Lecture Notes in Differential Equations - Bruce E. Shapiro

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210 LESSON 24. VARIATION OF PARAMETERS Next, we need to find a particular solution; we can do this using the variation of parameters formula. Since a(t) = t 2 and f(t) = 2t, we have ∫ ∫ y2 (t)f(t) y P = −y 1 (t) a(t)W (t) dt + y y1 (t)f(t) 2(t) dt (24.49) a(t)W (t) We previously had W = C; but here we need an exact value. To get the exact value of C, we calculate the Wronskina from the now-known homogeneous solutions, C = W = ∣ y ∣ ∣ 1 y 2 ∣∣∣ y 1 ′ y 2 ′ ∣ = t 2 1/t ∣∣∣ 2t −1/t 2 = −3 (24.50) Using y 1 = t 2 , y 2 = 1/t, f(t) = 2t, and a(t) = t 2 , ∫ y P = −t 2 2t t · t 2 · −3 dt + 1 ∫ t2 · 2t t t 2 dt (24.51) · −3 ∫ = 2t2 t −2 dt − 2 ∫ tdt (24.52) 3 3t = 2t2 3 · −1 − 2 t 3t · t2 (24.53) 2 = − 2t 3 − t = −t (24.54) 3 Hence y P = −t and therefore y = y H + y P = C 1 t 2 + C 2 t − 1 t (24.55) From the first initial condition we have 0 = C 1 + C 2 − 1 or To use the second condition we need the derivative, C 1 + C 2 = 1. (24.56) y ′ = 2C 1 t − C 2 t 2 − 1 (24.57) hence the second condition gives 1 = 2C 1 − C 2 − 1 or 2C 1 − C 2 = 2. (24.58) Solving for C 1 and C 2 gives C 1 = 1, hence C 2 = 0, so that the complete solution of the initial value problem is y = t 2 − 1 t (24.59)
Lesson 25 Harmonic Oscillations If a spring is extended from its resting length by an amount y then a restoring force, opposite in direction from but proportional to the displacement will attempt to pull the spring back; the subsequent motion of an object of mass m attached to the end of the spring is described by Newton’s laws of motion: my ′′ = −ky (25.1) where k is a positive constant that is determined by the mechanical properties of the spring. The right-hand side of quation (25.1) – that the force is proportional to the displacement – is known as Hooke’s Law. Simple Harmonic Motion Rearranging equation (25.1), y ′′ + ω 2 y = 0 (25.2) where ω = √ k/m is called the oscillation frequency. Equation (25.2) is called the simple harmonic oscillator equation because there are no additional drag or forcing functions. The oscillator, once started, continues to oscillate forever in this model. There are no physical realizations of (25.2) in nature because there is always some amount of drag. To keep a spring moving we need to add a motor. Before we see how to describe drag and forcing functions we will study the simple oscillator. 211
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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 −
Page 167 and 168: 159 and its roots are given by The
Page 169 and 170: 161 The motivation for equation 18.
Page 171 and 172: Lesson 19 Method of Undetermined Co
Page 173 and 174: 165 3. If f(t) = e rt and r is a ro
Page 175 and 176: 167 Example 19.4. Solve ⎫ y ′
Page 177 and 178: 169 Adding the two equations gives
Page 179 and 180: Lesson 20 The Wronskian We have see
Page 181 and 182: 173 Definition 20.1. The Wronskian
Page 183 and 184: 175 Example 20.3. Show that y = sin
Page 185 and 186: 177 and therefore the system of equ
Page 187 and 188: Lesson 21 Reduction of Order The me
Page 189 and 190: 181 The method of reduction of orde
Page 191 and 192: 183 Plugging these into Bessel’s
Page 193 and 194: 185 Example 21.5. Find a second sol
Page 195 and 196: Lesson 22 Non-homogeneous Equations
Page 197 and 198: 189 where r 1 and r 2 are the roots
Page 199 and 200: 191 This is a first order linear eq
Page 201 and 202: 193 Theorem 22.5. Properties of the
Page 203 and 204: 195 where (∫ ν(t) = exp ) −r 2
Page 205 and 206: 197 The characteristic equation is
Page 207 and 208: Lesson 23 Method of Annihilators In
Page 209 and 210: 201 Theorem 23.5. (D 2 − 2aD + (a
Page 211 and 212: 203 The method of annihilators is r
Page 213 and 214: Lesson 24 Variation of Parameters T
Page 215 and 216: 207 Substituting into equation (24.
Page 217: 209 Example 24.3. Solve the initial
Page 221 and 222: 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*
Page 227 and 228: 219 In the case just proven, there
Page 229 and 230: 221 Theorem 26.5. Under the same co
Page 231 and 232: 223 Since K n /(1 − K) → 0 as n
Page 233 and 234: 225 for any φ ∈ V. Let g, h be f
Page 235 and 236: Lesson 27 Higher Order Linear Equat
Page 237 and 238: 229 L n+1 (e rt y) = e rt a n (D +
Page 239 and 240: 231 Example 27.2. Find the general
Page 241 and 242: 233 Differentiating, u ′ (t) = d
Page 243 and 244: 235 Integrating, − 2K |t − t 0
Page 245 and 246: 237 a closed form expression for a
Page 249 and 250: 241 The characteristic equation is
Page 251 and 252: 243 The Wronskian In this section w
Page 253 and 254: 245 Certainly every φ(t) given by
Page 255 and 256: 247 the differential equation. Over
Page 257 and 258: 249 By the lemma, to obtain the der
Page 261 and 262: 253 So that f(t) = a n (t)y[ (n) +
Page 263 and 264: 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
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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
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289 Thus a Frobenius solution is y
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291 Example 30.6. Find the form of
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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
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301 Example 30.8. In example 30.4 w
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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
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345 can be written as as illustrate
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347 Translations in the Laplace Var
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349 Summary of Translation Formulas
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351 The inverse transform is [ ] f(
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353 Example 32.18. Find the Laplace
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355 Similarly, we can express a uni
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357 Figure 32.7: Solution of exampl
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Lesson 33 Numerical Methods Euler
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361 Figure 33.1: Illustration of Eu
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363 y 4 = y 3 + hf(t 3 , y 3 ) (33.
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365 Figure 33.3: Illustration of th
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367 result with a smaller step size
Page 377 and 378:
369 Expanding the final term in a T
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371 k 2 = y 0 + h 2 f(t 0, k 1 ) (3
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Lesson 34 Critical Points of Autono
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375 Since both f and g are differen
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377 Using the cos π/4 = √ 2/2 an
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379 values, of the matrix. We find
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381 Distinct Real Nonzero Eigenvalu
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383 eigendirection {λ 1 , v 1 }dom
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385 Figure 34.5: Phase portraits ty
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387 Complex Conjugate Pair with non
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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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