant coefficient second order equation with an unspecified. The next three examples illustrate this. Example 8.6.2 Find a formula... | Elementary Differential Equations with Boundary Value Problems_Page_454_Chunk2858 |
Section 8.6 Convolution 445 Since 1 (s2 + 4) ↔1 2 sin 2t and F (s) ↔f(t), the convolution theorem implies that L−1 1 (s2 + 4)F (s) = 1 2 Z t 0 f(t −τ) sin 2τ dτ. Therefore the solution of (8.6.8) is y(t) = k0 cos 2t + k1 2 sin 2t + 1 2 Z t 0 f(t −τ) sin 2τ dτ. Example 8.6.4 Find a formula for the solution of the in... | Elementary Differential Equations with Boundary Value Problems_Page_455_Chunk2859 |
446 Chapter 8 Laplace Transforms Example 8.6.5 Evaluate the convolution integral h(t) = Z t 0 (t −τ)5τ 7dτ. Solution We could evaluate this integral by expanding (t −τ)5 in powers of τ and then integrating. However, the convolution theorem provides an easier way. The integral is the convolution of f(t) = t5 and g(t) = ... | Elementary Differential Equations with Boundary Value Problems_Page_456_Chunk2860 |
Section 8.6 Convolution 447 is a Volterra integral equation. Here f and k are given functions and y is unknown. Since the integral on the right is a convolution integral, the convolution theorem provides a convenient formula for solving (8.6.11). Taking Laplace transforms in (8.6.11) yields Y (s) = F (s) + K(s)Y (s), a... | Elementary Differential Equations with Boundary Value Problems_Page_457_Chunk2861 |
448 Chapter 8 Laplace Transforms Proof Taking Laplace transforms in (8.6.13) yields p(s)Y (s) = F (s) + a(k1 + k0s) + bk0, where p(s) = as2 + bs + c. Hence, Y (s) = W(s)F (s) + V (s) (8.6.18) with W(s) = 1 p(s) (8.6.19) and V (s) = a(k1 + k0s) + bk0 p(s) . (8.6.20) Taking Laplace transforms in (8.6.15) and (8.6.16) sho... | Elementary Differential Equations with Boundary Value Problems_Page_458_Chunk2862 |
Section 8.6 Convolution 449 independent of the values of f(t −τ) for large τ, since limτ→∞w(τ) = 0. In this case we say that v and h are transient and steady state components, respectively, of the solution y of (8.6.13). These definitions apply to the initial value problem of Example 8.6.4, where the zeros of p(s) = s2 ... | Elementary Differential Equations with Boundary Value Problems_Page_459_Chunk2863 |
450 Chapter 8 Laplace Transforms (i) s −1 s2(s2 −2s + 2) (j) s(s + 3) (s2 + 4)(s2 + 6s + 10) (k) 1 (s −3)5s6 (l) 1 (s −1)3(s2 + 4) (m) 1 s2(s −2)3 (n) 1 s7(s −2)6 2. Find the Laplace transform. (a) Z t 0 sin aτ cos b(t −τ) dτ (b) Z t 0 eτ sin a(t −τ) dτ (c) Z t 0 sinh aτ cosh a(t −τ) dτ (d) Z t 0 τ(t −τ) sinωτ cos ω(t ... | Elementary Differential Equations with Boundary Value Problems_Page_460_Chunk2864 |
Section 8.6 Convolution 451 (f) y(t) = cos t −sin t + Z t 0 y(τ) sin(t −τ) dτ 5. Use the convolution theorem to evaluate the integral. (a) Z t 0 (t −τ)7τ 8 dτ (b) Z t 0 (t −τ)13τ 7 dτ (c) Z t 0 (t −τ)6τ 7 dτ (d) Z t 0 e−τ sin(t −τ) dτ (e) Z t 0 sin τ cos 2(t −τ) dτ 6. Show that Z t 0 f(t −τ)g(τ) dτ = Z t 0 f(τ)g(t −τ) ... | Elementary Differential Equations with Boundary Value Problems_Page_461_Chunk2865 |
452 Chapter 8 Laplace Transforms 11. Let w = L−1 1 as2 + bs + c , where a, b, and c are constants and a ̸= 0. (a) Show that w is the solution of aw′′ + bw′ + cw = 0, w(0) = 0, w′(0) = 1 a. (b) Let f be continuous on [0, ∞) and define h(t) = Z t 0 w(t −τ)f(τ) dτ. Use Leibniz’s rule for differentiating an integral wit... | Elementary Differential Equations with Boundary Value Problems_Page_462_Chunk2866 |
Section 8.7 Constant Coefficient Equations with Impulses 453 for t > 0, and y′′(t) = f(t) a + Z t 0 w′′(t −τ)f0(τ) dτ + u(t −t1) Z t−t1 0 w′′(t −t1 −τ)g(τ) dτ for 0 < t < t1 and t > t1. Also, show y satisfies the differential equation in (A) on(0, t1) and (t1, ∞). (d) Show that y and y′ are continuous on [0, ∞). 13. Supp... | Elementary Differential Equations with Boundary Value Problems_Page_463_Chunk2867 |
454 Chapter 8 Laplace Transforms where f is continuous or piecewise continuous on [0, ∞). In this section we consider initial value prob- lems where f represents a force that’s very large for a short time and zero otherwise. We say that such forces are impulsive. Impulsive forces occur, for example, when two objects co... | Elementary Differential Equations with Boundary Value Problems_Page_464_Chunk2868 |
Section 8.7 Constant Coefficient Equations with Impulses 455 1/h t0 t0+h t y Figure 8.7.1 y = fh(t) Therefore, (8.7.2) implies that yh(t) = 0, 0 ≤t < t0, 1 h Z t t0 w(t −τ) dτ, t0 ≤t ≤t0 + h, 1 h Z t0+h t0 w(t −τ) dτ, t > t0 + h. (8.7.4) Since yh(t) = 0 for all h if 0 ≤t ≤t0, it follows that li... | Elementary Differential Equations with Boundary Value Problems_Page_465_Chunk2869 |
456 Chapter 8 Laplace Transforms From this and (8.7.7), yh(t) −w(t −t0) = 1 h Z t0+h t0 (w(t −τ) −w(t −t0)) dτ. Therefore |yh(t) −w(t −t0)| ≤1 h Z t0+h t0 |w(t −τ) −w(t −t0)| dτ. (8.7.9) Now let Mh be the maximum value of |w(t −τ) −w(t −t0)| as τ varies over the interval [t0, t0 + h]. (Remember that t and t0 are fixed.)... | Elementary Differential Equations with Boundary Value Problems_Page_466_Chunk2870 |
Section 8.7 Constant Coefficient Equations with Impulses 457 t0 y t Figure 8.7.2 An illustration of Theorem 8.7.1 (remember that y′ −(t0) and y′ +(t0) are derivatives from the right and left, respectively) and y′(t0) does not exist. Thus, even though we defined y = u(t−t0)w(t−t0) to be the solution of (8.7.11), this func... | Elementary Differential Equations with Boundary Value Problems_Page_467_Chunk2871 |
458 Chapter 8 Laplace Transforms 0.1 0.2 0.3 0.4 t0 t0 + 1 t0 + 2 t0 + 3 t0 + 4 t0 + 5 t0 + 6 t0 + 7 t y Figure 8.7.3 y = u(t −t0)(t −t0)e−(t−t0) so Definition 8.7.2 yields y = u(t −t0)(t −t0)e−(t−t0) as the solution of (8.7.15) if t0 > 0. If t0 = 0, then (8.7.15) doesn’t have a solution; however, y = u(t)te−t (which we... | Elementary Differential Equations with Boundary Value Problems_Page_468_Chunk2872 |
Section 8.7 Constant Coefficient Equations with Impulses 459 Solution We leave it to you to show that the solution of y′′ + 6y′ + 5y = 3e−2t, y(0) = −3, y′(0) = 2 is ˆy = −e−2t + 1 2e−5t −5 2e−t. Since w(t) = L−1 1 s2 + 6s + 5 = L−1 1 (s + 1)(s + 5) = 1 4L−1 1 s + 1 − 1 s + 5 = e−t −e−5t 4 , the solution of ... | Elementary Differential Equations with Boundary Value Problems_Page_469_Chunk2873 |
460 Chapter 8 Laplace Transforms the solution of (8.7.18) is y = 1 −2 cos t + 2 sin t + 2u(t −π) sin(t −π) −3u(t −2π) sin(t −2π) = 1 −2 cos t + 2 sin t −2u(t −π) sin t −3u(t −2π) sin t, or y = 1 −2 cos t + 2 sint, 0 ≤t < π, 1 −2 cos t, π ≤t < 2π, 1 −2 cos t −3 sint, t ≥2π (8.7.19) (Figure 8.7.5). | Elementary Differential Equations with Boundary Value Problems_Page_470_Chunk2874 |
Section 8.7 Constant Coefficient Equations with Impulses 461 8.7 Exercises In Exercises 1–20 solve the initial value problem. Where indicated by C/G , graph the solution. 1. y′′ + 3y′ + 2y = 6e2t + 2δ(t −1), y(0) = 2, y′(0) = −6 2. C/G y′′ + y′ −2y = −10e−t + 5δ(t −1), y(0) = 7, y′(0) = −9 3. y′′ −4y = 2e−t + 5δ(t −1), ... | Elementary Differential Equations with Boundary Value Problems_Page_471_Chunk2875 |
462 Chapter 8 Laplace Transforms where t0 > 0 and h > 0. Then find w = L−1 1 as2 + bs + c and verify Theorem 8.7.1 by graphing w and yh on the same axes, for small positive values of h. 26. L y′′ + 2y′ + 2y = fh(t), y(0) = 0, y′(0) = 0 27. L y′′ + 2y′ + y = fh(t), y(0) = 0, y′(0) = 0 28. L y′′ + 3y′ + 2y = fh(t), y(... | Elementary Differential Equations with Boundary Value Problems_Page_472_Chunk2876 |
Section 8.8 A Brief Table of Laplace Transforms 463 8.8 A BRIEF TABLE OF LAPLACE TRANSFORMS f(t) F (s) 1 1 s (s > 0) tn n! sn+1 (s > 0) (n = integer > 0) tp, p > −1 Γ(p + 1) s(p+1) (s > 0) eat 1 s −a (s > a) tneat n! (s −a)n+1 (s > 0) (n = integer > 0) cos ωt s s2 + ω2 (s > 0) sinωt ω s2 + ω2 (s > 0) eλt cos ωt s −λ (s... | Elementary Differential Equations with Boundary Value Problems_Page_473_Chunk2877 |
464 Chapter 8 Laplace Transforms t sin ωt 2ωs (s2 + ω2)2 (s > 0) sin ωt −ωt cos ωt 2ω3 (s2 + ω2)2 (s > 0) ωt −sin ωt ω3 s2(s2 + ω2)2 (s > 0) 1 t sinωt arctan ω s (s > 0) eatf(t) F (s −a) tkf(t) (−1)kF (k)(s) f(ωt) 1 ωF s ω , ω > 0 u(t −τ) e−τs s (s > 0) u(t −τ)f(t −τ) (τ > 0) e−τsF (s) Z t o f(τ)g(t −τ) dτ F (s)... | Elementary Differential Equations with Boundary Value Problems_Page_474_Chunk2878 |
CHAPTER 9 Linear Higher Order Equations IN THIS CHAPTER we extend the results obtained in Chapter 5 for linear second order equations to linear higher order equations. SECTION 9.1 presents a theoretical introduction to linear higher order equations. SECTION 9.2 discusses higher order constant coefficient homogeneous equ... | Elementary Differential Equations with Boundary Value Problems_Page_475_Chunk2879 |
466 Chapter 9 Linear Higher Order Equations 9.1 INTRODUCTION TO LINEAR HIGHER ORDER EQUATIONS An nth order differential equation is said to be linear if it can be written in the form y(n) + p1(x)y(n−1) + · · · + pn(x)y = f(x). (9.1.1) We considered equations of this form with n = 1 in Section 2.1 and with n = 2 in Chap... | Elementary Differential Equations with Boundary Value Problems_Page_476_Chunk2880 |
Section 9.1 Introduction to Linear Higher Order Equations 467 Theorem 9.1.2 If Ly = 0 is normal on (a, b), then a set {y1, y2, . . ., yn} of n solutions of Ly = 0 on (a, b) is a fundamental set if and only if it’s linearly independent on (a, b). Example 9.1.1 The equation x3y′′′ −x2y′′ −2xy′ + 6y = 0 (9.1.5) is normal ... | Elementary Differential Equations with Boundary Value Problems_Page_477_Chunk2881 |
468 Chapter 9 Linear Higher Order Equations Example 9.1.2 The equation y(4) + y′′′ −7y′′ −y′ + 6y = 0 (9.1.9) is normal and has the solutions y1 = ex, y2 = e−x, y3 = e2x, and y4 = e−3x on (−∞, ∞). (Verify.) Show that {y1, y2, y3, y4} is linearly independent on (−∞, ∞). Then find the general solution of (9.1.9). Solution... | Elementary Differential Equations with Boundary Value Problems_Page_478_Chunk2882 |
Section 9.1 Introduction to Linear Higher Order Equations 469 for c1, c2, ..., cn. For a fixed x, the determinant of this system is W(x) = y1(x) y2(x) · · · yn(x) y′ 1(x) y′ 2(x) · · · y′ n(x) ... ... ... ... y(n−1) 1 (x) y(n−1) 2 (x) · · · y(n−1) n (x) . We call this determinant the Wronskian of {y1, y2, . . ., yn}. If... | Elementary Differential Equations with Boundary Value Problems_Page_479_Chunk2883 |
470 Chapter 9 Linear Higher Order Equations where we factored x2, x, and 2 out of the first, second, and third rows of W(x), respectively. Adding the second row of the last determinant to the first and third rows yields W(x) = 2x3 3 4x 0 2 3x −1 x3 3 6x 0 = 2x3 1 x3 3 4x 3 6x = 12x. Therefore W(x) ̸= 0 on (−∞, 0) and... | Elementary Differential Equations with Boundary Value Problems_Page_480_Chunk2884 |
Section 9.1 Introduction to Linear Higher Order Equations 471 General Solution of a Nonhomogeneous Equation The next theorem is analogous to Theorem 5.3.2. It shows how to find the general solution of Ly = F if we know a particular solution of Ly = F and a fundamental set of solutions of the complementary equation Ly = ... | Elementary Differential Equations with Boundary Value Problems_Page_481_Chunk2885 |
472 Chapter 9 Linear Higher Order Equations 2. Solve the initial value problem x3y′′′ −x2y′′ −2xy′ + 6y = 0, y(−1) = −4, y′(−1) = −14, y′′(−1) = −20. HINT: See Example 9.1.1. 3. Solve the initial value problem y(4) + y′′′ −7y′′ −y′ + 6y = 0, y(0) = 5, y′(0) = −6, y′′(0) = 10, y′′′(0) −36. HINT: See Example 9.1.2. 4. Fi... | Elementary Differential Equations with Boundary Value Problems_Page_482_Chunk2886 |
Section 9.1 Introduction to Linear Higher Order Equations 473 8. Find the Wronskian W of a set of four solutions of y(4) + (tan x)y′′′ + x2y′′ + 2xy = 0, given that W(π/4) = K. 9. (a) Evaluate the Wronskian W {ex, xex, x2ex}. Evaluate W(0). (b) Verify that y1, y2, and y3 satisfy y′′′ −3y′′ + 3y′ −y = 0. (A) (c) Use W(0... | Elementary Differential Equations with Boundary Value Problems_Page_483_Chunk2887 |
474 Chapter 9 Linear Higher Order Equations 15. Suppose {y1, y2, . . ., yn} is a fundamental set of solutions of P0(x)y(n) + P1(x)y(n−1) + · · · + Pn(x)y = 0 on (a, b), and let z1 = a11y1 + a12y2 + · · · + a1nyn z2 = a21y1 + a22y2 + · · · + a2nyn ... ... ... ... zn = an1y1 + an2y2 + · · · + annyn, where the {aij} are c... | Elementary Differential Equations with Boundary Value Problems_Page_484_Chunk2888 |
Section 9.1 Introduction to Linear Higher Order Equations 475 18. Let F = f11 f12 · · · f1n f21 f22 · · · f2n ... ... ... ... fn1 fn2 · · · fnn , where fij (1 ≤i, j ≤n) is differentiable. Show that F ′ = F1 + F2 + · · · + Fn, where Fi is the determinant obtained by differentiating the ith row of F . 19. Use Exercise 18... | Elementary Differential Equations with Boundary Value Problems_Page_485_Chunk2889 |
476 Chapter 9 Linear Higher Order Equations (e) {x, x2, 1/x} (f) {x + 1, ex, e3x} (g) {x, x3, 1/x, 1/x2} (h) {x, x lnx, 1/x, x2} (i) {ex, e−x, x, e2x} (j) {e2x, e−2x, 1, x2} 9.2 HIGHER ORDER CONSTANT COEFFICIENT HOMOGENEOUS EQUATIONS If a0, a1, ..., an are constants and a0 ̸= 0, then a0y(n) + a1y(n−1) + · · · + any = F... | Elementary Differential Equations with Boundary Value Problems_Page_486_Chunk2890 |
Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 477 Example 9.2.1 (a) Find the general solution of y′′′ −6y′′ + 11y′ −6y = 0. (9.2.3) (b) Solve the initial value problem y′′′ −6y′′ + 11y′ −6y = 0, y(0) = 4, y′(0) = 5, y′′(0) = 9. (9.2.4) Solution The characteristic polynomial of (9.2.3) is p(r) = r3 ... | Elementary Differential Equations with Boundary Value Problems_Page_487_Chunk2891 |
478 Chapter 9 Linear Higher Order Equations 0.5 1 1.5 2 .0 .0 100 200 300 400 500 600 700 x y Figure 9.2.1 y = 4ex −e2x + e3x Since p(r) can be factored as p(r) = (r −1)(r2 + 1) = (r2 + 1)(r −1), it’s reasonable to expect that p(D) can be factored as p(D) = (D −1)(D2 + 1) = (D2 + 1)(D −1). (9.2.7) However, before we ca... | Elementary Differential Equations with Boundary Value Problems_Page_488_Chunk2892 |
Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 479 From (b), (D2 + 1)(D −1)y = (D2 + 1)[(D −1)y] = (D2 + 1)(y′ −y) = D2(y′ −y) + (y′ −y) = (y′′′ −y′′) + (y′ −y) = y′′′ −y′′ + y′ −y = (D3 −D2 + D −1)y, (9.2.9) (D2 + 1)(D −1) = (D3 −D2 + D −1), which completes the justification of (9.2.7). Example 9.2.... | Elementary Differential Equations with Boundary Value Problems_Page_489_Chunk2893 |
480 Chapter 9 Linear Higher Order Equations or (D2 + 4)(D −2)(D + 2)y = 0 or (D −2)(D + 2)(D2 + 4)y = 0. Therefore y is a solution of (9.2.11) if it’s a solution of any of the three equations (D −2)y = 0, (D + 2)y = 0, (D2 + 4)y = 0. Hence, {e2x, e−2x, cos 2x, sin2x} is a set of solutions of (9.2.11). The Wronskian of ... | Elementary Differential Equations with Boundary Value Problems_Page_490_Chunk2894 |
Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 481 problem of finding solutions of p(D)y = 0 with p as in (9.2.14) reduces to finding solutions of each of these equations pj(D)y = 0, 1 ≤j ≤k, where pj is a power of a first degree term or of an irreducible quadratic. To find a fundamental set of solution... | Elementary Differential Equations with Boundary Value Problems_Page_491_Chunk2895 |
482 Chapter 9 Linear Higher Order Equations Example 9.2.5 Find the general solution of y′′′ + 3y′′ + 3y′ + y = 0. (9.2.18) Solution The characteristic polynomial of (9.2.18) is p(r) = r3 + 3r2 + 3r + 1 = (r + 1)3. Therefore (9.2.18) can be written as (D + 1)3y = 0, so Theorem 9.2.1 implies that the general solution of ... | Elementary Differential Equations with Boundary Value Problems_Page_492_Chunk2896 |
Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 483 Therefore (9.2.20) can be written as [(D + 1)2 + 1](D + 2)Dy = 0. Fundamental sets of solutions of (D + 1)2 + 1 y = 0, (D + 2)y = 0, and Dy = 0. are given by {e−x cos x, e−x sin x}, {e−2x}, and {1}, respectively. Therefore the general solution of ... | Elementary Differential Equations with Boundary Value Problems_Page_493_Chunk2897 |
484 Chapter 9 Linear Higher Order Equations In Exercises 15–27 solve the initial value problem. Where indicated by C/G , graph the solution. 15. y′′′ −2y′′ + 4y′ −8y = 0, y(0) = 2, y′(0) = −2, y′′(0) = 0 16. y′′′ + 3y′′ −y′ −3y = 0, y(0) = 0, y′(0) = 14, y′′(0) = −40 17. C/G y′′′ −y′′ −y′ + y = 0, y(0) = −2, y′(0) = 9,... | Elementary Differential Equations with Boundary Value Problems_Page_494_Chunk2898 |
Section 9.2 Higher Order Constant Coefficient Homogeneous Equations 485 39. It can be shown that 1 1 · · · 1 a1 a2 · · · an a2 1 a2 2 · · · a2 n ... ... ... ... an−1 1 an−1 2 · · · an−1 n = Y 1≤i<j≤n (aj −ai), (A) where the left side is the Vandermonde determinant and the right side is the product of all factors of the ... | Elementary Differential Equations with Boundary Value Problems_Page_495_Chunk2899 |
486 Chapter 9 Linear Higher Order Equations 41. (a) Show that if z = p(x) cos ωx + q(x) sin ωx, (A) where p and q are polynomials of degree ≤k, then (D2 + ω2)z = p1(x) cos ωx + q1(x) sin ωx, where p1 and q1 are polynomials of degree ≤k −1. (b) Apply (a) m times to show that if z is of the form (A) where p and q are pol... | Elementary Differential Equations with Boundary Value Problems_Page_496_Chunk2900 |
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