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S.l.dr.ing.mat. Alina BogoiDifferential Equations
POLITEHNICA University of Bucharest
Faculty of Aerospace Engineering
CHAPTER 5
Systems of First Order
Linear Equations
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Diff_Eq_7_2011 2
Outline Eigenvalues and Eigenvectors
Homogeneous Systems of Equationswith Constant Coefficients
Nonhomogeneous Systems ofEquations: Method of Variation of
Parameters
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Diff_Eq_7_2011 3
Eigenvalues and Eigenvectors
DefinitionLetAbe an n
n matrix. A scalar
is called an eigenvalue
ofA
if there exists a nonzero vector x
in Rn such that
Ax =
x.The vector x
is called an eigenvector
corresponding to .
Figure 6.1
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Eigenvalue and Eigenvector:
Ch6: Eigenvalues and Eigenvectors
:Example
The number is said to be an eigenvalue of the nxn matrix Aprovided there exists a nonzero vector v such that
v is called an eigenvector of the matrix A. v is associated with theeigenvalue
vAv =
=
=
1
2
v,22
65
:consider 1A
= 2
3
v2
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Diff_Eq_7_2011 5
Characteristic Equation:
Eigenvalues and Eigenvectors
( )P A I =
:Example
It is a polynomial of order n. ( A is nxn)
= 42
75
)Aa
=
02
80
)Ab
=
20
32)Ac
=
51516
264
003
)Ad
Find the characteristic ploynomial
=
322
102
124
)Ae
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Diff_Eq_7_2011 6
Ch6: Eigenvalues and Eigenvectors
:Example
Eigenvalues: Eigenvalues of A are the roots of the characteristic equation
=
42
75A
=
02
80A
=
20
32A
=
51516
264
003
A
Find all eigenvalues
Eigenvector:
:Example
=42
75A
=02
80A
=
20
32A
=
51516
264
003
A
Find all eigenvectors
6)( 2 = P 16)( 2 +=P 2)2()( =P )1)(3()( = P
:theosolution tzero-nonaisitif
eigenvaluewith theassociatedeigevectoranis
v [ ]0IA
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Diff_Eq_7_2011 7
Ch6: Eigenvalues and Eigenvectors
:Example
=
322
102
124
A
Find the charact. Poly.
Remark:
Find all eigenvalues
For higher-degree polynomial is not easy to factor. DONOT
expand but try to find a common factor
Remarks:Aofeigenvalueanis0= 0=A
singular
A
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Diff_Eq_7_2011 8
Eigenvalues and Eigenvectors
Eigenspace:
[ ]0-AsystemtheofspacesolutionThe =
Let be an eigenvalue of the matrix A .W = eigenspace of A associated with
= { all eigenvector of A associated with } U { zero vector}
=
322
102124
A
:Example 2=
[ ][ ]201
011
2
1
==
v
v
3=
[ ]1113=v
},{ 212 vvspanW = }{ 33 vspanW =2)dim( 2 =W
1)dim( 3 =W
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Diff_Eq_7_2011 9
Example 1Find the eigenvalues and eigenvectors of the matrix
=
53
64A
Let us first derive the characteristic polynomial of A.We getSolution
=
=
53
64
10
01
53
642IA
218)5)(4( 22 =+= IA
We now solve the characteristic equation ofA.
1or20)1)(2(022 ==+=
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Diff_Eq_7_2011 10
= 20=
2
1
33
66
x
x
This leads to the system of equations
Thus the eigenvectors ofA corresponding to
= 2 are
nonzero vectors of the form
033
066
21
21
=+
=
xx
xx
1
1
r
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Diff_Eq_7_2011 11
= 10=
2
1
63
63
x
x
This leads to the system of equations
Thus the eigenvectors ofA corresponding to
= 1
are
nonzero vectors of the form
063
063
21
21
=+
=
xx
xx
1
2s
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Diff_Eq_7_2011 12
Example 2Find the eigenvalues and eigenvectors of the matrix
=
222
254
245
A
The matrixA
I3
is obtained by subtracting
from
the diagonal elements ofA.Thus
Solution
=
222
254245
3IA
2
2
)1)(10()1)(10)(1(
]1011)[1(]8)2)(9)[(1(
242
294
001
==+==
=
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Diff_Eq_7_2011 13
= 10
We get
0
0x
=
=
3
2
1
3
822254
245
)10(
xx
x
I
1
2
2
r
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Diff_Eq_7_2011 14
= 1
0
0x
=
=
3
2
1
3
122
244
244
)1(
x
x
x
IA
The solution to this system of equations can be shown to be
x1
=
s
t,x2
= s, andx3
= 2t, where s and t are scalars.
Thus the eigenspace of = 1 is the space of vectors of theform.
ts
ts
2
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Diff_Eq_7_2011 15
Separating the parameters s and t, we can write
Thus the eigenspace of
= 1 is a two-dimensional subspace of
R2
with basis
+
=
2
0
1
0
1
1
2
ts
t
s
ts
0
0
1
,
0
1
1
.
Thus =10 has multiplicity 1, while =1 has multiplicity 2
in this example.
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Diff_Eq_7_2011 16
Theorem
LetAbe an n
n symmetric matrix.
(a) All the eigenvalues ofA are real numbers.(b)
The dimensional of an eigenspace ofA is the multiplicity of
the eigenvalues as a root of the characteristic equation.
(c) The eigenspaces ofA are orthogonal.(d)
A has n linearly independent eigenvectors.
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Diff_Eq_7_2011 17
Matrix Functions
The elements of a matrix can be functions of a real
variable. In this case, we write
Such a matrix is continuous at a point, or on an
interval
(a, b), if each element is continuous there. Similarly
with differentiation and integration:
=
=
)()()(
)()()(
)()()(
)(,
)(
)(
)(
)(
21
22221
11211
2
1
tatata
tatata
tatata
t
tx
tx
tx
t
mnmm
n
n
m L
MOMM
L
L
M Ax
=
= b
a ij
b
a
ij
dttadttdt
da
dt
d)()(, A
A
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Diff_Eq_7_2011 19
Introduction to Systems of First OrderLinear Equations
A system of simultaneous first order ordinary
differential equations has the general form
eachxk =xk( t).
),,,(
),,,(
),,,(
21
2122
2111
nnn
n
n
xxxtFx
xxxtFx
xxxtFx
K
MK
K
=
=
=
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Solutions of First Order Systems
A system of first order ordinary differential equations
It has a solution on I: < t < if there exists n functions
that are differentiable on I and satisfy the system ofequations at all points t in I.
Initial conditions may also be prescribed to give an IVP:
).,,,(
),,,(
21
2111
nnn
n
xxxtFx
xxxtFx
K
M
K
=
=
)(,),(),( 2211 txtxtx nn === K
00
0202
0101 )(,,)(,)( nn xtxxtxxtx === K
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Diff_Eq_7_2011 21
Theorem
Suppose F1,, Fn and F1/x1,, F1/xn,, Fn/ x1,, Fn/xn, are continuous in the region R definedby < t < ,
1
< x1
< 1
, , n
< xn
< n
, and let the
point
be contained in R. Then in some interval (t0 - h, t0 +h) there exists a unique solution
that satisfies the IVP.
00
2
0
10 ,,,, nxxxt K
)(,),(),( 2211 txtxtx nn === K
),,,(
),,,(
),,,(
21
2122
2111
nnn
n
n
xxxtFx
xxxtFx
xxxtFx
K
M
K
K
=
=
=
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Diff_Eq_7_2011 22
Linear Systems
If each Fk is a linear function ofx1,x2,
,xn, then the system of equations hasthe general form
Otherwise it is nonlinear.
)()()()(
)()()()()()()()(
2211
222221212
112121111
tgxtpxtpxtpx
tgxtpxtpxtpxtgxtpxtpxtpx
nnnnnnn
nn
nn
++++=
++++=++++=
KM
KK
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Diff_Eq_7_2011 23
General Solution The general solution of x' = P(t)x + g(t) on I: < t <
has the form
where
is the general solution of the homogeneous system
x' = P(t)x
and v(t) is a particular solution of thenonhomogeneous system x' = P(t)x + g(t).
)()()()( )()2(2)1(
1 ttctctc n
n vxxxx ++++= K
)()()( )()2(2
)1(
1
tctctc n
n
xxx +++ K
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Nth Order ODEs reduces to a Linear 1st
Order Systems An arbitrary nth order equation
Is transformed into a system of n first order
equations, by defining
)1()( ,,,,, = nn yyyytFy K
)1(
321 ,,,, ==== nn yxyxyxyx K
1 2
2 3
1 2( , , , )n n
x x
x x
x F t x x x
==
=M
K
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Practical Importance:
High-order
System Converted
First-order
System
Example: tyyyy sin5'2''3''' =++
'''
3
2
1
yxyx
yx
==
=
'''''''
''
3
2
1
yxyx
yx
==
=
txxxxxx
xx
sin523''
'
1233
32
21
++==
=First-order System
Linear Systems of Differential Equations
Li S t f Diff ti l E ti
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Diff_Eq_7_2011 26
Example: System of DE
)3sin(4022''
26''
tyxy
yxx
+=
+=
Independent variable: tdependent variables: x, y
2nd order
=
LL
LL
)(
)(
ty
tx
Solutions
Example:
yxy
yxx
22'
26'
=
+=
First-order
ytxy
ytxx
222'''
26''
=
+=
3rd-order zxz
zyxy
zyxx
=+=
+=
'
'
2'
First-order
Order of the system
Linear Systems of Differential Equations
Li S t f Diff ti l E ti
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Linear Systems of Differential Equations
Example:
'
'
4
3
2
1
yx
yx
xx
xx
=
=
=
=
Transform into first-order system
yxy
xyx
)1(''
)1(''
=
=
314
43
132
21
)1('
'
)1('
'
xxx
xx
xxx
xx
=
=
=
=
Example:
'
'
4
3
2
1
yx
yx
xxxx
=
=
==
Transform into first-order system
tyxy
yxx
+=
+=
''
''
txxx
xx
xxx
xx
+=
=
+==
314
43
312
21
'
'
'
'
+
=
tx
x
x
x
x
x
x
x
0
0
0
0101
1000
0101
0010
'
'
'
'
4
3
2
1
4
3
2
1
Li S t f Diff ti l E ti
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Diff_Eq_7_2011 28
Example:
yxyyxx
22'26'
=+=
3
2
22'
26'
yxy
yxx
=+=
zxz
zyxy
zyxx
=
+=
+=
'
'
2'
Linear system
yzxzzyxy
zxyxx
=+=
+=
''
2'
zexz
zyxy
zytxx
t=+=
+=
''
2' 2
Linear Systems of Differential Equations
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Linear Systems of Differential Equations
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Linear Systems of Differential Equations
1 11 1 12 2 1 1
2 21 1 22 2 2 2
1 1 2 2
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
n n
n n
n n n nn n n
x t a t x a t x a t x f t
x t a t x a t x a t x f t
x t a t x a t x a t x f t
= + + + +
= + + + +
= + + + +
KK
KK
M
KK
Matrix Form:
1 11 12 1 1 1
1 21 22 2 1 2
1 1 2 1 1
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
n
n
n n nn
x a t a t a t x f t
x a t a t a t x f t
x a t a t a t x f t
= +
K
K
M M M M M M
L
System of linear first-order DE
'X A X F= +
Linear Systems of Differential Equations
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Linear Systems of Differential Equations
'X A X F= +
If F=0 homogeneous system
If F 0 non-homogeneous system
Therorem ( Existence of a Unique Solution)
0
all entries of ( ) are cont on
all entries of ( ) are cont on
t I
t I
F t I
0 0
' ( ) ( ) (*)
( )
X A t X F t
X t X
= +
=
There exists a uniquesolution of IVP(*)
Linear Systems of Differential Equations
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Example:
yxytyxx
22'26'
=++=
zxz
zyxy
zyxx
=
+=
+=
'
'
2'
Homog and Non-homg
ttezexz
zyxy
tzytxx
2
22
'
'
2'
+=
+=
+=
+
=
022
26
'
' t
y
x
y
x
=
zy
x
zy
x
101111
121
''
'
+
=
tt e
t
zy
x
e
t
zy
x
2
22
001
111
121
''
'
homognon
'
+= FAXX
homog
' AXX =
homognon
'
+= FAXX
Linear Systems of Differential Equations
System of Equations: First-Order Linear
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System of Equations: First-Order Linear
Differential Equations - Substitution
Consider the 2x2 system of linear homogeneous differential
equations (with constant coefficients)
x'(t) = ax(t) + by(t)
y'(t) = cx(t) + dy(t)
We can solve this system using what we know:
1. Isolatey(t) in the first equation =>y(t) =x'(t)/b
ax(t)/b.
2. Differentiate thisy(t) equation =>y'(t) =x"(t)/b
ax'(t)/b.
3. Solve forx(t) andy'(t)
=>x"(t)
(a + d)x'(t) + (ad
bc)x(t) = 0.
4. Go back to step 1. Solve fory(t) in terms ofx'(t) andx(t).
System of Equations: First-Order Linear
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System of Equations: First-Order Linear
Differential Equations - SubstitutionExample:
x'(t) = 2x(t) +y(t)
y'(t) = 4x(t)
3y(t).
Solution:
x"(t) +x'(t) 2x(t) = 0.
x(t) =Aet +Be2t.
y(t) = Aet
4Be2t.
Linear Systems of Differential Equations
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Therorem ( Principle of Superposition)
solutionsnareLet 21 n,X,, XX L
XX ='Consider the sys of DE: (*)
1 1 2 2 also solutionn nc X c X c X + + +L
Example:
=
y
x
y
x
13
24
'
'
=
=
t
t
t
t
e
etX
e
etX
5
5
22
2
1
2)(,
3)(
are both
solutions of (*)
(*)
+
=
t
t
t
t
e
ec
e
ectX
5
5
22
2
23
2
3)( solution of (*)
DEF ( Wronskian)
solutionsnareLet 21 n,X,, XX L
XX ='Consider the sys of DE: (*)
nnn
n
n
xx
xx
XXXW
L
MOM
L
L
1
111
21
),,,( =
their wronskian is the nxn determinant
Example:
=
y
x
y
x
13
24
'
'
=
=
t
t
t
t
e
etXe
etX5
5
22
2
12)(,
3)(
Find W(X1,X2)
(*)
Linear Systems of Differential Equations
Linear Systems of Differential Equations
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Example: Consider the first-order linearsystem of DE
=
y
x
y
x
13
24
'
'
=
=
t
t
t
t
e
etX
e
etX
5
5
22
2
1
2)(,
3)(
Verify that the vector functions
are both solutions of (*)
(*)
Linear Systems of Differential Equations
First-order Systems
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First order Systems
THM ( Wronskian)
solutionsnareLet21 n
,X,, XX L
XX ='Consider the sys of DE: (*)
1 2( , , , ) 0nW X X X L
Example:
=
y
x
y
x
13
24
'
'
=
=
t
t
t
t
e
etX
e
etX
5
5
22
2
1
2)(,
3)(
Linearly dependent or independent ??
(*)
tindependenlinearly
21 n,X,, XX L
First-order Systems
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Diff_Eq_7_2011 38
THM ( general solution for Homog)
indeplinare21 n,X,, XX L
AXX ='Consider the sys of DE: (*)
Example:
=
y
x
y
x
13
24
'
'
=
=
t
t
t
t
e
etX
e
etX
5
5
22
2
1
2)(,
3)(
Find the general solution for (*)
(*)
nnXcXcXctX +++= L2211)(solutionsare
21 n
,X,, XX LThe general sol for (*) is
Example:
=
y
x
y
x
13
24
'
'
Solve IVP
(*)
=
1
1
)0(
)0(
y
x
First order Systems
First-order Systems
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Diff_Eq_7_2011 39
THM ( general solution for non-Homog)
indeplinare21 n,X,, XX L
FAXX +='Consider the sys of DE: (*)
Example:
++
=
t
t
y
x
y
x
3
14
13
24
'
'
=
=
t
t
t
t
e
etX
e
etX
5
5
22
2
1
2)(,
3)(
Find the general solution for (*)
(*)
nnc XcXcXctXwhere +++= L2211)(
(**)solutionsare21 n,X,, XX L The general sol for (*) is
Example: Solve IVP
=
1
1
)0(
)0(
y
x
AXX =' (**)
(*)forsolparticularpX)()()( tXtXtX pc +=
Sol for Homog
=
0)(
ttXP Particular sol for non-Homog
++
=
t
t
y
x
y
x
3
14
13
24
'
'
y
How to solve the system of ODE
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How to solve the system of ODE
System of Linear First-Order DE
(constant Coeff) 'X AX=
Distinct realEigenvalues
repeated realEigenvalues
complexEigenvalues
System of Linear First-Order DE
(Non-homog)'X AX F= +
Variation of
ParametersEigenv
alueM
ethod
Eigenvalue Method
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Diff_Eq_7_2011 41
g
Our Goal:
Example: Solve:
'X AX=Solve the Homog linear system
=
y
x
y
x
22
26
'
'
=
z
y
x
z
y
x
101
111
121
'
'
'
Method:Find all eigenvalues of the matrix A1 n ,,, 21 L
Distinct real
Eigenvalues
repeated real
Eigenvalues
complex
Eigenvalues
Solution: 1) Find n linearly independent solutions
2) The general solution is: their linear combination
nXXX ,,, 21 L
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Eigenvalue Method
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Diff_Eq_7_2011 43
1 ,123 11 1 veX
t=22
2 veX t= n
t
n veX n=
4 nnXcXcXctX +++= L2211)(
n,2 LLL
,1v ,2v nvLLL
LLL
:Example
=
4275A
2
7,
1
1
3,2 ==
Solve: 'X AX=
=1
1)( 21t
etX
=2
7)( 32t
etX
2211)( XcXctX +=The general solution:
+
= t
t
t
t
e
e
ce
e
ctX 3
3
22
2
12
7
)(
g
Eigenvalue Method
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Diff_Eq_7_2011 44
1 ,1
23 11 1 veX t= 22 2 veX t= ntn veX n=
4 nn
XcXcXctX +++= L2211
)(
n,2 LLL,1v ,2v nvLLL
LLL
:Example Solve: 'X AX=
=
3
1
0
)(1tX
=
5
2
0
)(2t
etX
332211)( XcXcXctX ++=The general solution:
+
+
=t
t
t
t
e
e
c
e
ecctX3
3
321
2
0
5
2
0
3
1
0
)(
=
51516
264
003
A
20
1
,5
2
0
,3
1
0
3,1,0 ===
=
2
0
1
)( 32t
etX
g
Eigenvalue Method
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Diff_Eq_7_2011 45
'X AX=
Method:Find all eigenvalues of the matrix A1
i =2,1
Complex conjugate eigenvalues
Find an eigenvector for21v
solution is:
31
1 veX t=
Two-lin independent solutions are:4
i +=1Complex vector
1)]sin()[cos( vtite t +=1
)(ve
ti+=
)real(1 XX = )Imag(2 XX =
Two-lin independent solutions are:52211)( XcXctX +=
Eigenvalue Method
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Diff_Eq_7_2011 46
:Example Solve: 'X AX=
)(1
tX )(2
tX
2211)( XcXctX +=
The general solution:
=
02
80A i4=
i4=
=
1
2iv
=
1
2)( 4
ietX
ti
+=
1
2)4sin4(cos
itit
+
+=
tit
tti
4sin4cos
4sin24cos2i
t
t
t
t
+
=
4sin
4cos2
4cos
4sin2
+
=
t
tc
t
tctX
4sin
4cos2
4cos
4sin2)( 21
Eigenvalue Method
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Diff_Eq_7_2011 47
'X AX=Method:
Find all eigenvalues of the matrix A1i =2,1
Complex conjugate
eigenvalues
Find an eigenvector for2iBBv 211 +=
Two-lin independent solutions are:
3
i +=1
Two-lin independent solutions are:42211)( XcXctX +=
1 1 2
2 2 1
[ cos sin ]
[ cos sin ]
t
t
X B t B t e
X B t B t e
=
= +
Eigenvalue Method
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Diff_Eq_7_2011 48
:Example Solve: 'X AX=
)(1
tX )(2
tX
=
02
80A i4=
i4=
=
1
2iv
=
1
2)( 4
ietX
ti
+=
1
2)4sin4(cos
itit
+
+=
tit
tti
4sin4cos
4sin24cos2i
t
t
t
t
+
=
4sin
4cos2
4cos
4sin2
i4=
=
1
2iv i
+
=
0
2
1
0
iBB 21+=
1 1 2
2 2 1
[ cos sin ]
[ cos sin ]
t
t
X B t B t e
X B t B t e
=
= +
ttX 4sin
0
24cos
1
01
= ttX 4sin
1
04cos
0
21
+
=
Multiple Eigenvalue Solution
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Diff_Eq_7_2011 49
, , , repeated real eigenvalues K 'X A X=
:Example 1, 1,5= Solve:
1 2 2
' 2 1 2
2 2 1
X X
=
3
1
1 for =5
1
K
=
Lin. indep eigenvectors
nvvv ,,, 21 L
One single eigenvector
v
n
t
n
tt veXveXveX === ,,, 2211 L
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Another method to compute: Generalized eigenvectors
Another method to compute: Generalized eigenvectors
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Diff_Eq_7_2011 52
, , , repeated real eigenvalues K
KKKKKKKKKKKKKK
[ ] 10 vIA [ ] 21 vvIA
[ ] 32 vvIA
[ ] kk vvIA 1
{ }kvvv ,,, 21 L
Chain of generalized
eigenvectors
Compute:)( IA 2)( IA LL kIA )(
k
k
vIA 0)(
Solve:
Compute:
kk vIAv )(1 =
LL
32 )( vIAv =
21 )( vIAv =
{ }kvvv ,,, 21 L
Chain of
generalized
eigenvectors
:Example Find all generalized eigenvectors:
1,11,=
12 )( = kk vIAv
=
010
122
001
A
Another method to compute: Generalized eigenvectors
Another method to compute: Generalized eigenvectors
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Diff_Eq_7_2011 53
Compute:)( IA 2)( IA LL
kIA )(
k
kvIA 0)(
Solve:
Compute:
kk vIAv )(1 =
LL
32 )( vIAv =
21 )( vIAv =
{ }kvvv ,,, 21 LChain of
generalized
eigenvectors
:Example Find all generalized eigenvectors:
1,11,=
12 )( = kk vIAv
=
010
122
001
A
:Solution[ ]
=
0110
0112
0000
0IA
0110
0001
000023 RR+
22
1R
1 lin indep
eigenvector
Length of
chain =3{ }321 ,, vvv
=
110
112
000
)( IA
=
002
002
000
)( 2IA
=
000
000
000
)( 3IA
=00
1
3v
=
=
02
0
00
1
110112
000
2v
=
=
22
0
02
0
110112
000
1v
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Diff_Eq_7_2011 54
Fundamental Matrices Suppose that x(1)(t),, x(n)(t) form a fundamental set of
solutions for x' = P(t)x on < t < .
The matrix
whose columns are x(1)(t),, x(n)(t), is a fundamental
matrix for the system x' = P(t)x. This matrix isnonsingular since its columns are linearly independent,
and hence det
0.
,
)()(
)()(
)()()1(
)(
1
)1(
1
=
txtx
txtx
tn
nn
n
L
MOM
L
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Diff_Eq_7_2011 55
Example 1: Consider the homogeneous equation x' = Ax below.
We found the following fundamental solutions for
this system:
Thus a fundamental matrix for this system is
xx
= 14
11
ttetet
=
=
2
1)(,
2
1)( )2(3)1( xx
=
tt
tt
ee
eet
22)(
3
3
Fundamental Matrices and General
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Diff_Eq_7_2011 56
Fundamental Matrices and General
Solution The general solution of x' = P(t)x
can be expressed x = (t)c, where c is a constantvector with components c1,, cn:
)()1(
1 )(
n
nctc xxx ++= L
==n
n
nn
n
c
c
txtx
txtx
t ML
MOM
L 1
)()1(
)(
1
)1(
1
)()(
)()(
)( c
x
Fundamental Matrix & Initial Value
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Diff_Eq_7_2011 57
Fundamental Matrix & Initial Value
Problem Consider an initial value problem
x' = P(t)x, x(t0
) = x0
where < t0 < and x0 is a given initial vector.
Now the solution has the form x = (t)c, hence wechoose c so as to satisfy x(t
0) = x0.
Recalling (t0) is nonsingular, it follows that
Thus our solution x = (t)c can be expressed as
0
0
10
0 )()( xcxc tt ==
0
0
1 )()( xx tt =
Nonhomogeneous Linear Systems
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Diff_Eq_7_2011 58
The general theory of a nonhomogeneous system of
equations
parallels that of a single nth order linear equation.
This system can be written as x' = P(t)x + g(t), where
)()()()(
)()()()(
)()()()(
2211
222221212
112121111
tgxtpxtpxtpx
tgxtpxtpxtpx
tgxtpxtpxtpx
nnnnnnn
nn
nn
++++=
++++=
++++=
K
M
K
K
=
=
=
)()()(
)()()(
)()()(
)(,
)(
)(
)(
)(,
)(
)(
)(
)(
21
22221
11211
2
1
2
1
tptptp
tptptp
tptptp
t
tg
tg
tg
t
tx
tx
tx
t
nnnn
n
n
nn L
MOMM
L
L
MM Pgx
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Variation of Parameters: Solution
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Diff_Eq_7_2011 60
We assume a particular solution of the form x =(t)u(t).
Substituting this into x' = P(t)x + g(t), we obtain
'(t)u(t) + (t)u'(t) = P(t)(t)u(t) + g(t) Since '(t) = P(t)(t), the above equation simplifies
to
u'(t) = -1(t)g(t)
Thus
where the vector c is an arbitrary constant of
integration.
cu += dttgtt )()()( 1
( ) arbitrary,,)()()()( 111
+= tdssstt
t
tgcx
Variation of Parameters: Initial Value
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Diff_Eq_7_2011 61
Problem For an initial value problem
x' = P(t)x + g(t),x(t0) = x
(0),
the general solution to x' = P(t)x + g(t)is
+=
t
tdsssttt
0
)()()()()( 1)0(01 gxx
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Diff_Eq_7_2011 62
Example :Variation of Parameters Consider
Fundamental matrix:
(t)u'(t) = g(t), or
2 1 2
1 2 3
te
t
= + x x
=
tt
tt
ee
eet3
3)(
=
t
e
u
u
ee
ee t
tt
tt
3
2
2
1
3
3
22 121
21 11
-1
-det
a aAa aA
=
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Diff_Eq_7_2011 63
Example 3: Solving for u(t) Solving (t)u'(t) = g(t) by row reduction,
It follows that
2 3
1
2
3 / 2
1 3 / 2
t t
t
u e te
u te
=
= +
++
++=
=
2
1
332
2
1
2/32/3
6/2/2/)(
cetet
cetee
u
ut
tt
ttt
u
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Diff_Eq_7_2011 64
Example 3: Solving for x(t) (3 of 3) Now x(t) = (t)u(t), and hence we multiply
to obtain, after collecting terms and simplifying,
Note that this is the same solution as in Example 1.
++++
=
2
1
332
3
3
2/32/36/2/2/
cetetcetee
eeee
tt
ttt
tt
tt
x
+
+
+
+
=
5
4
3
1
2
1
1
1
2
1
1
1
1
1
1
12
3
1 teteecec tttt
x
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Diff_Eq_7_2011 65
Example 3: Solving for x(t) (3 of 3)
t
egt
=
1 06 -1
A =
t
e
t3-3 1
2 -4
A =
2
4
4
4
t
t
eg
e
=
3 -1
-1 3A
=
1(0)
1x
=
Solve the nonhomogeneous system: X'=AX+g
(c)
(a)
(b)
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