hw6-solutions

# 2 3t 2 problem 13 the eigenvalues of a are given by

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Unformatted text preview: − λt) y (0) + t Ay (0) = e−t (1 + t) = e−t 3 −9 +t −1 −5 3 − 6t . −1 − 6t Problem 9: Let us consider the diﬀerential equation y ′ (t) = 2 3 −2 with initial condition y (0) = 3 2 −1 y (t) 3 . −2 We ﬁrst compute the eigenvalues of A: 0 = det 1 9 −3 2 = (λ − 2)(λ + 1) + = λ2 − λ + . λ+1 4 4 λ−2 3 2 1 Therefore, A has a single eigenvalue of multiplicity two (λ = 2 ). Applying the solution formula discussed in lecture gives y (t) = eλt (1 − λt) y (0) + t Ay (0) t = e− 2 t = e− 2 t 3 3 (1 − ) +t −5 2 −2 2 3 + 3t 2 . −2 − 3t 2 Problem 13: The eigenvalues of A are given by λ= a+d ± 2 (a + d)2 − (ad − bc). 4 (The calculation is the same as for Problem 23 in Section 3.4.) The statement that all solutions of the ODE approach the origin as t → ∞ is equivalent to the statement that both λ1 and λ2 have negative real part. It is useful to break the discussion into two cases: 2 Case 1: Let us ﬁrst consider the case that (a+d) ≥ ad − bc. In this 4 case, the eigenvalues of A are real numbers. I...
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## This note was uploaded on 02/09/2014 for the course MATH 53 taught by Professor Staff during the Winter '08 term at Stanford.

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