
˙x
1
(t)=f
i
(x
1
,...,x
n
),i=1, 2,...,p;
˙x
j
(t)=f
j
(x
1
,...,x
j
)+a
j+1
x
j+1
,j= p +1,...,n− 2;
............................................................;
˙x
n−2
(t)=f
n−2
(x
1
,...,x
n−2
) −
a
n−1
β
2,n−1
n−2
k=1
β
2k
x
k
−
a
n−1
β
2,n−1
ϕ
2
(x
1
,...,x
n−2
).
β
2,n−1
u
3
= a
n−1
ϕ
2
(x
1
,...,x
n−2
)
u
3
(x
1
,...,x
n−2
)
ψ
3
=
n−2
k=1
β
3k
x
k
+ ϕ
3
(x
1
,...,x
n−3
)=0.
ψ
4
=0 ψ
5
=0
ψ
l
=0
u
l
(x
1
,...,x
n−l+1
)=
a
n−l+1
β
l−1,n−l+2
ϕ
l−1
(x
1
,...,x
n−l+1
),l=1, 2,...,r;
u
l
=
1
β
l,n−l+1
p
i=1
β
li
+
∂ϕ
l
∂x
i
f
i
+ f
n−l+1
+
1
β
l,n−l+1
n−l
j=p+1
β
li
+
∂ϕ
l
∂x
i
×
× (f
j
+ a
j+1
x
j+1
) −
a
n−l+2
β
l−1,n−l+2
n−l+1
k=1
β
l−1,k
x
k
+
1
T
l
β
l,n−l+1
ψ
l
.
ψ
l
=
n−l+1
k=1
β
lk
x
k
+ ϕ
l
(x
1
,...,x
n−l
)=0
˙x
i
(t)=f
i
(x
1
,...,x
n
),i=1, 2,...,p;
˙x
j
(t)=f
j
(x
1
,...,x
j
)+a
j+1
x
j+1
,j= p +1,p+2,...,n− l +1;
..................................................................;
˙x
n−l
(t)=f
n−l
(x
1
,...,x
n−l
) −
a
n−l+1
β
l,n−l+2
n−l
k=1
β
lk
x
k
−
−
a
n−l+1
β
l,n−l+2
ϕ
l
(x
1
,...,x
n−l
),l=1, 2,...,r, r n − 1.
u
1
(x
1
,...,x
n
)
ψ
l
=0 l =2,...,r
u
l
(x
1
,...,x
n−l+1
)