Lecture Notes in Differential Equations - Bruce E. Shapiro

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396 APPENDIX A. TABLE OF INTEGRALS ∫ ∫ ∫ ∫ ∫ ∫ 1 1 + x 2 dx = tan−1 x (A.8) 1 a 2 + x 2 dx = 1 a tan−1 x a x a 2 + x 2 dx = 1 2 ln |a2 + x 2 | + c x 2 a 2 + x 2 dx = x − a tan−1 x a x 3 a 2 + x 2 dx = 1 2 x2 − 1 2 a2 ln |a 2 + x 2 | 1 ax 2 + bx + c dx = 2 2ax + b √ 4ac − b 2 tan−1 √ 4ac − b 2 ∫ 1 (x + a)(x + b) dx = 1 b − a ln a + x b + x , a ≠ b ∫ x (x + a) 2 dx = a + ln |a + x| a + x (A.15) (A.9) (A.10) (A.11) (A.12) (A.13) (A.14) ∫ x ax 2 + bx + c dx = 1 2a ln |ax2 + bx + c| − b a √ 2ax + b 4ac − b 2 tan−1 √ 4ac − b 2 Integrals with Roots (A.16) ∫ √x − adx = 2 3 (x − a)3/2 (A.17) ∫ 1 √ x ± a dx = 2 √ x ± a (A.18) ∫ 1 √ a − x dx = −2 √ a − x (A.19)
397 ∫ x √ x − adx = 2 3 a(x − a)3/2 + 2 (x − a)5/2 (A.20) 5 ∫ ( √ax 2b + bdx = 3a + 2x ) √ax + b (A.21) 3 ∫ ∫ (ax + b) 3/2 dx = 2 (ax + b)5/2 (A.22) 5a x √ x ± a dx = 2 3 (x ∓ 2a)√ x ± a (A.23) ∫ √ √ x x(a − x) a − x dx = −√ x(a − x) − a tan −1 x − a (A.24) ∫ √ x a + x dx = √ x(a + x) − a ln [√ x + √ x + a ] (A.25) ∫ x √ ax + bdx = 2 15a 2 (−2b2 + abx + 3a 2 x 2 ) √ ax + b (A.26) ∫ √x(ax 1 [ + b)dx = (2ax + b) √ ax(ax + b) 4a 3/2 −b 2 ∣ ln ∣a √ x + √ ] a(ax + b) ∣ (A.27) ∫ √x3 (ax + b)dx = [ b 12a − b2 8a 2 x + x ] √x3 (ax + b) 3 + b3 8a 5/2 ln ∣ ∣∣a √ x + √ a(ax + b) ∣ ∣∣ (A.28) ∫ √x2 ± a 2 dx = 1 2 x√ x 2 ± a 2 ± 1 2 a2 ln ∣ ∣x + √ x 2 ± a 2 ∣ ∣∣ (A.29)
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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,
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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
Page 201 and 202:
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
Page 219 and 220:
Lesson 25 Harmonic Oscillations If
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213 It is standard to define a new
Page 223 and 224:
215 As with the unforced case, we c
Page 225 and 226:
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
Page 235 and 236:
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
Page 251 and 252:
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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Page 261 and 262:
253 So that f(t) = a n (t)y[ (n) +
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Lesson 28 Series Solutions In many
Page 265 and 266:
257 Changing the index of the secon
Page 267 and 268:
259 Since the first two terms (corr
Page 269 and 270:
261 Hence ∞∑ ∞∑ ∞∑ 0 =
Page 271 and 272:
263 has an analytic solution at t =
Page 273 and 274:
265 By the triangle inequality, |(k
Page 275 and 276:
267 Table 28.1: Table of Special Fu
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269 Thus y = a 0 ( 1 + 1 6 t3 + 1 +
Page 279 and 280:
271 into (28.114) and collect terms
Page 281 and 282:
273 Summary of Power series method.
Page 283 and 284:
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 ′
Page 291 and 292:
Lesson 30 The Method of Frobenius I
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285 This is a homogeneous linear eq
Page 295 and 296:
287 Example 30.4. Find a Frobenius
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289 Thus a Frobenius solution is y
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291 Example 30.6. Find the form of
Page 301 and 302:
293 term by term to (30.97). Starti
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295 Let j = n − k. Then |n − 1
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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
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305 is where λ 2 − T λ + ∆ =
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307 (b) If λ 1 ≠ λ 2 ∈ R, i.e
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309 We will verify (31.54) by induc
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311 The Jordan Form Let A be a squa
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313 y = 1 [( ) ( ) ] ( ) 4 1 e 2t 1
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315 Theorem 31.8. (Abel’s Formula
Page 325 and 326:
317 By a similar argument, the seco
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319 we can replace (31.118) with a
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321 Corollary 31.12. The generalize
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323 where (A − λI)w 2 = w 1 , i.
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325 so that λ = 3, −5. The eigen
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327 Non-constant Coefficients We ca
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329 we find that ∫ M(t)g(t)dt = (
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Lesson 32 The Laplace Transform Bas
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333 Figure 32.1: A piecewise contin
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335 Example 32.4. From integral A.1
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337 apply this result iteratively.
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L [ t x−1] [ ] 1 d = L x dt tx =
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341 Equating numerators and expandi
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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
Page 399 and 400: 391 Figure 34.8: Topological instab
Page 401 and 402: 393 Figure 34.10: phase portraits f
Page 403: Appendix A Table of Integrals Basic
Page 407 and 408: 399 ∫ x √ ax2 + bx + c dx = 1 a
Page 409 and 410: 401 ∫ ∫ ∫ ∫ e ax2 dx = −
Page 411 and 412: 403 ∫ tan 3 axdx = 1 a ln cos ax
Page 413 and 414: 405 Products of Trigonometric Funct
Page 415 and 416: Appendix B Table of Laplace Transfo
Page 417 and 418: 409 e at cosh kt t sin kt t cos kt
Page 419 and 420: Appendix C Summary of Methods First
Page 421 and 422: 413 The resulting equation is linea
Page 423 and 424: 415 for y once z is known. Method o
Page 425 and 426: Bibliography [1] Bear, H.S. Differe
Page 427 and 428: BIBLIOGRAPHY 419
Page 429 and 430: BIBLIOGRAPHY 421
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