6. Phuong trinh vi phan.pdf

7/29/2019 6. Phuong trinh vi phan.pdf

1/93

Mc lc

1 Cc khi nim c bn. Cch gii mt s phng trnh cp mt ngin 21.1 Bi ton m u. Cc khi nim c bn . . . . . . . . . . . . . . . 21.2 Mt s phng trnh vi phn cp mt gii c . . . . . . . . . . 4

1.2.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Phng trnh tch bin . . . . . . . . . . . . . . . . . . . . 51.2.3 Phng trnh ng cp . . . . . . . . . . . . . . . . . . . . 71.2.4 Phng trnh tuyn tnh cp 1 . . . . . . . . . . . . . . . . 101.2.5 Phng trnh Bernoulli . . . . . . . . . . . . . . . . . . . . 111.2.6 Phng trnh Ricati . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Phng trnh vi phn ton phn. Tha s tch phn . . . . . . . . . 131.3.1 Phng trnh vi phn ton phn . . . . . . . . . . . . . . . 13

1.3.2 Tha s tch phn . . . . . . . . . . . . . . . . . . . . . . 161.3.3 Cch tm tha s tch phn . . . . . . . . . . . . . . . . . 17

1.4 Mt s phng trnh vi phn cp mt cha gii i vi o hm . 191.4.1 Phng trnh F(x, y) = 0 hay F(y, y) = 0 . . . . . . . . 19

1.5 Phng trnh Lagrange v phng trnh Clairaut . . . . . . . . . . 221.5.1 Phng trnh Lagrange . . . . . . . . . . . . . . . . . . . . 221.5.2 Phng trnh Clairaut . . . . . . . . . . . . . . . . . . . . 24

1


2/93

Chng 1

Cc khi nim c bn. Cch gii mt sphng trnh cp mt n gin

1.1 Bi ton m u. Cc khi nim c bn

Xt chuyn ng thng ca mt cht im c vn tc a khng i xut pht t vtr O v chuyn ng trn trc Ox. Tm quy lut chuyn ng ca cht im.

Do ngha c hc ca o hm cp mt th vn tc a =dx

dt, trong x l on

ng m cht im i c trong thi gian t. Nh vy, vic tm quy lut chuynng ca cht im a v vic gii phng trnh

dx

dt= a, (x = x(t)).

l phng trnh vi phn. Gii bi ton ta c

x = at + C, C = x(0).

Xt chuyn ng ri t do trong chn khng ca mt vt. Hy tm quy lut vnng ca vt .

Ta chn hng dng theo chiu thng ng t di ln. Nh ta bit, nuvt ri t do trong chn khng th gia tc g ca n l khng i v hng xungdi. Do n mang du m. Mt khc, theo ngha ca o hm cp hai, gia

tc bngd2x

dt2. V vy ta c

d2x

dt2= g.

Gii phng trnh ta c x =

gt2

2+ C

1t + C

2, trong C

1, C

2l cc hng

s xc nh bi C1 = v0 = v|t=0, C2 = x0 = x|t=0.2


3/93

1.1. BI TON M U. CC KHI NIM C BN 3

Mt cht im khi lng m chuyn ng theo trc Ox di tc dng ca mtlc bx(b > 0) hng v gc ta . Hy tm quy lut chuyn ng ca chtim , bit rng lc t = 0 cht im v tr x = x0 v c vn tc v = v0.

Theo nh lut Newton, ta c

md2x

dt2= bx

hayd2x

dt2+

b

mx = 0.

t 2 =b

m> 0. Gii phng trnh trn ta c

x = C1 cos t + C2 sin t,

trong C1, C2 l hai hng s c xc nh bi iu kin u.

C1 = x0 = x(0), C2 =v0

, v0 = v(0).

nh ngha 1.1.1. Phng trnh vi phn l mt phng trnh c cha cc bin clp, hm phi tm v cc o hm ca n.

V d 1.1.1.1. y = x2 + 1.

2. y 2y + 2y = 2x + 3.

3.2z

x2+

2z

y2= 0.

Trong phng trnh vi phn, nu n hm l hm theo mt bin c lp th tagi l phng trnh vi phn thng hay gi tt l phng trnh vi phn. Nu nhm l hm theo nhiu bin c lp th ta gi l phng trnh vi phn o hmring hay gi tt l phng trnh o hm ring.

Trong gio trnh ny, ta ch nghin cu v phng trnh vi phn thng. Nhvy, nu ta gi x l bin c lp, y(x) l n hm th phng trnh vi phn thngl phng trnh c dng

F

x, y, y, y, . . . , y(n)

= 0. (1.1)

nh ngha 1.1.2. Cp ca phng trnh l o hm cp cao nht c mt trongphng trnh .

Bin son: GVC.ThS. Phan vn Danh


4/93

4 Chng 1. Cc khi nim c bn. Cch gii mt s phng trnh cp mt n gin

V d 1.1.2.

1. yy 2 + 3x + 4y = 0 l phng trnh cp 1.

2. x2

y + x4

= y l phng trnh cp 2.

3.2z

x2+

2z

y2= 0 l phng trnh cp 2.

nh ngha 1.1.3. Hm sy = (x) kh vi n cp n trn khong (a, b) c gil mtnghim ca phng trnh (1.1) nu khi thay y = (x), y = (x), . . . , y(n) =(n)(x) vo (1.1) th ta c ng nht thc trn (a, b).

Ch 1.1.1. Thng thng khi tm nghim ca phng trnh cp n, nghim c

dng y = (x, C1, . . . , C n), trong C1, . . . , C n l cc hng s. Vi C1, . . . , C nc gi tr c th, ta tm c mt nghim ring ca phng trnh.

Ni chung, ngi ta khng bit c cch gii ca mt phng trnh vi phntrong trng hp tng qut ngay c vi phng trnh vi phn cp mt m ch bitcch gii mt s phng trnh vi phn trong mt s trng hp c bit.

Bi ton Cauchy. Bi ton tm nghim ca phng trnh vi phn

y(n) = f(x, y, y, y. . . . , y(n)) (1.2)

tha mn iu kin u

y(x0) = y0, y(x0) = y0, y

(x0) = y0 , . . . , y(n1)(x0) = y

(n1)0 (1.3)

trong y0, y0, . . . , y(n1)0 l cc gi tr cho trc, c gi l bi ton Cauchy.

1.2 Mt s phng trnh vi phn cp mt gii c

1.2.1 nh ngha

nh ngha 1.2.1. Phng trnh vi phn cp mt l phng trnh c dng

F(x, y, y) = 0 (1.4)

trong F l mt hm s xc nh trong min D

R

3.

Ta xt mt s trng hp phng trnh cp 1 gii c nh sau.



5/93

1.2.2. Phng trnh tch bin 5

1.2.2 Phng trnh tch bin

a) Phng trnh dng y = f(x), trong f l hm lin tc.

Phng trnh ny tng ng vi phng trnh dy = f(x)dx. Ly tch phn

hai v ta c nghim tng qut ca phng trnh l

y =

f(x)dx + C.

V d 1.2.1. Cho phng trnhdy

dx=

1

x. y l phng trnh tch bin. Gii ra

ta c y =

f(x)dx + C = ln |x| + C.

b) Phng trnh dng y = f(y), trong f(y) = 0 v l hm lin tc.Xem x l hm ca y, khi phng trnh trn tng ng vi phng trnh

dx

dy=

1

f(y)hay dx =

1

f(y)dy. Ly tch phn hai v ta c nghim tng qut ca

phng trnh l

x =

dy

f(y)+ C.


dx = y2.Gii. Vi y = 0 th y l mt nghim ca phng trnh.Vi y = 0, gii ta c nghim tng qut ca phng trnh l

x =

dy

y2+ C = 1

y+ C hay y =

1

x + C1 .

c) Phng trnh dngdy

dx= f1(x) f2(y) trong f1(x), f2(y) l cc hm lin tc.

Gii. Vi f2(y) = 0 () ta tm nghim ca phng trnh (*) y = y0 cng lnghim ca phng trnh cho.

Vi f2(y) = 0, khi phng trnh cho tng ng vi phng trnhdy

f2(y)= f1(x)dx.

Ly tch phn hai v ta c

dy

f2(y)=

f1(x)dx + C.



6/93


Tng qut phng trnh c dng

M1(x)N1(y)dx + M2(x)N2(y)dy = 0

trong M1(x), N1(y), M2(x), N2(y) l cc hm lin tc.Gii. Nu y = y0 l nghim ca phng trnh N1(y) = 0 th y = y0 cng l nghimca phng trnh cho.

Nu x = x0 l nghim ca phng trnh M2(x) = 0 th x = x0 cng l nghimca phng trnh cho.

Nu N1(y) M2(x) = 0 th phng trnh cho tng ng vi phng trnhM1(x)

M2(x)

dx =

N2(y)

N1(y)

dy.

Ly tch phn hai v ta cM1(x)

M2(x)dx =

N2(y)

N1(y)dy + C.

V d 1.2.3. Cho phng trnh x(y2 1)dx + y(x2 1)dy = 0.Ta c x = 1 v y = 1 l cc nghim ca phng trnh cho.Nu (y2

1)(x2

1)= 0 th phng trnh cho tng ng vi phng trnh

x

x2 1dx +y

y2 1dy = 0.


ln |x2 1| + ln |y2 1| = ln |C1|, C1 = 0.

Suy ra (x2 1)(y2 1) = C, C= 0 l nghim ca phng trnh cho.

V d 1.2.4. Cho phng trnh x2(y + 1)dx + (x3 1)(y 1)dy = 0 vi (y +1)(x3 1) = 0.

Vi (y + 1)(x3 1) = 0, phng trnh cho tng ng vi phng trnhx2

x3 1dx +y 1y + 1

dy = 0.

Ly tch phn hai v ta c nghim ca phng trnh

13

ln |x3 1| + y 2 ln |y + 1| = C.



7/93

1.2.3. Phng trnh ng cp 7

c) Phng trnh c dngdy

dx= f(ax + by + c). y l phng trnh c th a v

dng tch bin bng cch t z = ax + by + c. Khi ta cdz

dx= a + b

dy

dxhay

dy

dx=

dzdx a

b.

Thay vo phng trnh cho ta cdz

dx= a + bf(z) hay

dz

a + bf(z)= dx.

y l phng trnh tch bin.


dx= x y + 5.

Gii. t z = x y + 5 ta c dzdx

= 1 dydx

. Thay vo phng trnh cho, ta

thu c phng trnh 1 dzdx

= z haydz

1 z = dx. y l phng trnh tchbin. Gii phng trnh ny ta c nghim

z = 1 Cex

.

Suy ra nghim ca phng trnh cho l

y = Cex + x + 4.

1.2.3 Phng trnh ng cp

a) Phng trnh dngdy

dx= f(x, y), trong f(tx, ty) = f(x, y) vi mi t R.

Gii phng trnh by bng cch t u = yx

, ta c th a phng trnh trn

v dng tch bin. Tht vy, vi u =y

xta suy ra y = ux. Do

dy

dx= u + x

du

dx.

Thay vo phng trnh cho ta c phng trnh u + xdu

dx= f(1, u) = (u)

hay (u) u = xdudx

.



8/93


Vi (u) u = 0 ta c dydx

=y

x. Suy ra y = Cx l nghim ca phng trnh

cho.Vi (u) u = 0, phng trnh trn tng ng vi phng trnh

dx

x=

du

(u) u.

y l phng trnh tch bin, gii phng trnh ny ta c

x = Ce(u), trong (u) =

du

(u) u.


dx=

2xy

x2 y2.

Gii. t u =y

x, khi phng trnh tr thnh

xdu

dx=

u(u2 + 1)

1 u2 .

Nu u = 0 th y = 0 l nghim ca phng trnh cho.Nu u = 0 th phng trnh cho tng ng vi phng trnh

dxx

= 1 u2

u(u2 + 1)du.

Gii phng trnh ny ta c

x(1 + u2) = Cu.

Thay u =y

x, ta suy ra y2 + x2 = Cy l nghim ca phng trnh cho.

b) Phng trnh dng dydx

= f

a1x + b1y + c1a2x + b2y + c2

.

(i) Nu a1b2 a2b1 = 0 th ta tx = u + h

y = v + k,

vi h, k l cc hng s. Thay vo phng trnh trn ta c

dydx

= f

a1u + b1v + a1h + b1k + c1a2u + b2v + a2h + b2k + c2

.



9/93

1.2.3. Phng trnh ng cp 9

Chn cc hng sh, k sao choa1h + b1k + c1 = 0

a2h + b2k + c2 = 0.

Khi phng trnh cho tr thnh phng trnh ng cp

dv

du= f

a1u + b1v

a2u + b2v

.


dx=x y + 2

x y + 4 .Gii. Ta c a1b2 a2b1 = 2 = 0. Khi ta tm h, k tha h phng trnh

h k + 2 = 0h k + 4 = 0.

Gii h ny ta c h = 1, k = 3. Thay bin mi vo phng trnh cho tac phng trnh

dv

du=u v

u v .y l mt phng trnh ng cp, gii phng trnh ny ta c nghim

u2 2uv v2 = C.Vy nghim ca phng trnh cho l

y2 x2 2xy 8y + 4x = C1.

(ii) Nu a1b2 a2b1 = 0 th ta t u = a2x + b2y.Khi phng trnh trn tr thnh

du

dx = a2 + b2 fu + c1u + c2

trong =a1

a2=

b1

b2. Phng trnh trn cng c vit li l

du

dx= (u). y

l phng trnh tch bin.


dx=

2x + y 14x + 2y + 5

.

Gii. Ta c a1b2 a2b1 = 0. t u = 2x + y, phng trnh cho tr thnhdudx

= 5u + 92u + 5

.



10/93


Gii phng trnh tch bin ny ta c nghim

10u + 7 ln |5u + 9| = 25x + C.Vy nghim ca phng trnh cho l

10y + 7 ln |10x + 5y + 9| 5x = C.

1.2.4 Phng trnh tuyn tnh cp 1

nh ngha 1.2.2. Phng trnh tuyn tnh cp 1 l phng trnh c dng

y +p(x)y = f(x). (1.5)

Nu f(x)

0 th (1.5) c gi l phng trnh tuyn tnh thun nht

y +p(x)y = 0. (1.6)

V d 1.2.9. Phng trnh y yx

= x2 l phng trnh tuyn tnh v phng trnh

y yx

= 0 c gi l phng trnh tuyn tnh thun nht tng ng.

Cch gii.

1. Bc 1. Gii phng trnh tuyn tnh thun nht (1.6). y l phng trnhtch bin, gii phng trnh ny ta c nghim tng qut sau

y = Cep(x)dx (1.7)

2. Bc 2. Gii phng trnh (1.5) bng phng php bin thin hng s.

Ta nhn thy rng trong cng thc nghim (1.7) vi C l hng s th (1.7)khng th l nghim ca phng trnh (1.5). Do cng thc dng (1.7) chc th l nghim ca phng trnh (1.5) khi C = C(x). Vy ta s i tm C(x)sao cho (1.7) l nghim ca (1.5).

Ly o hm (1.7) theo bin x v thay vo (1.5) ta suy ra

dC

dx= f(x)e

p(x)dx. (1.8)

Gii phng trnh (1.8) ta c nghim

C(x) =

f(x)e

p(x)dx + C1. (1.9)

Thay (1.9) vo (1.7) ta c nghim ca tng qut ca phng trnh (1.5)

y = C1ep(x)dx + e

p(x)dx

f(x)e

p(x)dx. (1.10)



11/93

1.2.5. Phng trnh Bernoulli 11

Ch 1.2.1. Nghim tng qut ca phng trnh (1.5) bng nghim tng qut caphng trnh (1.6) cng vi mt nghim ring ca phng trnh (1.5).

V d 1.2.10. Cho phng trnh y yx

= x2.Gii phng trnh thun nht y yx

= 0

ta c nghim y = Cx.Bng phng php bin thin hng s, ta tm C = C(x) l nghim ca phng

trnhdC

dx= x. Gii phng trnh ny ta c C =

x2

2+ C1.

Vy nghim ca phng trnh tuyn tnh khng thun nht l

y =x3

2+ Cx.

V d 1.2.11. Cho phng trnh y +y

x

= 3x.

p dng cng thc (1.10) ta c nghim

y =C

x+ x2.

1.2.5 Phng trnh Bernoulli

nh ngha 1.2.3. Phng trnh Bernoulli l phng trnh c dng

y +p(x)y = f(x)y

vi = 0, = 1. (1.11)Cch gii.

Ta c y = 0 l mt nghim ca phng trnh (1.11).Vi y = 0 ta t z = y1. Khi phng trnh (1.11) tng ng vi phng

trnhz + (1 )p(x)z = (1 )f(x). (1.12)

y l phng trnh tuyn tnh cp 1, gii phng trnh (1.12) ta c nghimz = z(x). Thay z = y1 ta suy ra nghim ca phng trnh (1.11).

V d 1.2.12. Cho phng trnh y 2xy = x3y2.Gii. y = 0 l mt nghim ca phng trnh.

Vi y = 0 ta t z = y1, khi phng trnh cho tng ng vi phngtrnh z + 2xz = x3. Gii phng trnh tuyn tnh ny ta tm c nghim

z = Cex2

+ 1 x2.Vy nghim ca phng trnh cho l

y = 1Cex2 + 1 x2 .



12/93


V d 1.2.13. Cho phng trnh y +y

x= x2y4.

Gii. y = 0 l mt nghim ca phng trnh.Vi y = 0 ta t z = y3. Khi phng trnh trn tng ng vi phng

trnh z 3x

z = 3x2.Gii phng trnh ny ta c

z = 3x3 lnC

x

.Vy nghim ca phng trnh l

y =1

x

33 ln

Cx.

V d 1.2.14. Cho phng trnh y 4 yx

= x

y.

Gii. y = 0 l mt nghim ca phng trnh.Vi y = 0 ta t z = y 12 . Khi phng trnh trn tng ng vi phng

trnh

z 2x

z =x

2.


z = x2

12

ln |x| + C.Vy nghim ca phng trnh l

y = x41

2ln |x| + C2.

1.2.6 Phng trnh Ricati

nh ngha 1.2.4. Phng trnh Ricati l phng trnh c dng

y = p(x)y2 + q(x)y + r(x) (1.13)

trong p(x), q(x), r(x) l cc hm lin tc.

Cch gii.1) Nu bit mt nghim ring y = y1(x) th ta t z = yy1(x). Khi phng

trnh (1.13) tng ng vi phng trnh

dz

dx

= p(x)z2 + 2p(x)y1(x) + q(x)z. (1.14)y l phng trnh Bernoulli vi n hm z.



13/93

1.3. PHNG TRNH VI PHN TON PHN. THA S TCH PHN 13

V d 1.2.15. Gii phng trnh x2y = x2y2 + xy + 1 khi bit mt nghim ring

y1(x) = 1x

.

Gii. t z = y +1

x

v thay vo phng trnh cho ta thu c phng trnh

z = z2 zx

.

y l phng trnh Bernoulli vi = 2 nn ta t u =1

z. Khi ta c phng

trnh tuyn tnh cp 1

u ux

= 1.

Gii phng trnh ny ta c nghim u = x(C

ln|x

|). Suy ra z =

1

x(C ln |x|).

Vy nghim ca phng trnh cho l

y =1

x(C ln |x|) 1

x.

2) Nu bit hai nghim ring y = y1(x) v y = y2(x) th ta t z = y y1(x).Khi phng trnh (1.13) a v dng (1.14). Sau ta li t u =

1

zv phng

trnh (1.14) tr thnh

du

dx+

2p(x)y1(x) + q(x)

u = p(x). (1.15)

y l phng trnh tuyn tnh cp 1 khng thun nht vi mt nghim ring

u(x) =1

y2(x) y1(x) .

1.3 Phng trnh vi phn ton phn. Tha s tch phn

1.3.1 Phng trnh vi phn ton phn

nh ngha 1.3.1. Phng trnh

M(x, y)dx + N(x, y)dy = 0 (1.16)

c gi l phng trnh vi phn ton phn nu tn ti hm U(x, y) kh vi trnmin D R2 sao cho

M(x, y)dx + N(x, y)dy = d(U(x, y)). (1.17)

Khi tch phn tng qut ca (1.16) l U(x, y) = C.



14/93


V d 1.3.1. Cho phng trnh 3(x2 +y

3)dx + (x + 2y)dy = 0. Phng trnh ny

tng ng vi phng trnh d(x3) + d(xy) + d(y2) = 0 hay d(x3 + xy + y2) = 0.Suy ra tch phn tng qut ca phng trnh l x3 + xy + y2 = C.

nh l 1.3.1. iu kin cn v biu thc vi phn

M(x, y)dx + N(x, y)dy, (1.18)

trong M(x, y), N(x, y) lin tc v khng ng thi trit tiu ti bt k im

no trong min n lin D R2 v c cc o hm ring My

,N

xlin tc trong

min , l mtvi phn ton phn l ng thc

M

y =N

x vi mi (x, y) D. (1.19)Chng minh. iu kin cn. Gi s (1.18) l mt vi phn ton phn ca hmU(x, y). Khi

M(x, y)dx + N(x, y)dy =U

xdx +

U

ydy.

Suy ra

M(x, y) =

U

x v N(x, y) =U

y . (1.20)

Ly o hm ng thc (1.20) ta c

M

y=

2U

xy,

N

x=

2U

yx.

Do gi thit cc o hm ring ny lin tc nn ta suy raM

y=

N

x.

iu kin . Xt phng trnh

U

x = M(x, y), n c nghim dng

U(x, y) =

xx0

M(x, y)dx + (y), (1.21)

vi P0(x0, y0) D. V D n lin nn biu thc U(x, y) xc inh bi (1.21) cngha. Ta chn (y) sao cho phng trnh

U

y= N(x, y) c tha mn.

Gi s kh vi, suy ra

Uy

(x, y) = y

x

x0

M(x, y)dx + (y).



15/93

1.3.1. Phng trnh vi phn ton phn 15

Theo tnh cht ca tch phn ph thuc tham s ta c

N(x, y) =

xx0

M

y(x, y)dx + (y).

VM

y=

N

xnn

(y) = N(x, y) xx0

N

x(x, y)dx.

Suy ra (y) = N(x0, y). Vy

(y) = y

y0

N(x0, y)dy + C1.

V vy tch phn tng qut c dng

U(x, y) =

xx0

M(x, y0)dx +

yy0

N(x, y)dy + C1.

V d 1.3.2. Cho phng trnh

(7x + 3y)dx + (3x 5y)dy = 0.Gii. V

M

y=

N

x= 3 nn y l mt phng trnh vi phn ton phn. V vy

ta cn tm hm U(x, y) sao choU

x= M(x, y). Do ta c th tm hm U(x, y)

nh sau.

U(x, y) =

(7x + 3y)dx + (y) =

7x2

2+ 3xy + (y).

Suy raUy

= 3x + (y) = 3x 5y

hay (y) = 5y do (y) = 52

y2 + C. Vy hm cn tm l

U(x, y) =7

2x2 + 3xy 5

2y2 + C

hay tch phn tng qut ca phng trnh l

72

x2 + 3xy 52

y2 = C1.



16/93


1.3.2 Tha s tch phn

Trong trng hp (1.16) khng phi l phng trnh vi phn ton phn, nhng nuta c th tm c mt hm (x, y) sao cho

(x, y)M(x, y)dx + (x, y)N(x, y)dy = 0 (1.22)

l mt phng trnh vi phn ton phn th hm (x, y) c gi l tha s tchphn ca phng trnh (1.16).

Vn t ra l c tn ti tha s tch phn cho mt phng trnh vi phn dngtng qut khng? nh l sau y s tr li cu hi .

nh l 1.3.2. Mi phng trnh vi phn cp mt tha mn iu kin tn ti v

duy nht nghim lun tn ti v s tha s tch phn.Chng minh. Theo nh l tn ti nghim, phng trnh (1.16) nhn tch phn tngqut l U(x, y) = C.


U

xdx +

U

ydy = 0

haydy

dx = U

xUy

.

Mt khc phng trnh (1.16) vi N(x, y) = 0 ta c th vit dydx

= M(x, y)N(x, y)

.

Suy raUxUy

=M(x, y)

N(x, y)

hay

Ux

M(x, y)=

Uy

N(x, y).

t

(x, y) =Ux

M(x, y)=

Uy

N(x, y).

Suy ra phng trnh

(x, y)M(x, y)dx + (x, y)N(x, y)dy = 0

l phng trnh vi phn ton phn.



17/93

1.3.3. Cch tm tha s tch phn 17

By gi ta chng minh c v s tha s tch phn. Gi s 1(x, y) l mt thas tch phn. Khi

1(x, y)M(x, y)dx + 1(x, y)N(x, y)dy = d(U(x, y)) = 0,

trong U(x, y) l mt hm kh vi, do U(x, y) = C.Gi (U) l hm kh vi ty v t (x, y) = (U)1(x, y). Ta c

(x, y)M(x, y)dx + (x, y)N(x, y)dy

= (U)1(x, y)M(x, y)dx + (U)1(x, y)N(x, y)dy

= (U)d(U(x, y)) = d

(U)dU = 0.

Vy (x, y) l mt tha s tch phn.

1.3.3 Cch tm tha s tch phn

Ni chung khng c phng php tng qut tm tha s tch phn, m ta ch cth tm c tha s tch phn trong mt s trng hp c bit. Tht vy gi s(x, y) l mt tha s tch phn ca phng trnh (1.16), ta c

(M)

y

=(N)

xhay

N

x M

y=

M

y N

x

.

Suy raM

y N

x= N

ln

x Mln

y. (1.23)

Ta thy rng vic gii phng trnh (1.23) li phc tp hn phng trnh (1.16). V

vy ta ch c th gii phng trnh (1.23) trong mt s trng hp c bit sau.1) Khi hm = (x), ngha l

y= 0. T phng trnh (1.23) ta suy ra

d ln

dx=

M

y N

x

N.

Ly tch phn theo bin x, ta c

ln (x) =

My NxN

+ C.



18/93


Vy(x) = Ce

(x)dx,

trong (x) =My

Nx

N

. Chn C = 1, ta c tha s tch phn cn tm.

V d 1.3.3. Gii phng trnh (x + y2)dx 2xydy = 0.Gii. Ta c

My

Nx

N= 2

x.

Suy ra (x) =1

x2. Nhn (x) vi phng trnh cho ta c phng trnh

1x

+y2

x2 dx

2y

x

dy = 0.

Ta tm c tch phn tng qut

ln |x| y2

x= C1 hay x = Ce

y2x .

2) Khi hm = (y) ngha l

x= 0. T phng trnh (1.23) ta suy ra

d ln

dy = My

Nx

M .

Ly tch phn theo bin y ta c

ln (y) = M

y N

x

M+ C.

Vy(y) = Ce

(y)dy,

trong (y) =M

y N

x

M. Chn C = 1, ta c tha s tch phn cn tm.

V d 1.3.4. Cho phng trnh (2xy2 3y3)dx + (7 3xy2)dy = 0.Gii. Ta c

My

Nx

M=

2

y.

Suy ra (y) =1

y2. Nhn (y) vi phng trnh cho ta c phng trnh

(2x 3y)dx + 7y2 3xdy = 0.



19/93

1.4. MT S PHNG TRNH VI PHN CP MT CHA GII I VI OHM 19

Ta tm c tch phn tng qut

x2 3xy 7y

= C.

1.4 Mt s phng trnh vi phn cp mt cha gii i vi o hm

1.4.1 Phng trnh F(x, y) = 0 hay F(y, y) = 0

1.- Phng trnh F(x, y) = 0

a) Trng hp 1. Phng trnh c a v dng y = f(x). Khi nghim caphng trnh c dng

y =

f(x)dx + C.

b) Trng hp 2. Phng trnh c a v dng x = (y). Khi ta t p = y

th phng trnh tr thnh x = (p).

Tdy

dx= p suy ra dy = pdx = p(p)dp. Do

y = p(p)dp + C.

Vy nghim tng qut ca phng trnh c dngx = (p)

y =

p(p)dp + C.

V d 1.4.1. Gii phng trnh

ey

+ y = x.

Gii. t p = y, ta c x = ep +p. Suy ra dx = (ep + 1)dp. Khi

dy = pdx = y =

p(ep + 1)dp + C

hay y = ep(p 1) + p2

2+ C. Vy nghim ca phng trnh l

x = ep +p

y = ep(p 1) + p22

+ C.



20/93


c) Trng hp 3. Phng trnh c a v dng tham s

x = (t)y = (t).

Khi

y =

ydx + C =

(t)(t)dt + C.

Vy nghim ca phng trnh lx = (t)

y = (t)(t)dt + C.V d 1.4.2. Gii phng trnh

x3 + y3 3xy = 0.Gii. t y = tx trong t l tham s. Khi phng trnh trn c a vdng

x =

3t

1 + t3

y =3t2

1 + t3 .

Tch phn ta c

y =

ydx + C =

3t2

1 + t3 3(1 + t

3) 3t 3t2(1 + t3)2

dt + C

hay

y = 92(1 + t3)2

+6

1 + t3+ C.

Vy nghim ca phng trnh l

x =3t

1 + t3

y = 92(1 + t3)2

+6

1 + t3+ C.

2.- Phng trnh F(y, y) = 0

a) Trng hp 1. Phng trnh c a v dng y = f(y). Khi nghim ca

phng trnh c dngx + C =

dy

f(y).



21/93

1.4.1. Phng trnh F(x, y) = 0 hay F(y, y) = 0 21

Ngoi ra vi y = y0 l nghim ca phng trnh f(y) = 0 cng l mt nghimca phng trnh cho.

b) Trng hp 2. Phng trnh c a v dng y = (y). Khi ta t p = yv phng trnh tr thnh y = (p). Suy ra dy = (p)dp.

Tx =

dy

y+ C suy ra x =

(p)

pdp + C.

Vy nghim tng qut ca phng trnh c dng

x =

(p)

pdp + C

y = (p).

c) Trng hp 3. Phng trnh c a v dng tham sy = (t)

y = (t).

Khi

x = dy

y+ C =

(t)(t)

dt + C.

Vy nghim ca phng trnh lx =

(t)(t)

dt + C

y = (t).


y3 y2(1 y) = 0.Gii. t y = ty trong t l tham s. Khi phng trnh trn c a v dng

y =

t3

1 + t2

y =t2

1 + t2.

Suy ra

x = dyy

+ C = 3t2(1 + t2) 2t t2

(1 + t2

)2

1 + t2

t3

dt + C

hayx = t + 2 arctan t + C.



22/93



x = t + 2 arctan t + C

y = t

3

1 + t2 .

1.5 Phng trnh Lagrange v phng trnh Clairaut

1.5.1 Phng trnh Lagrange

nh ngha 1.5.1. Phng trnh Lagrange l phng trnh c dng

y = (y)x + (y), (1.24)

trong , l cc hm kh vi lin tc.

Cch gii. t p = y, phng trnh (1.24) tr thnh

y = (p)x + (p). (1.25)

Suy rady

dx= p = (p)

dp

dxx + (p)

dp

dx+ (p)

hay

p (p) = dpdx

[(p)x + (p)].

Khi phng trnh (1.25) tr thnh

dx

dp=

(p)x + (p)p (p) .

a) Gi sp (p) = 0. Suy radx

dp (p)x

p (p)=

(p)

p (p). (1.26)

Xem x = x(p) khi phng trnh (1.26) l phng trnh tuyn tnh. Gii phngtrnh (1.26) ta c

x = C f(p) + g(p)v

y = (p)[Cf(p) + g(p)] + (p).

Nu p

(p) = 0 khi gii phng trnh ny ta c nghim p = p0. Suy ra

y = (p0)x + (p0).

Ty trng hp ta c nghim ring hay nghim k d.



23/93

1.5.1. Phng trnh Lagrange 23

V d 1.5.1. Cho phng trnh y = (y)2x + (y)2.Gii. t p = y khi y = p2x +p2.Suy ra dy = pdx = p2dx + (2px + 2p)dp hay p(p

1)dx + 2p(x + 1)dp = 0.

Nu p = 0 v p = 1 th phng trnh trn tng ng vi phng trnhdx

dp+

2x

p 1 =2

1 p. (1.27)y l phng trnh tuyn tnh khng thun nht. Gii phng trnh (1.27) ta cnghim

x =C

(p 1)2 1.Vy nghim ca bi ton l

x = C

(p 1)2 1

y = p2C

(p 1)2 .

Ta c th khp trong phng trnh trn v tm c nghim ca phng trnh didng y = (

x + 1 + C)2.

Nu p = 0 th y = 0 l mt nghim ca phng trnh.Nu p = 1 th y = x + 1 l mt nghim ca phng trnh.

V d 1.5.2. Cho phng trnh y = 2xy y3.Gii. t p = y, phng trnh trn tr thnh y = 2xp p3. Ly o hm theo xphng trnh trn, ta c

y = 2p + 2xp 3p2p = p. (1.28)Nu p = 0 th y = 0 l nghim ca phng trnh.Nu p = 0 th ta a phng trnh (1.28) v phng trnh tuyn tnh

dx

dp

+2

p

x = 3p.


x =C

p2+

3

4p2.

Suy ra nghim ca phng trnh l

x =C

p2+

3

4p2

y = 2C

p2

+1

2

p3.

c bit nu p (p) 0, khi phng trnh trn tr thnh phng trnhClairaut.



24/93


1.5.2 Phng trnh Clairaut

nh ngha 1.5.2. Phng trnh Clairaut l phng trnh c dng

y = yx + (y) (1.29)

trong l hm kh vi lin tc.

gii phng trnh Clairaut, ta cn nghin cu bao hnh ca h cc ngcong.

nh ngha 1.5.3. Bao hnh ca h cc ng cong trn mt phng l mt ngcong

Cm ti mi im ca n,

Ctip xc vi t nht mt ng cong ca h v

trn mi cung ca n, C tip xc vi v s ng cong ca h ng cong.Cch tm bao hnh. Gi s phng trnh ca h ng cong trn mt phng c

dng(x, y, C ) = 0. (1.30)

Ta xt h sau

(x, y, C ) = 0

C(x, y, C ) = 0.

(1.31)

Kh tham sC ta c bit tuyn ca h (1.30). Bit tuyn c th l bao hnhca h m cng c th l qu tch ca cc im k d, ngha l nhng im tha

mn ng thi

x= 0 v

y= 0. Trong hnh hc vi phn, ngi ta chng

minh c rng nu h (1.31) c nghim

x = x(C)y = y(C)

(1.32)

v nu x, y l cc hm kh vi lin tc tha iu kin xC2 + yC

2 = 0 th (1.32) lbao hnh ca h (1.30).

V d 1.5.3. Cho h ng trn

(x C)2 + y2 = R2.

Bit tuyn ca h l (x C)2 + y2 = R22(x C) = 0.



25/93

1.5.2. Phng trnh Clairaut 25

Kh C ta c y = R. y l bao hnh ca h ng trn cho. Tht vy, tac

x = Cy = R

l nhng hm kh vi lin tc v xC2 + yC

2 = 1 = 0.By gi ta tr li vi phng trnh Clairaut v ch ra cch gii phng trnh ny.t y = p, ta c y = px + (p), trong l hm kh vi phi tuyn theo p.

Ly o hm theo x ta c

y = px +p + (p)p = p.

Rt gn ta c

x + (p)p = 0.

Nu p = 0 th p = C. Vy nghim ca phng trnh l y = Cx + (C).Nu x + (p) = 0 th x = (p). T y ta suy ra nghim ca phng trnh

c dng x = (p)y = p(p) + (p).

c bit, nu gii c p = (x) th nghim ca phng trnh s l

y = (x)x +

(x)

.

Vi nghim ca phng trnh Clairaut nh trn ta tm c

y = Cx + (C). (1.33)

Vy

x = (p)

y = p(p) + (p).(1.34)

Ta s chng minh rng (1.34) l bao hnh ca h ng thng (1.33). Tht vy, taxt h sau

0 = x + (C)y = Cx + (C).

(1.35)

Vi C = p th h (1.35) trng vi h (1.34). H ny xc nh bit tuyn ca hng thng (1.33). T (1.35) ta suy ra

x = (C)y = C(C) + (C). (1.36)



26/93


Theo gi thit hm kh vi phi tuyn nn xC = (C) = 0, hn nayC = x +

(C) = 0. Suy ra xC2 + yC

2 = 0. Vy (1.34) l bao hnh ca h ngthng (1.33). Khi ti mi im ca ng cong (1.34), tnh duy nht nghim

b ph v v ti c hai ng cong i qua: mt l nghim (1.34) v mt ngthng ca h (1.33). Vy nghim (1.34) l nghim k d ca phng trnh Clairaut.

V d 1.5.4. Cho phng trnh y = xy + 2y.

Gii. t y = p khi phng trnh cho tr thnh y = px + 2p.

Nu p = 0 th phng trnh c nghim y = Cx + 2C vi C 0.

Nu x = (2p) th p = 1x2

. Khi nghim ca phng trnh l y =1

x.

Vy nghim ca phng trnh cho l

y = C1x + C2

y =1

x(nghim k d).

V d 1.5.5. Cho phng trnh y = xy y2.Gii. t p = y khi phng trnh cho tr thnh y = px p2.

Nu p = 0 th p = C. Suy ra nghim ca phng trnh ny l y = Cx C2.Nu x = (p)2 th p =

x

2. Khi

y = x x2 x

2

2=

x2

4.


y = C1x + C2

y =x2

4(nghim k d).



27/93

Mc lc

1 S tn ti v duy nht nghim 281.1 S tn ti nghim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.1.1 ng gp khc Euler . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.1.2 Nghim -xp x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.1.3 B Arzela-Ascoli . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.1.4 nh l Cauchy-Peano . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.1.5 nh l v s tn ti nghim ca phng trnh vi phn . . . . . . . . . . . 321.2 Tnh duy nht nghim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321.3 Thc trin nghim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.4 Tnh trn ca nghim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.5 S ph thuc lin tc ca nghim vo iu kin ban u v v phi . . . . . . . . 36

27
http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


28/93

Chng 1

S tn ti v duy nht nghim

1.1 S tn ti nghim

Xt phng trnhdy

dx= f(x, y). (1.1)

Khi bi ton tm nghim y = y(x) ca phng trnh (2.1) vi iu kin ban uy(x0) = y0 c gi l bi ton Cauchy.Vn c t ra l:

Bi ton Cauchy c nghim khi no?Nghim ca bi ton Cauchy c duy nht khng? tr li nhng cu hi ny, ta s gii quyt mt s vn sau.

1.1.1 ng gp khc Euler

Gi sf(x, y) lin tc trong hnh ch nht D = [x0 a, x0 + a] [y0 b, y0 + b].Khi hm f b chn trong hnh ch nht ny.

tM = max

(x,y)D|f(x, y)|

v

= min

a, bM

.

Vi > 0 cho trc, ta dng ng gp khc Euler nh sau.V hm f lin tc trn hnh ch nht D nn lin tc u. Do vi mi > 0

cho trc, tn ti > 0 sao cho vi mi (x, y), (x, y) D tha |x x| < v|y y| < th ta c |f(x, y) f(x, y)| < .

Chia on [x0, x0 + ] thnh n on nh bi cc im chia x0 < x1 < 0 cho trc ta c

|(x) f(x, (x))| < , x [a, b]

c gi l nghim -xp x.

T cch xy dng mc 1.1.1, ta c

(x) = (xk1) + f(xk1, (xk1))(x xk1)

xk1 < x < xk, k = 1, 2, . . . , n .

(x0) = y0

Khi ta thy hm xc nh trn [x0, x0 + ] v c o hm lin tc tng khctrn [x0, x0 + ].

Ngoi ra do cch xy dng, ta c

|(x) (x)| M|x x|, x, x [x0, x0 + ].

Nu x (xk1, xk) th |(x) (xk1)| < (v max1kn

|xk xk1| min

,

M

).

Suy ra

|(x) f(x, (x))| = |f(xk1, (xk1)) f(x, (x))| < .

Bt ng thc ny chng t l nghim -xp x ca phng trnh (2.1).

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


30/93

30 Chng 1. S tn ti v duy nht nghim

1.1.3 B Arzela-Ascoli

nh ngha 1.1.2. H F gm cc hm f : I R c gi l lin tc ng bctrn I nu vi mi s > 0 cho trc, tn ti mt s > 0 sao cho vi mi f F

v vi mi x, x I, vi |x x| < th ta u c |f(x) f(x)| < .B 1.1.1 (Arzela-Ascoli). Cho h F cc hm xc nh trn I. Nu F l h cchm b chn u v lin tc ng bc trn I th tn ti mt dy hm {fn}, fn Fhi t u v mt hm f xc nh trn I.

Chng minh. Gi s{rk} l dy s hu t trn I. Vi r1 I, tp hp {f(r1) : f F} b chn trong R nn tn ti dy cc hm {fn1} sao cho dy s {fn1(r1)} hit.

Tng t vi r2 I, dy {fn1(r2)} b chn trong

R

nn tn ti dy con {fn2}ca {fn1} sao cho dy s{fn2(r2)} hi t.

Tip tc qu trnh ny ta c dy {fnk}, fnk F sao cho dy {fnk(rk)} hi t. n gin ta k hiu fn = fnn, n N. Ta chng minh dy {fn} hi t u

trn I. Theo cch xy dng trn, dy {fn} hi t ti cc im hu t trn I nnvi rk I v mi > 0 cho trc, tn ti n0(rk) N sao cho vi mi s nguynm, n > n0(rk) ta c

|fn(rk) fm(rk)| 0 cho trc, tn ti > 0sao cho vi mi f F v vi mi x, x I m |x x| < ta c

|f(x) f(x)| n0 = max

1kpn0(rk),

ta c|fn(x)fm(x)| |fn(x)fn(rk)|+|fn(rk)fm(rk)|+|fm(rk)fm(x)| < . (1.4)

Vy dy hm {fn} hi t u trn I.

1.1.4 nh l Cauchy-Peano

nh l 1.1.1 (Cauchy-Peano). Nu f lin tc trn hnh ch nht D th tn ti mt

nghim kh vi lin tc trn [x0 , x0 + ] ca phng trnh (2.1) tha mniu kin u (x0) = y0.

http://-/?-http://-/?-


31/93

1.1.4. nh l Cauchy-Peano 31

Chng minh. Ly dy s thc dng {n} n iu gim v 0. Vi mi n v on[x0 , x0 + ], ta xy dng nghim n-xp x n cho phng trnh (2.1) tha mnn(x0) = y0. Theo cch xy dng ca n ta c

|n(x) n(x)| M|x x|. (1.5)

Suy ra

|n(x) n(x0)| M|x x0| M b

M= b, (1.6)

hay |n(x)| |n(x0)| + b = C (hng s), vi mi x [x0 , x0 + ].Vy {n} b chn u trn [x0 , x0 + ]. Hn na, t (1.5) suy ra h {n} lintc ng bc. Theo b Arzela-Ascoli, 1.1.1, tn ti dy con {nk} hi t u

v hm trn [x0 , x0 + ]. Do xc nh, lin tc trn [x0 , x0 + ].By gi ta chng minh l nghim ca phng trnh (2.1) v tha mn (x0) =y0. Ta c

n(x) = n(x0) +

xx0

n(t)dt

= y0 +

xx0

[f(t, n(t)) + n(t) f(t, n(t))] dt

= y0 +xx0

[f(t, n(t)) + n(t)] dt,

trong

n(t) =

n(t) f(f, n(t)) ti cc im c

n

0 ti cc im cn li.

V n l nghim n-xp x nn |n(t)| < n. Do f lin tc u trn D v dy

{nk} hi t u n trn [x0 , x0 + ] nn ta c th chuyn qua gii hnng thc sau

nk(x) = y0 +

xx0

[f(t, nk(t)) + nk(t)]dt

v ta c

(x) = y0 +

xx0

f(t, (t))dt. (1.7)

T (1.7) ta suy ra (x0) = y0 v

(x) = f(x, (x)).Vy l nghim ca phng trnh (2.1) v tha mn (x0) = y0.

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


32/93


1.1.5 nh l v s tn ti nghim ca phng trnh vi phn

nh l 1.1.2. Cho f l mt hm lin tc trn min D v (x0, y0) D. Khi tnti nghim ca phng trnh vi phn (2.1) tha iu kin u (x0) = y0.

1.2 Tnh duy nht nghim

nh ngha 1.2.1. Hm f : G R2 R t tng ng mi (x, y) vi mt sf(x, y) c gi l hm s Lipschitz i vi y nu tn ti hng s k dng saocho

|f(x, y) f(x, y)| k|y y|

vi mi (x, y), (x, y) G.

nh l 1.2.1 (Picard). Cho phng trnh vi phn (2.1)

dy

dx= f(x, y)

vi iu kin u y(x0) = y0. Gi s hm s f lin tc trn hnh ch nhtD = [x0 a, x0 + a] [y0 b, y0 + b] v Lipschitz i vi y trong D. Khi biton tn ti duy nht nghim C1[x0,x0+], trong

= min

a,

b

M

, vi M = max

D|f(x, y)|.

Chng minh. Tm hm y = (x) tha phng trnh (2.1) v tha mn iu kinu y(x0) = y0 tng ng vi bi ton tm hm y = (x) tha phng trnh

(x) = y0 +

xx0

f(t, (t))dt. (1.8)

Ta chn sao cho k < 1. Trong khng gian C = C1[x0,x0+] vi metric xcnh nh sau

d(y1, y2) = max|xx0|

|y1(x) y2(x)|.

Xt tp con C ca C sao cho

|y(x) y0| b, vi mi x [x0 , x0 + ].

Khi C l mt khng gian metric y . Tht vy, gi s dy {yn} C l dy

Cauchy, khi dy {yn} hi t v y C (v C vi metric trn l mt khng giany ).

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


34/93


By gi ta xt xem vi iu kin no th nghim ca phng trnh c th thctrin c, ngha l c th tm c xc nh trn (a1, b1) (a, b) v trng vihm trn (a, b).

Ta chng minh rng:

1) Nu hm f lin tc trn D, b chn trong D v (b 0) = limxb0

(x), trong (b, (b 0)) D th nghim c th thc trin c n im b.

Tht vy, theo gi thit

(x) = y0 +

xx0

f(t, (t))dt, x (a, b). (1.9)

T suy ra rng nu x1, x2 (a, b) th

|(x0) (x2)|

x2

x1

|f(t, (t))|dt M|x2 x1|.

Cho x1, x2 a + 0 (hay b 0). Khi theo tiu chun Cauchy, cc gii hn sautn ti.

(a + 0) = limxa+0

(x), (b 0) = limxb0

(x).

Nu (b, (b 0)) D th t

(x) =

(x) nu x (a, b)

(b 0) nu x = b.

Khi l nghim ca phng trnh vi phn. Hn na, C1 trn (a, b].Tht vy,

(x) =

xx0

f(t, (t))dt, x (a, b]. (1.10)

T suy ra s tn ti o hm tri ca ti b v

(b 0) = f(b, (b 0)).

Hm l nghim thc trin ca trn (a, b].2) Ta tip tc thc trin nghim . p dng nh l tn ti nghim i vi

phng trnh vi phn vi im u (b, (b 0)) ta c nghim C1 i quaim (b, (b 0)) v xc nh trn [b, b + ] ( > 0). Khi ta xt hm

=

nu x (a, b](x) nu x [b, b + ].



35/93

1.4. TNH TRN CA NGHIM 35

By gi ta s chng minh l nghim ca phng trnh vi phn trn (a, b + ],ngha l C1 v (x0) = y0. Mun vy ta phi chng minh rng

(x) = y0 + x

x0

f(t, (t))dt, x (a, b + ]. (1.11)

Vi a < x b (1.11) ng.Vi x > b theo cch xy dng ta c

(x) = (b 0) +

xb

f(t, (t))dt.

Nhng

(b 0) = y0bx0

f(t, (t))dt,

nn (1.11) ng vi x > b. V lin tc nn f(x, (x)) lin tc. Ly o hm haiv ca (1.11) ta c

(x) = f(x, (x)), x (a, b + ].

nh l 1.3.1. Gi s f lin tc v b chn trn D. Nu l nghim ca phngtrnh vi phn th tn ti gii hn (a + 0) v (b 0). Nu (a, (a + 0)) D th c th thc trin v bn tri a. Nu (b, (b 0)) D th c th thc trin vbn phi b.

1.4 Tnh trn ca nghim

Xt phng trnhdy

dx= f(x, y)

trn hnh ch nht D.nh l 1.4.1. Nu f c cc o hm ring lin tc theo x v y n cp k (k > 0)trn D th mi nghim ca phng trnh c o hm lin tc n cp k + 1 trn[x0 a, x0 + a].

Chng minh. Gi sy(x) l nghim bt k ca phng trnh. Khi ta c

y(x) = f(x, y(x)). (1.12)

V y(x) tha mn phng trnh ang xt nn c cc o hm lin tc theo x. Hnna do f lin tc theo x v y nn y(x) lin tc theo x.



36/93


Gi sk 1. Khi v phi ca (1.12) c o hm lin tc theo x. V vy vtri c o hm lin tc ngha l y lin tc v

y(x) = fx(x, y(x)) + fy(x, y(x)) y

(x).

Tip tc lp lun nh trn ta suy ra kt lun.

1.5 S ph thuc lin tc ca nghim vo iu kin ban u vv phi

nh l 1.5.1. Nu f(x, y) lin tc v gii ni trong min D v ti mi im trong(x0, y0) ca min c duy nht nghim th nghim ph thuc lin tc vo v

phi f(x, y) v im (x0, y0).Chng minh. Gi s ti im (x0, y0) D c nghim y0(x) ca phng trnh (1.12)i qua v xc nh trn on [a, b]. Khi vi mi > 0 tn ti > 0 sao cho khi|y0 y0| < , |x0 x0| < , |f(x, y) f(x, y)| < (f(x, y) l hm lin tc trnG D) th nghim y0(x) ca phng trnh

y = f(x, y)

i qua im (x0, y0) s tn ti trn ton [a, b] v tha mn

|y0(x) y0(x)| < , x [, ].

Bng phn chng, gi s iu trn khng ng. Khi tn ti mt s 0 > 0v mt dy {yn(x)} nghim ca phng trnh y = fn(x, y) vi iu kin uy(xn) = yn, trong xn x0, yn y0, sup |fn f| 0(n ) sao cho khithc trin nghim trn [a, b] th bt ng thc

|yn(x) y0(x)| 0 (1.13)

khng cn ng na.Gi sM l mt s thc no sao cho M sup |f(x, y)| trn G. Ta xt hnh

ch nhtR = {(x, y) : |x x0| , |y y0| M }

nm trong D. Khi vi n ln, cc hm y0(x) v yn(x) s xc nh trnon [x0 , x0 + ]. Ta chng minh rng vi mi > 0 v n ln, ta c|yn(x) y0(x)| < vi mi x [x0 , x0 + ]. Tht vy, gi s iu khngxy ra. Khi tn ti mt s > 0 v mt dy {kn} sao cho vi mi n 1 vvi xn no thuc on [x0 , x0 + ], ta c

|ykn(xn) yn(xn)| . (1.14)



37/93

1.5. S PH THUC LIN TC CA NGHIM VO IU KIN BAN U V V PHI 37

Dy hm {yn(x)} b chn u v khi n ln, yn c o hm b chn nn vinhng n ta c

|yn(x) yn(x) M|x x|.

V vy dy hm {ykn} ng lin tc v b chn u nn theo nh l Azela-Ascoli, tn ti dy con ykn hi t u v nghim ca phng trnh (1.12) v i quaim (x0, y0). Do tnh duy nht nghim, suy ra nghim ny chnh l nghim y0.iu ny mu thun vi bt ng thc (1.14). Vy |yn(x) y0(x)| < , vi mix [x0 , x0 + ] vi mi n ln.

By gi xt hnh ch nht Q1 = {(x, y) : |x x10| 1, |y y10| M 1}

nm trong D, y x10 = x0 , y10 = y0(x

10). Theo chng minh trn vi n ln

th im (x10, yn(x10)) kh gn im (x0, y0). L lun nh trn cho nghim y0(x) v

yn(x) trn on [x10 1, x10 + 1] ta c |yn(x) y0(x)| < , nu n ln. Gi

s [a, b] l on sao cho ng cong y = y0(x), x [a, b] nm trong D. Tip tclp lun nh trn ta lp c cc hnh ch nht Qn nm trong D (vi n ln),

Qn = {(x, y) : |x xn0 | n, |y y

n0 | M n},

trong xn0 = xn10 n1, y

n0 = y0(x

n0). Khi ta chng minh c bt ng thc

(1.13) nghim ng trn on [a, b] vi n ln. V qu trnh trn ta ch dng mts hu hn hnh ch nht Qn, iu ny mu thun vi gi thit phn chng. Vynh l c chng minh.



38/93

Mc lc

3 Phng trnh vi phn cp cao 393.1 nh l tn ti v duy nht nghim i vi phng trnh vi phn

cp cao . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.2 Phng trnh vi phn tuyn tnh thun nht . . . . . . . . . . . . . 41

3.2.1 Khi nim phng trnh vi phn tuyn tnh thun nht cpcao . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.2.2 Mt s tnh cht ca nghim phng trnh thun nht . . . 423.2.3 S ph thuc tuyn tnh v c lp tuyn tnh ca hm . . 433.2.4 nh thc Wronski . . . . . . . . . . . . . . . . . . . . . . 443.2.5 H nghim c bn. Nghim tng qut . . . . . . . . . . . 45

3.3 Phng trnh tuyn tnh khng thun nht . . . . . . . . . . . . . . 463.4 Phng trnh vi phn tuyn tnh vi h s hng . . . . . . . . . . . 47

3.4.1 Phng trnh vi phn tuyn tnh cp hai i vi h s hng 473.4.2 Phng trnh vi phn tuyn tnh thun nht cp cao vi hs hng . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.5 Phng trnh vi phn tuyn tnh cp hai khng thun nht . . . . . 523.5.1 Phng php h s bt nh . . . . . . . . . . . . . . . . . 523.5.2 Phng php bin thin hng s. . . . . . . . . . . . . . . 563.5.3 Tm h nghim c bn khi bit mt nghim . . . . . . . . 59

3.6 S dao ng nghim ca phng trnh tuyn tnh thun nht cp hai 65

3.6.1 a phng trnh v dng khng cha o hm cp mt . 653.6.2 a phng trnh v dng lin hp . . . . . . . . . . . . . 683.6.3 Nghim dao ng v nghim khng dao ng . . . . . . . 69

38
http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


39/93

Chng 3

Phng trnh vi phn cp cao

3.1 nh l tn ti v duy nht nghim i vi phng trnh vi

phn cp caonh ngha 3.1.1. Phng trnh vi phn cp n l phng trnh c dng

F(x , y , y, y, . . . , y(n) = 0, (3.1)

trong nht thit phi c mt y(n).

Gi s hm F lin tc theo cc bin v ti im (x0, y0, y0, . . . , y(n)0 ) tha mn

cc iu kin

i) F(x0, y0, y0, . . . , y(n)0 ) = 0;

ii)F

y(n)(x0, y0, y

0, . . . , y

(n)0 ) = 0;

iii)F

y(n1)lin tc.

Khi theo nh l hm n, ta c th gii c

y(n) = f(x , y , y, y, . . . , y(n1)). (3.2)

Nghim tng qut ca phng trnh (3.1) l mt biu thc

y = (x, C1, . . . , C n)

ph thuc vo n hng s c lp C1, . . . , C n m khi thay vo phng trnh (3.1)ta c mt ng nht thc.

Nhiu khi gii phng trnh (3.1) ta i n biu thc dng

(x, C1, . . . , C n) = 0.

39


40/93

40 Chng 3. Phng trnh vi phn cp cao

nh l 3.1.1. Gi s trong min D Rn+1, hm f(x, u1, u2, . . . , un) lin tc vtha mn iu kin Lipschitz theo u1, u2, . . . , un. Khi vi bt k im trong(x0, y0, y

0, . . . , y

(n1)0 ) D, phng trnh (3.2) tn ti duy nht nghim y = y(x)

tha mn iu kin u

y(x0) = y0, y(x0) = y0, . . . , y

(n1)0 (x0) = y

(n1)0 (3.3)

trn mt on [x0 , x0 + ].H qu 3.1.1. Nu hm f(x, u1, u2, . . . , un) cng cc o hm ring lin tc trnmin D Rn+1 no th phng trnh (3.2) tn ti duy nht nghim tha iukin u (3.3).

Nhn xt 3.1.1. Ta c th a phng trnh vi phn cp n v h phng trnh viphn cp 1 bng cch t

y = y1

y = dy1dx

= y2

y2 =dy2dx

= y3

. . . . . . . . .

yn1 =dyn1dx

= yn.

(3.4)

Thay vo (3.2) ta c

dy1dx = y2dy2dx

= y3

. . . . . .dyndx

= f(x, y1, y2, . . . , yn).

V d 3.1.1. Phng trnh vi phn cp 3

y = xy + y2 + y.

Ta c th a phng trnh trn v h ba phng trnh bng cch t

y = y1dydx

= dy1dx

= y2d2ydx2

= dy2dx

= y3

ta c

dy1dx

= y2dy2dx

= y3dy3dx

= xy3 + y22 + y1.

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


41/93

3.2. PHNG TRNH VI PHN TUYN TNH THUN NHT 41

H(3.4) l mt dng c bit ca h phng trnh

dy1dx

= f1(x, y1, y2, . . . , yn)dy2dx

= f2(x, y1, y2, . . . , yn)

. . . . . . . . . . . . . . . . . . . . .dyndx

= fn(x, y1, y2, . . . , yn).

3.2 Phng trnh vi phn tuyn tnh thun nht

3.2.1 Kh i nim phng trnh vi phn tuyn tnh thun nht cp cao

nh ngha 3.2.1. Phng trnh vi phn tuyn tnh cp n c dng tng qut l

a0(x)y(n) + a1(x)y

(n1) + + an(x)y = f(x) (3.5)trong y(k) l cc o hm cp k ca y theo bin x, cn a0(x), a1(x), . . . , an(x), f(x)l nhng hm s bt k ca x. Ta gi thit cc hm s ny lin tc trn mt khong(a, b) no .

Nu f(x) 0 th phng trnh

a0(x)y

(n)

+ a1(x)y

(n

1)

+ + an(x)y = 0c gi l phng trnh thun nht, cn khi f(x) 0 th phng trnh (3.5) cgi l phng trnh khng thun nht.

Khi cc hm s a0(x), a1(x), . . . , an(x) l nhng hm hng th ta gi phngtrnh (3.5) l phng trnh vi phn tuyn tnh vi h s hng.



42/93


Nhn xt 3.2.1. Khi a0(x) = 0 trn (a, b) th trn khong ny phng trnh (3.5)tng ng vi

y(n) + p1(x)y(n1) + + pn1y + pn(x)y = g(x) (3.6)

trong pi(x) = ai(x)a0(x) vg(x) = f(x)a0(x)

.

Khi cc hm s p1(x), p2(x), . . . , pn(x).g(x) lin tc trn (a, b) th bi tonCauchy ca phng trnh (3.6) s c nghim duy nht trn min (a, b) Rn. Vvy t nay v sau ta ch nghin cu phng trnh dng (3.6).

3.2.2 Mt s tnh cht ca nghim phng trnh thun nht

Trc tin ta nghin cu mt s tnh cht ca nghim phng trnh vi phn tuyn

tnh thun nhty(n) + p1(x)y

(n1) + + pn1(x)y + pn(x)y = 0 (3.7)trn mt khong (a, b).

Vi mi hm kh vi cp n trn (a, b), t

L[y] = y(n) + p1(x)y(n1) + + pn1(x)y + pn(x)y.

Ta cL[y1 + y2] = (y1 + y2)

(n) + p1(x)(y1 + y2)(n1) + +

+ pn1(x)(y1 + y2) + pn(x)(y1 + y2)L[y1 + y2] = y

(n)1 + y

(n)2 + p1(x)(y

(n1)1 + y

(n1)2 ) + +

+ pn1(x)(y1 + y2) + pn(x)(y1 + y2)

= L[y1] + L[y2].

Nu C R thL[Cy] = (Cy)(n) + p1(x)(Cy)

(n1) + + pn1(x)(Cy) + pn(x)(Cy) = CL[y].Vy L l mt nh x tuyn tnh t khng gian vect cc hm s kh vi cp n

trn khong (a, b) vo khng gian cc hm s xc nh trn [a, b].Da vo tnh tuyn tnh ca L, ta suy ra cc khng nh sau.

Mnh 3.2.1. a) Nu (x) l mt nghim ca phng trinh vi phn tuyn tnhthun nht (3.7) v C l mt hng s ty th C(x) cng l nghim ca (3.7).

Nu 1(x), 2(x), . . . , k(x) l cc nghim ca (3.7) th

C11(x) + C22(x) + + ckk(x)cng l nghim ca (3.7), trong C1, C2, . . . , C k l cc hng s ty .

Ni cch khc, tp hp cc nghim ca phng trnh (3.7) to thnh mt khnggian vect.

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


43/93

3.2.3. S ph thuc tuyn tnh v c lp tuyn tnh ca hm 43

3.2.3 S ph thuc tuyn tnh v c lp tuyn tnh ca hm

Gi s ta c mt h cc hm s cng xc nh trn khong (a, b).

nh ngha 3.2.2. H cc hm 1, 2, . . . , k c gi l ph thuc tuyn tnh trn

khong (a, b) nu tn ti cc hng s c1, c2, . . . , ck khng ng thi bng khngsao cho

c11(x) + c22(x) + + ckk(x) 0 (3.8)trn (a, b).

Mt h khng phi ph thuc tuyn tnh c gi l c lp tuyn tnh. Vymt h hm 1(x), 2(x), . . . , k(x) c lp tuyn tnh trn (a, b) nu v ch nu(??) xy ra khi c1 = c2 = = ck = 0.

V d 3.2.1. a) H {sin x, cos x} c lp tuyn tnh trn mi khong.b) H {1, x , x2, . . . , xk} c lp tuyn tnh trn mi khong (a, b) v nu c1, c2, . . . , ck+1khng ng thi bng 0 th c1 + c2x + + ck+1xk l mt a thc nn c hu hnnghim. Do

c1 + c2x + + ck+1xk 0trn mi khong (a, b).

c) Nu 1, 2, . . . , k l cc s i mt khc nhau th h {e1x, e2x, . . . , ekx}l c lp tuyn tnh trn mi khong (a, b). Tht vy gi s tn ti (, ) v cc

hng s (1, 2, . . . , k) sao cho1e

1x + 2e2x + + kekx 0 (3.9)

trn (, ). Chia hai v ca (3.9) cho e1x ri ly o hm ng nht thc ny, tac

2(2 1)e(21)x + + k(k 1)e(k1)x 0.Tip tc chia hai v ca ng nht thc ny cho e(21)x ri ly o hm, ta i

n

3(3 1)(3 2)e(32)x + + k(k 2)e(k2)x 0.C nh vy mi, cui cng ta c ng nht thc

k(k 1)(k 2) (k k1)e(kk1)x 0.T y suy ra k = 0 vi k.

H {1, sin2 x, cos2 x, ex} ph thuc tuyn tnh v nu chn c1 = c3 = 1, c2 =

1, c4 = 0 ta s c

c1 1 + c2 sin2 x + c3( cos2 x) + c4ex = 1 sin2 x cos2 x + 0 ex 0.

http://-/?-http://-/?-


44/93


3.2.4 nh thc Wronski

Vic nghin cu s c lp tuyn tnh ca mt h hm s n gin hn khi ta dngnh thc Wronski. Cho 1(x), 2(x), . . . , k(x) l cc hm kh vi ti cp k 1trn (a, b). Khi nh thc

W[1, 2, . . . , k](x) =

1(x) 2(x) . . . k(x)1(x)

2(x) . . .

k(x)

. . . . . . . . . . . .

(k1)1 (x)

(k1)2 (x) . . .

(k1)k (x)

c gi l nh thc Wronski ca h hm trn.

V d 3.2.2. H {x, sin x, cos x} c nh thc Wronski l

W =

x sin x cos x1 cos x sin x0 sin x cos x

= x.nh l 3.2.1. Gi s h hm {1(x), 2(x), . . . , k(x)} kh vi ti cp k 1 trn(a, b) v h ny ph thuc tuyn tnh trn khong ny. Khi nh thc Wronskica h s ng nht bng 0 trn (a, b).

H qu 3.2.1. Nu nh thc Wronski ca h hm {1(x), 2(x), . . . , k(x)} khc0 d ch ti mt im ca (a, b) th h hm trn c lp tuyn tnh trn (a, b).

V d 3.2.3. H {x, sin x, cos x} c lp tuyn tnh trn mi khong (a, b). (Xemv d 3.2.2.

Nhn xt 3.2.2. nh l 3.2.1 ch l iu kin cn ch khng phi l iu kin .Chng hn h hm

y1(x) =

(x 1)2 nu 0 x 10 nu 1 < x 2,

y2(x) =

0 nu 0 x 1(x 1)2 nu 1 < x 2

c W[y1, y2] 0 trn [0, 2] nhng y1(x), y2(x) c lp tuyn tnh trn [0, 2].Tuy nhin i vi h n nghim ca phng trnh vi phn tuyn tnh cp n, nh

thc Wronski ca h s ng nht bng 0 trn (a, b) cng chnh l iu kin cnv nh ta thy di y.

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


45/93

3.2.5. H nghim c bn. Nghim tng qut 45

nh l 3.2.2. Gi s y1, y2, . . . , yn l n nghim ca phng trnh vi phn tuyntnh thun nht cp n 3.7. iu kin cn v h hm trn ph thuc tuyntnh trn (a, b) l nh thc Wronski ca h ng nht bng khng trn khongny. Ni cch khc, nh thc Wronski ca n nghim ca mt phng trnh vi phn

tuyn tnh cp n s ng nht bng khng hoc khc khng ti mi im.

3.2.5 H nghim c bn. Nghim tng qut

nh ngha 3.2.3. H gm n nghim c lp tuyn tnh ca phng trnh tuyntnh thun nht cp n c gi l h nghim c bn ca phng trnh .

By gi ta kho st h nghim c bn ca phng trnh tuyn tnh thun nht

y(n) + p1(x)y(n1) + + pn1(x)y + pn(x)y = 0 (3.10)

trong cc h sp1(x), p2(x), . . . , pn(x) l nhng hm lin tc trn khong (a, b).

nh l 3.2.3. Vi mi ma trn vung khng suy bin A cp n, tn ti mt hnghim c bn ca phng trnh (3.10) c nh thc Wronski ti im x0 (a, b)cho trc bng det A. Do mi phng trnh vi phn tuyn tnh thun nht uc v s h nghim c bn.

Chng minh. Ly ty x0 (a, b). Do phng trnh (3.10) c duy nht nghimtrn (a, b)Rn nn tn ti nghim y1(x), y2(x), . . . , yn(x) tha mn iu kin u

y1(x0) = a11, y1(x0) = a21, . . . y

(n1)1 (x0) = an1

y2(x0) = a11, y2(x0) = a21, . . . y

(n1)2 (x0) = an2

. . . . . . . . . . . .

yn(x0) = a11, yn(x0) = a21, . . . y

(n1)n (x0) = ann.

R rng nh thc Wronski ca h nghim ny ti im x0 l

W[y1, y2, . . . , yn](x0) =

a11 a12 . . . a1na21 a22 . . . a2n. . . . . . . . . . . .

an1 an2 . . . ann

= det A = 0.

Do h nghim trn l h nghim c bn ca phng trnh (3.10). V c v s

nh thc A (cng nh im x0) nh vy nn phng trnh c v s h nghim cbn.



46/93


nh l 3.2.4. Gi s{y1, y2, . . . , yn} l mt h nghim c bn ca phng trnhvi phn tuyn tnh thun nht cp n (3.10). Khi biu thc

y(x) = C1y1(x) + C2y2(x) + + Cnyn(x)trong C1, C2, . . . , C n l cc hng s s cho ta nghim tng qut ca phngtrnh (3.10).

nh l 3.2.5. Mi h gm n + 1 nghim ca phng trnh tuyn tnh thun nhtcp n u ph thuc tuyn tnh.

Nhn xt 3.2.3. T cc nh l ny ta suy ra tp nghim ca phng trnh viphn tuyn tnh thun nht cp n to thnh mt khng gian vectn chiu. Mi hnghim c bn chnh l mt c s ca khng gian nghim.

3.3 Phng trnh tuyn tnh khng thun nht

Trong phn ny ta s xt mt s tnh cht v cu trc nghim ca phng trnh viphn tuyn tnh khng thun nht

L[y] = y(n) + p1(x)y(n1) + + pn1(x)y + pn(x)y = f(x) (3.11)

trong cc hm s p1(x), p2(x), . . . , f (x) lin tc trn (a, b). Da vo tnh chttuyn tnh ca ton t L ta d dng suy ra cc tnh cht sau.

a) Nu z(x) l nghim ca phng trnh tuyn tnh thun nht

L[y] = y(n) + p1(x)y(n1) + + pn1(x)y + pn(x)y = 0 (3.12)

cn y1 l nghim ca phng trnh khng thun nht (3.11) th y(x) = y1(x) + z(x)cng l nghim ca phng trnh khng thun nht (3.11) v

L[y] = L[y1 + z] = L[y1] + L[z] = f(x) + 0 = f(x).

Nu phng trnh L[y] = f1(x) c nghim y1(x) v phng trnh L[y] f2(x)c nghim y2(x) th phng trnh L[y] = f1(x) + f2(x) c nghim y1(x) + y2(x).Tnh cht ny gi l nguyn l chng cht nghim.

nh l 3.3.1. Gi s{y1(x), y2(x), . . . , yn(x)} l mt nghim c bn ca phngtrnh thun nht cp n L[y] = 0 v y(x) l mt nghim ring no ca phngtrnh L[y] = f(x). Khi

y(x) = C1y1(x) + C2y2(x) + + Cnyn(x) + y(x)Bin son: GVC.ThS. Phan vn Danh


47/93

3.4. PHNG TRNH VI PHN TUYN TNH VI H S HNG 47

s l nghim tng qut ca ca phng trnh khng thun nht L[y] = f(x). Nicch khc nghim tng qut ca phng trnh khng thun nht bng nghim tngqut ca phng trnh thun nht cng vi nghim ring ca phng trnh khngthun nht.

Nh vy, tm nghim tng qut ca phng trnh khng thun nht, ta tinhnh theo hai bc sau.

Bc 1. Tm mt nghim c bn ca phng trnh thun nht tng ng.

Bc 2. Tm mt nghim ring ca phng trnh khng thun nht.

Nghim tng qut ca phng trnh khng thun nht bng tng ca nghim tngqut ca phng trnh thun nht vi mt nghim ring ca phng trnh khng

thun nht.V d 3.3.1. Xt phng trnh

y + 4y = ex.

Gi s ta bit c rng phng trnh thun nht tng ng

y + 4y = 0

c cc nghim y1(x) = cos 2x, y2(x) = sin 2x. D dng kim tra c hai nghim

ny c lp tuyn tnh (v nh thc Wronski) ca chng khc 0) nn y l mth nghim c bn ca phng trnh thun nht v y(x) =

1

5ex l mt nghim

ring ca phng trnh khng thun nht. Do phng trnh khng thun nht cnghim tng qut l

y(x) = C1 cos2x + C2 sin2x +1

5ex, C1, C2 R.

Ch . Ging nh phng trnh khng thun nht, phng php tm nghim

tng qut vt ht tt c cc nghim ca n.

3.4 Phng trnh vi phn tuyn tnh vi h s hng

3.4.1 Phng trnh vi phn tuyn tnh cp hai i vi h s hng

nh ngha 3.4.1. Phng trnh vi phn tuyn tnh cp hai i vi h s hng lphng trnh dng

L[y] = y + a 1y + a2y = 0 (3.13)trong a1, a2 l cc hng s.



48/93


Ta ch cn tm h nghim c bn ca phng trnh (3.13). Ta tm chng l nhngnghim c dng y = ex. Ta c y = ex, y = 2ex nn

L[y] = (2 + a1 + a2)ex.

a thc () = 2 + a1 + a2 c gi l a thc c trng ca phng trnh(3.13). Nh vy ta s c

L[y] = ()ex.

Do ex = 0 vi mi x nn y = ex l nghim ca (3.13) khi v ch khi l nghimca phng trnh

() = 2 + a1 + a2 = 0. (3.14)

Phng trnh ny c gi l phng trnh c trng tng ng ca phng trnh

thun nht (3.13). Ta xt cc trng hp sau.

a) Trng hp 1. Phng trnh c trng (3.14) c hai nghim thc phn bit1, 2, (1 = 2). Khi e1x v e2x l hai nghim ca phng trnh thun nht(3.13). V vy nghim tng qut ca phng trnh (3.13) l

y = C1e1x + C2e

2x

trong C1, C2 l cc hng s ty .

V d 3.4.1. Tm nghim tng qut ca phng trnh

y 5y + 6y = 0.Phng trnh c trng tng ng

2 5 + 6 = 0c hai nghim phn bit 1 = 2 v 2 = 3. Do nghim tng qut ca phng

trnh thun nht ly = C1e

2x + C2e3x, C1, C2 R.

b) Trng hp 2. Phng trnh c trng khng c nghim thc. Khi =a21 4a2 < 0 v phng trnh c trng c hai nghim phc lin hp

+ i, i.

B 3.4.1. Nu phng trnh (3.13) c nghim phc y(x) = u(x) + iv(x) thphn thc u(x) v phn o v(x) cng l nghim thc ca phng trnh ny.

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


49/93

3.4.1. Phng trnh vi phn tuyn tnh cp hai i vi h s hng 49

Chng minh. Theo gi thit ta c

L[y(x)] = L[u(x) + iv(x)] = L[u(x)] + iL[v(x)] = 0.

Vy L[u(x)]

0 v L[v(x)]

0 hay u(x) v v(x) l nhng nghim thc caphng trnh (3.13).

T biu thc e(+i)x = ex cos x + iex sin x l nghim phc ca phngtrnh (3.13), theo b trn phng trnh thun nht c hai nghim l ex cos xv ex sin x. Vy nghim tng qut ca (3.13) lc ny l

y = C1ex sin x + C2e

x cos x



y + 4y = 0.

Phng trnh c trng 2 + 4 = 0 c nghim phc = 2i ( y = 0).Do phng trnh cho c h nghim c bn l e0x sin2x v e0x cos2x v cnghim tng qut l

y = C1 sin2x + C2 cos2x, C1, C2

R.


y + y + y = 0.

Phng trnh c trng 2 + + 1 = 0 c nghim 1 = 12

+ i

3

2, 2 =

12 i

3

2. Vy nghim tng qut ca phng trnh l

y = C1ex

2 sin 32

x + C2ex

2 cos 32

x, C1, C2 R.

c) Trng hp 3. Phng trnh c trng c nghim kp. Gi s phng trnhc trng

() = 2 + a1 + a2 = 0

c nghim kp = a12

(lc ny = a21 4a2 = 0). Khi y1(x) = exl mt nghim ca phng trnh (3.13). Ta ch cn tm mt nghim na thu

c mt h nghim c bn cho (3.13). Ta tm nghim th hai ny di dngy2(x) = u(x) y1(x) = u(x)ex. (Phng php Bernoulli).



50/93


Ta cy = uex, y = (u + 2u + 2u)ex.

Thay vo (3.13) ta c

L[y2] = (u + 2u + 2u)ex + a1(u + u)ex + a2ex

= uex + (a1 + 2)uex + (2 + a2u + a2)uex = 0

v L[y2] = 0.Do l nghim kp ca phng trnh c trng 2 + a1u + a2 = 0 nn

a1 + 2 = 0 v ta suy ra uex = 0 hay u = 0. T y ta c u = C1x + C2.do ch cn mt hm u nn c th chn u(x) = x. D dng kim tra c h hm

{ex, xex

}c lp tuyn tnh nn n l mt h nghim c bn ca phng trnh

(3.13).Vy nghim tng qut ca phng trnh (3.13) l

y(x) = C1ex + C2xe

x, C1, C2 R.


y + 4y + 4y = 0.

Phng trnh c trng 2 + 4 + 4 = 0 c nghim kp = 2. Vy nghimtng qut ca phng trnh l

y(x) = C1e2x + C2xe2x, C1, C2 R.

3.4.2 Phng trnh vi phn tuyn tnh thun nht cp cao vi h s hng

nh ngha 3.4.2. Phng trnh vi phn tuyn tnh thun nht cp n vi h shng l phng trnh dng

y(n) + a1y(n1) + + an1y + any (3.15)trong a1, a2, . . . , an l cc hng s.

Tng t nh phng trnh cp 2, ta tm nghim ca phng trnh dngy = ex. Khi phi l nghim ca phng trnh

() = n + a1n1 + + an1 + an = 0. (3.16)

Phng trnh ny c gi l phng trnh c trng ca phng trnh thun nht(3.15).



51/93

3.4.2. Phng trnh vi phn tuyn tnh thun nht cp cao vi h s hng 51

a) Trng hp 1. Phng trnh c trng (3.16) c n nghim thc phn bit1, 2, . . . , n th nghim tng qut ca phng trnh thun nht (3.16) s l

y(x) = C1e1x + C2e

2x + + Cnenx

trong C1, C2, . . . , C n l nhng hng s ty .b) Trng hp 2. Nu i l cp nghim phc lin hp ca phng trnh

c trng (3.16) (= 0) th ex sin(x) v ex cos(x) s l nghim ca phngtrnh thun nht (3.15).

c) Trng hp 3. Nu l nghim thc bi m ca phng trnh c trng thta s c cc nghim

ex, xex, , xm1ex

ca phng trnh thun nht.d) Trng hp 4. Nu i l cp nghim phc bi m ca phng trnh c

trng thex sin x xex sin x . . . xm1ex sin xex cos x xex cos x . . . xm1ex cos x

l cc nghim ca phng trnh thun nht.


y(4) 3y(3) + 3y y = 0.Phng trnh c trng tng ng

4 33 + 32 = 0c nghim n = 0 v nghim bi 3 l = 1. Do nghim tng qut caphng trnh cho l

y = C1 + ex(C2 + C3x + C4x

2), Ci R, 1 i 4.

V d 3.4.6. Tm nghim tng qut ca phng trnhy(5) y(4) + 8y(3) 8y + 16y 16y = 0.

Phng trnh c trng

5 4 + 83 82 + 16 16 = 0c mt nghim thc n 1 = 1 v cp nghim phc lin hp bi 2 l 2 = 3 =2i, 4 = 5 = 2i. Do nghim tng qut ca phng trnh cho l

y = C1ex + C2 cos2x + C3 sin2x + C4x cos2x + C5x sin2x,

trong Ci R, 1 i 5.



52/93


3.5 Phng trnh vi phn tuyn tnh cp hai khng thun nht

Di y ta nghin cu phng trnh vi phn tuyn tnh cp cao n gin nht l phng trnh cp hai.

3.5.1 Phng php h s bt nh

Phng php ny ch s dng cho phng trnh vi phn tuyn tnh vi h s hng, tng ca phng php: Vi trng hp v phi ca phng trnh

L[y] = f(x) (3.17)

c mt s dng no , ta s cho dng nghim ring y(x) tng ng (c cha

cc tha s hng s ty ). Cc tha s ny s c xc nh khi thay y(x) vophng trnh cho ri cn bng cc h s. Ni chung dng nghim ring y(x)c dng nh dng ca f(x) v cc o hm ca f(x).

a) Trng hp 1. f(x) l mt a thc vi f(x) = Pm(x) =mi=0

Aixmi.

Nu s 0 khng phi l nghim ca phng trnh c trng th ta gi thitnghim ring c dng

y(x) = Qm(x) =

mi=0

Bixmi

trong Bi l cc h s ta phi xc nh.Nu s 0 l nghim bi k ca phng trnh c trng th ta tm nghim ring

c dng

y(x) = xkQm(x) = xkmi=0

Bixmi.

V d 3.5.1. Tm nghim ring ca phng trnh

y 3y 4y = 4x2.Phng trnh c trng 234 = 0 c nghim 1 = 1, 2 = 4. S 0 khng

phi l nghim ca phng trnh c trng. Do f(x) = 4x2 l a thc bc hai nnta tm c nghim ring cng l a thc bc hai,

y(x) = B0x2 + B1x + B2.

Khi y

= 2B0x + B1 v y

= 2B0. Thay vo phng trnh cho ta c

4B0x2 (6B0 + 4B1)x + (2B0 3B1 4B2) = 4x2



53/93

3.5.1. Phng php h s bt nh 53

2B0 3B1 4B2 = 06B0 + 4B1 = 0

4B0 = 4

Gii h ny ta c B0 = 1, B1 = 32

, B2 = 138

. Vy ta c nghim ring ca

phng trnh vi phn l

y(x) = x2 + 32

x 138

.

b) Trng hp 2. f(x) = exPm(x) vi Pm(x) l a thc bc m.Nu khng l nghim ca phng trnh c trng th ta tm y dng y(x) =

exQm(x) trong Qm cng l a thc bc m.Nu l nghim bi k ca phng trnh c trng th ta tm y dng

y(x) = xkexQm(x).


y 3y + 2y = 5ex.Phng trnh c trng 2

3 + 2 = 0 c nghim 1 = 1, 2 = 2. V

f(x) = exP0(x), = 1 = 1 l nghim n nn ta tm y dng y(x) = xexQ0(x) =xexB. Thay vo phng trnh cho v ng nht h s, ta i n phng trnhBex = 5ex. T y suy ra B = 5 v nghim ring y(x) = 5xex.

c) Trng hp 3. f(x) = ex[Pm(x)cos x + Pn(x)sin x].Nu + i khng phi l nghim ca phng trnh c trng th

y(x) = ex[P(x)cos x + P(x)sin x]

trong = max{m, n}.



54/93


Nu + i l nghim bi k ca phng trnh c trng th

y(x) = xkex[P(x)cos x + P(x)sin x]

trong = max{

m, n

}.


y y = 2sin x.Phng trnh c trng 2 = 0 c nghim 1 = 0, 2 = 1. V phi ca

phng trnh c dng f(x) = 2 sin x + 0 cos x = e0xP0(x). S phc 0 + i khngphi l nghim ca phng trnh c trng nn tm nghim y dng y(x) =A cos x + B sin x.Ta c

y = A sin x + B cos x, y = A cos x B sin x.Thay vo phng trnh cho v ng nht cc h s, ta c

A + B = 0

A B = 2.T A = 1, B = 1 v nghim ring ca phng trnh l y(x) = cos x sin x.

V d 3.5.4. Gii phng trnhy + y = 2 sin x.

Phng trnh c trng 2+1 = 0 c nghim phc 1 = i, 2 = i. Vy phngtrnh thun nht c nghim tng qut

y = C1 cos x + C2 sin x.

Ta thy hm f(x) = 2 sin x c dng

f(x) = e0x[P0(x)cos x + Q0(x)sin x]

v 0 + i 1 l nghim n ca phng trnh c trng nn ta tm nghim ring dng

y(x) = x(A cos x + B sin x).

Thay vo phng trnh cho ta tm c A = 1, B = 0 v do y(x) =x cos x.

Vy nghim tng qut ca phng trnh cho l

y(x) = C1 cos x + C2 sin x x cos x, C1, C2 R.



55/93

3.5.1. Phng php h s bt nh 55

d) Trng hp 4. f(x) = f1(x) + f2(x) + + fn(x).Ta s tm nghim ring ca tng phng trnh L[y] = fi(x), i = 1, . . . , n.

Nghim ring ca phng trnh L[y] = f(x), theo nguyn l chng cht nghim,s l y

(x) = y1(x) + y2(x) +

+ yn(x).


y y = 5ex sin2x.

Phng trnh c trng 2 = 0 c nghim 1 = 0, 2 = 1 nn nghim tngqut ca phng trnh thun nht tng ng l

y = C1 + C2ex.

Ta tm nghim ring ca cc phng trnh

y y = 5ex, (3.18)

y y = sin2x. (3.19)i vi phng trnh (3.18), v phi c dng f1(x) = ex[P0(x)sin0x+Q0(x)cos0x]

vi = 1 l nghim ca phng trnh c trng nn nghim ring c dngy1(x) = Axe

x. i vi phng trnh (3.19), v phi f2(x) = e0x[P0(x)sin2x +Q0(x)cos2x] nn c nghim ring dng y2(x) = B cos2x + Csin2x. Thay ccnghim y1(x), y2(x) vo cc phng trnh tng ng ta tm c A = 5, B =

110

, C =1

5. Do y1(x) = 5xex, y2(x) = 1

10cos2x +

1

5sin2x v ta c nghim

ring ca phng trnh cho l

y(x) = y1(x) + y2(x) = 5xex

1

10 cos2x +

1

5 sin2x.

Vy phng trnh cho c nghim tng qut l

y = 5xex 110

cos2x +1

5sin2x + C1e

x + C2.

Kt lun. Phng php h s bt nh ni chung ch p dng c khi hm v phic dng f(x) = ex[Pm(x)sin x +Pn(x)cos x] hoc l tng qut ca nhng hm

loi . Khi f(x) khng c dng ny, ta phi dng mt phng php khc mnhhn, gi l phng php bin thin hng s hay cn gi l phng php Lagrange.

http://-/?-http://-/?-http://-/?-http://-/?-


56/93


3.5.2 Phng php bin thin hng s

Kho st phng trnh vi phn tuyn tnh cp hai

y + p1(x)y + p2(x)y = q(x). (3.20)

Gi s ta bit c phng trnh thun nht tng ng

y + p1(x)y + p2(x)y = 0. (3.21)

c mt h nghim c bn l y1(x), y2(x) v c c nghim tng qut ca phngtrnh thun nht

y(x) = C1y1(x) + C2y2(x)


Ta tm nghim ring ca (3.20) di dng

y(x) = C1(x)y1(x) + C2(x)y2(x) (3.22)

ngha l phi tm C1(x), C2(x) sao cho biu thc tha mn phng trnh (3.20).Nh vy c mt h thc lin h gia C1(x) v C2(x), ta cn tm mt h thcmi na mi c th xc nh c C1(x) v C2(x).

T (3.22) ta suy ra

y(x) = C1(x)y

1

(x) + C2(x)y2

(x) + C1

(x)y1(x) + C2

(x)y2(x).

Ta chn cc hm C1(x), C2(x) sao cho tng lin quan ti o hm cp 1 ca chngtrit tiu,

C1(x)y1(x) + C2(x)y2(x) = 0.

Di iu kin ny

y(x) = C1(x)y1(x) + C2(x)y

2(x)

nn

y(x) = C1(x)y1(x) + C1(x)y1(x) + C2(x)y2(x) + C2(x)y2(x).Thay y v y vo phng trnh (3.20) ta c

C1y1 + C1y

1 + C

2y2 + C2y

2 + p1(C1y

1 + C2y

2) + p2(C1y1 + C2y2) = q

hay

C1(y1 + p1y

1 + p2y1) + C2(y

2 + p1y

2 + p2y2) + C

1y1 + C

2y

2 = q.

Do y1, y2 l cc nghim ca phng trnh thun nht (3.21) nn

y1 + p1y1 + p2y1 = 0 v y

2 + p1y

2 + p2y2 = 0.



57/93

3.5.2. Phng php bin thin hng s 57

Ta suy raC1y

1 + C

2y2 = q.

nh thc h ny

y1 y2y1 y2 = W[y1, y2] (3.23)chnh l nh thc Wronski ca h {y1, y2}. V y l nghim c bn ca phngtrnh thun nht nn W[y1, y2] = 0 ti mi im thuc (a, b). Do (3.23) l hCrame v v vy lun c nghim. Gii C1, C

2 t h (3.23) ta s suy ra c C1 v

C2.


y + y =

1

cos x.

Phng trnh c trng c nghim phc 1 = i, 2 = i nn phng trnh thunnht tng ng c nghim tng qut

y(x) = C1 sin x + C2 cos x.

V vy ta tm mt nghim ring dng

y(x) = C1(x)sin x + C2(x)cos x

C1, C2 phi tha mn h phng trnh

C1 sin x + C2 cos x = 0

C1 cos x C2 sin x =1

cos x.

Gii h ny ta c C1 = 1, C2 = tg x nn ta c th chn C1(x) = x, C2 =

ln | cos x|. Vy ta c nghim ringy(x) = x sin x + cos x ln

|cos x

|v nghim tng qut ca phng trnh cho l

y(x) = x sin x + cos x ln | cos x| + C1 sin x + C2 cos x.V d 3.5.7. Gii phng trnh

xy y = x2.Ta a phng trnh v dng

y 1x

y = x.



58/93


Xt phng trnh thun nht tng ng

y 1x

y = 0.

t z = y, ta a n phng trnhz 1

xz = 0.

Tch bin ta cdz

z dx

x= 0

v suy raln |z| ln |x| = ln |C|.

T y, z = C1x v do y = C1x

2 + C2.

By gi ta tm nghim ca phng trnh cho di dng

y(x) = C1(x)x2 + C2(x)

trong C1, C2 tha mn h

C1x2 + C2 = 0C1x + C20 = x.

Gii h ny ta c C1(x) = 1 v C2(x) = x2 nn C1(x) = x, C2(x) =

1

3x3.

Vy ta c nghim ring y(x) = x3 13

x3 =2

3x3 v t c nghim tng qut

ca phng trnh cho l

y(x) =2

3x3 + C1x

2 + C2.

Nhn xt. Cch tm nghim ring ca phng trnh vi phn tuyn tnh cp n cngtng t nh trn. Gi s cn tm nghim ring ca phng trnh

y(n)p1(x)y(n1) + + pn1(x)y + pn(x)y = f(x)

vi gi thit nghim tng qut ca phng trnh thun nht tng ng

y(n)p1(x)y(n1) + + pn1(x)y + pn(x)y = 0

c bit l

y(x) = C1y1(x) + C2y2(x) + + Cnyn(x)Bin son: GVC.ThS. Phan vn Danh


59/93

3.5.3. Tm h nghim c bn khi bit mt nghim 59

trong C1, C2, . . . , C n l cc hng s ty . Khi ta tm mt nghim ring caphng trnh khng thun nht di dng

y(x) = C1(x)y1(x) + C2(x)y2(x) + + Cn(x)yn(x).

Ta c th tm c cc hm s C1(x), C2(x), . . . , C n(x) nh cc o hm cpmt ca chng l nghim ca h phng trnh

C1y1 + C2y2 + + Cnyn = 0

C1y1 + C

2y

2 + + Cnyn = 0

. . . . . . . . . . . . . . .

C1y(n2)1 + C

2y

(n2)2 + + Cny(n2)n = 0

C1y(n1)1 + C

2y

(n1)2 +

+ Cny

(n1)n = f(x).

y l h thc Crame v nh thc ca h lun khc 0 v n chnh l nh thcWronski W[y1, y2m . . . , yn] ca h nghim c bn y1, y2, . . . , yn ca phng trnhthun nht. Gii h ny ta tm c C1, C

2, . . . , C

n v t c C1(x), C2(x), . . . , C n(x).

3.5.3 Tm h nghim c bn khi bit mt nghim

Vi phng trnh vi phn tuyn tnh thun nht c h s khng phi l hng s,ni chung ta khng c phng php tng qut tm c mt h nghim c bn.

Tuy nhin i vi phng trnh cp hai, nu bit c mt nghim khc 0 th tac th tm nghim th hai c lp tuyn tnh vi nghim th nht bng mt trongnhng phng php sau.

1. Phng php th nht (Phng php Bernoulli).Gi s phng trnh thun nht

y + p1(x)y + p2(x)y = 0 (3.24)

c nghim y1(x) 0. Ta tm mt nghim th hai c lp tuyn tnh vi y1 dngy2(x) = u(x)y1(x). Thay vo (3.24) ta cuy1 + u(2y1 + p1y1) + u(y

1 + p1y

1 + p2y) = 0.

T uy1 + u(2y1 + p1y1) = 0

(v y1 l nghim ca phng trnh vi phn).t v = u ta c phng trnh tuyn tnh cp mt

v +

2y1y1

+ p1

v = 0.

http://-/?-http://-/?-


60/93


Sau khi tm c v ta s tm c u. Dng nh thc Wronski, ngi ta chngminh c hai nghim y1, y2 tm c theo cch trn l c lp tuyn tnh.


y + 3x

y + 1x2

y = 0

bit rng phng trnh c nghim l y1(x) =1

x.

t y2(x) =1

xu(x). Thay vo phng trnh cho, ta i n phng trnh

v +1

xv = 0.

Tch bin v tch phn ln ta c v(x) =1

x. Vy

u(x) =

1

xdx = ln |x|, y2(x) = 1

xln |x|

v ta c nghim tng qut

y = C11

x+ C2

1

xln |x|.

Phng php th haiTrc tin ta ch ra rng nh thc Wronski

W[y1, y2] =

y1 y2y1 y2 = y1y2 y1y2

ca hai nghim y1, y2 bt k ca phng trnh (3.24) tha mn phng trnh

W(x) + p1(x)W(x) = 0. (3.25)

Tht vy

W(x) = (y1y2 y1y2) = y1y2 y1y2= y1(p1y2 p2y2) y2(p1y1 p2y1)= p1(y

1y2 y1y2) = p1(x)W(x)

nn ta c phng trnh (3.25).Do

W(x) = Cep1(x)dx, C1 = const

hocW(x) = W(x0)e

p1(x)dx.Bin son: GVC.ThS. Phan vn Danh


61/93


By gi ta bit mt nghim khc khng l y1(x) ca (3.24) v gi y(x) lmt nghim bt k khc vi y1(x). Theo chng minh trn ta c

W[y1, y] = y1y y1y = C1e

p1(x)dx.

Chia hai v cho y21(x) = 0 ta c phng trnhd

dx

y1

=1

y21C1e

p1(x)dx.T y ta c nghim tng qut ca phng trnh (3.24) l

y = y1

C1

ep1(x)dx

y21(x)dx + C2

.

V d 3.5.9. Gii phng trnh sau(1 x2)y 2xy + 2y = 0.

Ta d dng kim tra c y = x l mt nghim ca phng trnh. Chuyn v

dng (3.24) ta c p1(x) = 2x1 x2 nn nghim tng qut ca phng trnh cho

l

y = x

C1

e

2x1x2

dx

x2dx + C2

= x

C1

dx

x2(1

x2)

+ C2= x

1

x+ 1

2ln 1 + x

1 x + C2

= C2x + C1

x2

ln 1 + x1 x 1

.

3. Phng trnh Euler

nh ngha 3.5.1. Phng trnh tuyn tnh dng

xny(n) + an1xn1y(n1) + + a1xy + a0y = 0 (3.26)trong a0, a1, . . . , an

1 l cc hng s c gi l phng trnh Euler.

Cch gii. C hai phng php gii phng trnh ny.

1) Phng php th nhtTa tm nghim ca phng trnh di dng y = xr. Thay vo phng trnh

cho ta suy ra

r(r 1)(r 2) (r n + 1) + an1r(r 1)(r 2) (r n) + +a2r(r

1) + a1r + a0 = 0

y l phng trnh bc n v c gi l phng trnh c trng.



62/93


a Trng hp 1. Phng trnh c trng c n nghim thc phn bit th phngtrnh Euler c nghim tng qut dng

y = C1xr1 + C2x

r2 + + Cnxrn

trong C1, C2, . . . , C n l cc hng s ty .b Trng hp 2. Nu r l nghim bi m ca phng trnh c trng th

xr, xr ln x , . . . , xr(ln x)m1

l cc nghim ca phng trnh Euler.

c Trng hp 3. Mi cp nghim phc lin hp i ca phng trnh ctrng c th nhm li v tham gia vo cp nghim

x cos(ln x), x sin(ln x).

d Trng hp 4. Mi cp nghim phc lin i bi m ca phng trnh ctrng s cho 2m nghim ca phng trnh Euler

x cos(ln x), x cos(ln x) ln x , . . . , x cos(ln x)(ln x)m1

x sin(ln x), x sin(ln x) ln x , . . . , x sin(ln x)(ln x)m1

V d 3.5.10. Gii phng trnhx2y + 5xy 2y = 0.

Ta tm nghim dng y = xr. Thay vo phng trnh ta c

r(r 1)xr + 5rxr 2xr = 0hay r2 + 4r 2 = 0. Phng trnh ny c hai nghim l r = 26 nn phngtrnh vi phn c h nghim c bn l y1 = x2+

6, y2 = x

26.Vy nghim tng qut ca phng trnh l

y = C1x2+6 + C2x2

6.


x2y + 5xy + 4y = 0.

Phng trnh c trngr2 + 4r + 4 = 0

c nghim kp r = 2. Vy nghim tng qut ca phng trnh vi phn l

y = x2(C1 + C2 ln x).



63/93



x2y + 3xy + 6y = 0.

Phng trnh c trngr2 + 3r + 6 = 0

c nghim phc r = 3i15

2 nn nghim tng qut vi x > 0 s l

y = C1x3

2 cos15

2ln x

+ C2x

32 sin

152

ln x

.


x3y(3)

5x2y + 18xy

26y = 0.

Phng trnh c trng

r(r 1)(r 2) 5r(r 1) + 18r 26 = r3 8r2 + 25r 26 = 0c nghim r = 2, r = 3 2i. Vy phng trnh cho c nghim tng qut l

y = C1x2 + C2x

3 cos(2 ln x) + C3x3 sin(2 ln x).


x2y 4xy + 4y = x2 + x4.Trc ht ta gii phng trnh thun nht tng ng

x2y 4xy + 4y = 0.Thay y = xr vo phng trnh ny ta c r(r1)4r+4 = 0. Tc l r = 1, r = 4.

Do nghim tng qut ca phng trnh thun nht tng ng l

y = C1y1 + C2y2 = C1x + C2x4.

tm nghim ring, trc ht ta a phng trnh v dng c h s bc caonht bng 1

y 4x

y +4

x2y = x2 + 1.

By gi xem C1, C2 l cc hm ca x. Gii phng trnh

y1C1 + y2C2 = xC1 + x

4

C2 = 0y1C

1 + y

2C

2 = C

1 + 4x

3C2 = x2 + 1



64/93


ta c

C1 = x2 + 1

3, C2 =

1

3

1

x+

1

x3

.

T

C1(x) = x3

9 x

3, C2(x) = 1

3ln x 1

6x3.

Vy nghim tng qut ca phng trnh cho l

y = C1x + C2x4 x

4

9 x

2

2+

x4

3ln x


Vi phng php ny kh x < 0 th th no?

2) Phng php th haiTa thc hin php i bin mi

z = ln x.

Theo quy tc o hm hm hp

y =dy

dx=

dydz

dzdx=

1

x

dy

dz

y = ddx

(y) = ddx

1x

dydz

= 1

x2dydz

+ 1x

ddx

dydz

=

1

x2dy

dz+

1

x

d2y

dz2dz

dx= 1

x2dy

dz+

1

x2d2y

dz2.

Thay vo phng trnhx2y + axy + by = 0

ta cd2y

dz2 + (a 1)dy

dz + by = 0.y l phng trnh tuyn tnh cp hai vi h s hng. Sau khi gii y theo z ta tmc biu thc ca y theo x.


x2y 4xy + 6y = 0.t z = ln x, phng trnh cho c a v dng

d2ydz2

5dydz

+ 6y = 0. (3.27)



65/93

3.6. S DAO NG NGHIM CA PHNG TRNH TUYN TNH THUN NHT CP HAI65

Phng trnh c trng tng ng 2 5 + 6 = 0 c nghim 1 = 2, 2 = 3. Do phng trnh (3.27) c nghim tng qut l

y = C1e2z + C2e

3z.

Tr v bin x ta c nghim tng qut ca phng trnh cho l

y = C1x2 + C2x

3.

Ch . Phng trnh tuyn tnh dng

(ax + b)ny(n) + an1(ax + b)n1y(n1) + + a1(ax + b)y + a0y = f(x) (3.28)vi a = 0 cng c gi l phng trnh Euler. R rng ta c th chuyn phngtrnh (3.28) v phng trnh (3.26) bng php bin i t = ax + b.

3.6 S dao ng nghim ca phng trnh tuyn tnh thun nhtcp hai

3.6.1 a phng trnh v dng khng cha o hm cp mt

Gi s phng trnh

y + p(x)y + q(x) = 0 (3.29)c q(x) lin tc v p(x) kh vi lin tc trn khong (a, b). Ta s a phng trnhny v dng khng cha o hm cp mt

z + I(x)z = 0 (3.30)

bng php bin iy = u(x)z

trong u(x) l cc hm s c chn thch hp.

Tht vy, thay y = u(x)z vo phng trnh (3.29) ta c

u(x) + 2u(x)z+ u(x)z + p(x)[u(x)z+ u(x)z] + q(x)u(x)z = 0

hay

z +

2u(x)u(x)

+ p(x)

z +

u(x)u(x)

+ p(x)u(x)u(x)

+ q(x)

z = 0. (3.31)

Ta chn u(x) sao cho

2u(x)u(x)

+ p(x) = 0



66/93


tc lu(x) = e

p(x)2

dx.

Khi

u(x) = p(x)

2 ep(x)2 dx

u(x) =p(x)

2+

p2(x)

4

e

p(x)2

dx.

V vy phng trnh (3.31) c a v dng

z + I(x)z = 0

trong

I(x) = p(x)

2 p2(x)

4 + q(x).

Hm I(x) c gi l ci bt bin ca phng trnh (3.31).Phng trnh (3.31) s c tch phn nu I(x) l hng s hoc c dng I(x) =c

(x a)2 vi a, c l cc hng s.


y +2

x

y + y = 0.

Vi phng trnh ny ta c p(x) =2

x, q(x) = 1 nn

I(x) =1

x2 4

4x2+ 1 = 1.

Vy vi php th

y = e

1xdxz =

1

xz

ta a phng trnh cho v dng

z + z = 0.

Phng trnh ny c nghim tng qut

z = C1 cos x + C2 sin x

nn phng trnh cho c nghim tng qut l

y = C1cos xx

+ C2 sin xx

.



67/93

3.6.1. a phng trnh v dng khng cha o hm cp mt 67

V d 3.6.2. Xt phng trnh Bessel

x2y + xy + (x2 n2)y = 0.Phng trnh trn c vit li

y +1

xy +

1 n

2

x2

y = 0.

Thc hin php th

y = e

12xdxz =

zx

th phng trnh Bessel c a v dng

z +

1 +1

4n2

4x2

z = 0.

c bit, vi n = 12

th phng trnh Bessel

y +1

xy +

1 1

4x2

y = 0 (3.32)

c a v phng trnh vi h s hng z + z = 0.Phng trnh ny c cc nghim c lp tuyn tnh z1 = cos x, z2 = sin x nn

(3.32) c cc nghim c lp tuyn tnh

y1 =cos x

x,

sin xx

.

nh l 3.6.1. cc phng trnh tuyn tnh thun nht cp hai

y + p1(x)y + q1(x)y = 0 (1)y + p2(x)y + q2(x)y = 0 (2)

(3.33)

c th a v ln nhau qua php th y = u(x)z, iu kin cn v l chngcng bt bin I(x).

Chng minh. Gi s I1(x) v I2(x) ln lt l ci bt bin ca cc phng trnh(1) v (2) trong (3.33).

Gi s php th y1 = (x)y2 a phng trnh (1) v (2), php th y2 = u(x)za (2) v dng

z + I2(x)z = 0.

Khi php th y1 = (x)u(x)z a (1) v dng z + I2(x)z = 0. Do tnh btbin ca I1(x) i vi cc bin i dng y = u(x)z nn ta c c I1(x) I2(x).

Bin son: GV

Documents

6. Phuong trinh vi phan.pdf