2130_Chapter7_Notes

# 11 a 3iv 0 11 11 0 v1 v2 0 v1 v2 0 either

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Unformatted text preview: 0 −1 =⇒ x = (vt + w)e2t = 1 0 t+ −1 −1 e2t = t e2t . −t − 1 7/9 4 Exa (solution) 1 −1 x 1 3 x′ = Eigenvalue: λ = 2 (repeated); eigenvectors: v = c Solutions of x′ = Ax : x(1) = 1 −1 1 −1 e2t and x(2) = (vt + w)e2t = General solution: , c = 0. 1 −1 0 −1 t+ x = c1 x(1) + c2 x(2) = c1 1 −1 = c1 e2t = e2t + c2 1 −1 + c2 t −t−1 t −t−1 t −t−1 e2t . e2t e2t . 8/9 Summary of Results If λ is a repeated eigenvalue, there are two cases to consider: 1. Two independent eigenvectors v(1) and v(2) : x(1) = v(1) eλt and x(2) = v(2) eλt . This is just business as usual. 2. Only one independent eigenvector v : x(1) = veλt and x(2) = (cvt + w)eλt , c = 0 where w is a generalized eigenvector: (A − λI)w = cv, c = 0. 9/9 5 APMA 2130 Section 7.9: Undetermined Coefﬁcients Post-Lecture Handout Dr. Bernard Fulgham University of Virginia July 29, 2013 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....
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