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Unformatted text preview: igenvalues, wrote them as a + bi and pulled out
common real factors. You are not allowed to scale the eigenvalues
so you are stuck with the 1⁄2. For the eigenvectors, I wrote them
as a + bi and eliminated fractions by scaling the vectors. You are
allowed to scale eigenvectors.
b) Note that Re(λ1 ) = 1 > 0 so this is a source spiral (or spiral source).
2
To determine the direction of the spin, notice that when (x, y ) =
(1, 0) we have y (t) = 5 so we are moving in an upward direction at
(1, 0). That means that the spiral must turn counterclockwise. c) spiral source (or source spiral)
d) see b)
e)
√
√
√
√
−3 + 11i
−3 − 11i
1/2(1+ 11i)t
1/2(1− 11i)t
x(t) = c1 e
+ c2 e
10
10
—4— Winter 2014 Math 377 Homework III Solution f) Now things get messy. We just work on the ﬁrst solution (the c1
part) since all the information is buried in either of the solutions.
√
√
√
11t
11t
−3
11
t/2
+ i sin
+i
x1 (t) = e
cos
10
0
2
2
√
√
√
11t
11t
−3
11
t/2
cos
− sin
Re(x1 (t)) = e
10
0
2
2
√
√
√
11t
11t
−3
11
t/2
sin
Im(x1 (t)) = e
+ cos
10
0
2
2
√
t/2 x(t) = c1 e cos
t/2 + c2 e 11t
2
√ sin −3...
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 Spring '14
 Montgomery
 Math, Differential Equations, Equations

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