an1k1 an2k2 (ann )kn 0
Thus to find a nontrival solution X of (2) we must
find a nontrival solution of the foregoing system.
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Find a nontrival vector K 0 that satisfies (3),
X c1K1e1t c2K2e2t cn Knent .
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EXAMPLE 1 Distinct Eigenvalues
Solve
dx 4x y z dt
dy x 5 y z
(6)
dt
dz dt
y 3z
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Solution we find
system
X ' AX (2)
Where A is an n n matrix of constants.
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Eigenvalues（特征值） and Eigenvectors（特征向量）
If X Ket is a solution vector of the linear system X ' AX , then we have
eigenvalues 1, 2,n , then a set of n linearly
independent eigenvectorsK1, K2,Kn, can always
be found and
X1
K1e1t ,
X2
K
e2t
2
,
,
Xn
Knent ,
is a fundamental set of solutions of (2) on (, ).
eigenvalue is called an eigenvector of A.
a solution of the homogeneous system (2) is then
X Ket
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In the discussion that follows, We examine three cases:
c1 11e2t
c2
3 5
e6t
Is the general solution of the homogeneous system
X ' 15 33X
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We guess that a solution of the form
k1
X
k2
et
Ket
(1)
kn
For the general homogeneous linear first-order
we must have
det(A I ) 0
This polynomial equation in is called the
characteristic equation of the matrix A; its solutions are the eigenvalues of A.
A solution K 0 of (3) corresponding to an
x1 x2 xn1 0
xnKn 0, Kn 0 xn 0
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THEOREM 8.7 General solution-homogeneous systems
Let 1, 2,n be n distinct real eigenvalues of
the coefficient matrix A of the homogeneous system (2), and let K1, K2,Kn be the corresponding eigenvectors. Then the general solution of (2) on the interval (, ) is given by
1, real and distinct eigenvalues 2, repeated eigenvalues
3, complex eigenvalues
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8.2.1 DISTINCT REAL EIGENVALUES
When the nxn matrix A possesses n distinct real
1 8
1 1
0 0
row
operations
1 0
0 1
1 0 0 0.
0100 Nhomakorabea0
0
0
0
Therefore K1 K3 and K2 0. The choiceK3 1
gives an eigenvector an corresponding solution
Ket AKet (et 0) AK K ( K IK ) (A I )K 0 (3)
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The matrix equation (3) is equivalent to the simultaneous algebraic equations
(a11 )k1 a12k2 a1nkn 0 a21k1 (a22 )k2 a2nkn 0
4 1
1
det(A I ) 1 5 1
0
1 3
( 3)( 4)( 5) 0,
and so the eigenvalues are
1 3, 2 4, and 3 5.
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For 1 3 Gauss-Jordan elimination gives
1 ( A 3I 0) 1
8 systems of linear first-order differential equations
8.2 homogeneous linear systems with constant coefficients
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We have known that
X
c1 X 1
c2 X 2
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若特征值1, 2,...n互不相同，
则特征向量K1,
K2
,
,
K
线性无关
n
x1K1 x2K2 xnKn 0 (1)
用A乘两端, x11K1 x22K2 xnnKn 0 (2)
(2) (1)n x1(1 n )K1 x2 (2 n )K2 xn1(n1 n )Kn1 0