Lecture Notes in Differential Equations - Bruce E. Shapiro

312 LESSON 31. LINEAR SYSTEMS
( ) 1 1
The eigenvalues of A =
are λ = 2, −3, which we find by solving
4 −2
the the characteristic equation. Let x 1 and x 2 be the eigenvectors of A.
From the last theorem we know that
e At = ( ) ( ) ( )
e
x 1 x 2t 0 x1
2
0 e −3t (31.77)
x 2
where x 1 is the eigenvector with eigenvalue 2 and x 2 is the eigenvector with
eigenvalue -3.
( )
ui
Let x i = .
v i
Then since Ax i = λ i x i ,
( ) ( ) ( )
1 1 u1 u1
= 2
4 −2 v 1 v 1
( ) ( ) ( )
1 1 u2 u2
= −3
4 −2 v 2 v 2
=⇒ u 1 = v 1 (31.78)
=⇒ v 2 = −4u 2 (31.79)
Since one component of each eigenvector is free, we are free to choose the
following eigenvectors:
( ( 1 1
x 1 = and x
1)
2 =
(31.80)
−4)
Let
Then
Therefore
U = ( ( )
) 1 1
x 1 x 2 =
1 −4
U −1 = 1
det U [cof(U)]T = − 1 5
(
−4
)
−1
−1 1
e At = 1 ( ) ( ) ( )
1 1 e
2t
0 4 1
5 1 −4 0 e −3t 1 −1
= 1 ( ) ( )
1 1 4e
2t
e 2t
5 1 −4 e −3t −e −3t
= 1 ( )
4e 2t + e −3t e 2t − e −3t
5 4e 2t − 4e −3t e 2t + 4e −3t
= 1 [( ) ( ) ]
4 1
e 2t 1 −1
+
e −3t
5 4 1 −4 4
(31.81)
(31.82)
(31.83)
(31.84)
(31.85)
(31.86)
Hence the solution is