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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 . Therefore a′ 2m(r) = a2m(r) n X j=1 1 2j + r − 1 2j + r −3 . Setting r = 3 here and recalling (7.7.28) yields a′ 2m(3) = (−1)m Qm j=1(2j + 3) 2mm! m X j=1 1 2j + 3 −1 2j . (7.7.30) Since 1 2j + 3 −1 2j = − 3 2j(2j + 3), we can rewrite (7.7.30) as a′ 2m(3) = −3 2 (−1)n Qm j=1(2j + 3) 2mm!   n X j=1 1 j(2j + 3)  . Substituting this into (7.7.29) yields y2 = x−5 1 −1 2x2 + 1 8x4 + 1 16x6 −3 16y1 ln x + 9 32x3 ∞ X m=1 (−1)m Qm j=1(2j + 3) 2mm!   m X j=1 1 j(2j + 3)  x2m. Example 7.7.4 Find a fundamental set of Frobenius solutions of x2(1 −2x2)y′′ + x(7 −13x2)y′ −14x2y = 0. Give explicit formulas for the coefﬁcients in the solutions. Solution For the given equation, the polynomials deﬁned in Theorem 7.7.2 are p0(r) = r(r −1) + 7r = r(r + 6), p2(r) = −2r(r −1) −13r −14 = −(r + 2)(2r + 7).	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)) form a fundamental set of solutions of Ly = 0. The recurrence formulas (7.7.22) are a0(r) = 1, a2m(r) = −p2(2m + r −2) p0(2m + r) a2m−2(r) = (2m + r)(4m + 2r + 3) (2m + r)(2m + r + 6) a2m−2(r) = 4m + 2r + 3 2m + r + 6 a2m−2(r), m ≥1, (7.7.33) which implies that a2m(r) = m Y j=1 4j + 2r + 3 2j + r + 6 . Setting r = 0 yields a2m(0) = 6 Qm j=1(4j + 3) 2m(m + 3)! . Substituting this into (7.7.31) yields y1 = 6 ∞ X m=0 Qm j=1(4j + 3) 2m(m + 3)! x2m. To compute the coefﬁcients a0(−6), a2(−6), and a4(−6) in y2, we set r = −6 in (7.7.33) and apply the resulting recurrence formula for m = 1, 2; thus, a0(−6) = 1, a2m(−6) = 4m −9 2m a2m−2(−6), m = 1, 2. The last formula yields a2(−6) = −5 2 and a4(−6) = 5 8. Since p2(−2) = 0, the constant C in (7.7.23) is zero. Therefore (7.7.32) reduces to y2 = x−6 1 −5 2x2 + 5 8x4 . 7.7 Exercises In Exercises 1–40 ﬁnd a fundamental set of Frobenius solutions. Give explicit formulas for the coefﬁ- cients.	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′′ + 4x(3 + 8x)y′ −(5 −49x)y = 0 10. x2(1 + x)y′′ −x(3 + 10x)y′ + 30xy = 0 11. x2y′′ + x(1 + x)y′ −3(3 + x)y = 0 12. x2y′′ + x(1 −2x)y′ −(4 + x)y = 0 13. x(1 + x)y′′ −4y′ −2y = 0 14. x2(1 + 2x)y′′ + x(9 + 13x)y′ + (7 + 5x)y = 0 15. 4x2y′′ −2x(4 −x)y′ −(7 + 5x)y = 0 16. 3x2(3 + x)y′′ −x(15 + x)y′ −20y = 0 17. x2(1 + x)y′′ + x(1 −10x)y′ −(9 −10x)y = 0 18. x2(1 + x)y′′ + 3x2y′ −(6 −x)y = 0 19. x2(1 + 2x)y′′ −2x(3 + 14x)y′ + (6 + 100x)y = 0 20. x2(1 + x)y′′ −x(6 + 11x)y′ + (6 + 32x)y = 0 21. 4x2(1 + x)y′′ + 4x(1 + 4x)y′ −(49 + 27x)y = 0 22. x2(1 + 2x)y′′ −x(9 + 8x)y′ −12xy = 0 23. x2(1 + x2)y′′ −x(7 −2x2)y′ + 12y = 0 24. x2y′′ −x(7 −x2)y′ + 12y = 0 25. xy′′ −5y′ + xy = 0 26. x2y′′ + x(1 + 2x2)y′ −(1 −10x2)y = 0 27. x2y′′ −xy′ −(3 −x2)y = 0 28. 4x2y′′ + 2x(8 + x2)y′ + (5 + 3x2)y = 0 29. x2y′′ + x(1 + x2)y′ −(1 −3x2)y = 0 30. x2y′′ + x(1 −2x2)y′ −4(1 + 2x2)y = 0 31. 4x2y′′ + 8xy′ −(35 −x2)y = 0 32. 9x2y′′ −3x(11 + 2x2)y′ + (13 + 10x2)y = 0 33. x2y′′ + x(1 −2x2)y′ −4(1 −x2)y = 0 34. x2y′′ + x(1 −3x2)y′ −4(1 −3x2)y = 0 35. x2(1 + x2)y′′ + x(5 + 11x2)y′ + 24x2y = 0 36. 4x2(1 + x2)y′′ + 8xy′ −(35 −x2)y = 0 37. x2(1 + x2)y′′ −x(5 −x2)y′ −(7 + 25x2)y = 0	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 ln x + xr1 ∞ X n=1 a′ n(r1)xn ! are linearly independent. HINT: Show that if c1 and c2 are constants such that c1y1 +c2y2 ≡ 0 on an interval (0, ρ), then x−r2(c1y1(x) + c2y2(x)) = 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 7.7.1. 42. Find a fundamental set of Frobenius solutions of Bessel’s equation x2y′′ + xy′ + (x2 −ν2)y = 0 in the case where ν is a positive integer. 43. Prove Theorem 7.7.2. 44. Under the assumptions of Theorem 7.7.1, show that C = 0 if and only if p1(r2 + ) = 0 for some integer in {0, 1, . . ., k −1}. 45. Under the assumptions of Theorem 7.7.2, show that C = 0 if and only if p2(r2 + 2) = 0 for some integer ℓin {0, 1, . . ., k −1}. 46. Let Ly = α0x2y′′ + β0xy′ + (γ0 + γ1x)y and deﬁne p0(r) = α0r(r −1) + β0r + γ0. Show that if p0(r) = α0(r −r1)(r −r2) where r1 −r2 = k, a positive integer, then Ly = 0 has the solutions y1 = xr1 ∞ X n=0 (−1)n n! Qn j=1(j + k) γ1 α0 n xn and y2 = xr2 k−1 X n=0 (−1)n n! Qn j=1(j −k) γ1 α0 n xn − 1 k!(k −1)! γ1 α0 k  y1 lnx −xr1 ∞ X n=1 (−1)n n! Qn j=1(j + k) γ1 α0 n   n X j=1 2j + k j(j + k)  xn  .	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 (−1)m 4mm! Qm j=1(j −k) γ2 α0 m x2m − 2 4kk!(k −1)! γ2 α0 k  y1 lnx −xr1 2 ∞ X m=1 (−1)m 4mm! Qm j=1(j + k) γ2 α0 m   m X j=1 2j + k j(j + k)  x2m  . 48. Let L be as in Exercises 7.5.57 and 7.5.58, and suppose the indicial polynomial of Ly = 0 is p0(r) = α0(r −r1)(r −r2), with k = r1 −r2, where k is a positive integer. Deﬁne a0(r) = 1 for all r. If r is a real number such that p0(n + r) is nonzero for all positive integers n, deﬁne an(r) = − 1 p0(n + r) n X j=1 pj(n + r −j)an−j(r), n ≥1, and let y1 = xr1 ∞ X n=0 an(r1)xn. Deﬁne an(r2) = − 1 p0(n + r2) n X j=1 pj(n + r2 −j)an−j(r2) if n ≥1 and n ̸= k, and let ak(r2) be arbitrary. (a) Conclude from Exercise 7.6..66 that L y1 ln x + xr1 ∞ X n=1 a′ n(r1)xn ! = kα0xr1. (b) Conclude from Exercise 7.5..57 that L xr2 ∞ X n=0 an(r2)xn ! = Axr1,	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 adds a multiple of y1 to y2. Conclude that we may as well take ak(r2) = 0.	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 discussed here transforms an initial value problem for a constant coefﬁcient equation into an algebraic equation whose solution can then be used to solve the initial value problem. In some cases this method is merely an alternative procedure for solving problems that can be solved equally well by methods that we considered previously; however, in other cases the method of Laplace transforms is more efﬁcient than the methods previously discussed. This is especially true in physical problems dealing with discontinuous forcing functions. SECTION 8.1 deﬁnes the Laplace transform and developes its properties. SECTION 8.2 deals with the problem of ﬁnding a function that has a given Laplace transform. SECTION 8.3 applies the Laplace transform to solve initial value problems for constant coefﬁcient second order differential equations on (0, ∞). SECTION 8.4 introduces the unit step function. SECTION 8.5 uses the unit step function to solve constant coefﬁcient equations with piecewise continu- ous forcing functions. SECTION 8.6 deals with the convolution theorem, an important theoretical property of the Laplace trans- form. SECTION 8.7 introduces the idea of impulsive force, and treats constant coefﬁcient equations with im- pulsive forcing functions. SECTION 8.8 is a brief table of Laplace transforms. 393	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 g(t) dt = lim T →∞ Z T a g(t) dt. (8.1.1) We say that the improper integral converges if the limit in (8.1.1) exists; otherwise, we say that the improper integral diverges or does not exist. Here’s the deﬁnition of the Laplace transform of a function f. Deﬁnition 8.1.1 Let f be deﬁned for t ≥0 and let s be a real number. Then the Laplace transform of f is the function F deﬁned by F (s) = Z ∞ 0 e−stf(t) dt, (8.1.2) for those values of s for which the improper integral converges. It is important to keep in mind that the variable of integration in (8.1.2) is t, while s is a parameter independent of t. We use t as the independent variable for f because in applications the Laplace transform is usually applied to functions of time. The Laplace transform can be viewed as an operator L that transforms the function f = f(t) into the function F = F (s). Thus, (8.1.2) can be expressed as F = L(f). The functions f and F form a transform pair, which we’ll sometimes denote by f(t) ↔F (s). It can be shown that if F (s) is deﬁned for s = s0 then it’s deﬁned for all s > s0 (Exercise 14(b)). Computation of Some Simple Laplace Transforms Example 8.1.1 Find the Laplace transform of f(t) = 1. Solution From (8.1.2) with f(t) = 1, F (s) = Z ∞ 0 e−st dt = lim T →∞ Z T 0 e−st dt. If s ̸= 0 then Z T 0 e−stdt = −1 se−st T 0 = 1 −e−sT s . (8.1.3) Therefore lim T →∞ Z T 0 e−stdt = ( 1 s, s > 0, ∞, s < 0. (8.1.4)	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 transform pair 1 ↔1 s, s > 0. REMARK: It is convenient to combine the steps of integrating from 0 to T and letting T →∞. Therefore, instead of writing (8.1.3) and (8.1.4) as separate steps we write Z ∞ 0 e−stdt = −1 se−st ∞ 0 = ( 1 s, s > 0, ∞, s < 0. We’ll follow this practice throughout this chapter. Example 8.1.2 Find the Laplace transform of f(t) = t. Solution From (8.1.2) with f(t) = t, F (s) = Z ∞ 0 e−stt dt. (8.1.5) If s ̸= 0, integrating by parts yields Z ∞ 0 e−stt dt = −te−st s ∞ 0 + 1 s Z ∞ 0 e−st dt = − t s + 1 s2 e−st ∞ 0 = ( 1 s2 , s > 0, ∞, s < 0. If s = 0, the integral in (8.1.5) becomes Z ∞ 0 t dt = t2 2 ∞ 0 = ∞. Therefore F (0) is undeﬁned and F (s) = 1 s2 , s > 0. This result can also be written as L(t) = 1 s2 , s > 0, or as the transform pair t ↔1 s2 , s > 0.	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 by s −a here shows that F (s) = 1 s −a, s > a. This can also be written as L(eat) = 1 s −a, s > a, or eat ↔ 1 s −a, s > a. Example 8.1.4 Find the Laplace transforms of f(t) = sin ωt and g(t) = cos ωt, where ω is a constant. Solution Deﬁne F (s) = Z ∞ 0 e−st sin ωt dt (8.1.6) and G(s) = Z ∞ 0 e−st cos ωt dt. (8.1.7) If s > 0, integrating (8.1.6) by parts yields F (s) = −e−st s sinωt ∞ 0 + ω s Z ∞ 0 e−st cos ωt dt, so F (s) = ω s G(s). (8.1.8) If s > 0, integrating (8.1.7) by parts yields G(s) = −e−st cos ωt s ∞ 0 −ω s Z ∞ 0 e−st sin ωt dt, so G(s) = 1 s −ω s F (s). Now substitute from (8.1.8) into this to obtain G(s) = 1 s −ω2 s2 G(s).	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 transforms in the Appendix will be adequate for our purposes. Example 8.1.5 Use the table of Laplace transforms to ﬁnd L(t3e4t). Solution The table includes the transform pair tneat ↔ n! (s −a)n+1 . Setting n = 3 and a = 4 here yields L(t3e4t) = 3! (s −4)4 = 6 (s −4)4 . We’ll sometimes write Laplace transforms of speciﬁc functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms. Linearity of the Laplace Transform The next theorem presents an important property of the Laplace transform. Theorem 8.1.2 [Linearity Property] Suppose L(fi) is deﬁned for s > si, 1 ≤i ≤n). Let s0 be the largest of the numbers s1, s2, ...,sn, and let c1, c2,..., cn be constants. Then L(c1f1 + c2f2 + · · · + cnfn) = c1L(f1) + c2L(f2) + · · · + cnL(fn) for s > s0. Proof We give the proof for the case where n = 2. If s > s0 then L(c1f1 + c2f2) = Z ∞ 0 e−st (c1f1(t) + c2f2(t))) dt = c1 Z ∞ 0 e−stf1(t) dt + c2 Z ∞ 0 e−stf2(t) dt = c1L(f1) + c2L(f2). Example 8.1.6 Use Theorem 8.1.2 and the known Laplace transform L(eat) = 1 s −a to ﬁnd L(cosh bt) (b ̸= 0).	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 s > \|b\|. Simplifying the last expression in (8.1.9) yields L(cosh bt) = s s2 −b2 , s > \|b\|. The First Shifting Theorem The next theorem enables us to start with known transform pairs and derive others. (For other results of this kind, see Exercises 6 and 13.) Theorem 8.1.3 [First Shifting Theorem] If F (s) = Z ∞ 0 e−stf(t) dt (8.1.10) is the Laplace transform of f(t) for s > s0, then F (s −a) is the Laplace transform of eatf(t) for s > s0 + a. PROOF. Replacing s by s −a in (8.1.10) yields F (s −a) = Z ∞ 0 e−(s−a)tf(t) dt (8.1.11) if s −a > s0; that is, if s > s0 + a. However, (8.1.11) can be rewritten as F (s −a) = Z ∞ 0 e−st	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 Transforms Not every function has a Laplace transform. For example, it can be shown (Exercise 3) that Z ∞ 0 e−stet2dt = ∞ for every real number s. Hence, the function f(t) = et2 does not have a Laplace transform. Our next objective is to establish conditions that ensure the existence of the Laplace transform of a function. We ﬁrst review some relevant deﬁnitions from calculus. Recall that a limit lim t→t0 f(t) exists if and only if the one-sided limits lim t→t0−f(t) and lim t→t0+ f(t) both exist and are equal; in this case, lim t→t0 f(t) = lim t→t0−f(t) = lim t→t0+ f(t). Recall also that f is continuous at a point t0 in an open interval (a, b) if and only if lim t→t0 f(t) = f(t0), which is equivalent to lim t→t0+ f(t) = lim t→t0−f(t) = f(t0). (8.1.12) For simplicity, we deﬁne f(t0+) = lim t→t0+ f(t) and f(t0−) = lim t→t0−f(t), so (8.1.12) can be expressed as f(t0+) = f(t0−) = f(t0). If f(t0+) and f(t0−) have ﬁnite but distinct values, we say that f has a jump discontinuity at t0, and f(t0+) −f(t0−) is called the jump in f at t0 (Figure 8.1.1).	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 function f with a removable discontinuity at t0 can be made continuous at t0 by deﬁning (or redeﬁning) f(t0) = f(t0+) = f(t0−). REMARK: We know from calculus that a deﬁnite integral isn’t affected by changing the values of its integrand at isolated points. Therefore, redeﬁning a function f to make it continuous at removable dis- continuities does not change L(f). Deﬁnition 8.1.4 (i) A function f is said to be piecewise continuous on a ﬁnite closed interval [0, T] if f(0+) and f(T−) are ﬁnite and f is continuous on the open interval (0, T) except possibly at ﬁnitely many points, where f may have jump discontinuities or removable discontinuities. (ii) A function f is said to be piecewise continuous on the inﬁnite interval [0, ∞) if it’s piecewise continuous on [0, T] for every T > 0. Figure 8.1.3 shows the graph of a typical piecewise continuous function. It is shown in calculus that if a function is piecewise continuous on a ﬁnite closed interval then it’s integrable on that interval. But if f is piecewise continuous on [0, ∞), then so is e−stf(t), and therefore Z T 0 e−stf(t) dt	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) converges for s in some interval (s0, ∞). For example, we noted earlier that (8.1.13) diverges for all s if f(t) = et2. Stated informally, this occurs because et2 increases too rapidly as t →∞. The next deﬁnition provides a constraint on the growth of a function that guarantees convergence of its Laplace transform for s in some interval (s0, ∞) . Deﬁnition 8.1.5 A function f is said to be of exponential order s0 if there are constants M and t0 such that \|f(t)\| ≤Mes0t, t ≥t0. (8.1.14) In situations where the speciﬁc value of s0 is irrelevant we say simply that f is of exponential order. The next theorem gives useful sufﬁcient conditions for a function f to have a Laplace transform. The proof is sketched in Exercise 10. Theorem 8.1.6 If f is piecewise continuous on [0, ∞) and of exponential order s0, then L(f) is deﬁned for s > s0. REMARK: We emphasize that the conditions of Theorem 8.1.6 are sufﬁcient, but not necessary, for f to have a Laplace transform. For example, Exercise 14(c) shows that f may have a Laplace transform even though f isn’t of exponential order. Example 8.1.8 If f is bounded on some interval [t0, ∞), say \|f(t)\| ≤M, t ≥t0, then (8.1.14) holds with s0 = 0, so f is of exponential order zero. Thus, for example, sin ωt and cos ωt are of exponential order zero, and Theorem 8.1.6 implies that L(sin ωt) and L(cos ωt) exist for s > 0. This is consistent with the conclusion of Example 8.1.4.	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, ∞). Therefore Exercise 9 and Theorem 8.1.6 imply that L(tα) exists for s ≥s0. However, since s0 is an arbitrary positive number, this really implies that L(tα) exists for all s > 0. This is consistent with the results of Example 8.1.2 and Exercises 6 and 8. Example 8.1.10 Find the Laplace transform of the piecewise continuous function f(t) = 1, 0 ≤t < 1, −3e−t, t ≥1. Solution Since f is deﬁned by different formulas on [0, 1) and [1, ∞), we write F (s) = Z ∞ 0 e−stf(t) dt = Z 1 0 e−st(1) dt + Z ∞ 1 e−st(−3e−t) dt. Since Z 1 0 e−st dt =    1 −e−s s , s ̸= 0, 1, s = 0, and Z ∞ 1 e−st(−3e−t) dt = −3 Z ∞ 1 e−(s+1)t dt = −3e−(s+1) s + 1 , s > −1, it follows that F (s) =      1 −e−s s −3e−(s+1) s + 1 , s > −1, s ̸= 0, 1 −3 e , s = 0. This is consistent with Theorem 8.1.6, since \|f(t)\| ≤3e−t, t ≥1, and therefore f is of exponential order s0 = −1. REMARK: In Section 8.4 we’ll develop a more efﬁcient method for ﬁnding Laplace transforms of piece- wise continuous functions. Example 8.1.11 We stated earlier that Z ∞ 0 e−stet2dt = ∞ for all s, so Theorem 8.1.6 implies that f(t) = et2 is not of exponential order, since lim t→∞ et2 Mes0t = lim t→∞ 1 M et2−s0t = ∞, so et2 > Mes0t for sufﬁciently large values of t, for any choice of M and s0 (Exercise 3).	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 functions. (a) cosh t sin t (b) sin2 t (c) cos2 2t (d) cosh2 t (e) t sinh 2t (f) sin t cos t (g) sin t + π 4 (h) cos 2t −cos 3t (i) sin 2t + cos 4t 3. Show that Z ∞ 0 e−stet2dt = ∞ for every real number s. 4. Graph the following piecewise continuous functions and evaluate f(t+), f(t−), and f(t) at each point of discontinuity. (a) f(t) =    −t, 0 ≤t < 2, t −4, 2 ≤t < 3, 1, t ≥3. (b) f(t) =    t2 + 2, 0 ≤t < 1, 4, t = 1, t, t > 1. (c) f(t) =    sin t, 0 ≤t < π/2, 2 sin t, π/2 ≤t < π, cos t, t ≥π. (d) f(t) =            t, 0 ≤t < 1, 2, t = 1, 2 −t, 1 ≤t < 2, 3, t = 2, 6, t > 2. 5. Find the Laplace transform: (a) f(t) = e−t, 0 ≤t < 1, e−2t, t ≥1. (b) f(t) = 1, 0 ≤t < 4, t, t ≥4. (c) f(t) = t, 0 ≤t < 1, 1, t ≥1. (d) f(t) = tet, 0 ≤t < 1, et, t ≥1. 6. Prove that if f(t) ↔F (s) then tkf(t) ↔(−1)kF (k)(s). HINT: Assume that it’s permissible to differentiate the integral R ∞ 0 e−stf(t) dt with respect to s under the integral sign. 7. Use the known Laplace transforms L(eλt sin ωt) = ω (s −λ)2 + ω2 and L(eλt cos ωt) = s −λ (s −λ)2 + ω2 and the result of Exercise 6 to ﬁnd L(teλt cos ωt) and L(teλt sin ωt). 8. Use the known Laplace transform L(1) = 1/s and the result of Exercise 6 to show that L(tn) = n! sn+1 , n = integer. 9. (a) Show that if limt→∞e−s0tf(t) exists and is ﬁnite then f is of exponential order s0. (b) Show that if f is of exponential order s0 then limt→∞e−stf(t) = 0 for all s > s0.	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 ≥0) such that \|g(t)\| ≤w(t) for t ≥τ and R ∞ τ w(t) dt converges. Then R ∞ 0 g(t) dt converges. Use Theorem A to show that if f is piecewise continuous on [0, ∞) and of exponential order s0, then f has a Laplace transform F (s) deﬁned for s > s0. 11. Prove: If f is piecewise continuous and of exponential order then lims→∞F (s) = 0. 12. Prove: If f is continuous on [0, ∞) and of exponential order s0 > 0, then L Z t 0 f(τ) dτ = 1 sL(f), s > s0. HINT: Use integration by parts to evaluate the transform on the left. 13. Suppose f is piecewise continuous and of exponential order, and that limt→0+ f(t)/t exists. Show that L f(t) t = Z ∞ s F (r) dr. HINT: Use the results of Exercises 6 and 11. 14. Suppose f is piecewise continuous on [0, ∞). (a) Prove: If the integral g(t) = R t 0 e−s0τf(τ) dτ satisﬁes the inequality \|g(t)\| ≤M (t ≥0), then f has a Laplace transform F (s) deﬁned for s > s0. HINT: Use integration by parts to show that Z T 0 e−stf(t) dt = e−(s−s0)T g(T) + (s −s0) Z T 0 e−(s−s0)tg(t) dt. (b) Show that if L(f) exists for s = s0 then it exists for s > s0. Show that the function f(t) = tet2 cos(et2) has a Laplace transform deﬁned for s > 0, even though f isn’t of exponential order. (c) Show that the function f(t) = tet2 cos(et2) has a Laplace transform deﬁned for s > 0, even though f isn’t of exponential order. 15. Use the table of Laplace transforms and the result of Exercise 13 to ﬁnd the Laplace transforms of the following functions. (a) sin ωt t (ω > 0) (b) cos ωt −1 t (ω > 0) (c) eat −ebt t (d) cosh t −1 t (e) sinh2 t t 16. The gamma function is deﬁned by Γ(α) = Z ∞ 0 xα−1e−x dx, which can be shown to converge if α > 0.	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. HINT: Change the variable of integration in the integral for Γ(α + 1). 17. Suppose f is continuous on [0, T] and f(t + T) = f(t) for all t ≥0. (We say in this case that f is periodic with period T.) (a) Conclude from Theorem 8.1.6 that the Laplace transform of f is deﬁned for s > 0. HINT: Since f is continuous on [0, T] and periodic with period T, it’s bounded on [0, ∞). (b) (b) Show that F (s) = 1 1 −e−sT Z T 0 e−stf(t) dt, s > 0. HINT: Write F (s) = ∞ X n=0 Z (n+1)T nT e−stf(t) dt. Then show that Z (n+1)T nT e−stf(t) dt = e−nsT Z T 0 e−stf(t) dt, and recall the formula for the sum of a geometric series. 18. Use the formula given in Exercise 17(b) to ﬁnd the Laplace transforms of the given periodic functions: (a) f(t) = t, 0 ≤t < 1, 2 −t, 1 ≤t < 2, f(t + 2) = f(t), t ≥0 (b) f(t) = 1, 0 ≤t < 1 2, −1, 1 2 ≤t < 1, f(t + 1) = f(t), t ≥0 (c) f(t) = \| sin t\| (d) f(t) = sin t, 0 ≤t < π, 0, π ≤t < 2π, f(t + 2π) = f(t) 8.2 THE INVERSE LAPLACE TRANSFORM Deﬁnition of the Inverse Laplace Transform In Section 8.1 we deﬁned the Laplace transform of f by F (s) = L(f) = Z ∞ 0 e−stf(t) dt. We’ll also say that f is an inverse Laplace Transform of F , and write f = L−1(F ).	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 to ﬁnd inverse transforms that we’ll need. Example 8.2.1 Use the table of Laplace transforms to ﬁnd (a) L−1 1 s2 −1 and (b) L−1 s s2 + 9 . SOLUTION(a) Setting b = 1 in the transform pair sinh bt ↔ b s2 −b2 shows that L−1 1 s2 −1 = sinh t. SOLUTION(b) Setting ω = 3 in the transform pair cos ωt ↔ s s2 + ω2 shows that L−1 s s2 + 9 = cos 3t. The next theorem enables us to ﬁnd inverse transforms of linear combinations of transforms in the table. We omit the proof. Theorem 8.2.1 [Linearity Property] If F1, F2, ..., Fn are Laplace transforms and c1, c2, ..., cn are constants, then L−1(c1F1 + c2F2 + · · · + cnFn) = c1L−1(F1) + c2L−1(F2) + · · · + cnL−1Fn. Example 8.2.2 Find L−1 8 s + 5 + 7 s2 + 3 . Solution From the table of Laplace transforms in Section 8.8„ eat ↔ 1 s −a and sinωt ↔ ω s2 + ω2 . Theorem 8.2.1 with a = −5 and ω = √ 3 yields L−1 8 s + 5 + 7 s2 + 3 = 8L−1 1 s + 5 + 7L−1 1 s2 + 3 = 8L−1 1 s + 5 + 7 √ 3L−1 √ 3 s2 + 3 ! = 8e−5t + 7 √ 3 sin √ 3t.	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 + 1)2 + 4, and write L−1 3s + 8 (s + 1)2 + 4 = L−1 3s + 3 (s + 1)2 + 4 + L−1 5 (s + 1)2 + 4 = 3L−1 s + 1 (s + 1)2 + 4 + 5 2L−1 2 (s + 1)2 + 4 = e−t(3 cos 2t + 5 2 sin2t). REMARK: We’ll often write inverse Laplace transforms of speciﬁc functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms in Section 8.8. Inverse Laplace Transforms of Rational Functions Using the Laplace transform to solve differential equations often requires ﬁnding the inverse transform of a rational function F (s) = P (s) Q(s), where P and Q are polynomials in s with no common factors. Since it can be shown that lims→∞F (s) = 0 if F is a Laplace transform, we need only consider the case where degree(P ) < degree(Q). To obtain L−1(F ), we ﬁnd the partial fraction expansion of F , obtain inverse transforms of the individual terms in the expansion from the table of Laplace transforms, and use the linearity property of the inverse transform. The next two examples illustrate this. Example 8.2.4 Find the inverse Laplace transform of F (s) = 3s + 2 s2 −3s + 2. (8.2.1) Solution (METHOD 1) Factoring the denominator in (8.2.1) yields F (s) = 3s + 2 (s −1)(s −2). (8.2.2) The form for the partial fraction expansion is 3s + 2 (s −1)(s −2) = A s −1 + B s −2. (8.2.3)	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) by (s −1)(s −2) to compute A and B. We can obtain A by simply ignoring the factor s −1 in the denominator of (8.2.2) and setting s = 1 elsewhere; thus, A = 3s + 2 s −2 s=1 = 3 · 1 + 2 1 −2 = −5. (8.2.4) Similarly, we can obtain B by ignoring the factor s −2 in the denominator of (8.2.2) and setting s = 2 elsewhere; thus, B = 3s + 2 s −1 s=2 = 3 · 2 + 2 2 −1 = 8. (8.2.5) To justify this, we observe that multiplying (8.2.3) by s −1 yields 3s + 2 s −2 = A + (s −1) B s −2, and setting s = 1 leads to (8.2.4). Similarly, multiplying (8.2.3) by s −2 yields 3s + 2 s −1 = (s −2) A s −2 + B and setting s = 2 leads to (8.2.5). (It isn’t necesary to write the last two equations. We wrote them only to justify the shortcut procedure indicated in (8.2.4) and (8.2.5).) The shortcut employed in the second solution of Example 8.2.4 is Heaviside’s method. The next theo- rem states this method formally. For a proof and an extension of this theorem, see Exercise 10. Theorem 8.2.2 Suppose F (s) = P (s) (s −s1)(s −s2) · · · (s −sn), (8.2.6) where s1, s2, ..., sn are distinct and P is a polynomial of degree less than n. Then F (s) = A1 s −s1 + A2 s −s2 + · · · + An s −sn , where Ai can be computed from (8.2.6) by ignoring the factor s −si and setting s = si elsewhere. Example 8.2.5 Find the inverse Laplace transform of F (s) = 6 + (s + 1)(s2 −5s + 11) s(s −1)(s −2)(s + 1) . (8.2.7)	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 coefﬁcients are given by B = 6 + (2)(7) (1)(−1)(2) = −10, C = 6 + 3(5) 2(1)(3) = 7 2, and D = 6 (−1)(−2)(−3) = −1. Therefore F (s) = 17 2 1 s − 10 s −1 + 7 2 1 s −2 − 1 s + 1 and L−1(F ) = 17 2 L−1 1 s −10L−1 1 s −1 + 7 2L−1 1 s −2 −L−1 1 s + 1 = 17 2 −10et + 7 2e2t −e−t. REMARK: We didn’t “multiply out” the numerator in (8.2.7) before computing the coefﬁcients in (8.2.8), since it wouldn’t simplify the computations. Example 8.2.6 Find the inverse Laplace transform of F (s) = 8 −(s + 2)(4s + 10) (s + 1)(s + 2)2 . (8.2.9) Solution The form for the partial fraction expansion is F (s) = A s + 1 + B s + 2 + C (s + 2)2 . (8.2.10) Because of the repeated factor (s + 2)2 in (8.2.9), Heaviside’s method doesn’t work. Instead, we ﬁnd a common denominator in (8.2.10). This yields F (s) = A(s + 2)2 + B(s + 1)(s + 2) + C(s + 1) (s + 1)(s + 2)2 . (8.2.11) If (8.2.9) and (8.2.11) are to be equivalent, then A(s + 2)2 + B(s + 1)(s + 2) + C(s + 1) = 8 −(s + 2)(4s + 10). (8.2.12)	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 = −2 to obtain C = −8, and s = −1 to obtain A = 2. We can now choose any third value of s to determine B. Taking s = 0 yields 4A + 2B + C = −12. Since A = 2 and C = −8 this implies that B = −6. Therefore F (s) = 2 s + 1 − 6 s + 2 − 8 (s + 2)2 and L−1(F ) = 2L−1 1 s + 1 −6L−1 1 s + 2 −8L−1 1 (s + 2)2 = 2e−t −6e−2t −8te−2t. Example 8.2.7 Find the inverse Laplace transform of F (s) = s2 −5s + 7 (s + 2)3 . Solution The form for the partial fraction expansion is F (s) = A s + 2 + B (s + 2)2 + C (s + 2)3 . The easiest way to obtain A, B, and C is to expand the numerator in powers of s + 2. This yields s2 −5s + 7 = [(s + 2) −2]2 −5[(s + 2) −2] + 7 = (s + 2)2 −9(s + 2) + 21. Therefore F (s) = (s + 2)2 −9(s + 2) + 21 (s + 2)3 = 1 s + 2 − 9 (s + 2)2 + 21 (s + 2)3 and L−1(F ) = L−1 1 s + 2 −9L−1 1 (s + 2)2 + 21 2 L−1 2 (s + 2)3 = e−2t 1 −9t + 21 2 t2 . Example 8.2.8 Find the inverse Laplace transform of F (s) = 1 −s(5 + 3s) s [(s + 1)2 + 1]. (8.2.13)	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 inverse transforms e−t cos t and e−t sin t of s + 1 (s + 1)2 + 1 and 1 (s + 1)2 + 1 respectively. Therefore, instead of (8.2.14) we write F (s) = A s + B(s + 1) + C (s + 1)2 + 1 . (8.2.15) Finding a common denominator yields F (s) = A (s + 1)2 + 1 + B(s + 1)s + Cs s [(s + 1)2 + 1] . (8.2.16) If (8.2.13) and (8.2.16) are to be equivalent, then A (s + 1)2 + 1 + B(s + 1)s + Cs = 1 −s(5 + 3s). This is true for all s if it’s true for three distinct values of s. Choosing s = 0, −1, and 1 yields the system 2A = 1 A −C = 3 5A + 2B + C = −7. Solving this system yields A = 1 2, B = −7 2, C = −5 2. Hence, from (8.2.15), F (s) = 1 2s −7 2 s + 1 (s + 1)2 + 1 −5 2 1 (s + 1)2 + 1. Therefore L−1(F ) = 1 2L−1 1 s −7 2L−1 s + 1 (s + 1)2 + 1 −5 2L−1 1 (s + 1)2 + 1 = 1 2 −7 2e−t cos t −5 2e−t sin t. Example 8.2.9 Find the inverse Laplace transform of F (s) = 8 + 3s (s2 + 1)(s2 + 4). (8.2.17)	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 the denom- inator of (8.2.17), there’s an easier way: the expansion of F1(s) = 1 (s2 + 1)(s2 + 4) can be obtained quickly by using Heaviside’s method to expand 1 (x + 1)(x + 4) = 1 3 1 x + 1 − 1 x + 4 and then setting x = s2 to obtain 1 (s2 + 1)(s2 + 4) = 1 3 1 s2 + 1 − 1 s2 + 4 . Multiplying this by 8 + 3s yields F (s) = 8 + 3s (s2 + 1)(s2 + 4) = 1 3 8 + 3s s2 + 1 −8 + 3s s2 + 4 . Therefore L−1(F ) = 8 3 sin t + cos t −4 3 sin2t −cos 2t. USING TECHNOLOGY Some software packages that do symbolic algebra can ﬁnd partial fraction expansions very easily. We recommend that you use such a package if one is available to you, but only after you’ve done enough partial fraction expansions on your own to master the technique. 8.2 Exercises 1. Use the table of Laplace transforms to ﬁnd the inverse Laplace transform. (a) 3 (s −7)4 (b) 2s −4 s2 −4s + 13 (c) 1 s2 + 4s + 20 (d) 2 s2 + 9 (e) s2 −1 (s2 + 1)2 (f) 1 (s −2)2 −4 (g) 12s −24 (s2 −4s + 85)2 (h) 2 (s −3)2 −9 (i) s2 −4s + 3 (s2 −4s + 5)2 2. Use Theorem 8.2.1 and the table of Laplace transforms to ﬁnd the inverse Laplace transform.	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 method to ﬁnd the inverse Laplace transform. (a) 3 −(s + 1)(s −2) (s + 1)(s + 2)(s −2) (b) 7 + (s + 4)(18 −3s) (s −3)(s −1)(s + 4) (c) 2 + (s −2)(3 −2s) (s −2)(s + 2)(s −3) (d) 3 −(s −1)(s + 1) (s + 4)(s −2)(s −1) (e) 3 + (s −2)(10 −2s −s2) (s −2)(s + 2)(s −1)(s + 3) (f) 3 + (s −3)(2s2 + s −21) (s −3)(s −1)(s + 4)(s −2) 4. Find the inverse Laplace transform. (a) 2 + 3s (s2 + 1)(s + 2)(s + 1) (b) 3s2 + 2s + 1 (s2 + 1)(s2 + 2s + 2) (c) 3s + 2 (s −2)(s2 + 2s + 5) (d) 3s2 + 2s + 1 (s −1)2(s + 2)(s + 3) (e) 2s2 + s + 3 (s −1)2(s + 2)2 (f) 3s + 2 (s2 + 1)(s −1)2 5. Use the method of Example 8.2.9 to ﬁnd the inverse Laplace transform. (a) 3s + 2 (s2 + 4)(s2 + 9) (b) −4s + 1 (s2 + 1)(s2 + 16) (c) 5s + 3 (s2 + 1)(s2 + 4) (d) −s + 1 (4s2 + 1)(s2 + 1) (e) 17s −34 (s2 + 16)(16s2 + 1) (f) 2s −1 (4s2 + 1)(9s2 + 1) 6. Find the inverse Laplace transform. (a) 17s −15 (s2 −2s + 5)(s2 + 2s + 10) (b) 8s + 56 (s2 −6s + 13)(s2 + 2s + 5) (c) s + 9 (s2 + 4s + 5)(s2 −4s + 13) (d) 3s −2 (s2 −4s + 5)(s2 −6s + 13) (e) 3s −1 (s2 −2s + 2)(s2 + 2s + 5) (f) 20s + 40 (4s2 −4s + 5)(4s2 + 4s + 5) 7. Find the inverse Laplace transform. (a) 1 s(s2 + 1) (b) 1 (s −1)(s2 −2s + 17) (c) 3s + 2 (s −2)(s2 + 2s + 10) (d) 34 −17s (2s −1)(s2 −2s + 5) (e) s + 2 (s −3)(s2 + 2s + 5) (f) 2s −2 (s −2)(s2 + 2s + 10) 8. Find the inverse Laplace transform.	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) If s1, s2, ..., sn are distinct and P is a polynomial of degree less than n, then P (s) (s −s1)(s −s2) · · · (s −sn) = A1 s −s1 + A2 s −s2 + · · · + An s −sn . Multiply through by s −si to show that Ai can be obtained by ignoring the factor s −si on the left and setting s = si elsewhere. (b) Suppose P and Q1 are polynomials such that degree(P ) ≤degree(Q1) and Q1(s1) ̸= 0. Show that the coefﬁcient of 1/(s −s1) in the partial fraction expansion of F (s) = P (s) (s −s1)Q1(s) is P (s1)/Q1(s1). (c) Explain how the results of (a) and (b) are related. 8.3 SOLUTION OF INITIAL VALUE PROBLEMS Laplace Transforms of Derivatives In the rest of this chapter we’ll use the Laplace transform to solve initial value problems for constant coefﬁcient second order equations. To do this, we must know how the Laplace transform of f′ is related to the Laplace transform of f. The next theorem answers this question. Theorem 8.3.1 Suppose f is continuous on [0, ∞) and of exponential order s0, and f′ is piecewise continuous on [0, ∞). Then f and f′ have Laplace transforms for s > s0, and L(f′) = sL(f) −f(0). (8.3.1) Proof We know from Theorem 8.1.6 that L(f) is deﬁned for s > s0. We ﬁrst consider the case where f′ is continuous on [0, ∞). Integration by parts yields Z T 0 e−stf′(t) dt = e−stf(t) T 0 + s Z T 0 e−stf(t) dt = e−sT f(T) −f(0) + s Z T 0 e−stf(t) dt (8.3.2) for any T > 0. Since f is of exponential order s0, limT →∞e−sT f(T) = 0 and the last integral in (8.3.2) converges as T →∞if s > s0. Therefore Z ∞ 0 e−stf′(t) dt = −f(0) + s Z ∞ 0 e−stf(t) dt = −f(0) + sL(f),	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) dt = e−stif(ti) −e−sti−1f(ti−1) + s Z ti ti−1 e−stf(t) dt. Summing both sides of this equation from i = 1 to n and noting that	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 implies that L(f′) exists and satisﬁes (8.3.4) for s > s0. To prove that L(f′′) exists and satisﬁes (8.3.5) for s > s0, we ﬁrst apply Theorem 8.3.1 to g = f′. Since g satisﬁes the hypotheses of Theorem 8.3.1, we conclude that L(g′) is deﬁned and satisﬁes L(g′) = sL(g) −g(0) for s > s0. However, since g′ = f′′, this can be rewritten as L(f′′) = sL(f′) −f′(0). Substituting (8.3.4) into this yields (8.3.5). Solving Second Order Equations with the Laplace Transform We’ll now use the Laplace transform to solve initial value problems for second order equations. Example 8.3.2 Use the Laplace transform to solve the initial value problem y′′ −6y′ + 5y = 3e2t, y(0) = 2, y′(0) = 3. (8.3.6) Solution Taking Laplace transforms of both sides of the differential equation in (8.3.6) yields L(y′′ −6y′ + 5y) = L	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 necessary to write all the steps that we used to obtain (8.3.8). To see how to avoid this, let’s apply the method of Example 8.3.2 to the general initial value problem ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1. (8.3.9) Taking Laplace transforms of both sides of the differential equation in (8.3.9) yields aL(y′′) + bL(y′) + cL(y) = F (s). (8.3.10) Now let Y (s) = L(y). Theorem 8.3.2 and the initial conditions in (8.3.9) imply that L(y′) = sY (s) −k0 and L(y′′) = s2Y (s) −k1 −k0s. Substituting these into (8.3.10) yields a	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, so the solution of (8.3.14) is y = L−1(Y (s)) = 4 3e−t/2 −8e−t + 8 3e−2t (Figure 8.3.1). 1 2 3 4 5 6 7 −1 −2 −3 −4 t y Figure 8.3.1 y = 4 3e−t/2 −8e−t + 8 3e−2t 1 2 3 4 5 6 7 1 −1 −2 −3 −4 .5 t y Figure 8.3.2 y = 1 2 −7 2e−t cos t −5 2e−t sin t Example 8.3.4 Solve the initial value problem y′′ + 2y′ + 2y = 1, y(0) = −3, y′(0) = 1. (8.3.15) Solution The characteristic polynomial is p(s) = s2 + 2s + 2 = (s + 1)2 + 1 and F (s) = L(1) = 1 s,	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 therefore the solution of (8.3.15). REMARK: In our examples we applied Theorems 8.3.1 and 8.3.2 without verifying that the unknown function y satisﬁes their hypotheses. This is characteristic of the formal manipulative way in which the Laplace transform is used to solve differential equations. Any doubts about the validity of the method for solving a given equation can be resolved by verifying that the resulting function y is the solution of the given problem. 8.3 Exercises In Exercises 1–31 use the Laplace transform to solve the initial value problem. 1. y′′ + 3y′ + 2y = et, y(0) = 1, y′(0) = −6 2. y′′ −y′ −6y = 2, y(0) = 1, y′(0) = 0 3. y′′ + y′ −2y = 2e3t, y(0) = −1, y′(0) = 4 4. y′′ −4y = 2e3t, y(0) = 1, y′(0) = −1 5. y′′ + y′ −2y = e3t, y(0) = 1, y′(0) = −1 6. y′′ + 3y′ + 2y = 6et, y(0) = 1, y′(0) = −1 7. y′′ + y = sin2t, y(0) = 0, y′(0) = 1 8. y′′ −3y′ + 2y = 2e3t, y(0) = 1, y′(0) = −1 9. y′′ −3y′ + 2y = e4t, y(0) = 1, y′(0) = −2 10. y′′ −3y′ + 2y = e3t, y(0) = −1, y′(0) = −4 11. y′′ + 3y′ + 2y = 2et, y(0) = 0, y′(0) = −1 12. y′′ + y′ −2y = −4, y(0) = 2, y′(0) = 3 13. y′′ + 4y = 4, y(0) = 0, y′(0) = 1 14. y′′ −y′ −6y = 2, y(0) = 1, y′(0) = 0 15. y′′ + 3y′ + 2y = et, y(0) = 0, y′(0) = 1 16. y′′ −y = 1, y(0) = 1, y′(0) = 0 17. y′′ + 4y = 3 sin t, y(0) = 1, y′(0) = −1 18. y′′ + y′ = 2e3t, y(0) = −1, y′(0) = 4 19. y′′ + y = 1, y(0) = 2, y′(0) = 0 20. y′′ + y = t, y(0) = 0, y′(0) = 2	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 t, y(0) = 1, y′(0) = 0 27. y′′ −5y′ + 6y = 10et cos t, y(0) = 2, y′(0) = 1 28. y′′ + 2y′ + 2y = 2t, y(0) = 2, y′(0) = −7 29. y′′ −2y′ + 2y = 5 sin t + 10 cost, y(0) = 1, y′(0) = 2 30. y′′ + 4y′ + 13y = 10e−t −36et, y(0) = 0, y′(0) = −16 31. y′′ + 4y′ + 5y = e−t(cos t + 3 sint), y(0) = 0, y′(0) = 4 32. 2y′′ −3y′ −2y = 4et, y(0) = 1, y′(0) = −2 33. 6y′′ −y′ −y = 3e2t, y(0) = 0, y′(0) = 0 34. 2y′′ + 2y′ + y = 2t, y(0) = 1, y′(0) = −1 35. 4y′′ −4y′ + 5y = 4 sin t −4 cos t, y(0) = 0, y′(0) = 11/17 36. 4y′′ + 4y′ + y = 3 sin t + cos t, y(0) = 2, y′(0) = −1 37. 9y′′ + 6y′ + y = 3e3t, y(0) = 0, y′(0) = −3 38. Suppose a, b, and c are constants and a ̸= 0. Let y1 = L−1 as + b as2 + bs + c and y2 = L−1 a as2 + bs + c . Show that y1(0) = 1, y′ 1(0) = 0 and y2(0) = 0, y′ 2(0) = 1. HINT: Use the Laplace transform to solve the initial value problems ay′′ + by′ + cy = 0, y(0) = 1, y′(0) = 0 ay′′ + by′ + cy = 0, y(0) = 0, y′(0) = 1. 8.4 THE UNIT STEP FUNCTION In the next section we’ll consider initial value problems ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1, where a, b, and c are constants and f is piecewise continuous. In this section we’ll develop procedures for using the table of Laplace transforms to ﬁnd Laplace transforms of piecewise continuous functions, and to ﬁnd the piecewise continuous inverses of Laplace transforms.	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) To relate the ﬁrst term to a Laplace transform, we add and subtract Z ∞ 2 e−st(2t + 1) dt in (8.4.2) to obtain L(f) = Z ∞ 0 e−st(2t + 1) dt + Z ∞ 2 e−st(3t −2t −1) dt = Z ∞ 0 e−st(2t + 1) dt + Z ∞ 2 e−st(t −1) dt = L(2t + 1) + Z ∞ 2 e−st(t −1) dt. (8.4.3) To relate the last integral to a Laplace transform, we make the change of variable x = t −2 and rewrite the integral as Z ∞ 2 e−st(t −1) dt = Z ∞ 0 e−s(x+2)(x + 1) dx = e−2s Z ∞ 0 e−sx(x + 1) dx. Since the symbol used for the variable of integration has no effect on the value of a deﬁnite integral, we can now replace x by the more standard t and write Z ∞ 2 e−st(t −1) dt = e−2s Z ∞ 0 e−st(t + 1) dt = e−2sL(t + 1). This and (8.4.3) imply that L(f) = L(2t + 1) + e−2sL(t + 1). Now we can use the table of Laplace transforms to ﬁnd that L(f) = 2 s2 + 1 s + e−2s 1 s2 + 1 s .	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 piecewise continuous function. It is convenient to introduce the unit step function, deﬁned as u(t) = 0, t < 0 1, t ≥0. (8.4.4) Thus, u(t) “steps” from the constant value 0 to the constant value 1 at t = 0. If we replace t by t −τ in (8.4.4), then u(t −τ) = 0, t < τ, 1, t ≥τ ; that is, the step now occurs at t = τ (Figure 8.4.2). The step function enables us to represent piecewise continuous functions conveniently. For example, consider the function f(t) = ( f0(t), 0 ≤t < t1, f1(t), t ≥t1, (8.4.5) where we assume that f0 and f1 are deﬁned on [0, ∞), even though they equal f only on the indicated intervals. This assumption enables us to rewrite (8.4.5) as f(t) = f0(t) + u(t −t1) (f1(t) −f0(t)) . (8.4.6) To verify this, note that if t < t1 then u(t −t1) = 0 and (8.4.6) becomes f(t) = f0(t) + (0) (f1(t) −f0(t)) = f0(t). If t ≥t1 then u(t −t1) = 1 and (8.4.6) becomes f(t) = f0(t) + (1) (f1(t) −f0(t)) = f1(t). We need the next theorem to show how (8.4.6) can be used to ﬁnd L(f).	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 −τ)g(t)) = Z τ 0 e−st(0) dt + Z ∞ τ e−stg(t) dt. The ﬁrst integral on the right equals zero. Introducing the new variable of integration x = t −τ in the second integral yields L (u(t −τ)g(t)) = Z ∞ 0 e−s(x+τ)g(x + τ) dx = e−sτ Z ∞ 0 e−sxg(x + τ) dx. Changing the name of the variable of integration in the last integral from x to t yields L (u(t −τ)g(t)) = e−sτ Z ∞ 0 e−stg(t + τ) dt = e−sτL(g(t + τ)). Example 8.4.2 Find L	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(t) =      f0(t), 0 ≤t < t1, f1(t), t1 ≤t < t2, f2(t), t ≥t2, as f(t) = f0(t) + u(t −t1) (f1(t) −f0(t)) + u(t −t2) (f2(t) −f1(t)) if f0, f1, and f2 are all deﬁned on [0, ∞). Example 8.4.4 Find the Laplace transform of f(t) =          1, 0 ≤t < 2, −2t + 1, 2 ≤t < 3, 3t, 3 ≤t < 5, t −1, t ≥5 (8.4.7) (Figure 8.4.3). Solution In terms of step functions, f(t) = 1 + u(t −2)(−2t + 1 −1) + u(t −3)(3t + 2t −1) +u(t −5)(t −1 −3t), or f(t) = 1 −2u(t −2)t + u(t −3)(5t −1) −u(t −5)(2t + 1). Now Theorem 8.4.1 implies that L(f) = L(1) −2e−2sL(t + 2) + e−3sL (5(t + 3) −1) −e−5sL (2(t + 5) + 1) = L(1) −2e−2sL(t + 2) + e−3sL(5t + 14) −e−5sL(2t + 11) = 1 s −2e−2s 1 s2 + 2 s + e−3s 5 s2 + 14 s −e−5s 2 s2 + 11 s . The trigonometric identities sin(A + B) = sinA cos B + cos A sin B (8.4.8) cos(A + B) = cos A cos B −sin A sin B (8.4.9) are useful in problems that involve shifting the arguments of trigonometric functions. We’ll use these identities in the next example.	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 functions, f(t) = sin t + u(t −π/2)(cos t −4 sin t) + u(t −π)(2 cos t + 3 sin t). Now Theorem 8.4.1 implies that L(f) = L(sint) + e−π 2 sL	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)) exists for s > s0 and L(u(t −τ)g(t −τ)) = e−sτL(g(t)), or, equivalently, if g(t) ↔G(s), then u(t −τ)g(t −τ) ↔e−sτG(s). (8.4.12) REMARK: Recall that the First Shifting Theorem (Theorem 8.1.3 states that multiplying a function by eat corresponds to shifting the argument of its transform by a units. Theorem 8.4.2 states that multiplying a Laplace transform by the exponential e−τs corresponds to shifting the argument of the inverse transform by τ units. Example 8.4.6 Use (8.4.12) to ﬁnd L−1 e−2s s2 . Solution To apply (8.4.12) we let τ = 2 and G(s) = 1/s2. Then g(t) = t and (8.4.12) implies that L−1 e−2s s2 = u(t −2)(t −2).	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 + 1. Hence, (8.4.12) and the linearity of L−1 imply that h(t) = L−1 (G0(s)) −L−1	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, 0 ≤t < 1, e−2t, t ≥1. 11. f(t) =        −t, 0 ≤t < 2, t −4, 2 ≤t < 3, 1, t ≥3. 12. f(t) =        0, 0 ≤t < 1, t, 1 ≤t < 2, 0, t ≥2. 13. f(t) =        t, 0 ≤t < 1, t2, 1 ≤t < 2, 0, t ≥2. 14. f(t) =        t, 0 ≤t < 1, 2 −t, 1 ≤t < 2, 6, t > 2. 15. C/G f(t) =          sin t, 0 ≤t < π 2 , 2 sint, π 2 ≤t < π, cos t, t ≥π. 16. C/G f(t) =        2, 0 ≤t < 1, −2t + 2, 1 ≤t < 3, 3t, t ≥3. 17. C/G f(t) =        3, 0 ≤t < 2, 3t + 2, 2 ≤t < 4, 4t, t ≥4. 18. C/G f(t) = ( (t + 1)2, 0 ≤t < 1, (t + 2)2, t ≥1. In Exercises 19–28 use Theorem 8.4.2 to express the inverse transforms in terms of step functions, and then ﬁnd distinct formulas the for inverse transforms on the appropriate intervals, as in Example 8.4.7. Where indicated by C/G , graph the inverse transform. 19. H(s) = e−2s s −2 20. H(s) = e−s s(s + 1) 21. C/G H(s) = e−s s3 + e−2s s2 22. C/G H(s) = 2 s + 1 s2 + e−s 3 s −1 s2 + e−3s 1 s + 1 s2	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) = 1 s −2 s3 + e−2s 3 s −1 s3 + e−4s s2 29. Find L (u(t −τ)). 30. Let {tm}∞ m=0 be a sequence of points such that t0 = 0, tm+1 > tm, and limm→∞tm = ∞. For each nonnegative integer m, let fm be continuous on [tm, ∞), and let f be deﬁned on [0, ∞) by f(t) = fm(t), tm ≤t < tm+1 (m = 0, 1, . . .). Show that f is piecewise continuous on [0, ∞) and that it has the step function representation f(t) = f0(t) + ∞ X m=1 u(t −tm) (fm(t) −fm−1(t)) , 0 ≤t < ∞. How do we know that the series on the right converges for all t in [0, ∞)? 31. In addition to the assumptions of Exercise 30, assume that \|fm(t)\| ≤Mes0t, t ≥tm, m = 0, 1, . . ., (A) and that the series ∞ X m=0 e−ρtm (B) converges for some ρ > 0. Using the steps listed below, show that L(f) is deﬁned for s > s0 and L(f) = L(f0) + ∞ X m=1 e−stmL(gm) (C) for s > s0 + ρ, where gm(t) = fm(t + tm) −fm−1(t + tm). (a) Use (A) and Theorem 8.1.6 to show that L(f) = ∞ X m=0 Z tm+1 tm e−stfm(t) dt (D) is deﬁned for s > s0.	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 ∞ X m=0 Z ∞ tm+1 e−stfm(t) dt both converge (absolutely) if s > s0 + ρ. (d) Show that (E) can be rewritten as L(f) = L(f0) + ∞ X m=1 Z ∞ tm e−st (fm(t) −fm−1(t)) dt if s > s0 + ρ. (e) Complete the proof of (C). 32. Suppose {tm}∞ m=0 and {fm}∞ m=0 satisfy the assumptions of Exercises 30 and 31, and there’s a positive constant K such that tm ≥Km for m sufﬁciently large. Show that the series (B) of Exercise 31 converges for any ρ > 0, and conclude from this that (C) of Exercise 31 holds for s > s0. In Exercises 33–36 ﬁnd the step function representation of f and use the result of Exercise 32 to ﬁnd L(f). HINT: You will need formulas related to the formula for the sum of a geometric series. 33. f(t) = m + 1, m ≤t < m + 1 (m = 0, 1, 2, . . .) 34. f(t) = (−1)m, m ≤t < m + 1 (m = 0, 1, 2, . . .) 35. f(t) = (m + 1)2, m ≤t < m + 1 (m = 0, 1, 2, . . .) 36. f(t) = (−1)mm, m ≤t < m + 1 (m = 0, 1, 2, . . .) 8.5 CONSTANT COEEFFICIENT EQUATIONS WITH PIECEWISE CONTINUOUS FORCING FUNC- TIONS We’ll now consider initial value problems of the form ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1, (8.5.1) where a, b, and c are constants (a ̸= 0) and f is piecewise continuous on [0, ∞). Problems of this kind occur in situations where the input to a physical system undergoes instantaneous changes, as when a switch is turned on or off or the forces acting on the system change abruptly. It can be shown (Exercises 23 and 24) that the differential equation in (8.5.1) has no solutions on an open interval that contains a jump discontinuity of f. Therefore we must deﬁne what we mean by a solution of (8.5.1) on [0, ∞) in the case where f has jump discontinuities. The next theorem motivates our deﬁnition. We omit the proof.	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) y(0) = k0 and y′(0) = k1. (b) y and y′ are continuous on [0, ∞). (c) y′′ is deﬁned on every open subinterval of [0, ∞) that does not contain any of the points t1, ..., tn, and ay′′ + by′ + cy = f(t) on every such subinterval. (d) y′′ has limits from the right and left at t1, ..., tn. We deﬁne the function y of Theorem 8.5.1 to be the solution of the initial value problem (8.5.1). We begin by considering initial value problems of the form ay′′ + by′ + cy = ( f0(t), 0 ≤t < t1, f1(t), t ≥t1, y(0) = k0, y′(0) = k1, (8.5.2) where the forcing function has a single jump discontinuity at t1. We can solve (8.5.2) by the these steps: Step 1. Find the solution y0 of the initial value problem ay′′ + by′ + cy = f0(t), y(0) = k0, y′(0) = k1. Step 2. Compute c0 = y0(t1) and c1 = y′ 0(t1). Step 3. Find the solution y1 of the initial value problem ay′′ + by′ + cy = f1(t), y(t1) = c0, y′(t1) = c1. Step 4. Obtain the solution y of (8.5.2) as y = ( y0(t), 0 ≤t < t1 y1(t), t ≥t1. It is shown in Exercise 23 that y′ exists and is continuous at t1. The next example illustrates this procedure. Example 8.5.1 Solve the initial value problem y′′ + y = f(t), y(0) = 2, y′(0) = −1, (8.5.3) where f(t) =    1, 0 ≤t < π 2 , −1, t ≥π 2 .	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 yields y0(π/2) = 0 and y′ 0(π/2) = −1, so the second initial value problem is y′′ + y = −1, y π 2 = 0, y′ π 2 = −1. We leave it to you to verify that the solution of this problem is y1 = −1 + cos t + sin t. Hence, the solution of (8.5.3) is y =    1 + cos t −sint, 0 ≤t < π 2 , −1 + cos t + sint, t ≥π 2 (8.5.4) (Figure:8.5.1). If f0 and f1 are deﬁned on [0, ∞), we can rewrite (8.5.2) as ay′′ + by′ + cy = f0(t) + u(t −t1) (f1(t) −f0(t)) , y(0) = k0, y′(0) = k1, and apply the method of Laplace transforms. We’ll now solve the problem considered in Example 8.5.1 by this method.	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, transforming (8.5.5) yields (s2 + 1)Y (s) = 1 −2e−πs/2 s −1 + 2s, so Y (s) = (1 −2e−πs/2)G(s) + 2s −1 s2 + 1 , (8.5.6) with G(s) = 1 s(s2 + 1). The form for the partial fraction expansion of G is 1 s(s2 + 1) = A s + Bs + C s2 + 1 . (8.5.7) Multiplying through by s(s2 + 1) yields A(s2 + 1) + (Bs + C)s = 1, or (A + B)s2 + Cs + A = 1. Equating coefﬁcients of like powers of s on the two sides of this equation shows that A = 1, B = −A = −1 and C = 0. Hence, from (8.5.7), G(s) = 1 s − s s2 + 1. Therefore g(t) = 1 −cos t. From this, (8.5.6), and Theorem 8.4.2, y = 1 −cos t −2u t −π 2 1 −cos t −π 2 + 2 cos t −sin t. Simplifying this (recalling that cos(t −π/2) = sin t) yields y = 1 + cos t −sint −2u t −π 2 (1 −sin t),	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 function y with the properties stated in Theorem 8.5.1; that is, such that y and y′ are continuous on [0, ∞) and y′′ has limits from the right and left at t1. However, this is true if f0 and f1 are continuous and of exponential order on [0, ∞). A proof is sketched in Exercises 8.6.11–8.6.13. Example 8.5.3 Solve the initial value problem y′′ −y = f(t), y(0) = −1, y′(0) = 2, (8.5.8) where f(t) = t, 0 ≤t < 1, 1, t ≥1. Solution Here f(t) = t −u(t −1)(t −1), so L(f) = L(t) −L (u(t −1)(t −1)) = L(t) −e−sL(t) (from Theorem 8.4.1) = 1 s2 −e−s s2 . Since transforming (8.5.8) yields (s2 −1)Y (s) = L(f) + 2 −s, we see that Y (s) = (1 −e−s)H(s) + 2 −s s2 −1, (8.5.9) where H(s) = 1 s2(s2 −1) = 1 s2 −1 −1 s2 ; therefore h(t) = sinh t −t. (8.5.10) Since L−1 2 −s s2 −1 = 2 sinh t −cosh t, we conclude from (8.5.9), (8.5.10), and Theorem 8.4.1 that y = sinh t −t −u(t −1) (sinh(t −1) −t + 1) + 2 sinh t −cosh t, or y = 3 sinh t −cosh t −t −u(t −1) (sinh(t −1) −t + 1) (8.5.11) We leave it to you to verify that y and y′ are continuous and y′′ has limits from the right and left at t1 = 1.	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 (cos 2(t + π/4)) −e−πsL (cos 2(t + π)) = −e−πs/4L(sin 2t) −e−πsL(cos 2t) = −2e−πs/4 s2 + 4 −se−πs s2 + 4. Since transforming (8.5.12) yields (s2 + 1)Y (s) = L(f), we see that Y (s) = e−πs/4H1(s) + e−πsH2(s), (8.5.13) where H1(s) = − 2 (s2 + 1)(s2 + 4) and H2(s) = − s (s2 + 1)(s2 + 4). (8.5.14) To simplify the required partial fraction expansions, we ﬁrst write 1 (x + 1)(x + 4) = 1 3 1 x + 1 − 1 x + 4 . Setting x = s2 and substituting the result in (8.5.14) yields H1(s) = −2 3 1 s2 + 1 − 1 s2 + 4 and H2(s) = −1 3 s s2 + 1 − s s2 + 4 . The inverse transforms are h1(t) = −2 3 sin t + 1 3 sin2t and h2(t) = −1 3 cos t + 1 3 cos 2t. From (8.5.13) and Theorem 8.4.2, y = u t −π 4 h1 t −π 4 + u(t −π)h2(t −π). (8.5.15) Since h1 t −π 4 = −2 3 sin t −π 4 + 1 3 sin 2 t −π 4 = − √ 2 3 (sin t −cos t) −1 3 cos 2t	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)(cos t + cos 2t) or y =              0, 0 ≤t < π 4 , − √ 2 3 (sin t −cos t) −1 3 cos 2t, π 4 ≤t < π, − √ 2 3 sint + 1 + √ 2 3 cos t, t ≥π. (8.5.16) We leave it to you to verify that y and y′ are continuous and y′′ has limits from the right and left at t1 = π/4 and t2 = π (Figure 8.5.2). 8.5 Exercises In Exercises 1–20 use the Laplace transform to solve the initial value problem. Where indicated by C/G , graph the solution.	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, 1 ≤t < 2, −1, t ≥2, y(0) = −3, y′(0) = 1 6. C/G y′′ + 4y = ( \| sin t\|, 0 ≤t < 2π, 0, t ≥2π, y(0) = −3, y′(0) = 1 7. y′′ −5y′ + 4y =        1, 0 ≤t < 1 −1, 1 ≤t < 2, 0, t ≥2, y(0) = 3, y′(0) = −5 8. y′′ + 9y =      cos t, 0 ≤t < 3π 2 , sin t, t ≥3π 2 , y(0) = 0, y′(0) = 0 9. C/G y′′ + 4y =    t, 0 ≤t < π 2 , π, t ≥π 2 , y(0) = 0, y′(0) = 0 10. y′′ + y = ( t, 0 ≤t < π, −t, t ≥π, y(0) = 0, y′(0) = 0 11. y′′ −3y′ + 2y = 0, 0 ≤t < 2, 2t −4, t ≥2, , y(0) = 0, y′(0) = 0 12. y′′ + y = t, 0 ≤t < 2π, −2t, t ≥2π, y(0) = 1, y′(0) = 2 13. C/G y′′ + 3y′ + 2y = 1, 0 ≤t < 2, −1, t ≥2, y(0) = 0, y′(0) = 0 14. y′′ −4y′ + 3y = −1, 0 ≤t < 1, 1, t ≥1, y(0) = 0, y′(0) = 0 15. y′′ + 2y′ + y = et, 0 ≤t < 1, et −1, t ≥1, y(0) = 3, y′(0) = −1 16. y′′ + 2y′ + y = 4et, 0 ≤t < 1, 0, t ≥1, y(0) = 0, y′(0) = 0 17. y′′ + 3y′ + 2y = e−t, 0 ≤t < 1, 0, t ≥1, y(0) = 1, y′(0) = −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′(0) = −1 21. Solve the initial value problem y′′ = f(t), y(0) = 0, y′(0) = 0, where f(t) = m + 1, m ≤t < m + 1, m = 0, 1, 2, . . .. 22. Solve the given initial value problem and ﬁnd a formula that does not involve step functions and represents y on each interval of continuity of f. (a) y′′ + y = f(t), y(0) = 0, y′(0) = 0; f(t) = m + 1, mπ ≤t < (m + 1)π, m = 0, 1, 2, . . .. (b) y′′ + y = f(t), y(0) = 0, y′(0) = 0; f(t) = (m + 1)t, 2mπ ≤t < 2(m + 1)π, m = 0, 1, 2, . . . HINT: You’ll need the formula 1 + 2 + · · · + m = m(m + 1) 2 . (c) y′′ + y = f(t), y(0) = 0, y′(0) = 0; f(t) = (−1)m, mπ ≤t < (m + 1)π, m = 0, 1, 2, . . .. (d) y′′ −y = f(t), y(0) = 0, y′(0) = 0; f(t) = m + 1, m ≤t < (m + 1), m = 0, 1, 2, . . .. HINT: You will need the formula 1 + r + · · · + rm = 1 −rm+1 1 −r (r ̸= 1). (e) y′′ + 2y′ + 2y = f(t), y(0) = 0, y′(0) = 0; f(t) = (m + 1)(sin t + 2 cos t), 2mπ ≤t < 2(m + 1)π, m = 0, 1, 2, . . .. (See the hint in (d).) (f) y′′ −3y′ + 2y = f(t), y(0) = 0, y′(0) = 0; f(t) = m + 1, m ≤t < m + 1, m = 0, 1, 2, . . .. (See the hints in (b) and (d).) 23. (a) Let g be continuous on (α, β) and differentiable on the (α, t0) and (t0, β). Suppose A = limt→t0−g′(t) and B = limt→t0+ g′(t) both exist. Use the mean value theorem to show that lim t→t0− g(t) −g(t0) t −t0 = A and lim t→t0+ g(t) −g(t0) t −t0 = B. (b) Conclude from (a) that g′(t0) exists and g′ is continuous at t0 if A = B.	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 y′ are continuous on (α, β) and y′′ on (α, t0) and (t0, β). Suppose also that ay′′ + by′ + cy = f(t) (A) on (α, t0) and (t0, β). Show that y′′(t0+) −y′′(t0−) = f(t0+) −f(t0−) a ̸= 0. (b) Use (a) and Exercise 23(c) to show that (A) does not have solutions on any interval (α, β) that contains a jump discontinuity of f. 25. Suppose P0, P1, and P2 are continuous and P0 has no zeros on an open interval (a, b), and that F has a jump discontinuity at a point t0 in (a, b). Show that the differential equation P0(t)y′′ + P1(t)y′ + P2(t)y = F (t) has no solutions on (a, b).HINT: Generalize the result of Exercise 24 and use Exercise 23(c). 26. Let 0 = t0 < t1 < · · · < tn. Suppose fm is continuous on [tm, ∞) for m = 1, . . ., n. Let f(t) = fm(t), tm ≤t < tm+1, m = 1, . . ., n −1, fn(t), t ≥tn. Show that the solution of ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1, as deﬁned following Theorem 8.5.1, is given by y =                z0(t), 0 ≤t < t1, z0(t) + z1(t), t1 ≤t < t2, ... z0 + · · · + zn−1(t), tn−1 ≤t < tn, z0 + · · · + zn(t), t ≥tn, where z0 is the solution of az′′ + bz′ + cz = f0(t), z(0) = k0, z′(0) = k1 and zm is the solution of az′′ + bz′ + cz = fm(t) −fm−1(t), z(tm) = 0, z′(tm) = 0 for m = 1, . . ., n. 8.6 CONVOLUTION	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′(0) = 0. Taking Laplace transforms yields (as2 + bs + c)Y (s) = F (s), so Y (s) = F (s)G(s), (8.6.1) where G(s) = 1 as2 + bs + c. Until now wen’t been interested in the factorization indicated in (8.6.1), since we dealt only with differ- ential equations with speciﬁc forcing functions. Hence, we could simply do the indicated multiplication in (8.6.1) and use the table of Laplace transforms to ﬁnd y = L−1(Y ). However, this isn’t possible if we want a formula for y in terms of f, which may be unspeciﬁed. To motivate the formula for L−1(F G), consider the initial value problem y′ −ay = f(t), y(0) = 0, (8.6.2) which we ﬁrst solve without using the Laplace transform. The solution of the differential equation in (8.6.2) is of the form y = ueat where u′ = e−atf(t). Integrating this from 0 to t and imposing the initial condition u(0) = y(0) = 0 yields u = Z t 0 e−aτf(τ) dτ. Therefore y(t) = eat Z t 0 e−aτf(τ) dτ = Z t 0 ea(t−τ)f(τ) dτ. (8.6.3) Now we’ll use the Laplace transform to solve (8.6.2) and compare the result to (8.6.3). Taking Laplace transforms in (8.6.2) yields (s −a)Y (s) = F (s), so Y (s) = F (s) 1 s −a, which implies that y(t) = L−1 F (s) 1 s −a . (8.6.4) If we now let g(t) = eat, so that G(s) = 1 s −a, then (8.6.3) and (8.6.4) can be written as y(t) = Z t 0 f(τ)g(t −τ) dτ	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; that is, Z t 0 f(t −τ)g(τ) dτ = Z t 0 f(τ)g(t −τ) dτ. Eqn. (8.6.5) shows that L−1(F G) = f ∗g in the special case where g(t) = eat. This next theorem states that this is true in general. Theorem 8.6.2 [The Convolution Theorem] If L(f) = F and L(g) = G, then L(f ∗g) = F G. A complete proof of the convolution theorem is beyond the scope of this book. However, we’ll assume that f ∗g has a Laplace transform and verify the conclusion of the theorem in a purely computational way. By the deﬁnition of the Laplace transform, L(f ∗g) = Z ∞ 0 e−st(f ∗g)(t) dt = Z ∞ 0 e−st Z t 0 f(τ)g(t −τ) dτ dt. This iterated integral equals a double integral over the region shown in Figure 8.6.1. Reversing the order of integration yields L(f ∗g) = Z ∞ 0 f(τ) Z ∞ τ e−stg(t −τ) dt dτ. (8.6.6) However, the substitution x = t −τ shows that Z ∞ τ e−stg(t −τ) dt = Z ∞ 0 e−s(x+τ)g(x) dx = e−sτ Z ∞ 0 e−sxg(x) dx = e−sτG(s). Substituting this into (8.6.6) and noting that G(s) is independent of τ yields L(f ∗g) = Z ∞ 0 e−sτf(τ)G(s) dτ = G(s) Z ∞ 0 e−stf(τ) dτ = F (s)G(s).	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 . Since eat ↔ 1 s −a and ebt ↔ 1 s −b, it follows that L(f ∗g) = 1 a −b 1 s −a − 1 s −b = 1 (s −a)(s −b) = L(eat)L(ebt) = L(f)L(g).	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 for the solution of the initial value problem y′′ −2y′ + y = f(t), y(0) = k0, y′(0) = k1. (8.6.7) Solution Taking Laplace transforms in (8.6.7) yields (s2 −2s + 1)Y (s) = F (s) + (k1 + k0s) −2k0. Therefore Y (s) = 1 (s −1)2 F (s) + k1 + k0s −2k0 (s −1)2 = 1 (s −1)2 F (s) + k0 s −1 + k1 −k0 (s −1)2 . From the table of Laplace transforms, L−1 k0 s −1 + k1 −k0 (s −1)2 = et (k0 + (k1 −k0)t) . Since 1 (s −1)2 ↔tet and F (s) ↔f(t), the convolution theorem implies that L−1 1 (s −1)2 F (s) = Z t 0 τeτf(t −τ) dτ. Therefore the solution of (8.6.7) is y(t) = et (k0 + (k1 −k0)t) + Z t 0 τeτf(t −τ) dτ. Example 8.6.3 Find a formula for the solution of the initial value problem y′′ + 4y = f(t), y(0) = k0, y′(0) = k1. (8.6.8) Solution Taking Laplace transforms in (8.6.8) yields (s2 + 4)Y (s) = F (s) + k1 + k0s. Therefore Y (s) = 1 (s2 + 4)F (s) + k1 + k0s s2 + 4 . From the table of Laplace transforms, L−1 k1 + k0s s2 + 4 = k0 cos 2t + k1 2 sin 2t.	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 initial value problem y′′ + 2y′ + 2y = f(t), y(0) = k0, y′(0) = k1. (8.6.9) Solution Taking Laplace transforms in (8.6.9) yields (s2 + 2s + 2)Y (s) = F (s) + k1 + k0s + 2k0. Therefore Y (s) = 1 (s + 1)2 + 1F (s) + k1 + k0s + 2k0 (s + 1)2 + 1 = 1 (s + 1)2 + 1F (s) + (k1 + k0) + k0(s + 1) (s + 1)2 + 1 . From the table of Laplace transforms, L−1 (k1 + k0) + k0(s + 1) (s + 1)2 + 1 = e−t ((k1 + k0) sin t + k0 cos t) . Since 1 (s + 1)2 + 1 ↔e−t sin t and F (s) ↔f(t), the convolution theorem implies that L−1 1 (s + 1)2 + 1F (s) = Z t 0 f(t −τ)e−τ sin τ dτ. Therefore the solution of (8.6.9) is y(t) = e−t ((k1 + k0) sin t + k0 cos t) + Z t 0 f(t −τ)e−τ sin τ dτ. (8.6.10) Evaluating Convolution Integrals We’ll say that an integral of the form R t 0 u(τ)v(t −τ) dτ is a convolution integral. The convolution theorem provides a convenient way to evaluate convolution integrals.	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) = t7. Since t5 ↔5! s6 and t7 ↔7! s8 , the convolution theorem implies that h(t) ↔5!7! s14 = 5!7! 13! 13! s14 , where we have written the second equality because 13! s14 ↔t13. Hence, h(t) = 5!7! 13! t13. Example 8.6.6 Use the convolution theorem and a partial fraction expansion to evaluate the convolution integral h(t) = Z t 0 sin a(t −τ) cos bτ dτ (\|a\| ̸= \|b\|). Solution Since sin at ↔ a s2 + a2 and cos bt ↔ s s2 + b2 , the convolution theorem implies that H(s) = a s2 + a2 s s2 + b2 . Expanding this in a partial fraction expansion yields H(s) = a b2 −a2 s s2 + a2 − s s2 + b2 . Therefore h(t) = a b2 −a2 (cos at −cos bt) . Volterra Integral Equations An equation of the form y(t) = f(t) + Z t 0 k(t −τ)y(τ) dτ (8.6.11)	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), and solving this for Y (s) yields Y (s) = F (s) 1 −K(s). We then obtain the solution of (8.6.11) as y = L−1(Y ). Example 8.6.7 Solve the integral equation y(t) = 1 + 2 Z t 0 e−2(t−τ)y(τ) dτ. (8.6.12) Solution Taking Laplace transforms in (8.6.12) yields Y (s) = 1 s + 2 s + 2Y (s), and solving this for Y (s) yields Y (s) = 1 s + 2 s2 . Hence, y(t) = 1 + 2t. Transfer Functions The next theorem presents a formula for the solution of the general initial value problem ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1, where we assume for simplicity that f is continuous on [0, ∞) and that L(f) exists. In Exercises 11–14 it’s shown that the formula is valid under much weaker conditions on f. Theorem 8.6.3 Suppose f is continuous on [0, ∞) and has a Laplace transform. Then the solution of the initial value problem ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1, (8.6.13) is y(t) = k0y1(t) + k1y2(t) + Z t 0 w(τ)f(t −τ) dτ, (8.6.14) where y1 and y2 satisfy ay′′ 1 + by′ 1 + cy1 = 0, y1(0) = 1, y′ 1(0) = 0, (8.6.15) and ay′′ 2 + by′ 2 + cy2 = 0, y2(0) = 0, y′ 2(0) = 1, (8.6.16) and w(t) = 1 ay2(t). (8.6.17)	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) shows that p(s)Y1(s) = as + b and p(s)Y2(s) = a. Therefore Y1(s) = as + b p(s) and Y2(s) = a p(s). (8.6.21) Hence, (8.6.20) can be rewritten as V (s) = k0Y1(s) + k1Y2(s). Substituting this into (8.6.18) yields Y (s) = k0Y1(s) + k1Y2(s) + 1 aY2(s)F (s). Taking inverse transforms and invoking the convolution theorem yields (8.6.14). Finally, (8.6.19) and (8.6.21) imply (8.6.17). It is useful to note from (8.6.14) that y is of the form y = v + h, where v(t) = k0y1(t) + k1y2(t) depends on the initial conditions and is independent of the forcing function, while h(t) = Z t 0 w(τ)f(t −τ) dτ depends on the forcing function and is independent of the initial conditions. If the zeros of the character- istic polynomial p(s) = as2 + bs + c of the complementary equation have negative real parts, then y1 and y2 both approach zero as t →∞, so limt→∞v(t) = 0 for any choice of initial conditions. Moreover, the value of h(t) is essentially	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 + 2s + 2 = (s + 1)2 + 1 are −1 ± i. From (8.6.10), we see that the solution of the general initial value problem of Example 8.6.4 is y = v + h, where v(t) = e−t ((k1 + k0) sin t + k0 cos t) is the transient component of the solution and h(t) = Z t 0 f(t −τ)e−τ sin τ dτ is the steady state component. The deﬁnitions don’t apply to the initial value problems considered in Examples 8.6.2 and 8.6.3, since the zeros of the characteristic polynomials in these two examples don’t have negative real parts. In physical applications where the input f and the output y of a device are related by (8.6.13), the zeros of the characteristic polynomial usually do have negative real parts. Then W = L(w) is called the transfer function of the device. Since H(s) = W(s)F (s), we see that W(s) = H(s) F (s) is the ratio of the transform of the steady state output to the transform of the input. Because of the form of h(t) = Z t 0 w(τ)f(t −τ) dτ, w is sometimes called the weighting function of the device, since it assigns weights to past values of the input f. It is also called the impulse response of the device, for reasons discussed in the next section. Formula (8.6.14) is given in more detail in Exercises 8–10 for the three possible cases where the zeros of p(s) are real and distinct, real and repeated, or complex conjugates, respectively. 8.6 Exercises 1. Express the inverse transform as an integral. (a) 1 s2(s2 + 4) (b) s (s + 2)(s2 + 9) (c) s (s2 + 4)(s2 + 9) (d) s (s2 + 1)2 (e) 1 s(s −a) (f) 1 (s + 1)(s2 + 2s + 2) (g) 1 (s + 1)2(s2 + 4s + 5) (h) 1 (s −1)3(s + 2)2	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 −τ) dτ (e) et Z t 0 sin ωτ cos ω(t −τ) dτ (f) et Z t 0 τ 2(t −τ)eτ dτ (g) e−t Z t 0 e−ττ cos ω(t −τ) dτ (h) et Z t 0 e2τ sinh(t −τ) dτ (i) Z t 0 τe2τ sin 2(t −τ) dτ (j) Z t 0 (t −τ)3eτ dτ (k) Z t 0 τ 6e−(t−τ) sin 3(t −τ) dτ (l) Z t 0 τ 2(t −τ)3 dτ (m) Z t 0 (t −τ)7e−τ sin2τ dτ (n) Z t 0 (t −τ)4 sin 2τ dτ 3. Find a formula for the solution of the initial value problem. (a) y′′ + 3y′ + y = f(t), y(0) = 0, y′(0) = 0 (b) y′′ + 4y = f(t), y(0) = 0, y′(0) = 0 (c) y′′ + 2y′ + y = f(t), y(0) = 0, y′(0) = 0 (d) y′′ + k2y = f(t), y(0) = 1, y′(0) = −1 (e) y′′ + 6y′ + 9y = f(t), y(0) = 0, y′(0) = −2 (f) y′′ −4y = f(t), y(0) = 0, y′(0) = 3 (g) y′′ −5y′ + 6y = f(t), y(0) = 1, y′(0) = 3 (h) y′′ + ω2y = f(t), y(0) = k0, y′(0) = k1 4. Solve the integral equation. (a) y(t) = t − Z t 0 (t −τ)y(τ) dτ (b) y(t) = sin t −2 Z t 0 cos(t −τ)y(τ) dτ (c) y(t) = 1 + 2 Z t 0 y(τ) cos(t −τ) dτ (d) y(t) = t + Z t 0 y(τ)e−(t−τ) dτ (e) y′(t) = t + Z t 0 y(τ) cos(t −τ) dτ, y(0) = 4	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) dτ by introducing the new variable of integration x = t −τ in the ﬁrst integral. 7. Use the convolution theorem to show that if f(t) ↔F (s) then Z t 0 f(τ) dτ ↔F (s) s . 8. Show that if p(s) = as2 + bs + c has distinct real zeros r1 and r2 then the solution of ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1 is y(t) = k0 r2er1t −r1er2t r2 −r1 + k1 er2t −er1t r2 −r1 + 1 a(r2 −r1) Z t 0 (er2τ −er1τ)f(t −τ) dτ. 9. Show that if p(s) = as2 + bs + c has a repeated real zero r1 then the solution of ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1 is y(t) = k0(1 −r1t)er1t + k1ter1t + 1 a Z t 0 τer1τf(t −τ) dτ. 10. Show that if p(s) = as2 + bs + c has complex conjugate zeros λ ± iω then the solution of ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1 is y(t) = eλt k0(cos ωt −λ ω sin ωt) + k1 ω sin ωt + 1 aω Z t 0 eλtf(t −τ) sin ωτ dτ.	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 with respect to a parameter to show that h is the solution of ah′′ + bh′ + ch = f, h(0) = 0, h′(0) = 0. (c) Show that the function y in Eqn. (8.6.14) is the solution of Eqn. (8.6.13) provided that f is continuous on [0, ∞); thus, it’s not necessary to assume that f has a Laplace transform. 12. Consider the initial value problem ay′′ + by′ + cy = f(t), y(0) = 0, y′(0) = 0, (A) where a, b, and c are constants, a ̸= 0, and f(t) = ( f0(t), 0 ≤t < t1, f1(t), t ≥t1. Assume that f0 is continuous and of exponential order on [0, ∞) and f1 is continuous and of exponential order on [t1, ∞). Let p(s) = as2 + bs + c. (a) Show that the Laplace transform of the solution of (A) is Y (s) = F0(s) + e−st1G(s) p(s) where g(t) = f1(t + t1) −f0(t + t1). (b) Let w be as in Exercise 11. Use Theorem 8.4.2 and the convolution theorem to show that the solution of (A) is y(t) = Z t 0 w(t −τ)f0(τ) dτ + u(t −t1) Z t−t1 0 w(t −t1 −τ)g(τ) dτ for t > 0. (c) Henceforth, assume only that f0 is continuous on [0, ∞) and f1 is continuous on [t1, ∞). Use Exercise 11 (a) and (b) to show that y′(t) = Z t 0 w′(t −τ)f0(τ) dτ + u(t −t1) Z t−t1 0 w′(t −t1 −τ)g(τ) dτ	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. Suppose f(t) =                f0(t), 0 ≤t < t1, f1(t), t1 ≤t < t2, ... fk−1(t), tk−1 ≤t < tk, fk(t), t ≥tk, where fm is continuous on [tm, ∞) for m = 0, . . ., k (let t0 = 0), and deﬁne gm(t) = fm(t + tm) −fm−1(t + tm), m = 1, . . ., k. Extend the results of Exercise 12 to show that the solution of ay′′ + by′ + cy = f(t), y(0) = 0, y′(0) = 0 is y(t) = Z t 0 w(t −τ)f0(τ) dτ + k X m=1 u(t −tm) Z t−tm 0 w(t −tm −τ)gm(τ) dτ. 14. Let {tm}∞ m=0 be a sequence of points such that t0 = 0, tm+1 > tm, and limm→∞tm = ∞. For each nonegative integer m let fm be continuous on [tm, ∞), and let f be deﬁned on [0, ∞) by f(t) = fm(t), tm ≤t < tm+1 m = 0, 1, 2 . . .. Let gm(t) = fm(t + tm) −fm−1(t + tm), m = 1, . . ., k. Extend the results of Exercise 13 to show that the solution of ay′′ + by′ + cy = f(t), y(0) = 0, y′(0) = 0 is y(t) = Z t 0 w(t −τ)f0(τ) dτ + ∞ X m=1 u(t −tm) Z t−tm 0 w(t −tm −τ)gm(τ) dτ. HINT: See Exercise30. 8.7 CONSTANT COEFFICIENT EQUATIONS WITH IMPULSES So far in this chapter, we’ve considered initial value problems for the constant coefﬁcient equation ay′′ + by′ + cy = f(t),	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 collide. Since it isn’t feasible to represent such forces as continuous or piecewise continuous functions, we must construct a different mathematical model to deal with them. If f is an integrable function and f(t) = 0 for t outside of the interval [t0, t0 + h], then R t0+h t0 f(t) dt is called the total impulse of f. We’re interested in the idealized situation where h is so small that the total impulse can be assumed to be applied instantaneously at t = t0. We say in this case that f is an impulse function. In particular, we denote by δ(t −t0) the impulse function with total impulse equal to one, applied at t = t0. (The impulse function δ(t) obtained by setting t0 = 0 is the Dirac δ function.) It must be understood, however, that δ(t −t0) isn’t a function in the standard sense, since our “deﬁnition” implies that δ(t −t0) = 0 if t ̸= t0, while Z t0 t0 δ(t −t0) dt = 1. From calculus we know that no function can have these properties; nevertheless, there’s a branch of mathematics known as the theory of distributions where the deﬁnition can be made rigorous. Since the theory of distributions is beyond the scope of this book, we’ll take an intuitive approach to impulse functions. Our ﬁrst task is to deﬁne what we mean by the solution of the initial value problem ay′′ + by′ + cy = δ(t −t0), y(0) = 0, y′(0) = 0, where t0 is a ﬁxed nonnegative number. The next theorem will motivate our deﬁnition. Theorem 8.7.1 Suppose t0 ≥0. For each positive number h, let yh be the solution of the initial value problem ay′′ h + by′ h + cyh = fh(t), yh(0) = 0, y′ h(0) = 0, (8.7.1) where fh(t) =        0, 0 ≤t < t0, 1/h, t0 ≤t < t0 + h, 0, t ≥t0 + h, (8.7.2) so fh has unit total impulse equal to the area of the shaded rectangle in Figure 8.7.1. Then lim h→0+ yh(t) = u(t −t0)w(t −t0), (8.7.3) where w = L−1 1 as2 + bs + c . Proof Taking Laplace transforms in (8.7.1) yields (as2 + bs + c)Yh(s) = Fh(s), so Yh(s) = Fh(s) as2 + bs + c. The convolution theorem implies that yh(t) = Z t 0 w(t −τ)fh(τ) dτ.	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 lim h→0+ yh(t) = 0 if 0 ≤t ≤t0. (8.7.5) We’ll now show that lim h→0+ yh(t) = w(t −t0) if t > t0. (8.7.6) Suppose t is ﬁxed and t > t0. From (8.7.4), yh(t) = 1 h Z t0+h t0 w(t −τ)dτ if h < t −t0. (8.7.7) Since 1 h Z t0+h t0 dτ = 1, (8.7.8) we can write w(t −t0) = 1 hw(t −t0) Z t0+h t0 dτ = 1 h Z t0+h t0 w(t −t0) dτ.	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.) Then (8.7.8) and (8.7.9) imply that \|yh(t) −w(t −t0)\| ≤1 hMh Z t0+h t0 dτ = Mh. (8.7.10) But limh→0+ Mh = 0, since w is continuous. Therefore (8.7.10) implies (8.7.6). This and (8.7.5) imply (8.7.3). Theorem 8.7.1 motivates the next deﬁnition. Deﬁnition 8.7.2 If t0 > 0, then the solution of the initial value problem ay′′ + by′ + cy = δ(t −t0), y(0) = 0, y′(0) = 0, (8.7.11) is deﬁned to be y = u(t −t0)w(t −t0), where w = L−1 1 as2 + bs + c . In physical applications where the input f and the output y of a device are related by the differential equation ay′′ + by′ + cy = f(t), w is called the impulse response of the device. Note that w is the solution of the initial value problem aw′′ + bw′ + cw = 0, w(0) = 0, w′(0) = 1/a, (8.7.12) as can be seen by using the Laplace transform to solve this problem. (Verify.) On the other hand, we can solve (8.7.12) by the methods of Section 5.2 and show that w is deﬁned on (−∞, ∞) by w = er2t −er1t a(r2 −r1), w = 1 ater1t, or w = 1 aω eλt sin ωt, (8.7.13) depending upon whether the polynomial p(r) = ar2 + br + c has distinct real zeros r1 and r2, a repeated zero r1, or complex conjugate zeros λ±iω. (In most physical applications, the zeros of the characteristic polynomial have negative real parts, so limt→∞w(t) = 0.) This means that y = u(t −t0)w(t −t0) is deﬁned on (−∞, ∞) and has the following properties: y(t) = 0, t < t0, ay′′ + by′ + cy = 0 on (−∞, t0) and (t0, ∞), and y′ −(t0) = 0, y′ +(t0) = 1/a (8.7.14)	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 function doesn’t satisfy the differential equation in (8.7.11) at t0, since it isn’t differentiable there; in fact (8.7.14) indicates that an impulse causes a jump discontinuity in velocity. (To see that this is reasonable, think of what happens when you hit a ball with a bat.) This means that the initial value problem (8.7.11) doesn’t make sense if t0 = 0, since y′(0) doesn’t exist in this case. However y = u(t)w(t) can be deﬁned to be the solution of the modiﬁed initial value problem ay′′ + by′ + cy = δ(t), y(0) = 0, y′ −(0) = 0, where the condition on the derivative at t = 0 has been replaced by a condition on the derivative from the left. Figure 8.7.2 illustrates Theorem 8.7.1 for the case where the impulse response w is the ﬁrst expression in (8.7.13) and r1 and r2 are distinct and both negative. The solid curve in the ﬁgure is the graph of w. The dashed curves are solutions of (8.7.1) for various values of h. As h decreases the graph of yh moves to the left toward the graph of w. Example 8.7.1 Find the solution of the initial value problem y′′ −2y′ + y = δ(t −t0), y(0) = 0, y′(0) = 0, (8.7.15) where t0 > 0. Then interpret the solution for the case where t0 = 0. Solution Here w = L−1 1 s2 −2s + 1 = L−1 1 (s −1)2 = te−t,	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 would usually write simply as y = te−t) is the solution of the modiﬁed initial value problem y′′ −2y′ + y = δ(t), y(0) = 0, y′ −(0) = 0. The graph of y = u(t −t0)(t −t0)e−(t−t0) is shown in Figure 8.7.3 Deﬁnition 8.7.2 and the principle of superposition motivate the next deﬁnition. Deﬁnition 8.7.3 Suppose α is a nonzero constant and f is piecewise continuous on [0, ∞). If t0 > 0, then the solution of the initial value problem ay′′ + by′ + cy = f(t) + αδ(t −t0), y(0) = k0, y′(0) = k1 is deﬁned to be y(t) = ˆy(t) + αu(t −t0)w(t −t0), where ˆy is the solution of ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1. This deﬁnition also applies if t0 = 0, provided that the initial condition y′(0) = k1 is replaced by y′ −(0) = k1. Example 8.7.2 Solve the initial value problem y′′ + 6y′ + 5y = 3e−2t + 2δ(t −1), y(0) = −3, y′(0) = 2. (8.7.16)	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 (8.7.16) is y = −e−2t + 1 2e−5t −5 2e−t + u(t −1)e−(t−1) −e−5(t−1) 2 (8.7.17) (Figure 8.7.4) . t y 1 2 3 4 −1 −2 −3 t = 1 Figure 8.7.4 Graph of (8.7.17) 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 3 4 5 −1 −2 −3 t = π t = 2π t y Figure 8.7.5 Graph of (8.7.19) Deﬁnition 8.7.3 can be extended in the obvious way to cover the case where the forcing function contains more than one impulse. Example 8.7.3 Solve the initial value problem y′′ + y = 1 + 2δ(t −π) −3δ(t −2π), y(0) = −1, y′(0) = 2. (8.7.18) Solution We leave it to you to show that ˆy = 1 −2 cos t + 2 sin t is the solution of y′′ + y = 1, y(0) = −1, y′(0) = 2. Since w = L−1 1 s2 + 1 = sin t,	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), y(0) = −1, y′(0) = 2 4. C/G y′′ + y = sin 3t + 2δ(t −π/2), y(0) = 1, y′(0) = −1 5. y′′ + 4y = 4 + δ(t −3π), y(0) = 0, y′(0) = 1 6. y′′ −y = 8 + 2δ(t −2), y(0) = −1, y′(0) = 1 7. y′′ + y′ = et + 3δ(t −6), y(0) = −1, y′(0) = 4 8. y′′ + 4y = 8e2t + δ(t −π/2), y(0) = 8, y′(0) = 0 9. C/G y′′ + 3y′ + 2y = 1 + δ(t −1), y(0) = 1, y′(0) = −1 10. y′′ + 2y′ + y = et + 2δ(t −2), y(0) = −1, y′(0) = 2 11. C/G y′′ + 4y = sin t + δ(t −π/2), y(0) = 0, y′(0) = 2 12. y′′ + 2y′ + 2y = δ(t −π) −3δ(t −2π), y(0) = −1, y′(0) = 2 13. y′′ + 4y′ + 13y = δ(t −π/6) + 2δ(t −π/3), y(0) = 1, y′(0) = 2 14. 2y′′ −3y′ −2y = 1 + δ(t −2), y(0) = −1, y′(0) = 2 15. 4y′′ −4y′ + 5y = 4 sin t −4 cos t + δ(t −π/2) −δ(t −π), y(0) = 1, y′(0) = 1 16. y′′ + y = cos 2t + 2δ(t −π/2) −3δ(t −π), y(0) = 0, y′(0) = −1 17. C/G y′′ −y = 4e−t −5δ(t −1) + 3δ(t −2), y(0) = 0, y′(0) = 0 18. y′′ + 2y′ + y = et −δ(t −1) + 2δ(t −2), y(0) = 0, y′(0) = −1 19. y′′ + y = f(t) + δ(t −2π), y(0) = 0, y′(0) = 1, and f(t) = ( sin 2t, 0 ≤t < π, 0, t ≥π. 20. y′′ + 4y = f(t) + δ(t −π) −3δ(t −3π/2), y(0) = 1, y′(0) = −1, and f(t) = ( 1, 0 ≤t < π/2, 2, t ≥π/2 21. y′′ + y = δ(t), y(0) = 1, y′ −(0) = −2 22. y′′ −4y = 3δ(t), y(0) = −1, y′ −(0) = 7 23. y′′ + 3y′ + 2y = −5δ(t), y(0) = 0, y′ −(0) = 0 24. y′′ + 4y′ + 4y = −δ(t), y(0) = 1, y′ −(0) = 5 25. 4y′′ + 4y′ + y = 3δ(t), y(0) = 1, y′ −(0) = −6 In Exercises 26-28, solve the initial value problem ay′′ h + by′ h + cyh =        0, 0 ≤t < t0, 1/h, t0 ≤t < t0 + h, 0, t ≥t0 + h, yh(0) = 0, y′ h(0) = 0,	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(0) = 0, y′(0) = 0 29. Recall from Section 6.2 that the displacement of an object of mass m in a spring–mass system in free damped oscillation is my′′ + cy′ + ky = 0, y(0) = y0, y′(0) = v0, and that y can be written as y = Re−ct/2m cos(ω1t −φ) if the motion is underdamped. Suppose y(τ) = 0. Find the impulse that would have to be applied to the object at t = τ to put it in equilibrium. 30. Solve the initial value problem. Find a formula that does not involve step functions and represents y on each subinterval of [0, ∞) on which the forcing function is zero. (a) y′′ −y = ∞ X k=1 δ(t −k), y(0) = 0, y′(0) = 1 (b) y′′ + y = ∞ X k=1 δ(t −2kπ), y(0) = 0, y′(0) = 1 (c) y′′ −3y′ + 2y = ∞ X k=1 δ(t −k), y(0) = 0, y′(0) = 1 (d) y′′ + y = ∞ X k=1 δ(t −kπ), y(0) = 0, y′(0) = 0	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 −λ)2 + ω2 (s > λ) eλt sin ωt ω (s −λ)2 + ω2 (s > λ) cosh bt s s2 −b2 (s > \|b\|) sinh bt b s2 −b2 (s > \|b\|) t cos ωt s2 −ω2 (s2 + ω2)2 (s > 0)	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) · G(s) δ(t −a) e−as (s > 0)	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 equations. SECTION 9.3 presents the method of undetermined coefﬁcients for higher order equations. SECTION 9.4 extends the method of variation of parameters to higher order equations. 465	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 Chapter 5. In this chapter n is an arbitrary positive integer. In this section we sketch the general theory of linear nth order equations. Since this theory has already been discussed for n = 2 in Sections 5.1 and 5.3, we’ll omit proofs. For convenience, we consider linear differential equations written as P0(x)y(n) + P1(x)y(n−1) + · · · + Pn(x)y = F (x), (9.1.2) which can be rewritten as (9.1.1) on any interval on which P0 has no zeros, with p1 = P1/P0, ..., pn = Pn/P0 and f = F/P0. For simplicity, throughout this chapter we’ll abbreviate the left side of (9.1.2) by Ly; that is, Ly = P0y(n) + P1y(n−1) + · · · + Pny. We say that the equation Ly = F is normal on (a, b) if P0, P1, ..., Pn and F are continuous on (a, b) and P0 has no zeros on (a, b). If this is so then Ly = F can be written as (9.1.1) with p1, ..., pn and f continuous on (a, b). The next theorem is analogous to Theorem 5.3.1. Theorem 9.1.1 Suppose Ly = F is normal on (a, b), let x0 be a point in (a, b), and let k0, k1, ..., kn−1 be arbitrary real numbers. Then the initial value problem Ly = F, y(x0) = k0, y′(x0) = k1, . . ., y(n−1)(x0) = kn−1 has a unique solution on (a, b). Homogeneous Equations Eqn. (9.1.2) is said to be homogeneous if F ≡0 and nonhomogeneous otherwise. Since y ≡0 is obviously a solution of Ly = 0, we call it the trivial solution. Any other solution is nontrivial. If y1, y2, ..., yn are deﬁned on (a, b) and c1, c2, ..., cn are constants, then y = c1y1 + c2y2 + · · · + cnyn (9.1.3) is a linear combination of {y1, y2 . . . , yn}. It’s easy to show that if y1, y2, ..., yn are solutions of Ly = 0 on (a, b), then so is any linear combination of {y1, y2, . . ., yn}. (See the proof of Theorem 5.1.2.) We say that {y1, y2, . . ., yn} is a fundamental set of solutions of Ly = 0 on (a, b) if every solution of Ly = 0 on (a, b) can be written as a linear combination of {y1, y2, . . ., yn}, as in (9.1.3). In this case we say that (9.1.3) is the general solution of Ly = 0 on (a, b). It can be shown (Exercises 14 and 15) that if the equation Ly = 0 is normal on (a, b) then it has in- ﬁnitely many fundamental sets of solutions on (a, b). The next deﬁnition will help to identify fundamental sets of solutions of Ly = 0. We say that {y1, y2, . . ., yn} is linearly independent on (a, b) if the only constants c1, c2, ..., cn such that c1y1(x) + c2y2(x) + · · · + cnyn(x) = 0, a < x < b, (9.1.4) are c1 = c2 = · · · = cn = 0. If (9.1.4) holds for some set of constants c1, c2, ..., cn that are not all zero, then {y1, y2, . . ., yn} is linearly dependent on (a, b) The next theorem is analogous to Theorem 5.1.3.	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 and has the solutions y1 = x2, y2 = x3, and y3 = 1/x on (−∞, 0) and (0, ∞). Show that {y1, y2, y3} is linearly independent on (−∞, 0) and (0, ∞). Then ﬁnd the general solution of (9.1.5) on (−∞, 0) and (0, ∞). Solution Suppose c1x2 + c2x3 + c3 x = 0 (9.1.6) on (0, ∞). We must show that c1 = c2 = c3 = 0. Differentiating (9.1.6) twice yields the system c1x2 + c2x3 + c3 x = 0 2c1x + 3c2x2 −c3 x2 = 0 2c1 + 6c2x + 2c3 x3 = 0. (9.1.7) If (9.1.7) holds for all x in (0, ∞), then it certainly holds at x = 1; therefore, c1 + c2 + c3 = 0 2c1 + 3c2 −c3 = 0 2c1 + 6c2 + 2c3 = 0. (9.1.8) By solving this system directly, you can verify that it has only the trivial solution c1 = c2 = c3 = 0; however, for our purposes it’s more useful to recall from linear algebra that a homogeneous linear system of n equations in n unknowns has only the trivial solution if its determinant is nonzero. Since the determinant of (9.1.8) is 1 1 1 2 3 −1 2 6 2 = 1 0 0 2 1 −3 2 4 0 = 12, it follows that (9.1.8) has only the trivial solution, so {y1, y2, y3} is linearly independent on (0, ∞). Now Theorem 9.1.2 implies that y = c1x2 + c2x3 + c3 x is the general solution of (9.1.5) on (0, ∞). To see that this is also true on (−∞, 0), assume that (9.1.6) holds on (−∞, 0). Setting x = −1 in (9.1.7) yields c1 −c2 −c3 = 0 −2c1 + 3c2 −c3 = 0 2c1 −6c2 −2c3 = 0. Since the determinant of this system is 1 −1 −1 −2 3 −1 2 −6 −2 = 1 0 0 −2 1 −3 2 −4 0 = −12, it follows that c1 = c2 = c3 = 0; that is, {y1, y2, y3} is linearly independent on (−∞, 0).	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 Suppose c1, c2, c3, and c4 are constants such that c1ex + c2e−x + c3e2x + c4e−3x = 0 (9.1.10) for all x. We must show that c1 = c2 = c3 = c4 = 0. Differentiating (9.1.10) three times yields the system c1ex + c2e−x + c3e2x + c4e−3x = 0 c1ex −c2e−x + 2c3e2x −3c4e−3x = 0 c1ex + c2e−x + 4c3e2x + 9c4e−3x = 0 c1ex −c2e−x + 8c3e2x −27c4e−3x = 0. (9.1.11) If (9.1.11) holds for all x, then it certainly holds for x = 0. Therefore c1 + c2 + c3 + c4 = 0 c1 −c2 + 2c3 −3c4 = 0 c1 + c2 + 4c3 + 9c4 = 0 c1 −c2 + 8c3 −27c4 = 0. The determinant of this system is 1 1 1 1 1 −1 2 −3 1 1 4 9 1 −1 8 −27 = 1 1 1 1 0 −2 1 −4 0 0 3 8 0 −2 7 −28 = −2 1 −4 0 3 8 −2 7 −28 = −2 1 −4 0 3 8 0 6 −24 = −2 3 8 6 −24 = 240, (9.1.12) so the system has only the trivial solution c1 = c2 = c3 = c4 = 0. Now Theorem 9.1.2 implies that y = c1ex + c2e−x + c3e2x + c4e−3x is the general solution of (9.1.9). The Wronskian We can use the method used in Examples 9.1.1 and 9.1.2 to test n solutions {y1, y2, . . ., yn} of any nth order equation Ly = 0 for linear independence on an interval (a, b) on which the equation is normal. Thus, if c1, c2 ,..., cn are constants such that c1y1 + c2y2 + · · · + cnyn = 0, a < x < b, then differentiating n −1 times leads to the n × n system of equations c1y1(x) + c2y2(x)+ · · · +cnyn(x) = 0 c1y′ 1(x) + c2y′ 2(x)+ · · · +cny′ n(x) = 0 ... c1y(n−1) 1 (x) + c2y(n−1) 2 (x)+ · · · +cny(n−1) n (x) = 0 (9.1.13)	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 W(x) ̸= 0 for some x in (a, b) then the system (9.1.13) has only the trivial solution c1 = c2 = · · · = cn = 0, and Theorem 9.1.2 implies that y = c1y1 + c2y2 + · · · + cnyn is the general solution of Ly = 0 on (a, b). The next theorem generalizes Theorem 5.1.4. The proof is sketched in (Exercises 17–20). Theorem 9.1.3 Suppose the homogeneous linear nth order equation P0(x)y(n) + P1(x)yn−1 + · · · + Pn(x)y = 0 (9.1.14) is normal on (a, b), let y1, y2, ..., yn be solutions of (9.1.14) on (a, b), and let x0 be in (a, b). Then the Wronskian of {y1, y2, . . ., yn} is given by W(x) = W(x0) exp − Z x x0 P1(t) P0(t) dt , a < x < b. (9.1.15) Therefore, either W has no zeros in (a, b) or W ≡0 on (a, b). Formula (9.1.15) is Abel’s formula. The next theorem is analogous to Theorem 5.1.6.. Theorem 9.1.4 Suppose Ly = 0 is normal on (a, b) and let y1, y2, ..., yn be n solutions of Ly = 0 on (a, b). Then the following statements are equivalent; that is, they are either all true or all false: (a) The general solution of Ly = 0 on (a, b) is y = c1y1 + c2y2 + · · · + cnyn. (b) {y1, y2, . . ., yn} is a fundamental set of solutions of Ly = 0 on (a, b). (c) {y1, y2, . . ., yn} is linearly independent on (a, b). (d) The Wronskian of {y1, y2, . . ., yn} is nonzero at some point in (a, b). (e) The Wronskian of {y1, y2, . . ., yn} is nonzero at all points in (a, b). Example 9.1.3 In Example 9.1.1 we saw that the solutions y1 = x2, y2 = x3, and y3 = 1/x of x3y′′′ −x2y′′ −2xy′ + 6y = 0 are linearly independent on (−∞, 0) and (0, ∞). Calculate the Wronskian of {y1, y2, y3}. Solution If x ̸= 0, then W(x) = x2 x3 1 x 2x 3x2 −1 x2 2 6x 2 x3 = 2x3 1 x 1 x3 2 3x −1 x3 1 3x 1 x3 ,	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 (0, ∞). Example 9.1.4 In Example 9.1.2 we saw that the solutions y1 = ex, y2 = e−x, y3 = e2x, and y4 = e−3x of y(4) + y′′′ −7y′′ −y′ + 6y = 0 are linearly independent on every open interval. Calculate the Wronskian of {y1, y2, y3, y4}. Solution For all x, W(x) = ex e−x e2x e−3x ex −e−x 2e2x −3e−3x ex e−x 4e2x 9e−3x ex −e−x 8e2x −27e−3x . Factoring the exponential common factor from each row yields W(x) = e−x 1 1 1 1 1 −1 2 −3 1 1 4 9 1 −1 8 −27 = 240e−x, from (9.1.12). REMARK: Under the assumptions of Theorem 9.1.4, it isn’t necessary to obtain a formula for W(x). Just evaluate W(x) at a convenient point in (a, b), as we did in Examples 9.1.1 and 9.1.2. Theorem 9.1.5 Suppose c is in (a, b) and α1, α2, ..., are real numbers, not all zero. Under the assump- tions of Theorem 10.3.3, suppose y1 and y2 are solutions of (5.1.35) such that αyi(c) + y′ i(c) + · · · + y(n−1) i (c) = 0, 1 ≤i ≤n. (9.1.16) Then {y1, y2, . . .yn} isn’t linearly independent on (a, b). Proof Since α1, α2, ..., αn are not all zero, (9.1.14) implies that y1(c) y′ 1(c) · · · y(n−1) 1 (c) y2(c) y′ 2(c) · · · y(n−1) 2 (c) ... ... ... ... yn(c) y′ n(c) · · · y(n−1) n (c) = 0, so y1(c) y2(c) · · · yn(c) y′ 1(c) y′ 2(c) · · · y′ n(c) ... ... ... ... y(n−1) 1 (c) y(n−1) 2 (c)(c) · · · y(n−1) n (c)(c) = 0 and Theorem 9.1.4 implies the stated conclusion.	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 = 0. Theorem 9.1.6 Suppose Ly = F is normal on (a, b). Let yp be a particular solution of Ly = F on (a, b), and let {y1, y2, . . ., yn} be a fundamental set of solutions of the complementary equation Ly = 0 on (a, b). Then y is a solution of Ly = F on (a, b) if and only if y = yp + c1y1 + c2y2 + · · · + cnyn, where c1, c2, . . ., cn are constants. The next theorem is analogous to Theorem 5.3.2. Theorem 9.1.7 [The Principle of Superposition] Suppose for each i = 1, 2, ..., r, the function ypi is a particular solution of Ly = Fi on (a, b). Then yp = yp1 + yp2 + · · · + ypr is a particular solution of Ly = F1(x) + F2(x) + · · · + Fr(x) on (a, b). We’ll apply Theorems 9.1.6 and 9.1.7 throughout the rest of this chapter. 9.1 Exercises 1. Verify that the given function is the solution of the initial value problem. (a) x3y′′′ −3x2y′′ + 6xy′ −6y = −24 x , y(−1) = 0, y′(−1) = 0, y′′(−1) = 0; y = −6x −8x2 −3x3 + 1 x (b) y′′′ −1 xy′′ −y′ + 1 xy = x2 −4 x4 , y(1) = 3 2, y′(1) = 1 2, y′′(1) = 1; y = x + 1 2x (c) xy′′′ −y′′ −xy′ + y = x2, y(1) = 2, y′(1) = 5, y′′(1) = −1; y = −x2 −2 + 2e(x−1) −e−(x−1) + 4x (d) 4x3y′′′ + 4x2y′′ −5xy′ + 2y = 30x2, y(1) = 5, y′(1) = 17 2 ; y′′(1) = 63 4 ; y = 2x2 ln x −x1/2 + 2x−1/2 + 4x2 (e) x4y(4) −4x3y′′′ + 12x2y′′ −24xy′ + 24y = 6x4, y(1) = −2, y′(1) = −9, y′′(1) = −27, y′′′(1) = −52; y = x4 lnx + x −2x2 + 3x3 −4x4 (f) xy(4) −y′′′ −4xy′′ + 4y′ = 96x2, y(1) = −5, y′(1) = −24 y′′(1) = −36; y′′′(1) = −48; y = 9 −12x + 6x2 −8x3	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. Find solutions y1, y2, ..., yn of the equation y(n) = 0 that satisfy the initial conditions y(j) i (x0) = ( 0, j ̸= i −1, 1, j = i −1, 1 ≤i ≤n. 5. (a) Verify that the function y = c1x3 + c2x2 + c3 x satisﬁes x3y′′′ −x2y′′ −2xy′ + 6y = 0 (A) if c1, c2, and c3 are constants. (b) Use (a) to ﬁnd solutions y1, y2, and y3 of (A) such that y1(1) = 1, y′ 1(1) = 0, y′′ 1 (1) = 0 y2(1) = 0, y′ 2(1) = 1, y′′ 2 (1) = 0 y3(1) = 0, y′ 3(1) = 0, y′′ 3 (1) = 1. (c) Use (b) to ﬁnd the solution of (A) such that y(1) = k0, y′(1) = k1, y′′(1) = k2. 6. Verify that the given functions are solutions of the given equation, and show that they form a fundamental set of solutions of the equation on any interval on which the equation is normal. (a) y′′′ + y′′ −y′ −y = 0; {ex, e−x, xe−x} (b) y′′′ −3y′′ + 7y′ −5y = 0; {ex, ex cos 2x, ex sin 2x}. (c) xy′′′ −y′′ −xy′ + y = 0; {ex, e−x, x} (d) x2y′′′ + 2xy′′ −(x2 + 2)y = 0; {ex/x, e−x/x, 1} (e) (x2 −2x + 2)y′′′ −x2y′′ + 2xy′ −2y = 0; {x, x2, ex} (f) (2x −1)y(4) −4xy′′′ + (5 −2x)y′′ + 4xy′ −4y = 0; {x, ex, e−x, e2x} (g) xy(4) −y′′′ −4xy′ + 4y′ = 0; {1, x2, e2x, e−2x} 7. Find the Wronskian W of a set of three solutions of y′′′ + 2xy′′ + exy′ −y = 0, given that W(0) = 2.	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) from (a) and Abel’s formula to calculate W(x). (d) What is the general solution of (A)? 10. Compute the Wronskian of the given set of functions. (a) {1, ex, e−x} (b) {ex, ex sinx, ex cos x} (c) {2, x + 1, x2 + 2} (d) x, x lnx, 1/x} (e) {1, x, x2 2! , x3 3! , · · · , xn n! } (f) {ex, e−x, x} (g) {ex/x, e−x/x, 1} (h) {x, x2, ex} (i) {x, x3, 1/x, 1/x2} (j) {ex, e−x, x, e2x} (k) {e2x, e−2x, 1, x2} 11. Suppose Ly = 0 is normal on (a, b) and x0 is in (a, b). Use Theorem 9.1.1 to show that y ≡0 is the only solution of the initial value problem Ly = 0, y(x0) = 0, y′(x0) = 0, . . ., y(n−1)(x0) = 0, on (a, b). 12. Prove: If y1, y2, ..., yn are solutions of Ly = 0 and the functions zi = n X j=1 aijyj, 1 ≤i ≤n, form a fundamental set of solutions of Ly = 0, then so do y1, y2, ..., yn. 13. Prove: If y = c1y1 + c2y2 + · · · + ckyk + yp is a solution of a linear equation Ly = F for every choice of the constants c1, c2 ,..., ck, then Lyi = 0 for 1 ≤i ≤k. 14. Suppose Ly = 0 is normal on (a, b) and let x0 be in (a, b). For 1 ≤i ≤n, let yi be the solution of the initial value problem Lyi = 0, y(j) i (x0) = ( 0, j ̸= i −1, 1, j = i −1, 1 ≤i ≤n, where x0 is an arbitrary point in (a, b). Show that any solution of Ly = 0 on (a, b), can be written as y = c1y1 + c2y2 + · · · + cnyn, with cj = y(j−1)(x0).	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 constants. Show that {z1, z2, . . ., zn} is a fundamental set of solutions of (A) if and only if the determinant a11 a12 · · · a1n a21 a22 · · · a2n ... ... ... ... an1 an2 · · · ann is nonzero.HINT: The determinant of a product of n × n matrices equals the product of the deter- minants. 16. Show that {y1, y2, . . ., yn} is linearly dependent on (a, b) if and only if at least one of the functions y1, y2, ..., yn can be written as a linear combination of the others on (a, b). Take the following as a hint in Exercises 17–19: By the deﬁnition of determinant, a11 a12 · · · a1n a21 a22 · · · a2n ... ... ... ... an1 an2 · · · ann = X ±a1i1a2i2, . . ., anin, where the sum is over all permutations (i1, i2, . . ., in) of (1, 2, . . ., n) and the choice of + or −in each term depends only on the permutation associated with that term. 17. Prove: If A(u1, u2, . . ., un) = a11 a12 · · · a1n a21 a22 · · · a2n ... ... ... ... an−1,1 an−1,2 · · · an−1,n u1 u2 · · · un , then A(u1 + v1, u2 + v2, . . ., un + vn) = A(u1, u2, . . ., un) + A(v1, v2, . . ., vn).	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 to show that if W is the Wronskian of the n-times differentiable functions y1, y2, ..., yn, then W ′ = y1 y2 · · · yn y′ 1 y′ 2 · · · y′ n ... ... ... ... y(n−2) 1 y(n−2) 2 · · · y(n−2) n y(n) 1 y(n) 2 · · · y(n) n . 20. Use Exercises 17 and 19 to show that if W is the Wronskian of solutions {y1, y2, . . ., yn} of the normal equation P0(x)y(n) + P1(x)y(n−1) + · · · + Pn(x)y = 0, (A) then W ′ = −P1W/P0. Derive Abel’s formula (Eqn. (9.1.15)) from this. HINT: Use (A) to write y(n) in terms of y, y′, . . ., y(n−1). 21. Prove Theorem 9.1.6. 22. Prove Theorem 9.1.7. 23. Show that if the Wronskian of the n-times continuously differentiable functions {y1, y2, . . ., yn} has no zeros in (a, b), then the differential equation obtained by expanding the determinant y y1 y2 · · · yn y′ y′ 1 y′ 2 · · · y′ n ... ... ... ... ... y(n) y(n) 1 y(n) 2 · · · y(n) n = 0, in cofactors of its ﬁrst column is normal and has {y1, y2, . . ., yn} as a fundamental set of solutions on (a, b). 24. Use the method suggested by Exercise 23 to ﬁnd a linear homogeneous equation such that the given set of functions is a fundamental set of solutions on intervals on which the Wronskian of the set has no zeros. (a) {x, x2 −1, x2 + 1} (b) {ex, e−x, x} (c) {ex, xe−x, 1} (d) {x, x2, ex}	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 (x) is said to be a constant coefﬁcient equation. In this section we consider the homogeneous constant coef- ﬁcient equation a0y(n) + a1y(n−1) + · · · + any = 0. (9.2.1) Since (9.2.1) is normal on (−∞, ∞), the theorems in Section 9.1 all apply with (a, b) = (−∞, ∞). As in Section 5.2, we call p(r) = a0rn + a1rn−1 + · · · + an (9.2.2) the characteristic polynomial of (9.2.1). We saw in Section 5.2 that when n = 2 the solutions of (9.2.1) are determined by the zeros of the characteristic polynomial. This is also true when n > 2, but the situation is more complicated in this case. Consequently, we take a different approach here than in Section 5.2. If k is a positive integer, let Dk stand for the k-th derivative operator; that is Dky = y(k). If q(r) = b0rm + b1rm−1 + · · · + bm is an arbitrary polynomial, deﬁne the operator q(D) = b0Dm + b1Dm−1 + · · · + bm such that q(D)y = (b0Dm + b1Dm−1 + · · · + bm)y = b0y(m) + b1y(m−1) + · · · + bmy whenever y is a function with m derivatives. We call q(D) a polynomial operator. With p as in (9.2.2), p(D) = a0Dn + a1Dn−1 + · · · + an, so (9.2.1) can be written as p(D)y = 0. If r is a constant then p(D)erx =	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 −6r2 + 11r −6 = (r −1)(r −2)(r −3). Therefore {ex, e2x, e3x} is a set of solutions of (9.2.3). It is a fundamental set, since its Wronskian is W(x) = ex e2x e3x ex 2e2x 3e3x ex 4e2x 9e3x = e6x 1 1 1 1 2 3 1 4 9 = 2e6x ̸= 0. Therefore the general solution of (9.2.3) is y = c1ex + c2e2x + c3e3x. (9.2.5) SOLUTION(b) We must determine c1, c2 and c3 in (9.2.5) so that y satisﬁes the initial conditions in (9.2.4). Differentiating (9.2.5) twice yields y′ = c1ex + 2c2e2x + 3c3e3x y′′ = c1ex + 4c2e2x + 9c3e3x. (9.2.6) Setting x = 0 in (9.2.5) and (9.2.6) and imposing the initial conditions yields c1 + c2 + c3 = 4 c1 + 2c2 + 3c3 = 5 c1 + 4c2 + 9c3 = 9. The solution of this system is c1 = 4, c2 = −1, c3 = 1. Therefore the solution of (9.2.4) is y = 4ex −e2x + e3x (Figure 9.2.1). Now we consider the case where the characteristic polynomial (9.2.2) does not have n distinct real zeros. For this purpose it is useful to deﬁne what we mean by a factorization of a polynomial operator. We begin with an example. Example 9.2.2 Consider the polynomial p(r) = r3 −r2 + r −1 and the associated polynomial operator p(D) = D3 −D2 + D −1.	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 can make this assertion we must deﬁne what we mean by saying that two operators are equal, and what we mean by the products of operators in (9.2.7). We say that two operators are equal if they apply to the same functions and always produce the same result. The deﬁnitions of the products in (9.2.7) is this: if y is any three-times differentiable function then (a) (D −1)(D2 + 1)y is the function obtained by ﬁrst applying D2 + 1 to y and then applying D −1 to the resulting function (b) (D2 + 1)(D −1)y is the function obtained by ﬁrst applying D −1 to y and then applying D2 + 1 to the resulting function. From (a), (D −1)(D2 + 1)y = (D −1)[(D2 + 1)y] = (D −1)(y′′ + y) = D(y′′ + y) −(y′′ + y) = (y′′′ + y′) −(y′′ + y) = y′′′ −y′′ + y′ −y = (D3 −D2 + D −1)y. (9.2.8) This implies that (D −1)(D2 + 1) = (D3 −D2 + D −1).	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 9.2.3 Use the result of Example 9.2.2 to ﬁnd the general solution of y′′′ −y′′ + y′ −y = 0. (9.2.10) Solution From (9.2.8), we can rewrite (9.2.10) as (D −1)(D2 + 1)y = 0, which implies that any solution of (D2 + 1)y = 0 is a solution of (9.2.10). Therefore y1 = cos x and y2 = sin x are solutions of (9.2.10). From (9.2.9), we can rewrite (9.2.10) as (D2 + 1)(D −1)y = 0, which implies that any solution of (D −1)y = 0 is a solution of (9.2.10). Therefore y3 = ex is solution of (9.2.10). The Wronskian of {ex, cos x, sinx} is W(x) = cos x sin x ex −sin x cos x ex −cos x −sin x ex . Since W(0) = 1 0 1 0 1 1 −1 0 1 = 2, {cos x, sinx, ex} is linearly independent and y = c1 cos x + c2 sin x + c3ex is the general solution of (9.2.10). Example 9.2.4 Find the general solution of y(4) −16y = 0. (9.2.11) Solution The characteristic polynomial of (9.2.11) is p(r) = r4 −16 = (r2 −4)(r2 + 4) = (r −2)(r + 2)(r2 + 4). By arguments similar to those used in Examples 9.2.2 and 9.2.3, it can be shown that (9.2.11) can be written as (D2 + 4)(D + 2)(D −2)y = 0	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 this set is W(x) = e2x e−2x cos 2x sin 2x 2e2x −2e−2x −2 sin 2x 2 cos 2x 4e2x 4e−2x −4 cos 2x −4 sin 2x 8e2x −8e−2x 8 sin2x −8 cos 2x . Since W(0) = 1 1 1 0 2 −2 0 2 4 4 −4 0 8 −8 0 −8 = −512, {e2x, e−2x, cos 2x, sin2x} is linearly independent, and y1 = c1e2x + c2e−2x + c3 cos 2x + c4 sin 2x is the general solution of (9.2.11). It is known from algebra that every polynomial p(r) = a0rn + a1rn−1 + · · · + an with real coefﬁcients can be factored as p(r) = a0p1(r)p2(r) · · ·pk(r), where no pair of the polynomials p1, p2, ..., pk has a commom factor and each is either of the form pj(r) = (r −rj)mj, (9.2.12) where rj is real and mj is a positive integer, or pj(r) = (r −λj)2 + ω2 j mj , (9.2.13) where λj and ωj are real, ωj ̸= 0, and mj is a positive integer. If (9.2.12) holds then rj is a real zero of p, while if (9.2.13) holds then λ + iω and λ −iω are complex conjugate zeros of p. In either case, mj is the multiplicity of the zero(s). By arguments similar to those used in our examples, it can be shown that p(D) = a0p1(D)p2(D) · · · pk(D) (9.2.14) and that the order of the factors on the right can be chosen arbitrarily. Therefore, if pj(D)y = 0 for some j then p(D)y = 0. To see this, we simply rewrite (9.2.14) so that pj(D) is applied ﬁrst. Therefore the	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 solutions {y1, y2, . . ., yn} of p(D)y = 0, we ﬁnd fundamental set of solutions of each of the equations and take {y1, y2, . . ., yn} to be the set of all functions in these separate fundamental sets. In Exercise 40 we sketch the proof that {y1, y2, . . ., yn} is linearly independent, and therefore a fundamental set of solutions of p(D)y = 0. To apply this procedure to general homogeneous constant coefﬁcient equations, we must be able to ﬁnd fundamental sets of solutions of equations of the form (D −a)my = 0 and (D −λ)2 + ω2m y = 0, where m is an arbitrary positive integer. The next two theorems show how to do this. Theorem 9.2.1 If m is a positive integer, then {eax, xeax, . . ., xm−1eax} (9.2.15) is a fundamental set of solutions of (D −a)my = 0. (9.2.16) Proof We’ll show that if f(x) = c1 + c2x + · · · + cmxm−1 is an arbitrary polynomial of degree ≤m −1, then y = eaxf is a solution of (9.2.16). First note that if g is any differentiable function then (D −a)eaxg = Deaxg −aeaxg = aeaxg + eaxg′ −aeaxg, so (D −a)eaxg = eaxg′. (9.2.17) Therefore (D −a)eaxf = eaxf′ (from (9.2.17) with g = f) (D −a)2eaxf = (D −a)eaxf′ = eaxf′′ (from (9.2.17) with g = f′) (D −a)3eaxf = (D −a)eaxf′′ = eaxf′′′ (from (9.2.17) with g = f′′) ... (D −a)meaxf = (D −a)eaxf(m−1) = eaxf(m) (from (9.2.17) with g = f(m−1)). Since f(m) = 0, the last equation implies that y = eaxf is a solution of (9.2.16) if f is any polynomial of degree ≤m −1. In particular, each function in (9.2.15) is a solution of (9.2.16). To see that (9.2.15) is linearly independent (and therefore a fundamental set of solutions of (9.2.16)), note that if c1eax + c2xeax + c · · · + cm−1xm−1eax = 0 for all x in some interval (a, b), then c1 + c2x + c · · · + cm−1xm−1 = 0 for all x in (a, b). However, we know from algebra that if this polynomial has more than m−1 zeros then c1 = c2 = · · · = cn = 0.	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 (9.2.18) is y = e−x(c1 + c2x + c3x2). The proof of the next theorem is sketched in Exercise 41. Theorem 9.2.2 If ω ̸= 0 and m is a positive integer, then {eλx cos ωx, xeλx cos ωx, . . ., xm−1eλx cos ωx, eλx sin ωx, xeλx sin ωx, . . ., xm−1eλx sin ωx} is a fundamental set of solutions of [(D −λ)2 + ω2]my = 0. Example 9.2.6 Find the general solution of (D2 + 4D + 13)3y = 0. (9.2.19) Solution The characteristic polynomial of (9.2.19) is p(r) = (r2 + 4r + 13)3 =	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 (9.2.20) is y = e−x(c1 cos x + c2 sin x) + c3e−2x + c4. Example 9.2.8 Find a fundamental set of solutions of [(D + 1)2 + 1]2(D −1)3(D + 1)D2y = 0. (9.2.21) Solution A fundamental set of solutions of (9.2.21) can be obtained by combining fundamental sets of solutions of (D + 1)2 + 1 2 y = 0, (D −1)3y = 0, (D + 1)y = 0, and D2y = 0. Fundamental sets of solutions of these equations are given by {e−x cos x, xe−x cos x, e−x sin x, xe−x sin x}, {ex, xex, x2ex}, {e−x}, and {1, x}, respectively. These ten functions form a fundamental set of solutions of (9.2.21). 9.2 Exercises In Exercises 1–14 ﬁnd the general solution. 1. y′′′ −3y′′ + 3y′ −y = 0 2. y(4) + 8y′′ −9y = 0 3. y′′′ −y′′ + 16y′ −16y = 0 4. 2y′′′ + 3y′′ −2y′ −3y = 0 5. y′′′ + 5y′′ + 9y′ + 5y = 0 6. 4y′′′ −8y′′ + 5y′ −y = 0 7. 27y′′′ + 27y′′ + 9y′ + y = 0 8. y(4) + y′′ = 0 9. y(4) −16y = 0 10. y(4) + 12y′′ + 36y = 0 11. 16y(4) −72y′′ + 81y = 0 12. 6y(4) + 5y′′′ + 7y′′ + 5y′ + y = 0 13. 4y(4) + 12y′′′ + 3y′′ −13y′ −6y = 0 14. y(4) −4y′′′ + 7y′′ −6y′ + 2y = 0	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, y′′(0) = 4 18. C/G y′′′ −2y′ −4y = 0, y(0) = 6, y′(0) = 3, y′′(0) = 22 19. C/G 3y′′′ −y′′ −7y′ + 5y = 0, y(0) = 14 5 , y′(0) = 0, y′′(0) = 10 20. y′′′ −6y′′ + 12y′ −8y = 0, y(0) = 1, y′(0) = −1, y′′(0) = −4 21. 2y′′′ −11y′′ + 12y′ + 9y = 0, y(0) = 6, y′(0) = 3, y′′(0) = 13 22. 8y′′′ −4y′′ −2y′ + y = 0, y(0) = 4, y′(0) = −3, y′′(0) = −1 23. y(4) −16y = 0, y(0) = 2, y′(0) = 2, y′′(0) = −2, y′′′(0) = 0 24. y(4) −6y′′′ + 7y′′ + 6y′ −8y = 0, y(0) = −2, y′(0) = −8, y′′(0) = −14, y′′′(0) = −62 25. 4y(4) −13y′′ + 9y = 0, y(0) = 1, y′(0) = 3, y′′(0) = 1, y′′′(0) = 3 26. y(4) + 2y′′′ −2y′′ −8y′ −8y = 0, y(0) = 5, y′(0) = −2, y′′(0) = 6, y′′′(0) = 8 27. C/G 4y(4) + 8y′′′ + 19y′′ + 32y′ + 12y = 0, y(0) = 3, y′(0) = −3, y′′(0) = −7 2, y′′′(0) = 31 4 28. Find a fundamental set of solutions of the given equation, and verify that it’s a fundamental set by evaluating its Wronskian at x = 0. (a) (D −1)2(D −2)y = 0 (b) (D2 + 4)(D −3)y = 0 (c) (D2 + 2D + 2)(D −1)y = 0 (d) D3(D −1)y = 0 (e) (D2 −1)(D2 + 1)y = 0 (f) (D2 −2D + 2)(D2 + 1)y = 0 In Exercises 29–38 ﬁnd a fundamental set of solutions. 29. (D2 + 6D + 13)(D −2)2D3y = 0 30. (D −1)2(2D −1)3(D2 + 1)y = 0 31. (D2 + 9)3D2y = 0 32. (D −2)3(D + 1)2Dy = 0 33. (D2 + 1)(D2 + 9)2(D −2)y = 0 34. (D4 −16)2y = 0 35. (4D2 + 4D + 9)3y = 0 36. D3(D −2)2(D2 + 4)2y = 0 37. (4D2 + 1)2(9D2 + 4)3y = 0 38. (D −1)4 −16 y = 0	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 form (aj −ai) with i and j between 1 and n and i < j. (a) Verify (A) for n = 2 and n = 3. (b) Find the Wronskian of {ea1x, ea2x, . . ., eanx}. 40. A theorem from algebra says that if P1 and P2 are polynomials with no common factors then there are polynomials Q1 and Q2 such that Q1P1 + Q2P2 = 1. This implies that Q1(D)P1(D)y + Q2(D)P2(D)y = y for every function y with enough derivatives for the left side to be deﬁned. (a) Use this to show that if P1 and P2 have no common factors and P1(D)y = P2(D)y = 0 then y = 0. (b) Suppose P1 and P2 are polynomials with no common factors. Let u1, ..., ur be linearly independent solutions of P1(D)y = 0 and let v1, ..., vs be linearly independent solutions of P2(D)y = 0. Use (a) to show that {u1, . . ., ur, v1, . . ., vs} is a linearly independent set. (c) Suppose the characteristic polynomial of the constant coefﬁcient equation a0y(n) + a1y(n−1) + · · · + any = 0 (A) has the factorization p(r) = a0p1(r)p2(r) · · ·pk(r), where each pj is of the form pj(r) = (r −rj)nj or pj(r) = [(r −λj)2 + w2 j]mj (ωj > 0) and no two of the polynomials p1, p2, ..., pk have a common factor. Show that we can ﬁnd a fundamental set of solutions {y1, y2, . . ., yn} of (A) by ﬁnding a fundamental set of solutions of each of the equations pj(D)y = 0, 1 ≤j ≤k, and taking {y1, y2, . . ., yn} to be the set of all functions in these separate fundamental sets.	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 polynomial of degree ≤m −1, then (D2 + ω2)mz = 0. (B) (c) Use Eqn. (9.2.17) to show that if y = eλxz then [(D −λ)2 + ω2]my = eλx(D2 + ω2)mz. (d) Conclude from (b) and (c) that if p and q are arbitrary polynomials of degree ≤m −1 then y = eλx(p(x) cos ωx + q(x) sin ωx) is a solution of [(D −λ)2 + ω2]my = 0. (C) (e) Conclude from (d) that the functions eλx cos ωx, xeλx cos ωx, . . . , xm−1eλx cos ωx, eλx sinωx, xeλx sinωx, . . . , xm−1eλx sin ωx (D) are all solutions of (C). (f) Complete the proof of Theorem 9.2.2 by showing that the functions in (D) are linearly inde- pendent. 42. (a) Use the trigonometric identities cos(A + B) = cos A cos B −sinA sin B sin(A + B) = cos A sin B + sin A cos B to show that (cos A + i sin A)(cos B + i sin B) = cos(A + B) + i sin(A + B). (b) Apply (a) repeatedly to show that if n is a positive integer then n Y k=1 (cos Ak + i sin Ak) = cos(A1 + A2 + · · · + An) + i sin(A1 + A2 + · · · + An). (c) Infer from (b) that if n is a positive integer then (cos θ + i sin θ)n = cos nθ + i sin nθ. (A) (d) Show that (A) also holds if n = 0 or a negative integer. HINT: Verify by direct calculation that (cos θ + i sin θ)−1 = (cos θ −i sin θ). Then replace θ by −θ in (A).	Elementary Differential Equations with Boundary Value Problems_Page_496_Chunk2900

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