
Review of Dynamic Systems 585
Therefore, the “forced” solution (including gravity) follows from Equation (A.53),
given by
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
x
1
(t)
x
2
(t)
x
3
(t)
x
4
(t)
x
5
(t)
x
6
(t)
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
100(t −t
0
) 00
010 0 (t −t
0
) 0
001 0 0 (t −t
0
)
000 1 0 0
000 0 1 0
000 0 0 1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
x
1
(t
0
)
x
2
(t
0
)
x
3
(t
0
)
x
4
(t
0
)
x
5
(t
0
)
x
6
(t
0
)
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
−
t
t
0
00(
τ
−t
0
) 001
T
gd
τ
or
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
x
1
(t)
x
2
(t)
x
3
(t)
x
4
(t)
x
5
(t)
x
6
(t)
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
x
1
(t
0
)+x
4
(t
0
)(t −t
0
)
x
2
(t
0
)+x
5
(t
0
)(t −t
0
)
x
3
(t
0
)+x
6
(t
0
)(t −t
0
) −1/2g(t −t
0
)
2
x
4
(t
0
)
x
5
(t
0
)
x
6
(t
0
) −g(t −t
0
)
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
which verify (again) the previous results and demonstrate the equivalence between
the preceding results of this section and conventional integration of the ...