Lecture Notes in Differential Equations - Bruce E. Shapiro
Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro
254 LESSON 27. HIGHER ORDER EQUATIONS Example 27.15. Solve y ′′′ + y ′ = tan t using variation of parameters. The characteristic equation is 0 = r 3 + r = r(r + i)(r − i); hence a fundamental set of solutions are y 1 = 1, y 2 = cos t, and y 3 = sin t. From either Abel’s formula or a direct calculation, W (t) = 1, since y P = y 1 ∫ t W = ∣ W 1 (s)f(s) W (s)a 3 (s) ds+y 2 1 cos t sin t 0 − sin t cos t 0 − cos t − sin t ∫ t W 2 (s)f(s) W (s)a 3 (s) ds+y 3 ∣ = 1 (27.207) ∫ W 3 (s)f(s) ds (27.208) W (s)a 3 (s) where a 3 (s) = 1, f(s) = tan s, and 0 cos t sin t W 1 = 0 − sin t cos t ∣ 1 − cos t − sin t ∣ = 1 (27.209) 1 0 sin t W 2 = 0 0 cos t = − cos t (27.210) ∣ 0 1 − sin t ∣ 1 cos t 0 W 3 = 0 − sin t 0 = − sin t (27.211) ∣ 0 − cos t 1 ∣ Therefore ∫ y P = t ∫ ∫ tan sds − cos t cos s tan sds − sin t sin s tan sds (27.212) t t Integrating the first term and substituting for the tangent in the third term, ∫ ∫ sin 2 s y P = − ln |cos t| − cos t sin sds − sin t ds (27.213) cos s t The second integral can not be integrated immediately, and the final integral can be solved by substituting sin 2 s = 1 − cos 2 s ∫ y P =− ln |cos t| + cos 2 1 − cos 2 s t − sin t ds (27.214) cos s Since the first term (the constant) is a solution of the homogeneous equation, we can drop it from the particular solution, giving and a general solution of t y P = ln |cos t| − sin t ln |sec t + tan t| (27.215) y = ln |cos t| − sin t ln |sec t + tan t| + C 1 + C 2 cos t + C 3 sin t. (27.216) t t
Lesson 28 Series Solutions In many cases all we can say about the solution of a(t)y ′′ + b(t)y ′ + c(t)y = f(t) y(t 0 ) = y 0 y ′ (t 0 ) = y 1 ⎫ ⎪⎬ ⎪ ⎭ (28.1) is a statement about whether or not a solution exists. So far, however, we do not have any generally applicable technique to actually find the solution. If a solution does exist, we know it must be twice differentiable (n times differentiable for an nth order equation). If the solution is not just twice, but infinitely, differentiable, we call it an analytic function. According to Taylor’s theorem, any function that is analytic at a point t = t 0 can be expanded in a power series where y(t) = ∞∑ a k (t − t 0 ) k (28.2) k=0 a k = y(k) (t 0 ) (28.3) k! Because of Taylor’s theorem, the term analytic is sometimes used to mean that a function can be expanded in a power series at a point (analytic ⇐⇒ infinitely differentiable ⇐⇒ power series exists). The method of series for solving differential equations solutions looks for analytic solutions to (28.1) by substituting equation (28.2) into the differential equation and solving for the coefficients a 0 , a 1 , ... 255
- 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
- Page 255 and 256: 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: 253 So that f(t) = a n (t)y[ (n) +
- 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
- 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 and 286: 277 ∑ ∞ ∞∑ ∞∑ 0 = t 2 a
- Page 287 and 288: 279 Case 2: Two equal real roots. S
- 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
- 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
- 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
Lesson 28<br />
Series Solutions<br />
In many cases all we can say about the solution of<br />
a(t)y ′′ + b(t)y ′ + c(t)y = f(t)<br />
y(t 0 ) = y 0<br />
y ′ (t 0 ) = y 1<br />
⎫<br />
⎪⎬<br />
⎪ ⎭<br />
(28.1)<br />
is a statement about whether or not a solution exists. So far, however, we<br />
do not have any generally applicable technique to actually f<strong>in</strong>d the solution.<br />
If a solution does exist, we know it must be twice differentiable (n times<br />
differentiable for an nth order equation).<br />
If the solution is not just twice, but <strong>in</strong>f<strong>in</strong>itely, differentiable, we call it an<br />
analytic function. Accord<strong>in</strong>g to Taylor’s theorem, any function that is<br />
analytic at a po<strong>in</strong>t t = t 0 can be expanded <strong>in</strong> a power series<br />
where<br />
y(t) =<br />
∞∑<br />
a k (t − t 0 ) k (28.2)<br />
k=0<br />
a k = y(k) (t 0 )<br />
(28.3)<br />
k!<br />
Because of Taylor’s theorem, the term analytic is sometimes used to mean<br />
that a function can be expanded <strong>in</strong> a power series at a po<strong>in</strong>t (analytic ⇐⇒<br />
<strong>in</strong>f<strong>in</strong>itely differentiable ⇐⇒ power series exists).<br />
The method of series for solv<strong>in</strong>g differential equations solutions looks<br />
for analytic solutions to (28.1) by substitut<strong>in</strong>g equation (28.2) <strong>in</strong>to the<br />
differential equation and solv<strong>in</strong>g for the coefficients a 0 , a 1 , ...<br />
255