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Homework 3 Solution

# 5 m 11 0 11t 2 1 and 0 0 1 1t xt e 2 cos 10 11t 2 3 t

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Unformatted text preview: − sin 10 11t 2 √ 11t 2 √ −3 + cos 10 √ 11 0 g) Plugging in t = 0 sets all the et/2 = 1, all of the cos √ all of the sin 11t 2 = c1 c1 = + III.5 M = 11 0 √ 11t 2 = 1 and = 0. 0 1 1t x(t) = e /2 cos 10 √ 11t 2 √ 3 t √ e /2 sin 10 11 −3 + c2 10 1 10 11t 2 √ c2 = −2 9 0 −2 —5— 11 0 3 √ 10 11 −3 − sin 10 11t 2 √ √ −3 + cos 10 11t 2 √ √ 11t 2 11 0 √ 11 0 Winter 2014 Math 377 Homework III Solution a) λ1 = −2 1 0 v1 = b) c) degenerate sink node d) see b) e) We need to ﬁnd a generalized eigenvector. I will use my method which ﬁxes the constant c1 at the beginning of the process. (M − λI ) w1 w2 9w2 0 = c1 c1 0 = w2 = w= 1 0 c1 9 c2 c1 9 x(t) = (e−2t + te−2t ) c1 0 x(t) = (e−2t + te−2t ) 9c1 c + e−2t 2 0 c1 —6— + e−2t c2 c1 9 Winter 2014 Math 377 Homework III Solution Note that I replaced c1 in the ﬁrst equation by 9c1 in the second equation to eliminate the fraction. f) see e) g) 0 1 = 9c1 + c1 = 1 x(t) = (e−2t + te−2t ) —7— c2 c1 c2 = −9 −9 9 + e−2t 1 0...
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