Unformatted text preview: ————————— CHAPTER 7. —— Therefore the general solution is . Imposing the initial conditions, we arrive at the equations
,
with and . Therefore the solution of the IVP is . Since , all solutions converge to the origin. 27 . Suppose that a
b
the vector
, respectively, a 0 . Since a and b are the real and imaginary parts of
and b
. Hence
0, which leads to
0
Now since and are linearly independent, we must have It follows that . . Recall that
u
v
Consider the equation u a
a
v b
b . 0 , for some . We can then write ________________________________________________________________________
page 400 —————————————————————————— CHAPTER 7. —— a b a b 0. Rearranging the terms, and dividing by the exponential,
a
From Part b 0. , since a and b are linearly independent, it follows that Without loss of generality, assume that the trigonometric factors are nonzero. Otherwise
proceed again from Equation
, above. We then conclude that
and , which leads to
. Thus u
and v
are linearly independent for some ,
and...
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 Spring '08
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 Linear Algebra, eigenvector

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