Download - A4 Bai 6 - Nguyen Quoc Lan
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BO MON TOAN NG DUNG - HBK-------------------------------------------------------------------------------------
TOAN 4
CHUOI VA PHNG TRNH VI PHAN BAI 6: HE PHNG TRNH VI PHAN CAP 1
TS. NGUYEN QUOC LAN (5/2006)
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NOI DUNG---------------------------------------------------------------------------------------------------------------------------
1 HE PHNG TRNH VI PHAN
2 PHNG PHAP KH
3 HE PHNG TRNH VI PHAN CAP 1 TUYEN TNH
THUAN NHAT HE SO HANG: PHNG PHAP TR
RIENG (CHEO HOA MA TRAN)
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KHAI NIEM (SGK, TRANG 165)---------------------------------------------------------------------------------------------------------------------------
He m phtrnh vi phan (cap n) vi m ham an: Minh
hoa m = 2
: dang chuan hoa( )( )
=
=
0',',,,0',',,,
yxyxtGyxyxtF ( )
( )
=
=
yxtgyyxtfx
,,'
,,'
VD: He cap 1
++=
+=teyxty
tyxtx
310)('sin3)('
VD: He cap 2
+=
+=
txtytytx
sin10)(''cos)(''
Van e: Bien oi he cap 1 tren ve 1 ptrnh vi phan va giai?
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PHNG PHAP KH (SGK, TRANG 166)-------------------------------------------------------------------------------------------------------------------------------------
a he n phng trnh vi phan cap 1 ve 1 phng trnh vi
phan cap n: ao ham len, lan lt kh (n 1) an khac
VD: Giai( )
( )
++=
+=
2310)('123)('
teyxty
yxtx
( ) ( )( ) ( ) )()(0
,,
31023
tbtAXdtdX
etb
tytx
tXAt
+=
=
=
=
Chu y: He phng trnh tuyen tnh Cach viet dang ma tran
He 2 phng trnh cap 1 : Xem nh tng ng 1 phng
trnh cap 2 Nghiem cha ung 2 hang so C1, C2
( ) ( ) ( ) tetxtxtx 211'6'' =
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HE PHNG TRNH TUYEN TNH (SGK, TRANG 170)--------------------------------------------------------------------------------------------------------------------------------
He n ham an, n phng trnh vi phan cap 1 tuyen tnh:
( )Etbxtaxtaxtatx
tbxtaxtaxtatxtbxtaxtaxtatx
nnnnnnn
nn
nn
++++=
++++=
++++=
)()()()()('............................................................................
)()()()()(')()()()()('
2211
222221212
112121111
L
L
L
Ma tran: )()( tbXtAdtdX
+= : he pt ttnh khong thuan nhat
=
)()()(.........................................
)()()()()()(
21
22221
11211
tatata
tatata
tatata
A
nnnn
n
n
L
L
L
=
)(.......
)()(
2
1
tx
tx
tx
X
n
=
)(.......
)()(
)( 21
tb
tbtb
tb
n
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HE PTVP TTNH THUAN NHAT (SGK, TRANG 170)--------------------------------------------------------------------------------------------------------------------------------------
He n ham an x1(t), x2(t) xn(t) & n phng trnh vi phan cap
1 tuyen tnh thuan nhat (khong co ve phai)
( ) ( ) ( )tXtAdtdXE
xtaxtaxtatx
xtaxtaxtatx
xtaxtaxtatx
nnnnnn
nn
nn
=
+++=
+++=
+++=
0
2211
22221212
12121111
)()()()('.................................................................
)()()()(')()()()('
L
L
L
Cau truc nghiem he thuan nhat: Nghiem tong quat cua he
thuan nhat: Xtq.tn(t) = c1X1(t) + c2X2(t) + + cnXn(t), ck vi he nghiem X1(t), X2(t) Xn(t) la he nghiem c s (tc n vect
{X1(t), X2(t) Xn(t)} oc lap tuyen tnh t (a, b))
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TTNH THUAN NHAT HE SO HANG (SGK, TRANG 173)-------------------------------------------------------------------------------------------------------------------------------
He p/trnh vi phan cap 1 tuyen tnh thuan nhat he so hang
( ) ( )02211
22221212
12121111
)('..................................................
)(')('
EtAXdtdX
xaxaxatx
xaxaxatx
xaxaxatx
nnnnnn
nn
nn
=
+++=
+++=
+++=
L
L
L
( )( )
=
=
2
1
2
1
2
1
1232
c
c
c
c
ec
ec
tytx
t
t
Vect v = [c1, c2]T: vect rieng ma tran A ng tr rieng !
Ma tran A (cap 2): 2 gia tr rieng thc 1, 2 & 2 vect rieng oc lap tuyen tnh: v1, v2 2211tq.tn 21 vecvecX
tt +=
VD: Giai
+=
+=
yxtyyxtx
2)('32)('
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NHAC LAI: TR RIENG, VECT RIENG-------------------------------------------------------------------------------------------------------------
Tr rieng: det(A I) = 0. Vect rieng v: (A I)v = 0
VD:
=
1232
A ( )
=
=
=
=
41
012
32det
2
1
IA
VTR v1 = [, ]T ng 1 = 1: Av1 = 1v1 (A 1I)v1 = 0
( )
==+=
=+
11
002233
0 11 vvIA
2 tr rieng thc, phan biet 2 VTR LTT Cheo hoa1
2131
4001
2131
1232
=
=A
Vect rieng v2 = [, ]T ng vi 2 = 4: v2 = [3, 2]T
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KET QUA TONG QUAT--------------------------------------------------------------------------------------------------------------------------------------
nh Ly: He X = AX(t), ma tran A n gia tr rieng thc 1, 2 n (khong bat buoc phan biet), tng ng n vect rieng v1, v2 vn oc lap tuyen tnh Nghiem tong quat thuan nhat:
( ) ( ) ( ) ( )[ ] =
==
n
kk
tk
Tn vectxtxtxtX k
121 ,,,
K
VD: Giai he
+=
+=
yxyyxx
2'32'
=
1232
A : 2 GTP thc, VTR LTT
( )( )
+=
+=
+
=
=
tt
tttt
ececty
ecectxecec
yx
tX4
21
4214
21 23
23
11)(
[ ] [ ]TT vv 23,4;11,1 2211 ==== Tr rieng, vect rieng:
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TONG QUAT: PHNG PHAP CHEO HOA MA TRAN--------------------------------------------------------------------------------------------------------------------------------
Ma tran A cua he c cheo hoa bi ma tran P: A = PDP-1
X(t) = AX(t) = (PDP-1)X(t) P-1X(t) = D.P-1X(t). oi bien
( )= tXPY 1 = )()(' tDYtY( )( )
( )
( )( )
( )
=
ty
tyty
ty
tyty
nnn
........
00.....................
0000
'
........
'
'
2
1
2
1
2
1
K
K
K
=
=
=
)()('........................
)()(')()('
222
111
tyty
tytytyty
nnn
({v1, vn}: vect rieng)
( )( )
( )
=
=
=
tnn
t
t
necty
ecty
ecty
....................
2
1
22
11
=
==n
kk
tk vecPYX k
1
Phai tnh cac vect rieng v1, vn; Khong tnh ma tran P1
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GIA TR RIENG PHC (THAM KHAO)--------------------------------------------------------------------------------------------------------------------------------
Cap gia tr rieng phc, lien hp = i tng ng cap vect rieng v = a ib (a, b: vect) 2 vect nghiem c s
[ ] [ ]tbtaetbtae tt cossin,sincos +
VD: Giai he
+=
=
yxyyxx25'
6'
=
2516
A
i= 42,1
+=
=i
vi
v21
,
21
21
=
=
10
,
21
ba
( )( )
+
+
=
tteCtteC
tytx tt cos
10
sin21
sin1
0cos
21 4
24
1
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HE KHONG THUAN NHAT (THAM KHAO) ------------------------------------------------------------------------------------------------------------------------------------------
Ma tran A cua X = AX + b(t) cheo hoa bi ma tran P: A =
PDP-1. He ban au X(t) = (PDP-1)X + b P-1X(t) = D.P-1X(t)
+ P-1b. oi bien:
( )( )
( )
+=
+=
+=
tbtyty
tbtytytbtyty
nnnn
~)()('..................................
~)()('
~)()('2222
1111
Phai tnh ma tran P
= [v1, vn] va P1:
e tnh vect P1b
( ) ( ) :tPYtX =
= XPY 1 ( ) += bDYtY ~'
+
=
nnnn b
bb
y
yy
y
yy
~
...
~
~
....
00.....................
0000
'
.....
'
'
2
1
2
1
2
1
2
1
K
K
K
Tnh tay: cong kenh! n gian hn nhieu: Phng phap kh!
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V DU GIAI HE KHONG THUAN NHAT (THAM KHAO)-------------------------------------------------------------------------------------------------------------------------------------
Giai he khong thuan nhat
++=
++=
tyxyeyxx t
sin52'532'
=
1232
A cheo hoa (2 VTR oc lap tuyen tnh) viMa tran he:
[ ] [ ]TT vv 23,4;11,1 2211 ==== Tr rieng, vect rieng:
11
4001
1232
51
51
53
52
2131
=
=
= PPPP
D43421
: Cheo hoa
ma tran
Y = P1X
He mi:
~
1~
'
sinsin32
sin55 bDYY
te
tebPb
t
ebt
tt
+=
+
==
=
-
V DU GIAI HE KHONG THUAN NHAT (TIEP THEO)-------------------------------------------------------------------------------------------------------------------------------------
Giai he khong thuan nhat
++=
++=
tyxyeyxx t
sin52'532'
He mi:
+
+
=
+=
=
te
te
v
u
v
ubDYY
v
uY
t
t
sinsin32
4001
'
'
'.
~
He
++=
+=
tevv
teuut
t
sin4'sin32' ( ) ( )
( ) ( )
+=
++=
tteeCtv
tteeCtu
tt
tt
sin4cos171
31
sincos23
42
1
Quay ve bien X: Y = P1X
( )( )
( )( )
+
++
==
=
17sin4cos32sincos3
2131
42
1
tteeCtteeC
PYtytx
Xtt
tt