Lecture Notes in Differential Equations - Bruce E. Shapiro

328 LESSON 31. LINEAR SYSTEMS
The corresponding eigenvectors are
( ) ( )
−1
1
v 1 = , v
1 2 =
1
The diagonalizing matrix is then
and its inverse is given by
so that
S = (v 1 v 2 ) =
S −1 = 1 2
D = S −1 PS = 1 2
( −1 1
1 1
( −1 1
1 1
)
)
( ) −t
2
0
0 t 2
is diagonal. Hence M = e P = Se D S −1 , which we calculate as follows.
( ) ( ) ( )
−1 1 e
M =
−t2 /2
0 1 −1 1
1 1 0 e t2 /2 2 1 1
( ) ( )
−1 1 −e −t2 /2
e −t2 /2
(31.209)
(31.210)
(31.211)
(31.212)
= 1 2
= 1 2
=
1 1 e t2 /2
e t2 /2
(
e −t2 /2 + e t2 /2
−e −t2 /2 + e t2 /2
−e −t2 /2 + e t2 /2
e −t2 /2 + e t2 /2
( )
cosh(t 2 /2) sinh(t 2 /2)
sinh(t 2 /2) cosh(t 2 /2)
)
(31.213)
and (using the identity cosh 2 x − sinh 2 x = 1),
( )
M −1 cosh(t
=
2 /2) − sinh(t 2 /2)
− sinh(t 2 /2) cosh(t 2 /2)
Furthermore,
( ) ( )
cosh(t
M(t)g(t) =
2 /2) sinh(t 2 /2) t
sinh(t 2 /2) cosh(t 2 /2) 3t
( )
t cosh(t
=
2 /2) + 3t sinh(t 2 /2)
t sinh(t 2 /2) + 3t cosh(t 2 /2)
(31.214)
(31.215)
(31.216)
Using the integral formulas
∫
t cosh(t 2 /2)dt = sinh(t 2 /2) (31.217)
∫
t sinh(t 2 /2)dt = cosh(t 2 /2) (31.218)