18.06 Problem Set 7
Due Wednesday, 16 April 2008 at 4 pm in 2106.
Problem 1:
Do problem 1 in section 6.3 (pg. 315) in the book.
Solution
(10 points)
We solve a linear system of differential equations by taking
u
(
t
) =
c
1
e
λ
1
t
x
1
+
c
2
e
λ
2
t
x
2
where
λ
1
, λ
2
are the eigenvalues,
x
1
, x
2
are the eigenvectors, and
c
1
, c
2
are constants
that satisfy
c
1
x
1
+
c
2
x
2
=
u
(0).
To write down the matrix exponential explicitly, we must find the eigenvalues
and eigenvectors of
A
. Since this
A
is diagonal, its eigenvalues are just the diagonal
entries, i.e.
λ
1
= 4 and
λ
2
= 1. The eigenvectors are
x
1
= (1
,
0) and
x
2
= (1
,

1).
Finally, if
u
(0) = (5
,

2) we must find how to write
u
(0) as a linear combination of
x
1
and
x
2
. We do this by solving the equation
1
1
0

1
c
1
c
2
=
5

2
We get
c
2
= 2 and
c
1
= 3. So, the final equation is
u
(
t
) = 3
e
4
t
1
0
+ 2
e
t
1

1
Problem 2:
Do problem 3 in section 6.3 (pg. 315).
Solution
(10 points)
To linearize this system, we identify
u
with the vector [
y, y
]
T
, so that we have
two equations
dy/dt
=
y
and
dy /dt
=
y
= 4
y
+ 5
y
. That is, we can “decouple”
the differential equation by adding
y
as a new variable, to obtain the system
dy/dt
dy /dt
=
0
1
4
5
y
y
We call the coefficient matrix
A
as usual. The eigenvalues of
A
satisfy the equation
λ
2

5
λ

4 = 0, so the eigenvalues are
λ
1
=
1
2
(5 +
√
41) and
λ
2
=
1
2
(5

√
41).
1
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Another way to find the eigenvalues is to substitute
y
=
e
λt
into the differential
equation. We obtain
λ
2
e
λt
= 5
λe
λt
+ 4
e
λt
Dividing by
e
λt
, we find the same relationship
λ
2

5
λ

4 = 0.
Problem 3:
a) Do problem 17 in section 6.3 (pg. 317).
b) Do problem 24 in section 6.3 (pg. 318).
Solution
(5+5 points)
a) The infinite series for
e
Bt
is
e
Bt
=
I
+
tB
+
1
2
t
2
B
2
+
. . .
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 Spring '08
 CHANILLO
 Differential Equations, Linear Algebra, Equations, Orthogonal matrix, eλt

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