System of First Order ODEs - 2 | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Example 2.4. Twoconjugatecomplexeigenvaluescase Find
the general solution to x
0
(t)=
·
2 ¡3
1 2
¸
x(t)
Solution Using Mathcad , functions eigenvals() and eigen-
vecs() we ?nd two conjugate complex eigenvalues, ¸
1
= 2+i
p
3 and
¸
2
=2¡i
p
3withassociatedeigenvectorv
1
=
· p
3
¡i
¸
withrespectto
¸
1
. Compare this with the Theorem 2.1, we have a = 2;b =
p
3;v
11
=
p
3;v
21
=0;v
12
=0; and v
22
=¡1:
So the fundamental matrix is
F(t)=e
2t
· p
3cos(bt) ¡sin(bt)
p
3sin(bt) ¡cos(bt)
¸
and the general solution is, c=
·
c
1
c
2
¸
,
x
c
(t) = F(t)c=e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
c
1
c
2
¸
= e
2t
· p
3c
1
cos(
p
3t)¡c
2
sin(
p
3t)
p
3c
1
sin(
p
3t)¡c
2
cos(
p
3t)
¸
Suppose we want to ?nd a solution such that x(0)=
·
1
2
¸
, then
x(t) = F(t)F(0)
-1
x(0)
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸· p
3 0
0 ¡1
¸
-1
·
1
2
¸
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
1
p
3
¡2
¸
= e
2t
·
cos(
p
3t)++2sin(
p
3t)
¡sin(
p
3t)+2cos(
p
3t)
¸
a
2.2. A is a 3£3 matrix. Suppose
A =
2
4
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
3
5
Then the characteristic polynomial p(¸) ofA given by
p(¸)=jA¡¸Ij;
Page 2


Example 2.4. Twoconjugatecomplexeigenvaluescase Find
the general solution to x
0
(t)=
·
2 ¡3
1 2
¸
x(t)
Solution Using Mathcad , functions eigenvals() and eigen-
vecs() we ?nd two conjugate complex eigenvalues, ¸
1
= 2+i
p
3 and
¸
2
=2¡i
p
3withassociatedeigenvectorv
1
=
· p
3
¡i
¸
withrespectto
¸
1
. Compare this with the Theorem 2.1, we have a = 2;b =
p
3;v
11
=
p
3;v
21
=0;v
12
=0; and v
22
=¡1:
So the fundamental matrix is
F(t)=e
2t
· p
3cos(bt) ¡sin(bt)
p
3sin(bt) ¡cos(bt)
¸
and the general solution is, c=
·
c
1
c
2
¸
,
x
c
(t) = F(t)c=e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
c
1
c
2
¸
= e
2t
· p
3c
1
cos(
p
3t)¡c
2
sin(
p
3t)
p
3c
1
sin(
p
3t)¡c
2
cos(
p
3t)
¸
Suppose we want to ?nd a solution such that x(0)=
·
1
2
¸
, then
x(t) = F(t)F(0)
-1
x(0)
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸· p
3 0
0 ¡1
¸
-1
·
1
2
¸
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
1
p
3
¡2
¸
= e
2t
·
cos(
p
3t)++2sin(
p
3t)
¡sin(
p
3t)+2cos(
p
3t)
¸
a
2.2. A is a 3£3 matrix. Suppose
A =
2
4
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
3
5
Then the characteristic polynomial p(¸) ofA given by
p(¸)=jA¡¸Ij;
is a cubic polynomial of ¸: From Algebra, we know that p(¸) = 0
has either 3 distinct real solutions, or 2 distinct solutions and one is a
doublesolution,oronerealsolutionand2conjugatecomplexsolutions,
or a triple solution. The following theorem summarize the solution to
the homogeneous system,
Theorem 2.2. Let p(¸) be the characteristic polynomial of A; for
x
0
(t)=Ax(t);
Case 1: p(¸)=0 has three distinct real solutions ¸
1
, ¸
2
, and ¸
3
:
Suppose v
1
=
2
4
v
11
v
21
v
31
3
5
, v
2
=
2
4
v
12
v
22
v
32
3
5
, and v
3
=
2
4
v
13
v
23
v
33
3
5
are associate eigenvector (i.e, Av
1
= ¸
1
v
1
, Av
2
= ¸
2
v
2
, and
Av
3
=¸
3
v
3
) Then the general solution is
x
c
(t)=c
1
v
1
e
¸
1
t
+c
2
v
2
e
¸
2
t
+c
3
v
3
e
¸
3
t
And the fundamental matrix is
F(t)=
2
4
v
11
e
¸
1
t
v
12
e
¸
2
t
v
13
e
¸
3
t
v
21
e
¸
1
t
v
22
e
¸
2
t
v
23
e
¸
3
t
v
31
e
¸
1
t
v
32
e
¸
2
t
v
33
e
¸
3
t
3
5
:
Case 2: p(¸)=0 has a double solutions ¸
0
:
So p(¸) = (¸¡¸
0
)
2
(¸¡¸
1
), and ¸
0
has multiplicity 2. Let
v
3
=
2
4
v
12
v
22
v
32
3
5
is the eigenvector associated with ¸
1
:
[1] ¸
0
has two linearly independent eigenvectors:
Supposev
1
=
·
v
11
v
21
¸
andv
2
=
·
v
12
v
22
¸
areassociatelinearly
independent eigenvectors. Then the general solution is
x
c
(t)=(c
1
v
1
+c
2
v
2
)e
¸
0
t
+c
3
v
3
e
¸
1
t
And
F(t)=
2
4
v
11
e
¸
0
t
v
12
e
¸
0
t
v
13
e
¸
1
t
v
21
e
¸
0
t
v
22
e
¸
0
t
v
23
e
¸
1
t
v
31
e
¸
0
t
v
32
e
¸
0
t
v
33
e
¸
1
t
3
5
Page 3


Example 2.4. Twoconjugatecomplexeigenvaluescase Find
the general solution to x
0
(t)=
·
2 ¡3
1 2
¸
x(t)
Solution Using Mathcad , functions eigenvals() and eigen-
vecs() we ?nd two conjugate complex eigenvalues, ¸
1
= 2+i
p
3 and
¸
2
=2¡i
p
3withassociatedeigenvectorv
1
=
· p
3
¡i
¸
withrespectto
¸
1
. Compare this with the Theorem 2.1, we have a = 2;b =
p
3;v
11
=
p
3;v
21
=0;v
12
=0; and v
22
=¡1:
So the fundamental matrix is
F(t)=e
2t
· p
3cos(bt) ¡sin(bt)
p
3sin(bt) ¡cos(bt)
¸
and the general solution is, c=
·
c
1
c
2
¸
,
x
c
(t) = F(t)c=e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
c
1
c
2
¸
= e
2t
· p
3c
1
cos(
p
3t)¡c
2
sin(
p
3t)
p
3c
1
sin(
p
3t)¡c
2
cos(
p
3t)
¸
Suppose we want to ?nd a solution such that x(0)=
·
1
2
¸
, then
x(t) = F(t)F(0)
-1
x(0)
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸· p
3 0
0 ¡1
¸
-1
·
1
2
¸
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
1
p
3
¡2
¸
= e
2t
·
cos(
p
3t)++2sin(
p
3t)
¡sin(
p
3t)+2cos(
p
3t)
¸
a
2.2. A is a 3£3 matrix. Suppose
A =
2
4
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
3
5
Then the characteristic polynomial p(¸) ofA given by
p(¸)=jA¡¸Ij;
is a cubic polynomial of ¸: From Algebra, we know that p(¸) = 0
has either 3 distinct real solutions, or 2 distinct solutions and one is a
doublesolution,oronerealsolutionand2conjugatecomplexsolutions,
or a triple solution. The following theorem summarize the solution to
the homogeneous system,
Theorem 2.2. Let p(¸) be the characteristic polynomial of A; for
x
0
(t)=Ax(t);
Case 1: p(¸)=0 has three distinct real solutions ¸
1
, ¸
2
, and ¸
3
:
Suppose v
1
=
2
4
v
11
v
21
v
31
3
5
, v
2
=
2
4
v
12
v
22
v
32
3
5
, and v
3
=
2
4
v
13
v
23
v
33
3
5
are associate eigenvector (i.e, Av
1
= ¸
1
v
1
, Av
2
= ¸
2
v
2
, and
Av
3
=¸
3
v
3
) Then the general solution is
x
c
(t)=c
1
v
1
e
¸
1
t
+c
2
v
2
e
¸
2
t
+c
3
v
3
e
¸
3
t
And the fundamental matrix is
F(t)=
2
4
v
11
e
¸
1
t
v
12
e
¸
2
t
v
13
e
¸
3
t
v
21
e
¸
1
t
v
22
e
¸
2
t
v
23
e
¸
3
t
v
31
e
¸
1
t
v
32
e
¸
2
t
v
33
e
¸
3
t
3
5
:
Case 2: p(¸)=0 has a double solutions ¸
0
:
So p(¸) = (¸¡¸
0
)
2
(¸¡¸
1
), and ¸
0
has multiplicity 2. Let
v
3
=
2
4
v
12
v
22
v
32
3
5
is the eigenvector associated with ¸
1
:
[1] ¸
0
has two linearly independent eigenvectors:
Supposev
1
=
·
v
11
v
21
¸
andv
2
=
·
v
12
v
22
¸
areassociatelinearly
independent eigenvectors. Then the general solution is
x
c
(t)=(c
1
v
1
+c
2
v
2
)e
¸
0
t
+c
3
v
3
e
¸
1
t
And
F(t)=
2
4
v
11
e
¸
0
t
v
12
e
¸
0
t
v
13
e
¸
1
t
v
21
e
¸
0
t
v
22
e
¸
0
t
v
23
e
¸
1
t
v
31
e
¸
0
t
v
32
e
¸
0
t
v
33
e
¸
1
t
3
5
[2] ¸
0
has one eigenvector:
Suppose v
1
=
2
4
v
11
v
21
v
31
3
5
is the associated eigenvector with re-
spect to ¸
0
and v
2
=
2
4
v
12
v
22
v
32
3
5
is a solution of
(¸
0
I¡A)v
2
=v
1
:
Then the general solution is,
x
c
(t)=(c
1
v
1
+c
2
(tv
1
+v
2
))e
¸
0
t
+c
3
v
3
e
¸
1
And
F(t)=
2
4
v
11
e
¸
0
t
(v
11
t+v
12
)e
¸
0
t
v
13
e
¸
1
v
21
e
¸
0
t
(v
21
t+v
22
)e
¸
0
t
v
23
e
¸
1
v
31
e
¸
0
t
(v
31
t+v
32
)e
¸
0
t
v
33
e
¸
1
3
5
is the fundamental solution matrix.
Case 3: p(¸)=0 has two conjugate complex solutions a§bi and a real
solution ¸
1
:
Suppose v =
2
4
v
11
+iv
12
v
21
+iv
22
v
31
+iv
32
3
5
is the associate complex eigen-
vector with respect to a+bi, then the general solution is, let
v
3
=
2
4
v
13
v
23
V
33
3
5
, are associated eigenvectors with respect to ¸
1
,
x
c
(t)=[c
1
(v
1
cos(bt)¡v
2
sin(bt))c
2
(v
2
cos(bt)+v
1
sin(bt))]e
at
+c
3
v
3
e
¸
1
:
And
F(t)=e
at
2
4
v
11
cos(bt)¡v
12
sin(bt) v
12
cos(bt)+v
11
sin(bt) v
13
e
¸
1
v
21
cos(bt)¡v
22
sin(bt) v
22
cos(bt)+v
21
sin(bt) v
23
e
¸
1
v
31
cos(bt)¡v
32
sin(bt) v
32
cos(bt)+v
31
sin(bt) v
33
e
¸
1
3
5
is the fundamental matrix.
Case 4: p(¸)=0 has solution ¸
0
with multiplicity 3.
In this case, p(¸)=(¸¡¸
0
)
3
:
[1]¸
0
hasthreelinearlyindependenteigenvectors.
Let v
1
=
2
4
v
11
v
21
v
31
3
5
, v
2
=
2
4
v
12
v
22
V
32
3
5
, and v
3
=
2
4
v
13
v
23
V
33
3
5
be
the three linearly independent eigenvectors. Then the general
Page 4


Example 2.4. Twoconjugatecomplexeigenvaluescase Find
the general solution to x
0
(t)=
·
2 ¡3
1 2
¸
x(t)
Solution Using Mathcad , functions eigenvals() and eigen-
vecs() we ?nd two conjugate complex eigenvalues, ¸
1
= 2+i
p
3 and
¸
2
=2¡i
p
3withassociatedeigenvectorv
1
=
· p
3
¡i
¸
withrespectto
¸
1
. Compare this with the Theorem 2.1, we have a = 2;b =
p
3;v
11
=
p
3;v
21
=0;v
12
=0; and v
22
=¡1:
So the fundamental matrix is
F(t)=e
2t
· p
3cos(bt) ¡sin(bt)
p
3sin(bt) ¡cos(bt)
¸
and the general solution is, c=
·
c
1
c
2
¸
,
x
c
(t) = F(t)c=e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
c
1
c
2
¸
= e
2t
· p
3c
1
cos(
p
3t)¡c
2
sin(
p
3t)
p
3c
1
sin(
p
3t)¡c
2
cos(
p
3t)
¸
Suppose we want to ?nd a solution such that x(0)=
·
1
2
¸
, then
x(t) = F(t)F(0)
-1
x(0)
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸· p
3 0
0 ¡1
¸
-1
·
1
2
¸
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
1
p
3
¡2
¸
= e
2t
·
cos(
p
3t)++2sin(
p
3t)
¡sin(
p
3t)+2cos(
p
3t)
¸
a
2.2. A is a 3£3 matrix. Suppose
A =
2
4
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
3
5
Then the characteristic polynomial p(¸) ofA given by
p(¸)=jA¡¸Ij;
is a cubic polynomial of ¸: From Algebra, we know that p(¸) = 0
has either 3 distinct real solutions, or 2 distinct solutions and one is a
doublesolution,oronerealsolutionand2conjugatecomplexsolutions,
or a triple solution. The following theorem summarize the solution to
the homogeneous system,
Theorem 2.2. Let p(¸) be the characteristic polynomial of A; for
x
0
(t)=Ax(t);
Case 1: p(¸)=0 has three distinct real solutions ¸
1
, ¸
2
, and ¸
3
:
Suppose v
1
=
2
4
v
11
v
21
v
31
3
5
, v
2
=
2
4
v
12
v
22
v
32
3
5
, and v
3
=
2
4
v
13
v
23
v
33
3
5
are associate eigenvector (i.e, Av
1
= ¸
1
v
1
, Av
2
= ¸
2
v
2
, and
Av
3
=¸
3
v
3
) Then the general solution is
x
c
(t)=c
1
v
1
e
¸
1
t
+c
2
v
2
e
¸
2
t
+c
3
v
3
e
¸
3
t
And the fundamental matrix is
F(t)=
2
4
v
11
e
¸
1
t
v
12
e
¸
2
t
v
13
e
¸
3
t
v
21
e
¸
1
t
v
22
e
¸
2
t
v
23
e
¸
3
t
v
31
e
¸
1
t
v
32
e
¸
2
t
v
33
e
¸
3
t
3
5
:
Case 2: p(¸)=0 has a double solutions ¸
0
:
So p(¸) = (¸¡¸
0
)
2
(¸¡¸
1
), and ¸
0
has multiplicity 2. Let
v
3
=
2
4
v
12
v
22
v
32
3
5
is the eigenvector associated with ¸
1
:
[1] ¸
0
has two linearly independent eigenvectors:
Supposev
1
=
·
v
11
v
21
¸
andv
2
=
·
v
12
v
22
¸
areassociatelinearly
independent eigenvectors. Then the general solution is
x
c
(t)=(c
1
v
1
+c
2
v
2
)e
¸
0
t
+c
3
v
3
e
¸
1
t
And
F(t)=
2
4
v
11
e
¸
0
t
v
12
e
¸
0
t
v
13
e
¸
1
t
v
21
e
¸
0
t
v
22
e
¸
0
t
v
23
e
¸
1
t
v
31
e
¸
0
t
v
32
e
¸
0
t
v
33
e
¸
1
t
3
5
[2] ¸
0
has one eigenvector:
Suppose v
1
=
2
4
v
11
v
21
v
31
3
5
is the associated eigenvector with re-
spect to ¸
0
and v
2
=
2
4
v
12
v
22
v
32
3
5
is a solution of
(¸
0
I¡A)v
2
=v
1
:
Then the general solution is,
x
c
(t)=(c
1
v
1
+c
2
(tv
1
+v
2
))e
¸
0
t
+c
3
v
3
e
¸
1
And
F(t)=
2
4
v
11
e
¸
0
t
(v
11
t+v
12
)e
¸
0
t
v
13
e
¸
1
v
21
e
¸
0
t
(v
21
t+v
22
)e
¸
0
t
v
23
e
¸
1
v
31
e
¸
0
t
(v
31
t+v
32
)e
¸
0
t
v
33
e
¸
1
3
5
is the fundamental solution matrix.
Case 3: p(¸)=0 has two conjugate complex solutions a§bi and a real
solution ¸
1
:
Suppose v =
2
4
v
11
+iv
12
v
21
+iv
22
v
31
+iv
32
3
5
is the associate complex eigen-
vector with respect to a+bi, then the general solution is, let
v
3
=
2
4
v
13
v
23
V
33
3
5
, are associated eigenvectors with respect to ¸
1
,
x
c
(t)=[c
1
(v
1
cos(bt)¡v
2
sin(bt))c
2
(v
2
cos(bt)+v
1
sin(bt))]e
at
+c
3
v
3
e
¸
1
:
And
F(t)=e
at
2
4
v
11
cos(bt)¡v
12
sin(bt) v
12
cos(bt)+v
11
sin(bt) v
13
e
¸
1
v
21
cos(bt)¡v
22
sin(bt) v
22
cos(bt)+v
21
sin(bt) v
23
e
¸
1
v
31
cos(bt)¡v
32
sin(bt) v
32
cos(bt)+v
31
sin(bt) v
33
e
¸
1
3
5
is the fundamental matrix.
Case 4: p(¸)=0 has solution ¸
0
with multiplicity 3.
In this case, p(¸)=(¸¡¸
0
)
3
:
[1]¸
0
hasthreelinearlyindependenteigenvectors.
Let v
1
=
2
4
v
11
v
21
v
31
3
5
, v
2
=
2
4
v
12
v
22
V
32
3
5
, and v
3
=
2
4
v
13
v
23
V
33
3
5
be
the three linearly independent eigenvectors. Then the general
solution is x
c
(t) = (c
1
v
1
+c
2
v
2
+c
3
v
3
)e
¸
0
t
and fundamental
matrix is F(t)=e
¸
0
t
2
6
6
6
6
4
v
11
v
12
v
13
v
21
v
22
V
23
v
31
v
32
V
33
3
7
7
7
7
5
[2] ¸
0
has two linearly independent eigenvectors.
Suppose v
1
=
2
4
v
11
v
21
v
31
3
5
, v
2
=
2
4
v
12
v
22
V
32
3
5
are the linearly inde-
pendent eigenvectors. Letv
3
=
2
4
v
13
v
23
V
33
3
5
; then only one of the
two equations, (A¡¸
0
I)v
3
= v
1
or (A¡¸
0
I)v
3
= v
2
can
has a solution that is linearly independent with v
1
;v
2
:
Suppose (A¡¸
0
I)v
3
=v
2
generates such a solution. Then
the general solution is x
c
(t)=[c
1
v
1
+c
2
v
2
+c
3
(tv
2
+v
3
)]e
¸
0
t
andfundamentalmatrixisF(t)=e
¸
0
t
2
4
v
11
v
12
tv
12
+v
13
v
21
v
22
tv
22
+V
23
v
31
v
32
tv
32
+V
33
3
5
[3] ¸
0
has only one eigenvector.
Let v
1
=
2
4
v
11
v
21
v
31
3
5
be the linearly independent eigenvectors.
Let v
2
=
2
4
v
12
v
22
V
32
3
5
and v
3
=
2
4
v
13
v
23
V
33
3
5
be two vectors that
satis?es
(A¡¸
0
I)v
2
=v
1
and (A¡¸
0
I)v
3
=v
2
:
Then the general solution is x
c
(t) = [c
1
v
1
+ c
2
(tv
1
+v
2
) +
c
3
(t
2
v
1
+ tv
2
+ v
3
)]e
¸
0
t
and fundamental matrix is F(t) =
e
¸
0
t
2
4
v
11
tv
11
+v
12
t
2
v
11
+tv
12
+v
13
v
21
tv
21
+v
22
t
2
v
21
+tv
22
+V
23
v
31
tv
31
+v
32
t
2
v
31
+tv
32
+V
33
3
5
Remark 2.1. Suppose A is an n£n matrix, for the homogeneous
system x
0
(t)=Ax(t); three general case would happen
Case 1: A has n distinct eigenvalues ¸
i
; i = 1;2;¢¢¢ ;n with linearly
independent eigenvectors v
i
; i = 1;2;¢¢¢ ;n then the general
solution will be x
c
(t)=c
1
v
1
e
¸
1
+c
2
v
2
e
¸
2
+¢¢¢+c
n
v
n
e
¸n
Page 5


Example 2.4. Twoconjugatecomplexeigenvaluescase Find
the general solution to x
0
(t)=
·
2 ¡3
1 2
¸
x(t)
Solution Using Mathcad , functions eigenvals() and eigen-
vecs() we ?nd two conjugate complex eigenvalues, ¸
1
= 2+i
p
3 and
¸
2
=2¡i
p
3withassociatedeigenvectorv
1
=
· p
3
¡i
¸
withrespectto
¸
1
. Compare this with the Theorem 2.1, we have a = 2;b =
p
3;v
11
=
p
3;v
21
=0;v
12
=0; and v
22
=¡1:
So the fundamental matrix is
F(t)=e
2t
· p
3cos(bt) ¡sin(bt)
p
3sin(bt) ¡cos(bt)
¸
and the general solution is, c=
·
c
1
c
2
¸
,
x
c
(t) = F(t)c=e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
c
1
c
2
¸
= e
2t
· p
3c
1
cos(
p
3t)¡c
2
sin(
p
3t)
p
3c
1
sin(
p
3t)¡c
2
cos(
p
3t)
¸
Suppose we want to ?nd a solution such that x(0)=
·
1
2
¸
, then
x(t) = F(t)F(0)
-1
x(0)
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸· p
3 0
0 ¡1
¸
-1
·
1
2
¸
= e
2t
· p
3cos(
p
3t) ¡sin(
p
3t)
p
3sin(
p
3t) ¡cos(
p
3t)
¸·
1
p
3
¡2
¸
= e
2t
·
cos(
p
3t)++2sin(
p
3t)
¡sin(
p
3t)+2cos(
p
3t)
¸
a
2.2. A is a 3£3 matrix. Suppose
A =
2
4
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
3
5
Then the characteristic polynomial p(¸) ofA given by
p(¸)=jA¡¸Ij;
is a cubic polynomial of ¸: From Algebra, we know that p(¸) = 0
has either 3 distinct real solutions, or 2 distinct solutions and one is a
doublesolution,oronerealsolutionand2conjugatecomplexsolutions,
or a triple solution. The following theorem summarize the solution to
the homogeneous system,
Theorem 2.2. Let p(¸) be the characteristic polynomial of A; for
x
0
(t)=Ax(t);
Case 1: p(¸)=0 has three distinct real solutions ¸
1
, ¸
2
, and ¸
3
:
Suppose v
1
=
2
4
v
11
v
21
v
31
3
5
, v
2
=
2
4
v
12
v
22
v
32
3
5
, and v
3
=
2
4
v
13
v
23
v
33
3
5
are associate eigenvector (i.e, Av
1
= ¸
1
v
1
, Av
2
= ¸
2
v
2
, and
Av
3
=¸
3
v
3
) Then the general solution is
x
c
(t)=c
1
v
1
e
¸
1
t
+c
2
v
2
e
¸
2
t
+c
3
v
3
e
¸
3
t
And the fundamental matrix is
F(t)=
2
4
v
11
e
¸
1
t
v
12
e
¸
2
t
v
13
e
¸
3
t
v
21
e
¸
1
t
v
22
e
¸
2
t
v
23
e
¸
3
t
v
31
e
¸
1
t
v
32
e
¸
2
t
v
33
e
¸
3
t
3
5
:
Case 2: p(¸)=0 has a double solutions ¸
0
:
So p(¸) = (¸¡¸
0
)
2
(¸¡¸
1
), and ¸
0
has multiplicity 2. Let
v
3
=
2
4
v
12
v
22
v
32
3
5
is the eigenvector associated with ¸
1
:
[1] ¸
0
has two linearly independent eigenvectors:
Supposev
1
=
·
v
11
v
21
¸
andv
2
=
·
v
12
v
22
¸
areassociatelinearly
independent eigenvectors. Then the general solution is
x
c
(t)=(c
1
v
1
+c
2
v
2
)e
¸
0
t
+c
3
v
3
e
¸
1
t
And
F(t)=
2
4
v
11
e
¸
0
t
v
12
e
¸
0
t
v
13
e
¸
1
t
v
21
e
¸
0
t
v
22
e
¸
0
t
v
23
e
¸
1
t
v
31
e
¸
0
t
v
32
e
¸
0
t
v
33
e
¸
1
t
3
5
[2] ¸
0
has one eigenvector:
Suppose v
1
=
2
4
v
11
v
21
v
31
3
5
is the associated eigenvector with re-
spect to ¸
0
and v
2
=
2
4
v
12
v
22
v
32
3
5
is a solution of
(¸
0
I¡A)v
2
=v
1
:
Then the general solution is,
x
c
(t)=(c
1
v
1
+c
2
(tv
1
+v
2
))e
¸
0
t
+c
3
v
3
e
¸
1
And
F(t)=
2
4
v
11
e
¸
0
t
(v
11
t+v
12
)e
¸
0
t
v
13
e
¸
1
v
21
e
¸
0
t
(v
21
t+v
22
)e
¸
0
t
v
23
e
¸
1
v
31
e
¸
0
t
(v
31
t+v
32
)e
¸
0
t
v
33
e
¸
1
3
5
is the fundamental solution matrix.
Case 3: p(¸)=0 has two conjugate complex solutions a§bi and a real
solution ¸
1
:
Suppose v =
2
4
v
11
+iv
12
v
21
+iv
22
v
31
+iv
32
3
5
is the associate complex eigen-
vector with respect to a+bi, then the general solution is, let
v
3
=
2
4
v
13
v
23
V
33
3
5
, are associated eigenvectors with respect to ¸
1
,
x
c
(t)=[c
1
(v
1
cos(bt)¡v
2
sin(bt))c
2
(v
2
cos(bt)+v
1
sin(bt))]e
at
+c
3
v
3
e
¸
1
:
And
F(t)=e
at
2
4
v
11
cos(bt)¡v
12
sin(bt) v
12
cos(bt)+v
11
sin(bt) v
13
e
¸
1
v
21
cos(bt)¡v
22
sin(bt) v
22
cos(bt)+v
21
sin(bt) v
23
e
¸
1
v
31
cos(bt)¡v
32
sin(bt) v
32
cos(bt)+v
31
sin(bt) v
33
e
¸
1
3
5
is the fundamental matrix.
Case 4: p(¸)=0 has solution ¸
0
with multiplicity 3.
In this case, p(¸)=(¸¡¸
0
)
3
:
[1]¸
0
hasthreelinearlyindependenteigenvectors.
Let v
1
=
2
4
v
11
v
21
v
31
3
5
, v
2
=
2
4
v
12
v
22
V
32
3
5
, and v
3
=
2
4
v
13
v
23
V
33
3
5
be
the three linearly independent eigenvectors. Then the general
solution is x
c
(t) = (c
1
v
1
+c
2
v
2
+c
3
v
3
)e
¸
0
t
and fundamental
matrix is F(t)=e
¸
0
t
2
6
6
6
6
4
v
11
v
12
v
13
v
21
v
22
V
23
v
31
v
32
V
33
3
7
7
7
7
5
[2] ¸
0
has two linearly independent eigenvectors.
Suppose v
1
=
2
4
v
11
v
21
v
31
3
5
, v
2
=
2
4
v
12
v
22
V
32
3
5
are the linearly inde-
pendent eigenvectors. Letv
3
=
2
4
v
13
v
23
V
33
3
5
; then only one of the
two equations, (A¡¸
0
I)v
3
= v
1
or (A¡¸
0
I)v
3
= v
2
can
has a solution that is linearly independent with v
1
;v
2
:
Suppose (A¡¸
0
I)v
3
=v
2
generates such a solution. Then
the general solution is x
c
(t)=[c
1
v
1
+c
2
v
2
+c
3
(tv
2
+v
3
)]e
¸
0
t
andfundamentalmatrixisF(t)=e
¸
0
t
2
4
v
11
v
12
tv
12
+v
13
v
21
v
22
tv
22
+V
23
v
31
v
32
tv
32
+V
33
3
5
[3] ¸
0
has only one eigenvector.
Let v
1
=
2
4
v
11
v
21
v
31
3
5
be the linearly independent eigenvectors.
Let v
2
=
2
4
v
12
v
22
V
32
3
5
and v
3
=
2
4
v
13
v
23
V
33
3
5
be two vectors that
satis?es
(A¡¸
0
I)v
2
=v
1
and (A¡¸
0
I)v
3
=v
2
:
Then the general solution is x
c
(t) = [c
1
v
1
+ c
2
(tv
1
+v
2
) +
c
3
(t
2
v
1
+ tv
2
+ v
3
)]e
¸
0
t
and fundamental matrix is F(t) =
e
¸
0
t
2
4
v
11
tv
11
+v
12
t
2
v
11
+tv
12
+v
13
v
21
tv
21
+v
22
t
2
v
21
+tv
22
+V
23
v
31
tv
31
+v
32
t
2
v
31
+tv
32
+V
33
3
5
Remark 2.1. Suppose A is an n£n matrix, for the homogeneous
system x
0
(t)=Ax(t); three general case would happen
Case 1: A has n distinct eigenvalues ¸
i
; i = 1;2;¢¢¢ ;n with linearly
independent eigenvectors v
i
; i = 1;2;¢¢¢ ;n then the general
solution will be x
c
(t)=c
1
v
1
e
¸
1
+c
2
v
2
e
¸
2
+¢¢¢+c
n
v
n
e
¸n
Case2: A has m<n distinct eigenvalues, in this case some eigenval-
ues would have multiplicity greater than 1.
Suppose ¸
r
has multiplicity r. Depending on how many
linearly independent eigenvectors are associated with ¸
r
the
situationcouldbeverycomplex. Let pbethenumberoflinearly
eigenvectors associated with ¸
r
; then d = r¡p is called the
de?cit of ¸
r
. The simply cases are either d=0 or d=r¡1.
When 0<d<r¡1 the situation could be very complex.
Suppose d=r¡1 and v
1
is the only eigenvector associate
with ¸
r
, then one will have to solve r ¡ 1 equations (A¡
¸
r
)
i
v
i+1
= v
i
; i = 1;2;¢¢¢ ;r¡1: And the general solution
would contains terms like [c
1
v
1
+c
2
(v
1
t+v
2
)+c
3
(v
1
t
2
+v
2
t+
v
3
)+¢¢¢+c
r
(v
r
1
+v
2
t
r¡1
+¢¢¢+v
r
)]e
¸r
:
Case 3: A complex root a+bi with associated eigenvectorv
a
+iv
b
, then
thegeneralsolutioncontainsterm, [c
1
(v
a
cos(bt)¡v
b
sin(bt))+
c
2
(v
a
sin(bt)+v
b
cos(bt))]e
at
:
Remark 2.2. Suppose x
1
(t);x
2
(t);x
3
(t);¢¢¢ ;x
n
(t) are n linearly
independent solution for n£n homogeneous system,x
0
(t)=Ax(t), the
fundamental matrix F(t) is a matrix whose columns are x
i
(t);i =
1;2;¢¢¢ ;n:
Example 2.5. (Two distinct eigenvalues) Find the general so-
lution to
x
0
1
= 3x
1
+4x
2
¡2x
3
x
0
2
= 2x
1
+x
2
¡4x
3
x
0
3
= x
1
+2x
2
Solution Let x(t) =
2
4
x
1
(t)
x
2
(t)
x
3
(t)
3
5
and
2
4
3 4 ¡2
2 1 ¡4
1 2 0
3
5
The equa-
tions can be written in matrix form x
0
(t)=Ax(t):
UsingMathcad,functionseigenvals()andeigenvecs()we?nd,¸
1
=
2 and ¸
2
= 1 with associated eigenvectors v
1
=
2
4
¡4
1
¡2
3
5
and v
2
=
2
4
1
0
1
3
5
respectively. Since ¸
1
has multiplicity 2 as 1 appeared twice
in the result of eigenvals() function, we need to solve the equation
(A¡¸
1
I)v
3
=v
2
: