Lecture Notes in Differential Equations - Bruce E. Shapiro

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340 LESSON 32. THE LAPLACE TRANSFORM Using Laplace Transforms to Solve Linear Differential Equations The idea is this: the Laplace Transform turns derivatives into algebraic functions. So if we convert an entire linear ODE into its Laplace Transform, since the transform is linear, all of the derivatives will go away. Then we can solve for F (s) as a function of s. A solution to the differential equation is given by any function f(t) whose Laplace transform is given by F (s). The process is illustrated with an example. Example 32.10. Solve y ′ + 2y = 3, y(0) = 1, using Laplace transforms. Let Y (s) = L[y(t)] and apply the Laplace Transform operator to the entire equation. From example 32.1, and equation 32.35, y ′ (t) + 2y(t) = 3t (32.66) L[y ′ (t) + 2y(t)] = L[3t] (32.67) L[y ′ (t)] + 2L[y(t)] = 3L[t] (32.68) sY (s) − y(0) + 2Y (s) = 3 s 2 (32.69) (s + 2)Y (s) = 3 s 2 + y(0) = 3 s 2 + 1 = 3 + s2 s 2 (32.70) (32.71) Solving for Y (s), Y (s) = 3 + s2 (s + 2)s 2 (32.72) The question then becomes the following: What function has a Laplace Transform that is equal to (3 + s 2 )/(s 2 (s + 2))? This is called the inverse Laplace Transform of Y (s) and gives y(t): [ ] 3 + s y(t) = L −1 2 (s + 2)s 2 (32.73) We typically approach this by trying to simplify the function into smaller parts until we recognize each part as a Laplace transform. For example, we can use the method of Partial Fractions to separated the expression: 3 + s 2 (s + 2)s 2 = A + Bs s 2 + C + Ds s + 2 = (A + Bs)(s + 2) s 2 (s + 2) + (C + Ds)s2 s 2 (s + 2) (32.74) (32.75)
341 Equating numerators and expanding the right hand side, 3 + s 2 = (A + Bs)(s + 2) + (C + Ds)s 2 (32.76) Equating coefficients of like powers of s, Hence = As + 2A + Bs 2 + 2Bs + Cs 2 + Ds 3 (32.77) = 2A + (A + 2B)s + (B + C)s 2 + Ds 3 (32.78) 3 = 2A =⇒ A = 3 2 0 = A + 2B =⇒ B = −A 2 = −3 4 (32.79) (32.80) 1 = B + C =⇒ C = 1 − B = 7 4 (32.81) 0 = D (32.82) 3 + s 2 3 (s + 2)s 2 = 2 − 3 4 s 7 4 s 2 + s + 2 = 3 2 · 1 s 2 − 3 4 · 1 s + 7 4 · 1 s + 2 (32.83) (32.84) From equations (32.4) and (32.8) we see that 3 + s 2 (s + 2)s 2 = 3 2 · L[t] − 3 4 · L[ e 0] + 7 4 · L[ e −2t] (32.85) [ 3 = L 2 t − 3 4 + 7 ] 4 · e−2t (32.86) [ ] 3 + s y(t) =L −1 2 (s + 2)s 2 = 3 2 t − 3 4 + 7 4 · e−2t (32.87) which gives us the solution to the initial value problem. Here is a summary of the procedure to solve the initial value problem for y ′ (t): 1. Apply the Laplace transform operator to both sides of the differential equation.
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Lecture Notes in Differential Equat
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Contents Front Cover . . . . . . .
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CONTENTS v Dedicated to the hundred
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Preface These lecture notes on diff
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Lesson 1 Basic Concepts A different
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3 Definition 1.2 (Solution, ODE). A
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5 1.22 is restricted to being a pos
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7 Figure 1.1 illustrates what this
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9 We will study linear equations in
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Lesson 2 A Geometric View One way t
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13 We can extend this geometric int
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15 that since the slope of the solu
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Lesson 3 Separable Equations An ODE
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19 Since it is not possible to solv
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21 where M(t) = −a(t) and N(y) =
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23 Example 3.10. Find a general sol
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Lesson 4 Linear Equations Recall th
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27 So far any function µ will work
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29 Example 4.1. Solve the different
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31 Since p(t) = 1 (the coefficient
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33 Multiplying equation 4.68 by µ
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35 Substituting y = t = 0 in this i
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37 ∫ t t 0 Evaluating the integra
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Lesson 5 Bernoulli Equations The Be
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41 This is a Bernoulli equation wit
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Lesson 6 Exponential Relaxation One
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45 Exponential Runaway First we con
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47 Figure 6.2: Illustration of the
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49 This is identical to with Theref
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51 this becomes a first-order ODE i
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Lesson 7 Autonomous Differential Eq
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55 Figure 7.1: A plot of the right-
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57 Figure 7.2: Solutions of the log
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59 Figure 7.4: Solutions of the thr
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Lesson 8 Homogeneous Equations Defi
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63 where z = y/t, the differential
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Lesson 9 Exact Equations We can re-
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67 Now compare equation (9.2) with
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69 Hence dg dy = 0 =⇒ g = C′ (9
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71 From the first of equations (9.5
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73 Differentiating equations (9.81)
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75 This has the form Mdt + Ndy = 0
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Lesson 10 Integrating Factors Defin
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79 Differentiating with respect to
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81 Proof. In each of the five cases
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83 as required by equation (10.31).
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85 Since M y ≠ N t , equation (10
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87 the revised equation (10.100) is
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89 Substituting (10.129) into (10.1
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Lesson 11 Method of Successive Appr
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93 because the integral is zero (th
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95 Example 11.1. Construct the Pica
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97 We can then plug this expression
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Lesson 12 Existence of Solutions* I
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101 • Interchangeability of Limit
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103 But on the square −1 ≤ t
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105 Thus lim φ n = φ 0 + lim n→
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107 because the right hand side doe
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Lesson 13 Uniqueness of Solutions*
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111 The proof of theorem (13.1) is
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113 But δ(t) is an absolute value,
Page 123 and 124:
115 Substituting (13.66) into (13.6
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Lesson 14 Review of Linear Algebra
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119 Definition 14.10. An m × n (or
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121 Definition 14.19. Matrix Multip
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123 In practical terms, computation
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125 Simplifying 4x − 2 + 3z = 0 (
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Lesson 15 Linear Operators and Vect
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129 Example 15.3. By a similar argu
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131 Therefore ‖y + z‖ 2 ≤ ‖
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133 Definition 15.5. Two vectors y,
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Lesson 16 Linear Equations With Con
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137 Hence both r = 1 and r = 3. Thi
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139 The second order linear initial
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141 The general solution to is give
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Lesson 17 Some Special Substitution
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145 Therefore since z = y ′ , Int
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147 Example 17.5. Solve yy ′′ +
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149 where I is the identity matrix.
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151 can be rewritten by solving a =
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Lesson 18 Complex Roots We know for
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155 Theorem 18.2. Euler’s Formula
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157 For k = 0, 1, 2, . . . , n −
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159 and its roots are given by The
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161 The motivation for equation 18.
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Lesson 19 Method of Undetermined Co
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165 3. If f(t) = e rt and r is a ro
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167 Example 19.4. Solve ⎫ y ′
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169 Adding the two equations gives
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Lesson 20 The Wronskian We have see
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173 Definition 20.1. The Wronskian
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175 Example 20.3. Show that y = sin
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177 and therefore the system of equ
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Lesson 21 Reduction of Order The me
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181 The method of reduction of orde
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183 Plugging these into Bessel’s
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185 Example 21.5. Find a second sol
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Lesson 22 Non-homogeneous Equations
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189 where r 1 and r 2 are the roots
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191 This is a first order linear eq
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193 Theorem 22.5. Properties of the
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195 where (∫ ν(t) = exp ) −r 2
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197 The characteristic equation is
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Lesson 23 Method of Annihilators In
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201 Theorem 23.5. (D 2 − 2aD + (a
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203 The method of annihilators is r
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Lesson 24 Variation of Parameters T
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207 Substituting into equation (24.
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209 Example 24.3. Solve the initial
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Lesson 25 Harmonic Oscillations If
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213 It is standard to define a new
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215 As with the unforced case, we c
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Lesson 26 General Existence Theory*
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219 In the case just proven, there
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221 Theorem 26.5. Under the same co
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223 Since K n /(1 − K) → 0 as n
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225 for any φ ∈ V. Let g, h be f
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Lesson 27 Higher Order Linear Equat
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229 L n+1 (e rt y) = e rt a n (D +
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231 Example 27.2. Find the general
Page 241 and 242:
233 Differentiating, u ′ (t) = d
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235 Integrating, − 2K |t − t 0
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237 a closed form expression for a
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Page 249 and 250:
241 The characteristic equation is
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243 The Wronskian In this section w
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245 Certainly every φ(t) given by
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247 the differential equation. Over
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249 By the lemma, to obtain the der
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253 So that f(t) = a n (t)y[ (n) +
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Lesson 28 Series Solutions In many
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257 Changing the index of the secon
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259 Since the first two terms (corr
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261 Hence ∞∑ ∞∑ ∞∑ 0 =
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263 has an analytic solution at t =
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265 By the triangle inequality, |(k
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267 Table 28.1: Table of Special Fu
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269 Thus y = a 0 ( 1 + 1 6 t3 + 1 +
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271 into (28.114) and collect terms
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273 Summary of Power series method.
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Lesson 29 Regular Singularities The
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277 ∑ ∞ ∞∑ ∞∑ 0 = t 2 a
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279 Case 2: Two equal real roots. S
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281 Example 29.6. Solve t 2 y ′
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Lesson 30 The Method of Frobenius I
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285 This is a homogeneous linear eq
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287 Example 30.4. Find a Frobenius
Page 297 and 298: 289 Thus a Frobenius solution is y
Page 299 and 300: 291 Example 30.6. Find the form of
Page 301 and 302: 293 term by term to (30.97). Starti
Page 303 and 304: 295 Let j = n − k. Then |n − 1
Page 305 and 306: 297 is a solution of (t − t 0 ) 2
Page 307 and 308: 299 Evaluation of the integral depe
Page 309 and 310: 301 Example 30.8. In example 30.4 w
Page 311 and 312: Lesson 31 Linear Systems The genera
Page 313 and 314: 305 is where λ 2 − T λ + ∆ =
Page 315 and 316: 307 (b) If λ 1 ≠ λ 2 ∈ R, i.e
Page 317 and 318: 309 We will verify (31.54) by induc
Page 319 and 320: 311 The Jordan Form Let A be a squa
Page 321 and 322: 313 y = 1 [( ) ( ) ] ( ) 4 1 e 2t 1
Page 323 and 324: 315 Theorem 31.8. (Abel’s Formula
Page 325 and 326: 317 By a similar argument, the seco
Page 327 and 328: 319 we can replace (31.118) with a
Page 329 and 330: 321 Corollary 31.12. The generalize
Page 331 and 332: 323 where (A − λI)w 2 = w 1 , i.
Page 333 and 334: 325 so that λ = 3, −5. The eigen
Page 335 and 336: 327 Non-constant Coefficients We ca
Page 337 and 338: 329 we find that ∫ M(t)g(t)dt = (
Page 339 and 340: Lesson 32 The Laplace Transform Bas
Page 341 and 342: 333 Figure 32.1: A piecewise contin
Page 343 and 344: 335 Example 32.4. From integral A.1
Page 345 and 346: 337 apply this result iteratively.
Page 347: L [ t x−1] [ ] 1 d = L x dt tx =
Page 351 and 352: 343 Derivatives of the Laplace Tran
Page 353 and 354: 345 can be written as as illustrate
Page 355 and 356: 347 Translations in the Laplace Var
Page 357 and 358: 349 Summary of Translation Formulas
Page 359 and 360: 351 The inverse transform is [ ] f(
Page 361 and 362: 353 Example 32.18. Find the Laplace
Page 363 and 364: 355 Similarly, we can express a uni
Page 365 and 366: 357 Figure 32.7: Solution of exampl
Page 367 and 368: Lesson 33 Numerical Methods Euler
Page 369 and 370: 361 Figure 33.1: Illustration of Eu
Page 371 and 372: 363 y 4 = y 3 + hf(t 3 , y 3 ) (33.
Page 373 and 374: 365 Figure 33.3: Illustration of th
Page 375 and 376: 367 result with a smaller step size
Page 377 and 378: 369 Expanding the final term in a T
Page 379 and 380: 371 k 2 = y 0 + h 2 f(t 0, k 1 ) (3
Page 381 and 382: Lesson 34 Critical Points of Autono
Page 383 and 384: 375 Since both f and g are differen
Page 385 and 386: 377 Using the cos π/4 = √ 2/2 an
Page 387 and 388: 379 values, of the matrix. We find
Page 389 and 390: 381 Distinct Real Nonzero Eigenvalu
Page 391 and 392: 383 eigendirection {λ 1 , v 1 }dom
Page 393 and 394: 385 Figure 34.5: Phase portraits ty
Page 395 and 396: 387 Complex Conjugate Pair with non
Page 397 and 398: 389 The angular change is described
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391 Figure 34.8: Topological instab
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393 Figure 34.10: phase portraits f
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Appendix A Table of Integrals Basic
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397 ∫ x √ x − adx = 2 3 a(x
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399 ∫ x √ ax2 + bx + c dx = 1 a
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401 ∫ ∫ ∫ ∫ e ax2 dx = −
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403 ∫ tan 3 axdx = 1 a ln cos ax
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405 Products of Trigonometric Funct
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Appendix B Table of Laplace Transfo
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409 e at cosh kt t sin kt t cos kt
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Appendix C Summary of Methods First
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413 The resulting equation is linea
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415 for y once z is known. Method o
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Bibliography [1] Bear, H.S. Differe
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BIBLIOGRAPHY 419
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BIBLIOGRAPHY 421
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Lecture Notes in Differential Equations - Bruce E. Shapiro

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