Lecture Notes in Differential Equations - Bruce E. Shapiro

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188 LESSON 22. NON-HOMOG. EQS.W/ CONST. COEFF. has the form y(t) = Ay H1 (t) + By H2 (t) + y P (t) (22.4) where y H1 and y H2 are linearly independent solutions of ay ′′ + by ′ + cy = 0 (22.5) and y P (t) is a solution of (22.3) that is linearly independent of y H1 and y H2 . Equation (22.4) is called the general solution of the ODE (22.3). The two linearly independent solutions of the homogeneous equation are called a fundamental set of solutions. General Concept: To solve the general equation with constant coefficients, we need to find two linearly independent solutions to the homogeneous equation as well as a particular solution to the non-homogeneous solutions. Theorem 22.3. Subtraction Principle. If y P 1 (t) and y P 2 (t) are two different particular solutions of then is a solution of the homogeneous equationLy = 0. Proof. Since Then Hence y H given by (22.7) satisfies Ly H = 0. Ly = f(t) (22.6) y H (t) = y P 1 (t) − y P 2 (t) (22.7) Ly P 1 = f(t) (22.8) Ly P 2 = f(t) (22.9) L(y P 1 − y P 2 ) = Ly P 1 − Ly P 2 (22.10) Theorem 22.4. The linear differential operator can be factored as = f(t) − f(t) (22.11) = 0 (22.12) Ly = aD 2 y + bDy + cy (22.13) Ly = (aD 2 + bD + c)y = a(D − r 1 )(D − r 2 )y (22.14)
189 where r 1 and r 2 are the roots of the characteristic polynomial ar 2 + br + c = 0 (22.15) Proof. Using the quadratic equation, the roots of the characteristic polynomial satisfy r 1 + r 2 = −b + √ b 2 − 4ac + −b − √ b 2 − 4ac = − b (22.16) ( 2a 2a a −b + √ ) ( b r 1 r 2 = 2 − 4ac −b − √ ) b 2 − 4ac = b2 − (b 2 − 4ac) 2a 2a 4a 2 = c a (22.17) hence b = −a(r 1 + r 2 ) (22.18) c = ar 1 r 2 (22.19) Thus Ly = ay ′′ + by ′ + cy = ay ′′ − a(r 1 + r 2 )y ′ + ar 1 r 2 y = a[y ′′ − (r 1 + r 2 )y ′ + r 1 r 2 y] = a(D − r 1 )(y ′ − r 2 y) = a(D − r 1 )(D − r 2 )y (22.20) Hence L = a(D − r 1 )(D − r 2 ) which is the desired factorization. Theorem (22.4) provides us with the following simple algorithm for solving any second order linear differential equation or initial value problem with constant coefficients.
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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
Page 7 and 8:
Preface These lecture notes on diff
Page 9 and 10:
Lesson 1 Basic Concepts A different
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3 Definition 1.2 (Solution, ODE). A
Page 13 and 14:
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
Page 39 and 40:
31 Since p(t) = 1 (the coefficient
Page 41 and 42:
33 Multiplying equation 4.68 by µ
Page 43 and 44:
35 Substituting y = t = 0 in this i
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37 ∫ t t 0 Evaluating the integra
Page 47 and 48:
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
Page 81 and 82:
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).
Page 93 and 94:
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
Page 121 and 122:
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 (
Page 135 and 136:
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,
Page 143 and 144:
Lesson 16 Linear Equations With Con
Page 145 and 146: 137 Hence both r = 1 and r = 3. Thi
Page 147 and 148: 139 The second order linear initial
Page 149 and 150: 141 The general solution to is give
Page 151 and 152: Lesson 17 Some Special Substitution
Page 153 and 154: 145 Therefore since z = y ′ , Int
Page 155 and 156: 147 Example 17.5. Solve yy ′′ +
Page 157 and 158: 149 where I is the identity matrix.
Page 159 and 160: 151 can be rewritten by solving a =
Page 161 and 162: Lesson 18 Complex Roots We know for
Page 163 and 164: 155 Theorem 18.2. Euler’s Formula
Page 165 and 166: 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: Lesson 22 Non-homogeneous Equations
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 and 218: 209 Example 24.3. Solve the initial
Page 219 and 220: Lesson 25 Harmonic Oscillations If
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 247 and 248:
239 Example 27.6. Find the general
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
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247 the differential equation. Over
Page 257 and 258:
249 By the lemma, to obtain the der
Page 259 and 260:
251 Example 27.14. Find the general
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
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
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
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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 + ∆ =
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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.
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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 = (
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 and 348:
L [ t x−1] [ ] 1 d = L x dt tx =
Page 349 and 350:
341 Equating numerators and expandi
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
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Lesson 33 Numerical Methods Euler
Page 369 and 370:
361 Figure 33.1: Illustration of Eu
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363 y 4 = y 3 + hf(t 3 , y 3 ) (33.
Page 373 and 374:
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
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
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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
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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 and 404:
Appendix A Table of Integrals Basic
Page 405 and 406:
397 ∫ x √ x − adx = 2 3 a(x
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
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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
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415 for y once z is known. Method o
Page 425 and 426:
Bibliography [1] Bear, H.S. Differe
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BIBLIOGRAPHY 419
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BIBLIOGRAPHY 421
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