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CHAPTER 10
Systems of Linear
Differential Equations
Contents
10.1 Preliminary Theory
10.2 Homogeneous Linear Systems
10.3 Solution by Diagonalization
10.4 Nonhomogeneous Linear Systems
10.5 Matrix Exponential
Ch10_2
10.1 Preliminary Theory
Introduction
Recall that in Sec 3.11, we have the following system
of linear DEs:
P11 ( D) x1  P12 ( D) x2    P1n ( D) xn  b1 (t )
P21 ( D) x1  P22 ( D) x2    P2 n ( D) xn  b2 (t )

(1)
Pn1 ( D) x1  Pn 2 ( D) x2    Pnn ( D) xn  bn (t )
Ch10_3
Consider the system of first-order DEs
dx1
 g1 (t , x1 , x2 , , xn )
dt
dx2
 g 2 (t , x1 , x2 , , xn )
dt

(2)
dxn
 g n (t , x1 , x2 , , xn )
dt
Ch10_4
Linear Systems
When (2) is linear, we have the normal form as
dx1
 a11 (t ) x1  a12 (t ) x2    a1n (t ) xn  f1 (t )
dt
dx2
 a21 (t ) x1  a22 (t ) x2    a2 n (t ) xn  f 2 (t )
(3)
dt

dxn
 an1 (t ) x1  an 2 (t ) x2    ann (t ) xn  f n (t )
dt
Ch10_5
Matrix Form of a Linear Systems
If we let
 x1 (t ) 
 a11 (t ) a12 (t )  a1n (t ) 






 x2 (t ) 
 a21 (t ) a22 (t )  a2 n (t ) 

X
,
A
(
t
)

,
F
(
t
)

 

 
 





 xn (t ) 
 an1 (t ) an 2 (t )  ann (t ) 

f1 (t ) 

f 2 (t ) 
 

f n (t ) 
then (3) becomes
X  AX  F
(4)
If it is homogeneous,
X  AX
(5)
Ch10_6
Example 1
 x
(a) If X    , the matrix form of
 y
dx
 3x  4 y
dt
dy
 5x  7 y
dt
is
3 4 
X  
X
5  7
Ch10_7
 x
 
(b) If X   y  , the matrix form of
z
 
dx
 6x  y  z  t
dt
dy
 8 x  7 y  z  10t is
dt
dz
 2 x  9 y  z  6t
dt
 6 1 1
 t


 
X   8 7  1 X  10t 
 2 9  1
 6t 


 
Ch10_8
DEFINITION 10.1
Solution Vector
A solution vector on an interval I is any column
matrix
 x1 (t ) 


 x2 (t ) 
X
 


 xn (t ) 
whose entries are differentiable functions satisfying (4)
on the interval.
Ch10_9
Example 2
Verify that on (−, )
 1  2 t  e  2 t 
X1   e   2t ,
  1
 e 
are solutions of
 3  6t  3e6t 
X 2   e   6t 
 5
 5e 
 1 3
X  
X
 5 3
(6)
Ch10_10
Example 2
Solution
From
we have
  2e2t 

X1  
 2t 
 2e 
(2)
 18e6t 

X2  
6t 
30
e


 1 3   e2t   e2t  3e2t    2e2t 

  X1
AX1  
  2t    2t

2
t

2
t



5
3

e
5
e

3
e
2
e


 
 

 1 3   3e6t   3e6t  15e6t   18e6t 

  X2
AX2  
  6t    6t
6
t
6
t



 5 3   5e  15e  15e   30e 
Ch10_11
Initial Value Problem (IVP)
 x1 (t0 ) 


 x2 (t0 ) 
X(t0 )  
,




 xn (t0 ) 
Then the problem
Let
 1 
 
 2 
X0   

 
 n 
X  A(t ) X  F(t )
with X(t0) = X0
(7)
is an IVP.
Ch10_12
THEOREM 10.1
Existence of a Unique Solution
Let the entries of A(t) and F(t) be continuous functions
on a common interval I that contains t0, Then there
exists a unique solution of (7) on I.
THEOREM 10.2
Superposition Principles
Let X1, X2,…, Xk be a set of solutions of the
homogeneous system (5) on I, then
X = c1X1 + c2X2 + … + ckXk
is also a solution on I.
Ch10_13
Example 3
Please verify that
cos t


X1    1/2 cos t  1/2 sin
  cos t  sin t

are solutions of
 1 0

X   1 1
 2 0



t ,


1

0 X
 1
0
 t
X2   e 
0
 
(8)
Ch10_14
Example 3 (2)
then
X  c1X1  c2 X 2
cos t


 c1   1/2 cos t  1/2 sin
  cos t  sin t


0

 t
t   c2  e 

0

 
is also a solution.
Ch10_15
DEFINITION 10.2
Linear Dependence/Independence
Let X1, X2, …, Xk be a set of solution vectors of the
homogeneous system (5) in an interval I. We say the
set is linear dependent in the interval if there exist
constants c1, c2, …, ck, not all zero, such that
c1X1 + c2X2 + … + ckXk = 0
for every t in the interval. If the set of vectors is not
linearly dependent on the interval, it is said to be
linearly independent.
Ch10_16
THEOREM 10.3
Let
Criterion for Linearly Independent
Solution
 x11 
 x12 
 x1n 
 
 
 
 x21 
 x22 
 x2 n 
X1    , X 2    , , X n   



 
 
 
 xn1 
 xn 2 
 xnn 
be n solution vectors of the homogeneous system (5) on
an interval I. Then the set of solution vectors is linearly
Independent on I if and only if the Wronskian
Ch10_17
(continued)
THEOREM 10.3
Criterion for Linearly Independent
Solution
W ( X1 , X 2 , , X n ) 
x11
x12  x1n
x21
x22  x2 n

xn1

0
(9)
xn 2  xnn
for every t in the interval.
Ch10_18
(continued)
Criterion for Linearly Independent
Solution
If X1 , X 2 , , X nare linearly dependent on I, then W=0
at every point of I.
If X1 , X 2 , , X n are linearly independent on I, then W 0
at every point of I.
Ch10_19
Example 4
 1 2t
 3  6t
We saw that X1   e , X 2   e
  1
 5
are solutions of (6). Since
W ( X1 , X 2 ) 
e
e
 2t
 2t
3e
6t
5e
6t
 8e 4t  0
they are linearly independent for all real t.
Ch10_20
DEFINITION 10.3
Fundamental Set of Solution
Any set of n linearly independent solution vectors
X1, X2, …, Xn of the homogeneous system (5) on an
Interval I is said to be a fundamental set of solutions
on the interval.
Ch10_21
THEOREM 10.4
Existence of a Fundamental Set
There exists a fundamental set of solutions for the
homogeneous system (5) on an interval I.
THEOREM 10.5
General Solution—Homogeneous
Systems
Let X1, X2, …, Xn be a fundamental set of solutions of
the homogeneous system (5) on an interval I. Then the
general solution of the system in the interval is
X = c1X1 + c2X2 + … + cnXn
where the ci, i = 1, 2,…, n are arbitrary constants.
Ch10_22
Example 5
We saw that X   1e 2t , X   3 e6t
1
2
  1
 5
are linearly independent solutions of (6) on (−, ).
Hence they form a fundamental set of solutions. Then
general solutions is then
 1 2t
 3  6t
X c1X1  c2 X1  c1   e  c2   e
  1
 5
(10)
Ch10_23
Example 6
Consider the vectors
cos t


0


  t
X1    1/2 cos t  1/2 sin t , X 2   1  e ,
  cos t  sin t 
0


 
sin t




X3    1/2 sin t  1/2 cos t 
  sin t  cos t 


Ch10_24
Example 6 (2)
Their Wronskian is
W ( X1 , X 2 , X3 )
cos t
0
sin t
  1/2 cos t  1/2 sin t et
 1/2 sin t  1/2 cos t
 cos t  sin t
 sin t  cos t
0
 et  0
Ch10_25
Example 6 (2)
Then the general solution is
cos t


0


  t
X  c1   1/2 cos t  1/2 sin t   c2  1  e
  cos t  sin t 
0


 
sin t



3
 c   1/2 sin t  1/2 cos t 
  sin t  cos t 


Ch10_26
THEOREM 10.6
General Solution—Nonhomogeneous
Systems
Let Xp be a given solution of the nonhomogeneous system (4)
on the interval I, and let
Xc = c1X1 + c2X2 + … + cnXn
denote the general solution on the same interval of the associated
homogeneous system (5). Then the general solution of the
nonhomogeneous system on the interval is
X = Xc + Xp.
The general solution Xc of the homogeneous system (5) is called
the complementary function of the nonhomogeneous system (4).
Ch10_27
Example 7
 3t  4 
The vector X p  
 is a particular solution of
  5t  6 
 1 3
12t  11
(11)

X 
 X

 5 3
 3 
on (−, ). In Example 5, we saw the solution of
 1 3
 1  2 t
 3  6t
X  
 X is Xc  c1   e  c2  e
 5 3
  1
 5
Thus the general solution of (11) on (−, ) is
 1 2t
 3  6t  3t  4 
X  Xc  X p  c1   e  c2   e  

  1
 5
  5t  6 
Ch10_28
10.2 Homogeneous Linear Systems
A Question
We are asked whether we can always find a solution
of the form
 k1 
 
 k2  t
X    e  Ke  t
(1)

 
 kn 
for the homogeneous linear first-order system
(2)
X  AX
Ch10_29
Eigenvalues and Eigenvectors
If (1) is a solution of (2), then X = Ket then (2)
becomes Ket = AKet . Thus we have AK = K or
AK – K = 0. Since K = IK, we have
(A – I)K = 0
(3)
The equation (3) is equivalent to
(a11   )k1  a12 k2   
a1n kn  0
a21k1  (a22   )k2   
a2 n kn  0


an1k1  an 2 k2    (ann   )kn  0
Ch10_30
If we want to find a nontrivial solution X, we must
have
det(A – I) = 0
The above discussions are similar to eigenvalues and
eigenvectors of matrices.
Ch10_31
8.8 The Eigenvalue Problems
DEFINITION 8.13
Eigenvalues and Eigenvectors
Let A be n  n matrix. A number  is said to be an
eigenvalue of A if there exists a nonzero solution
vector K of
AK = K
(1)
The solution vector K is said to be an eigenvector
corresponding to the eigenvalue .
Ch10_32
Procedure for finding a solution
Step 1: Determine the eigenvalue  .
Step 2: Determine the associated eigenvector by solving
AK=  K
t
e
Step 3: Form a solution K .
Ch10_33
THEOREM 10.7
General Solution—Homogeneous
Systems
Let 1, 2,…, n be n distinct eigenvalues of the matrix
A of (2), and let K1, K2,…, Kn be the corresponding
eigenvectors. Then the general solution of (2) is
X  c1K1e1t  c2K 2e2t  cnK nent
Ch10_34
THEOREM
Let 1, 2,…, n be n distinct eigenvalues of the matrix
A of (2), and let K1, K2,…, Kn be the corresponding
eigenvectors. Then K1, K2,…, Kn are linearly
Independent.
Ch10_35
Example 1
Solve
dx
 2x  3y
dt
dy
 2x  y
dt
(4)
Solution
det ( A   I ) 
2
3
2
1 
 2  3  4
 (  1)(  4)  0
we have 1 = −1, 2 = 4.
Ch10_36
Example 1
(2)
For 1 = −1, we have
3k1 + 3k2 = 0
2k1 + 2k2 = 0
 1
Thus k1 = – k2. When k2 = –1, then K1   
  1
For 1 = 4, we have
−2k1 + 3k2 = 0
2k1 − 2k2 = 0
 3
Thus k1 = 3k2/2. When k2 = 2, then K 2   
 2
Ch10_37
Example 1
(3)
We have
 1  t
 3  4t
X1    e , X 2    e
  1
 2
and the solution is
 1  t
 3  4t
X  c1X1  c2 X 2  c1   e  c2   e
  1
 2
(5)
Ch10_38
Example 2
Solve
dx
dy
dz
 4 x  y  z,
 x  5 y  z,  y  3z
dt
dt
dt
Solution
det ( A   I ) 
4
1
1
1
5
1
0
1
3
(6)
 (  3)(  4)(  5)  0
  3,  4, 5
Ch10_39
Example 2
(2)
For 1 = −3, we have
 1 1 1 0
 1 0 1 0




( A  3I | 0)   1 8  1 0    0 1 0 0 
 0 1 0 0
0 0 0 0




Thus k1 = k3, k2 = 0. When k3 = 1, then
 1
 
K1   0  ,
 1
 
 1
  3t
X1   0  e
 1
 
(7)
Ch10_40
Example 2
(3)
For 2 = −4, we have
0 1 1 0


( A  4I | 0)   1 9  1 0 
0 1 0 0



 1 0  10 0 


0 0
0 1
0 0

0
0


Thus k1 = 10k3, k2 = − k3. When k3 = 1, then
 10 
 
K 2    1 ,
 1
 
 10 
  4t
X 2    1 e
 1
 
(8)
Ch10_41
Example 2
(4)
For 3 = 5, we have
1 0
 9 1
 1 0 1 0




( A  5I | 0)   1 0  1 0    0 1  8 0 
 0 1  8 0
0 0

0
0




Then
 1
 1
 
  5t
K 3   8  , X3   8  e
(9)
 1
 1
 
 
Ch10_42
Example 2
(5)
Thus
 1
 10 
 1
  3t
  4t
  5t
X  c1  0  e  c2   1 e  c3  8  e
 1
 1
 1
 
 
 
Ch10_43
Example 3: Repeated Eigenvalues
Solve
 1 2 2 


X'    2 1  2  X
 2 2 1 


Solution
1 
2
det ( A   I )   2
1 
2
2
2
2 0
1 
Ch10_44
Example 3 (2) (repeated eigenvalue is complete)
We have – ( + 1)2(– 5) = 0, then 1 = 2 = – 1,
3 = 5.
For 1 = – 1,
2 0
 2 2
 1 1 1 0




( A  I | 0)    2
2  2 0  0 0 0 0
 2 2

0 0 0 0
2
0




k1 – k2 + k3 = 0 or k1 = k2 – k3.
Choosing k2 = 1, k3 = 0 and k2 = 1, k3 = 1, in turn we
have k1 = 1 and k1 = 0.
Ch10_45
Example 3 (3)
Thus the two linearly independent eigenvectors are
1
 
K1   1  ,
0
 
For 3 = 5,
K2
 0
 
 1
1
 
2 0
 4  2


( A  5I | 0)    2  4  2 0 
 2  2  4 0


 1 0 1 0


 0 1 1 0
0 0 0 0


Ch10_46
Example 3 (4)
Implies k1 = k3 and k2 = – k3. Choosing k3 = 1, then k1
= 1, k2 = –1, thus
 1
 
K 3    1
 1
 
Then the general solution is
 1
 0
 1
  t
  t
  5t
X  c1  1 e  c2  1 e  c3   1 e
0
 1
 1
 
 
 
Ch10_47
DEFINITION 7.7
Linear Independence
A set of vectors {x1, x2, …, xn} is said to be linearly
independent, if the only constants satisfying
k1x1 + k2x2 + …+ knxn = 0
(3)
are k1= k2 = … = kn = 0. If the set of vectors is not
linearly independent, it is linearly dependent.
Ch10_48
Repeated eigenvalue is defective
Suppose 1 is of multiplicity 2 and there is only one
eigenvector. A second solution will be of the form
X2  Kte1t  Pe1t
(12)
Thus X = AX becomes
( AK  1K )te1t  ( AP  1P  K )e1t  0
we have
( A  1I )K  0
( A  1I )P  K
(13)
(14)
Ch10_49
Example 4
Solve
 3  18 
X  
X
 2  9
Solution
First solve det (A – I) = 0 = ( + 3)2,  = －3, －3, and
then we get the first solution
 p1 
 3
Let K   , P   
 p2 
1
 3   3t
X1    e
1
Ch10_50
Example 4
(2)
From (14), we have (A + 3 I) P = K. Then
6 p1  18 p2  3
2 p1  6 p2  1
Choose p1 = 1, then p2 = 1/6. However, we choose
1/2 
p1 = ½, then p2 = 0. Thus P   
 0 
Ch10_51
Example 4
(3)
From (12), we have
 3  3t 1/2  3t
X 2    te    e
1
 0 
The general solution is
 3  3t 1/2  3t 
 3  3t
X  c1  e  c2   te    e 
1
 0 
 1 

Ch10_52
Eigenvalue of Multiplicity 3
When there is only one eigenvector associated with
an eigenvalue of multiplicity 3, we can find a third
solution as
2
t 1t
X3  K e  Pte 1t  Qe1t
2
and
( A  1I )K  0
(16)
( A  1I )P  K
(17)
( A  1I )Q  P
(18)
Ch10_53
Example 5
 2 1 6


X   0 2 5  X
 0 0 2
Solution


(1 – 2)3 = 0, 1 = 2 is of multiplicity 3. By solving
(A – 2I)K = 0, we have a single eigenvector
Solve
1
 
K  0
0
 
Ch10_54
Example 5 (2)
Next, solving (A – 2I) P = K and (A – 2I) Q = P,
then
0
 0 
 


P   1 ,
Q    6/5 
0
 1/5 
 


Thus
 1 

1
0
  2t
  2t   2t 

X  c1  0  e  c2  0  te   1  e


0




 0 

 
0
 1  2

0
 0 
  t 2t   2t 
 2t 

 c3  0  e   1  te    6/5  e


2






 0 

0
 1/5 
Ch10_55
THEOREM 10.8
Solution Corresponding to a
Complex Eigenvalue
Let A be the coefficient matrix having real entires of
the homogeneous system (2), and let K1 be an
eigenvector corresponding to the complex eigenvalue
1 =  + i ,  and  are real. Then
K1e1t and K1e1t
are two linearly independent solutions of (2).
Ch10_56
THEOREM 10.9
Real Solutions Corresponding to a
Complex Eigenvalue
Let 1 =  + i be a complex eigenvalue of the
coefficient matrix A in the homogeneous system (2),
and let B1=Re(K 1) and B2=Im(K 1). Then
X1  [B1 cos  t  B 2 sin  t ] e t
(23)
X2  [B 2 cos  t  B1 sin  t ] e t
are linearly independent solutions of (2) on (－, ).
Ch10_57
Example 6
Solve
8
 2
X  
X,
 1  2
 2
X(0)   
  1
Solution
First,
2
8
det ( A   I ) 
 2  4  0
1  2  
For 1 = 2i, (2 – 2i)k1 + 8k2 = 0,
– k1 + (–2 – 2i)k2 = 0, we get k1 = –(2 + 2i)k2.
Ch10_58
Example 6 (2)
Choosing k2 = –1, then
 2  2i   2   2 
K1  
  i  
  1    1  0 
2
 2
B1  Re(K1 )   ,
B 2  Im(K1 )   
  1
0
Since  = 0, then
 2 

 2 

 2
 2
X  c1   cos 2t    sin 2t   c2   cos 2t    sin 2t 
 0
  1
  1

 0 

 2 cos 2t  2 sin 2t 
 2 cos 2t  2 sin 2t 
 c1 
  c2 

 cos 2t
 sin 2t




Ch10_59
Fig 10.4
Ch10_60
10.3 Solution by Diagonalization
Formula
If A is diagonalizable, then there exists P, such that D
= P-1AP is diagonal. Letting X = PY then
X = AX becomes PY = APY, Y = P-1APY,
that is, Y = DY, the solution will be
 c1e1t 
 t
 c2e 2 
Y




 c ent 
 n 
Ch10_61
Example 1
Solve
  2  1 8


X   0  3 8  X
 0  4 9


Solution
From det (A – I) = – ( + 2)(– 1)(– 5), we get 1 = –
2, 2 = 1 and 3 = 5. Since they are distinct, the
eigenvectors are linearly independent. For i = 1, 2, 3,
solve (A –iI)K = 0, we have
1
 2
1
 
 
 
K1   0  , K 2   2  , K 3  1
0
1
1
 
 
 
Ch10_62
Example 1
 1 2 1


P   0 2 1
then
 0 1 1


Since Y = DY, then
Thus
(2)
  2 0 0


and D   0 1 0 
 0 0 5


 c1e 2t 
 t 
Y   c2e 
 c e 5t 
 3

2t
2t
t
5t




1
2
1
c
e
c
e

2
c
e

c
e

 1
1
2
3
 



t
t
5t
X  PY   0 2 1  c2e   
2c2e  c3e


t
5t
 0 1 1  c e5t  
c
e

c
e

 3
2
3
 

Ch10_63
10.4 Nonhomogeneous Linear Systems
Example 1
Solve
 1 2
  8
X'  
 X   , on (-, )
 1 1
 3
Solution
First solve X = AX,
det( A  I ) 
1 
2
1
1 
 2  1  0
 cos t  sin t 
 cos t  sin t 
  c2 

 = i, −i, Xc  c1
 cos t 
  sin t 
Ch10_64
Method of undetermined coefficients
Now let
Thus
 a1 
Xp   
 b1 
we have
0   a1  2b1  8
0   a1  b1  3
14 
Xp   
 11 
Finally, X = Xc + Xp
 cos t  sin t 
 cos t  sin t  14 
 c1 
  c2 
 
 cos t 
  sin t   11
Ch10_65
Example 2
Solve
 6 1
 6t 
 X  
, on (, )
X'  
 4 3
  10t 
Solution
First we solve X = AX. By similar procedures, we have
1 = 2, 2 = 7, and
Then
 1 
 1
K1   , K 2   
  2
 1
 1  2t
1 7t
Xc  c1  e  c2  e
  4
1
Ch10_66
Example 2
(2)
 a2   a1 
X p   t   
 b2   b1 
After substitution and simplification,
 a2   6 1   a2   a1   6   0 
 
  t     
t   
 b2   4 3   b2   b1    10   4 
or

 0   (6a2  b2  6)t  6a1  b1  a2

 
 0   (4a2  3b2  10)t  4a1  3b1  b2  4 
6a2  b2  6  0
6a1  b1  a2  0
and
4a2  3b2  10  0
4a1  3b1  b2  4  0
Now let
Ch10_67
Example 2
(3)
Solving the first two eqs. a2  2, b2  6 , and
substitute these values into the last two equations, we
4
10
obtain a1   7 , b1  7
then
4



2
   7 
X p   t  10
 6   7 
The general solution of the system on (－, ) is X =
Xc + Xp or
 4
 1  2t
1 7t   2   7 
X  c1  e  c2  e   t  

  4
1
 6   10 
 7 
Ch10_68
Example 3
Determine the form of Xp for
dx/dt =5x + 3y – 2e-t + 1
dy/dt =−x + y + e-t – 5t + 7
Solution
Since
  2  t  0   1 
F (t )   e   t   
 1 
5 7
then
 a3  t  a2   a1 
X p   e   t   
 b2   b1 
 b3 
Ch10_69
Fundamental Matrix
If X1, X2,…, Xn is a fundamental set of solutions of
X = AX on I, its general solution is the linear
combination X = c1X1 + c2X2 +…+ cnXn, or
 x11 
 x12 
 x1n 
 
 
 
 x21 
 x22 
 x2 n 
X  c1   c2      cn  



 
 
 
 xn1 
 xn 2 
 xnn 
 c1x11  c2 x12    cn x1n 


 c1x21  c2 x22    cn x2 n 





 c1xn1  c2 xn 2    cn xnn 
(1)
Ch10_70
(1) can be also written as
X = Φ(t)C
(2)
where C is the n  1 vector of arbitrary constants c1,
 x11 x12  x1n 
c2,…, cn, and


 x21 x22  x2 n 
Φ(t )  

 


 xn1 xn 2  xnn 
is called a fundamental matrix.
Two Properties of (t):
(i) nonsingular;
(ii) (t) = A(t)
(3)
Ch10_71
Variation of Parameters
In addition, we want to find a function
 u1 (t ) 


 u2 (t )  such that X = Φ(t)U(t)
U(t )  
p




 un (t ) 
is a particular solution of
X  AX  F(t )
Since
Xp  Φ(t )U(t )  Φ' (t )U(t )
then
Φ(t)U(t )  Φ' (t )U(t )  AΦ(t )U(t )  F(t )
(4)
(5)
(6)
(7)
Ch10_72
Using (t) = A(t), then
Φ(t )U(t )  AΦ(t )U(t )  AΦ(t )U(t )  F(t )
or
Φ(t )U(t )  F(t )
Thus
U(t )  Φ1 (t )F(t )
and so
U(t )    1 (t )F(t ) d t  C
(8)
Since Xp = Φ(t)U(t) by setting C=0, then
X p  (t )   1 (t )F(t ) dt
Finally, X = Xc + Xp
X  Φ(t )C  Φ(t )  Φ1 (t )F(t ) dt
(9)
(10)
Ch10_73
Example 4
Find the general solution of
1
 3
 3t 
X  
 X   t 
 2  4
e 
on (−, ).
Solution
First we solve the homogeneous system
 3 1 
X  
X
 2  4
The characteristic equation of the coefficient matrix is
3
1
det ( A  I ) 
 (  2)(  5)  0
2
4
Ch10_74
Example 4 (2)
We can get  = −2, −5, and the eigenvectors are
 1   1
 ,  
 1   2 
Thus the solutions are
 1  2 t  e  2 t 
 1  5 t  e  5 t 

X1    e   2t , X 2    e  
5 t 
 1
  2
e 
  2e 
Thus
 e 2t
Φ(t )   2t
e
5 t
e 
,
5 t 
 2e 
2 2t

e
1
3
Φ (t )   1 5t
 e
3
1
3
1

3
e 2t 

e5t 
Ch10_75
Example 4 (3)
Now
X p   (t )   1 (t )F (t ) dt
e 2t   3t 
   dt
t
5t 
e  e 
e 

5 t 
 2e 
2t
1 t
2t

e   2te  3 e 

dt
1 4t 
5t
5 t   
 2e   te  3 e 
1 2t
1 t
2t
5 t 
te

e

e 
e 
2
3


1 5t
1 4t 
5t
5 t   1
te

e

e 
 2e   5
25
12
e
  2t
e
e
  2t
e
 e 2t
  2t
e
5 t
 2 e 2t
3
  1 e5t
3
2t
1
3
1
3
5 t
 6 t  27  1 e t 
50
4

  53
21
1
 t   e t 
5

50
2
Ch10_76
Example 4 (4)
Hence
2t
e
X   2t
e
6
27
1 t

t


e 
c
e   1 5
50
4





3
21
1 t 
5 t 

c
t


e 
 2e   2   5
50
2
5 t
 1  2 t
 1  5 t
 c1   e  c2   e
 1
  2
 6   27   1  t
  4 e
  53  t   50
   21   1 
 5   50   2 
Ch10_77
Example 5
Solve
 3et 
 4 2
X  
 X   t 
 2 1
e 
Solution
From the same method, we have
 1
 2
1  0, 2  5, K1    , K 2   
  2
 1
Then
 1  2


 1 2
5
P
P 1   5
,
2
  2 1
2
5
5
Ch10_78
Example 5 (2)
Let X = PY,
 1  2
 1 et 

  3et  

5    5 
P 1F   5
 et 
2
2

    7 et 
5
5 
5
 1 et 


0 0
Y  
Y   5 
 0 5
 7 et 
5 
We have two DEs:
y1  1/5 et ,
y2  5 y2  7/5 et
Ch10_79
Example 5 (3)
The solutions are y1 = 1/5 et + c1 and y2 = –7/20 et + c2
e5t. Thus
t
1
2

1/5
e
 c1 



X  PY  
 
t
5t 

2
1

7/20
e

c
e



2
  1/2 et  c1  2c2e5t 

 
t
5t 
  3/4 e  2c1  c2e 
Ch10_80
10.5 Matrix Exponential
DEFINITION10.4
Matrix Exponential
For any n  n matrix A,
e At
2
k
k

t
t
t
 I  At  A 2    A k     A k
2!
k!
k!
k 0
(3)
Also A0 = I, A2 = AA, ….
Ch10_81
Derivative of eAt
d At
At
e  Ae
dt
(4)
Since
2
k

d At d 
t
t
2
k
e  I  At  A
  A
 
dt
dt 
2!
k!

1 32
2
 A  A t  A t 
2!
2


t
2
 A I  At  A
 
2!


 Ae At
Ch10_82
Besides,
d At
X  e C  Ae At C  A(e At C)  AX
dt
(5)
then eAt C is also the solution of X = AX.
Ch10_83
Computation of eAt
Ch10_84
Using the Laplace Transform

X = AX, X(0) = I
(7)
If x(s) = L{X(t)} = L{eAt}, then
sx(s) – X(0) = Ax(s) or (sI – A)x(s) = I
Now x(s) = (sI – A)–1I = (sI – A)–1. Thus
L{eAt} = (sI – A)–1
(8)
Ch10_85
Example 1
Use the Laplace Transform to compute eAt, where
 1  1
A

 2  2
Solution
1 
 s 1
sI  A  

  2 s  2
 s2
1

s

1
1


s ( s  1)
1

( sI  A)  
 
 2
  2 s  2
 s ( s  1)

1 

s ( s  1) 
s 1 
s ( s  1) 
Ch10_86
Example 1
(2)
1
1 
2  1
 


1
s s  1
( sI  A)   s s  1
1
2 
2  2

 s s 1 s s 1
e At
 2  e t
 
t
2

2
e

(9)
 1  e t 

t 
 1  2e 
Ch10_87
Using Powers Am

n 1
Ak   c j A j ,
j 0
n 1
k   c j  j
(10)
j 0
where the coefficient cj are the same in each and the
last expression is valid for the eigenvalues 1, 2, …,
n of A. Assume that the eigenvalues of A are distinct.
By setting  = 1, 2, …n in second expression in
(10), we found the cj in first expression by solving n
equations in n unknown. From (3) and (2), we have
e At

k
t
  Ak ,
k!
k 0

k
t
et   k
k!
k 0
Ch10_88
Replacing Ak and k as finite sums followed by an
interchange of the order of summations
e At
n 1
 k
n 1



t k  n1
t
    c j (k )A j    A j   c j (k )    A j b j (t ) (12)
k  0 k!  j  0
 j 0
 j  0  k  0 k!

k
n 1
n 1
 k
n 1




t
t
et     c j (k ) j     j   c j (k )     j b j (t ) (13)
k  0 k!  j  0
 j 0
 j  0  k  0 k!

Ch10_89
Example 2: Using Powers Am
Compute eAt, where
  2 4
A

  1 3
Solution
We already know 1= −1 and 2 = 2, then
eAt = b0I + b1A and et  b0  b1
Using the values of , then
e t  b0  b1
(14)
e 2t  b0  2b1
we have b0 = (1/3)[e2t + 2e– t], b1 = (1/3)[e2t – e–t].
Ch10_90
Example 2 (2)
Thus
e
At
  1/3 e 2t  4 /3 e t
 
2t
t

1/3
e

1/3
e

t
4 /3 e  4/3 e 

2t
t 
4 /3 e  1/3 e 
2t
(15)
Ch10_91