Lecture Notes in Differential Equations - Bruce E. Shapiro
Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro
324 LESSON 31. LINEAR SYSTEMS for some set of unknown functions u 1 (t), ..., u n (t), and u = (u 1 (t) u 2 (t) · · · u n (t)) T . Differentiating (31.166) gives y ′ P = (Wu) ′ = W ′ u + Wu ′ (31.167) Substitution into the differential equation (31.164) gives Since W ′ = AW, Subtracting the commong term AWu gives W ′ u + Wu ′ = AWu + g (31.168) AWu + Wu ′ = AWu + g (31.169) Multiplying both sides of the equation by W −1 gives Wu ′ = g (31.170) du dt = u′ = W −1 g (31.171) Since W = e At , W −1 = e −At , and therefore ∫ ∫ du u(t) = dt dt = e −At g(t)dt (31.172) Substition of (31.172) into (31.166) yields the desired result, using the fact that W = e At . Example 31.5. Find a particular solution to the system } x ′ = x + 3y + 5 y ′ = 4x − 3y + 6t The problem can be restated in the form y ′ = Ay + g where ( ) ( ) 1 3 5 A = , and g(t) = 4 −3 6t (31.173) (31.174) We first solve the homogeneous system. The eigenvalues of A satisfy 0 = (1 − λ)(−3 − λ) − 12 (31.175) = λ 2 + 2λ − 15 (31.176) = (λ − 3)(λ + 5) (31.177)
325 so that λ = 3, −5. The eigenvectors satisfy ( ) ( ) ( ) ( 1 3 a a 3 = 3 ⇒ 2a = 3b ⇒ v 4 −3 b b 3 = 2 ) (31.178) and ( 1 3 4 −3 ) ( c d ) ( c = −5 d ) ( 1 ⇒ d = −2c ⇒ v −5 = −2 ) (31.179) Hence y H = c 1 e 3t ( 3 2 ) + c 2 e −5t ( 1 −2 ) (31.180) where c 1 , c 2 are arbitrary constants. Define a matrix U whose columns are the eigenvectors of A. Then Then U = ( 3 1 2 −2 ) where D = diag(3, −5). Hence and U −1 = ( 1/4 1/8 1/4 −3/8 ) (31.181) e At = Ue Dt U −1 (31.182) ( ) ( ) ( ) e At 3 1 e 3t 0 1/4 1/8 = 2 −2 0 e −5t (31.183) 1/4 −3/8 ( ) ( ) 3 1 e = 3t /4 e 3t /8 2 −2 e −5t /4 −3e −5t (31.184) /8 ⎛ 3 ⎜ = ⎝4 e3t + 1 3 4 e−5t 8 e3t − 3 ⎞ 8 e−5t 1 2 e3t − 1 1 2 e−5t 4 e3t + 3 ⎟ ⎠ (31.185) 4 e−5t Since we have e −At = (Ue Dt U −1 ) −1 = Ue −Dt U −1 (31.186) ( ) ( ) ( ) e −At 3 1 e −3t 0 1/4 1/8 = 2 −2 0 e 5t (31.187) 1/4 −3/8 ⎛ 3 ⎜ = ⎝4 e−3t + 1 3 4 e5t 8 e−3t − 3 ⎞ 8 e5t 1 2 e−3t − 1 1 2 e5t 4 e−3t + 3 ⎟ ⎠ (31.188) 4 e5t
- Page 281 and 282: 273 Summary of Power series method.
- Page 283 and 284: Lesson 29 Regular Singularities The
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- Page 311 and 312: Lesson 31 Linear Systems The genera
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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
- Page 325 and 326: 317 By a similar argument, the seco
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- Page 331: 323 where (A − λI)w 2 = w 1 , i.
- Page 335 and 336: 327 Non-constant Coefficients We ca
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- Page 347 and 348: L [ t x−1] [ ] 1 d = L x dt tx =
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- Page 367 and 368: Lesson 33 Numerical Methods Euler
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324 LESSON 31. LINEAR SYSTEMS<br />
for some set of unknown functions u 1 (t), ..., u n (t), and u = (u 1 (t) u 2 (t) · · · u n (t)) T .<br />
Differentiat<strong>in</strong>g (31.166) gives<br />
y ′ P = (Wu) ′ = W ′ u + Wu ′ (31.167)<br />
Substitution <strong>in</strong>to the differential equation (31.164) gives<br />
S<strong>in</strong>ce W ′ = AW,<br />
Subtract<strong>in</strong>g the commong term AWu gives<br />
W ′ u + Wu ′ = AWu + g (31.168)<br />
AWu + Wu ′ = AWu + g (31.169)<br />
Multiply<strong>in</strong>g both sides of the equation by W −1 gives<br />
Wu ′ = g (31.170)<br />
du<br />
dt = u′ = W −1 g (31.171)<br />
S<strong>in</strong>ce W = e At , W −1 = e −At , and therefore<br />
∫ ∫ du<br />
u(t) =<br />
dt dt = e −At g(t)dt (31.172)<br />
Substition of (31.172) <strong>in</strong>to (31.166) yields the desired result, us<strong>in</strong>g the fact<br />
that W = e At .<br />
Example 31.5. F<strong>in</strong>d a particular solution to the system<br />
}<br />
x ′ = x + 3y + 5<br />
y ′ = 4x − 3y + 6t<br />
The problem can be restated <strong>in</strong> the form y ′ = Ay + g where<br />
( )<br />
( )<br />
1 3<br />
5<br />
A =<br />
, and g(t) =<br />
4 −3<br />
6t<br />
(31.173)<br />
(31.174)<br />
We first solve the homogeneous system. The eigenvalues of A satisfy<br />
0 = (1 − λ)(−3 − λ) − 12 (31.175)<br />
= λ 2 + 2λ − 15 (31.176)<br />
= (λ − 3)(λ + 5) (31.177)