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f08week12 - 2 = 3 , A-3 I = 6 6-10-10 1 1 , and v 2 = (-1 ,...

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Math 205, Fall 2008 B. Dodson Week 13: 1. 6.1 1st order systems of DE 2. 6.2 Diagonalizable Homog Systems 3. Supplement, Part B ————– Recall that Part A of the supplement had an extended discussion of a 2-by-2 non-diagonalizale example – A.3, then A.5 and in final form in A.9. A second problem type, with the same 2-by-2 Jordan block J λ 1 , was described in Example A.17. These were the two cases of an eigenvalue of (algebraic) multiplicity 2, m 1 = 2, with an eigenspace of dimension 1 (geometric multiplicity 1, d 1 = 1 ± = m 1 = 2). The new material from 6.1 is linear independence of vector functions (like the Wronskian); and non-coupled non-homog systems. In 6.2, we apply diagonalizability to solve systems of first order DE with diagonalizable coef matrix. Problem 7. Solve y ± 1 = 9 y 1 + 6 y 2 y ± 2 = - 10 y 1 - 7 y 2 using the method of 6.2. Solution. We first need the eigenvectors to find the matrix P . We have P ( λ ) = (9 - λ )( - 7 - λ ) + 60 = λ 2 - 2 λ - 63 + 60 = λ 2 - 2 λ - 3 = ( λ - 3)( λ + 1) , so λ 1 = - 1 , λ 2 = 3 .
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2 For λ 1 = - 1 , A + I = ± 10 6 - 10 - 6 ± 5 3 0 0 , ————– and ±v 1 = ( - 3 , 5) . For λ
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Unformatted text preview: 2 = 3 , A-3 I = 6 6-10-10 1 1 , and v 2 = (-1 , 1) . Writing the eigenvectors as columns, P = -3-1 5 1 will have P-1 AP = -1 3 = D . Now the method of Section 6.2 is to solve the new system of DE y = P-1 AP y, which is the uncoupled system y 1 =-y 1 y 2 = 3 y 2 , and has solution z 1 = c 1 e-t , z 2 = c 2 e 3 t so the original system had solution y = P z, which we write as y 1 y 2 = -3-1 5 1 c 1 e-t c 2 e 3 t = c 1 e-t -3 1 + c 2 e 3 t -1 1 . Notice that this solution is a linear combination of two LI solutions of the form e j t v j , where v j is an eigenvector with eigenvalue j . For non-diagonalizable systems, we recall that the method of the Supplement Part B is (except for the linear algebra, nding the matrix P so P-1 AP = J ) described in our text, 6.3, as in Example 2....
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f08week12 - 2 = 3 , A-3 I = 6 6-10-10 1 1 , and v 2 = (-1 ,...

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