Lecture Notes in Differential Equations - Bruce E. Shapiro
Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro
278 LESSON 29. REGULAR SINGULARITIES for some unknown number (integer, real, or possibly complex) r. To see what values of r might work, we substitute into the original ODE (29.4): t 2 r(r − 1)t r−2 + αtrt r−1 + βt r = 0 (29.16) =⇒ t r [r(r − 1) + αr + β] = 0 (29.17) Since this must hold for all values of t, the second factor must be identically equal to zero. Thus we obtain r(r − 1) + αr + β = 0 (29.18) without the restriction that r ∈ Z. We call this the indicial equation (even though there are no indices) due to its relationship (and similarity) to the indicial equation that we will have to solve in Frobenius’ method in chapter 30. We will now consider each of the three possible cases for the roots r 1,2 = 1 − α ± √ (1 − α) 2 − 4β 2 (29.19) Case 1: Two real roots. If r 1 , r 2 ∈ R and r 1 ≠ r 2 , then we have found two linearly independent solutions and the general solution is Example 29.4. Find the general solution of In standard from this becomes The indicial equation is y = C 1 t r1 + C 2 t r2 (29.20) 2t 2 y ′′ + 3ty ′ − y = 0 (29.21) t 2 y ′′ + 3 2 ty′ − 1 2 y = 0 (29.22) 0 = r(r − 1) − 3 2 r + 1 2 = 2 2 r2 + 1 2 r + 1 2 (29.23) 0 = 2r 2 + r + 1 = (2r − 1)(r + 1) =⇒ r = 1 , −1 (29.24) 2 Hence y = C 1 √ t + C 2 t (29.25)
279 Case 2: Two equal real roots. Suppose r 1 = r 2 = r. Then r = 1 − α 2 (29.26) because the square root must be zero (ie., (1 − α) 2 − 4β = 0) to give a repeated root. We have one solution is given by y 1 = t r = t (1−α)/2 . To get a second solution we can use reduction of order. From Abel’s formula the Wronskian is ( ∫ ) αtdt W = C exp − t 2 = C exp ( −α ∫ ) dt = Ce −α ln t = C t t α (29.27) By the definition of the Wronskian a second formula is given by W = y 1 y 2 ′ − y 2 y 1 ′ (29.28) ( = t (1−α)/2) ( ) 1 − α ( y 2 ′ − y 2 ′ t (−1−α)/2) (29.29) 2 Equating the two expressions for W , ( t (1−α)/2) ( ) 1 − α ( y 2 ′ − y 2 ′ t (−1−α)/2) = C 2 t α (29.30) ( ) 1 − α t y 2 ′ (−1−α)/2 C − 2 t (1−α)/2 y′ 2 = (29.31) t α (t (1−α)/2 ) ( ) α − 1 y 2 ′ + y ′ C 2 = (29.32) 2t t (1+α)/2 This is a first order linear equation; an integrating factor is (∫ ) ( ) α − 1 α − 1 µ(t) = exp dt = exp ln t = t (α−1)/2 (29.33) 2t 2 If we denote q(t) = Ct −(1+α)/2 then the general solution of (29.32) is y = 1 [∫ ] µ(t)q(t)dt + C 1 (29.34) µt [∫ ] = t (1−α)/2 t (α−1)/2 Ct −(1+α)/2 dt + C 1 (29.35) ∫ dt = Cy 1 t + C 1y 1 (29.36) = Cy 1 ln |t| + C 1 y 1 (29.37)
- 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 +
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- Page 241 and 242: 233 Differentiating, u ′ (t) = d
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- 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
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- Page 255 and 256: 247 the differential equation. Over
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- 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 =
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- Page 273 and 274: 265 By the triangle inequality, |(k
- Page 275 and 276: 267 Table 28.1: Table of Special Fu
- Page 277 and 278: 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
- Page 285: 277 ∑ ∞ ∞∑ ∞∑ 0 = t 2 a
- Page 289 and 290: 281 Example 29.6. Solve t 2 y ′
- Page 291 and 292: Lesson 30 The Method of Frobenius I
- Page 293 and 294: 285 This is a homogeneous linear eq
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- Page 317 and 318: 309 We will verify (31.54) by induc
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- Page 321 and 322: 313 y = 1 [( ) ( ) ] ( ) 4 1 e 2t 1
- Page 323 and 324: 315 Theorem 31.8. (Abel’s Formula
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- Page 329 and 330: 321 Corollary 31.12. The generalize
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278 LESSON 29. REGULAR SINGULARITIES<br />
for some unknown number (<strong>in</strong>teger, real, or possibly complex) r. To see<br />
what values of r might work, we substitute <strong>in</strong>to the orig<strong>in</strong>al ODE (29.4):<br />
t 2 r(r − 1)t r−2 + αtrt r−1 + βt r = 0 (29.16)<br />
=⇒ t r [r(r − 1) + αr + β] = 0 (29.17)<br />
S<strong>in</strong>ce this must hold for all values of t, the second factor must be identically<br />
equal to zero. Thus we obta<strong>in</strong><br />
r(r − 1) + αr + β = 0 (29.18)<br />
without the restriction that r ∈ Z. We call this the <strong>in</strong>dicial equation<br />
(even though there are no <strong>in</strong>dices) due to its relationship (and similarity)<br />
to the <strong>in</strong>dicial equation that we will have to solve <strong>in</strong> Frobenius’ method <strong>in</strong><br />
chapter 30.<br />
We will now consider each of the three possible cases for the roots<br />
r 1,2 = 1 − α ± √ (1 − α) 2 − 4β<br />
2<br />
(29.19)<br />
Case 1: Two real roots. If r 1 , r 2 ∈ R and r 1 ≠ r 2 , then we have found two<br />
l<strong>in</strong>early <strong>in</strong>dependent solutions and the general solution is<br />
Example 29.4. F<strong>in</strong>d the general solution of<br />
In standard from this becomes<br />
The <strong>in</strong>dicial equation is<br />
y = C 1 t r1 + C 2 t r2 (29.20)<br />
2t 2 y ′′ + 3ty ′ − y = 0 (29.21)<br />
t 2 y ′′ + 3 2 ty′ − 1 2 y = 0 (29.22)<br />
0 = r(r − 1) − 3 2 r + 1 2 = 2 2 r2 + 1 2 r + 1 2<br />
(29.23)<br />
0 = 2r 2 + r + 1 = (2r − 1)(r + 1) =⇒ r = 1 , −1 (29.24)<br />
2<br />
Hence<br />
y = C 1<br />
√<br />
t +<br />
C 2<br />
t<br />
(29.25)