A4 Bai 5 - Nguyen Quoc Lan

7/21/2019 A4 Bai 5 - Nguyen Quoc Lan

1/16

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)


2/16

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


3/16

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 ++=


4/16

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


5/16

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

=

=


6/16

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 ?


7/16

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


8/16

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 +=


9/16

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


10/16

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


11/16

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


12/16

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


13/16

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


14/16

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


15/16

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


16/16

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

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A4 Bai 5 - Nguyen Quoc Lan