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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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This document was uploaded on 03/30/2014 for the course MATH 377 at Central Washington University.
 Spring '14
 Montgomery
 Math, Differential Equations, Equations

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