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Unformatted text preview: 2 = 3 , A3 I = ± 6 61010 ¶ → ± 1 1 ¶ , and ±v 2 = (1 , 1) . Writing the eigenvectors as columns, P = ±31 5 1 ¶ will have P1 AP = ±1 3 ¶ = D . Now the “method of Section 6.2” is to solve the new system of DE ± y ± = P1 AP± y, which is the “uncoupled system” y ± 1 =y 1 y ± 2 = 3 y 2 , and has solution z 1 = c 1 et , z 2 = c 2 e 3 t so the original system had solution ± y = P± z, which we write as ± y 1 y 2 ¶ = ±31 5 1 ¶± c 1 et c 2 e 3 t ¶ = c 1 et ±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 λ j t ±v j , where ±v j is an eigenvector with eigenvalue λ j . For nondiagonalizable systems, we recall that the method of the Supplement Part B is (except for the linear algebra, ﬁnding the matrix P so P1 AP = J ) described in our text, 6.3, as in Example 2....
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This note was uploaded on 11/23/2011 for the course MATH 205 taught by Professor Zhang during the Fall '08 term at Lehigh University .
 Fall '08
 zhang
 Math

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