MA366 Final
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the problem.
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(1) The following vectors
X
1
and
Y
1
are eigenvectors for a certain
3
×
3 matrix
A
corresponding to the eigenvalues 2

i
and

4
respectively. Find the general solution to the system
X
0
=
AX
in real form
. No complex numbers allowed!
5 pts.
X
1
=
i
+ 1
i

2
i
,
Y
1
=
0
3
1
.
Solution:
Since
e
(2

i
)
t
= e
t
cos (2
t
) +
i
e
t
sin (2
t
)
we see that
(1 +
i
)
e
(2

i
)
t
=

e
t
cos (2
t
)

e
t
sin (2
t
) +
i
(
e
t
cos (2
t
)

e
t
sin (2
t
)
)
ie
(2

i
)
t
=

e
t
sin (2
t
) +
i
e
t
cos (2
t
)

2
e
(2

i
)
t
=

2 e
t
cos (2
t
)

2
i
e
t
sin (2
t
)
We form matrices from the real and imaginary parts:
x
1
(
t
) =

e
t
cos (2
t
)

e
t
sin (2
t
)

e
t
sin (2
t
)

2 e
t
cos (2
t
)
x
2
(
t
) =
e
t
cos (2
t
)

e
t
sin (2
t
)
e
t
cos (2
t
)

2 e
t
sin (2
t
)
The real eigenvector produces the solution
x
3
(
t
) =
e
4
t
0
3
1
.
The general solution is
X
(
t
) =
C
1
x
1
(
t
) +
c
2
x
2
(
t
) +
c
3
x
3
(
t
)
.
(2) Given that
X
1
,
X
2
and
X
3
are eigenvectors for the following
matrix, ﬁnd the general solution to
X
0
=
AX
.
Hint:
To ﬁnd
the eigenvalue, compute
AX
i
.
5 pts.
A
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 Fall '09
 EdrayGoins
 Differential Equations, Equations, Integrals, Complex differential equation, The Singularity Is Near

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