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BO MON TOAN NG DUNG - HBK-------------------------------------------------------------------------------------
TOAN 4CHUOI VA PHNG TRNH VI PHAN
BAI 5: PHNG TRNH VI PHAN CAP 2
TS. NGUYEN QUOC LAN (5/2006)
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NOI DUNG-----------------------------------------------------------------------------------------------------------------------------------
1 PHNG TRNH VI PHAN CAP 2. TRNG HPGIAM CAP
3 PHNG TRNH VI PHAN CAP 2 HE SO HAM2 PHNG TRNH VI PHAN CAP 2 HE SO HANG
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GIAM CAP PHNG TRNH VI PHAN CAP 2--------------------------------------------------------------------------------------------------------------------------------------
VD: Giai phng trnh vi phan cap 2: xxx
yy cos
''' +=
Phng trnh vi phan cap 2: F(x, y, y, y) = 0
BT Cosi: PT chuan hoa + K au ( )
( ) ( )
==
=
1000 ',
',,''
yxyyxy
yyxfy
Giam cap c ban: Phng trnh F(x, y, y) = 0
Nguyen tac: at u(x) = ao ham cap thap nhat cua an y
( ) ( ) ( ) ( ) ( ) ( ) 0',,:"''0'',', ==== uuxFxyxuxyxuyyxF
Nghiem tong quat PT vi phan cap 2 cha 2 hang so C1, C2
ap so: Nghiem xxxCxCy cossin2
2
1 ++=
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PHNG TRNH VI PHAN TUYEN TNH CAP 2--------------------------------------------------------------------------------------------------------------------------------------
Tuyen
tnh
(linear):y,y,y
bac 1
He so ham, k0 thuan nhat (ve phai): y + p(x)y + q(x)y = f(x)
V du:
He so hang, k0 thuan nhat (co ve phai): y + py + qy = f(x)
V du:
( )1sincossin''' 2 xxxyexyxy x +=+
( )3sincos4'3'' xxxyyy +=+
PT thuan nhat tng ng: y + p(x)y + q(x)y = 0
( )20sin''' 2 =+ yexyxy xV du: Tng ng (1):
PT thuan nhat tng ng: y + py + qy = 0
V du: Tng ng (3): ( )404'3'' =+ yyy
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GIAI PTVP TUYEN TNH C2 THUAN NHAT HE SO HANG----------------------------------------------------------------------------------------------------------------------------------------------
xkxk
tntq eCeCy 21
21. +=
xkxkee 21 ,scnghiem2PTVPC2 thuan
nhat he so hang
y + py + qy = 0
PTrnh ac trngk2 + pk + q = 0
> 0: k1 k2 R
< 0: N0 phc
= 0: k1 = k2 R
ik
im
=
=
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y y = 0
y 5y + 6y = 0
y 4y + 4y = 0
y 2y + 5y = 0
S O GIAI PTVP TUYEN TNH THUAN NHAT CAP n-----------------------------------------------------------------------------------------------------------------------------------------
PTVP t/tnh thuan nhat L[y] = 0 PT ac trng (ai so) an k
Tm u n ng. k1 knn ham c s y1 yn( )
==
n
i ii
xyCy1tq
k2 5k + 6 = 0: N0 2, 3 ytq= C1e3x + C2e2x
k2 4k + 4 = 0: 2 (kep) ytq= C1e2x + C2xe2x
k2
2k + 5 = 0 k1,2 = 1 2i: =1, = 2 Nghiem tong quat thuan nhat ytq.tn = ex(C1cos2x + C2sin2x)
k3 1 = 0 1 Nghiem k = 1 N0 CS ex, xex ?
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NGHIEM (HAM) C S TNG NG N0 PT AC TRNG------------------------------------------------------------------------------------------------------------------------------------------
K+= xktntq eCy 1
1.
xke 11 scnghiem
kxrkxkxexxeer
1,:NCS
K
K++= kxkx xeCeCy21
xexe xx sin,cosNCS2
( ) K++= xCxCey x sincos 21
Kxxexer xx cos,cos:NCS2
[ ]K++= xxCxCey x coscos 21
k1 R: Nghiem n
k R: boi cap r
i: phc
lien hp, n
i:
boi cap r 2r n0 n
PTT kn+p1kn-1
+ pn = 0
Tm n nghiem
thc phc.Nghiem boi cap
r r nghiem
n trung nhau
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PHNG TRNH TUYEN TNH KHONG THUAN NHAT--------------------------------------------------------------------------------------------------------------------------------
VD: Giai ptrnh y 3y + 2y = 2 bang cach ch ra 1
nghiem rieng yr ket hp vi nghiem tong quat thuan nhat
Nghiem rieng yr = 1 ytq= C1ex + C2e2x + 1
PTVP tuyen tnh khong thuan nhat cap n (he so tuy y):
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )Exfyxayxayxayxa nnnn =++++
'1
1
10 K
PTVP tuyen tnh thuan nhat cap n tng ng:
( ) ( ) ( ) ( ) ( ) ( ) ( )01
1
10 0' Eyxayxayxayxa nnnn =++++
K
Nghiem tong quat (E) = Tong quat (E0)+ Nghiem rieng (E)
nhatthuanngrieng.Khonhatnquat.Thuatongnhatthuanngquat.Khotong yyy +=
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TM NGHIEM RIENG VI VE PHAI AC BIET-----------------------------------------------------------------------------------------------------------------------
Ve phai: ex[Pn(x)cosx + Qm(x)sin x], Pn, Qm a thc
2/ Ve phai cha ex yr cha ex
3/ Ve phai cha lng giac yr cha 2 ham: sin x,cos x (du ve phai ch co 1 loai ham!)
1/ Ve phai cha a thc yr cha a thc (he so cha
xac nh) bac cao nhat. Hang so a thc bac 0
4/ + i (ve phai) nghiem boi cap r cua phng trnh
ac trng Nhan them xr vao yr can tm. Khong co
ham mu = 0; Khong co lng giac = 0
Tom lai: Ba cung Cung dang, cung bac, trung nghiem
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BA TRNG HP HAY GAP----------------------------------------------------------------------------------------------------------------------
y+py+qy=ex[Pn(x)cosx+Qm(x)sinx],NT: nghiem ac trng; H: a thc bac
( )mn,max=l
l
( )xP:phaiVe
Ng.rieng yr:
(*) khi 0
NT cap r.
( )
( ) ( )
*xHx
xH
r
VP: a thc
0=+ i
( )xPe x=K
Ng. rieng yr:
(*) khi
NT cap r
( )( ) ( )
*xHexxHe
xr
x
VP: mu
0=
Ve phai: Lng giac
( ) ( ) xxQxxP mn sincos +
Nghiem rieng yr co dang:
( ) ( )[ ] ( )
++
*sincossincosxHxRx
xxHxxRr
Bac R = Bac H. (*) khi
i NT boi cap r
ii =+=0
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NGUYEN LY CHONG CHAT (SGK, trang 150)--------------------------------------------------------------------------------------------------------------------------------
Nghiem tong quat ytqphng trnh vi phan tuyen tnh co ve
phai: y + p(x)y + q(x)y = f1(x) + f2(x) bieu dien qua:
Nghiem tong quat thuan nhat ytq.0: y + p(x)y + q(x)y = 0
Nghiem rieng yr.1 cua pt: y + p(x)y + q(x)y = f1(x)
Nghiem rieng yr.2 cua pt: y + p(x)y + q(x)y = f2(x)
Cong thc chong chat: ytq= ytq.0 + yr.1 + yr.2
Y ngha: Tach phng trnh co ve phai dang tong phc tap
thanh tong cac phng trnh co ve phai n gian
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VE PHAI TONG QUAT BIEN THIEN HANG SO-------------------------------------------------------------------------------------------------------------------
Ve phai: y + py + qy = f(x) Tm yr t ytq.tn: Bien
thien hang so C1 = C1(x), C2 = C2(x)
VD: y 3y + 2y = lnx
PTVP tuyen tnh k0 thuan nhat y + p(x)y + q(x)y = f(x)
& nghiem tong quat thuan nhat ytq.tn = C1y1(x) + C2y2(x).
( ) ( )
( ) ( ) ( ) ( ) ( ) K121
2211
2211','
''''
0''C
D
DxC
D
DxC
xfyxCyxC
yxCyxCyx ==
=+
=+
Tm nghiem rieng phng trnh khong thuan nhat: Xem
C1 = C1(x), C2 = C2(x) Ng. rieng yr = C1(x)y1 + C2(x)y2
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PTVP TUYEN TNH C2 HE SO HAM (THAM KHAO)--------------------------------------------------------------------------------------------------------------------------------
N0 c s th nh: y2(x) = C(x)y1(x)
( )
( )
2211.2
1
' yCyCyy
exC tntq
dxxp
+=
=
Nghiem tqy = C1y1 +
C2y2 + yr
PTVPC2 thuan nhat:
y + p(x)y+q(x)y = 0
Tm nghiem ac biet y1: oan
dang (x, a thc) hoac c gi y
PTVPC2TT tong
quat he so ham y +p(x)y + q(x)y = f(x)
Ng. rieng pt k0 tn:Bien thien hang so C1
= C1(x), C2 = C2(x)
( )
=+
=+
xfyCyC
yCyC
''''
0''
2211
2211
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PHNG TRNH EULER - COSI------------------------------------------------------------------------------------------------------------------
Thuan nhat ax2y + bxy + cy = 0 2 nghiem c s y = xm
PT he so ham: anxny(n) + an-1xn-1y(n-1) + a0y = f(x) De
tm nghiem c s thuan nhat hoac a ve he so hang
Dau hieu: He so xk
cua ao ham cap k y(k)
(0 k n)
2 nghiem thc phan biet m1 m2
Nghiem kep m
Phc: m1,2 = i
21
21
mm
tq xCxCy +=
xxCxCy mm
tq ln21 +=
( ) ( )[ ]xCxCxytq lnsinlncos
21
+=
( )0
2
=+
+
c
mabam
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PHNG PHAP OI BIEN---------------------------------------------------------------------------------------------------------------------
a/ x2yxy8y = 0 b/ 4x2y+y = 0 c/x2y3xy+13y = 0
PTrnh Euler: anxny(n) + an-1xn-1y(n-1) + a0y = f(x).oi bien x = et y(x) = y(t).t(x), y(x) =
VD: Giai phng trnh x2y 2xy + 2y = ln2x + ln(x2)
anxny(n) + + a
0y = 0
PTT theo m: g(m) = 0
n nghiem (thc, phc) n nghiem (ham) c s
m R: n NCS y =xm
m R: boi r xm,xmlnx m
xy=
( )
( )
=
=
xxy
xxyi
lnsin
lncos
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BAI TOAN BIEN--------------------------------------------------------------------------------------------------------------------------
Phan biet vi bai toan Cosi cap 2: ( )( ) ( )
==
=
21 ',
,',,"
ayay
axyyxfy
VD:( ) ( )
==
=+
Bbyy
bxyy
,00
0,0"
Bby =)( BbC = sin2
==
=
:0,0sin
:0,0sin
:0sin
Bb
Bb
kbb 1 nghiemvo nghiem
vo so nghiem
Bai toan bien cap 2, nghiem c s sin, cos Vo so nghiem
VD:( ) ( )
==
=
Bbyy
bxyy
,00
0,0"
( )
==
=
byay
bxayyxfy
,)(
),',,(''Bai toan bien: Tm y = y(x) thoa