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Quiz 4
Name:
Math 240  Calculus III
February 24, 2009
Note:
In order to receive full credit, you must show work that justiﬁes your answer.
1. (6 points) Find the general solution of the following system of diﬀerential equations:
X
0
=
±
1 3
3 1
²
X
+
±
2

3
²
e
3
t
Solution
: The matrix
±
1 3
3 1
²
has eigenvalues
λ
1
=

2 and
λ
2
= 4, with
corresponding eigenvectors
v
1
=
h
1
,
1
i
and
v
2
=
h
1
,
1
i
, respectively. The
complementary solution is then
X
c
=
c
1
±

1
1
²
e

2
t
+
c
2
±
1
1
²
e
4
t
.
We suppose a particular solution has the form
X
p
=
±
a
b
²
e
3
t
. Substituting
X
p
into
the diﬀerential equation, we ﬁnd that
a
= 1 and
b
= 0.
The general solution is thus
X
=
c
1
±

1
1
²
e

2
t
+
c
2
±
1
1
²
e
4
t
+
±
1
0
²
e
3
t
.
2. (2 points each) Find diﬀerential operators that “annihilate” the following functions:
(a)
y
(
x
) =
e
5
x
Solution
: The annihilator is
D

5 since
(
D

5)
y
=
y
0

5
y
= 5
e
5
x

5
e
5
x
= 0
.
(b)
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 Spring '07
 Cremins
 Differential Equations, Linear Algebra, Algebra, Equations

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