2/5/20151Chng 2PHNG TRNH VI PHN V NG DNGNi Dung1. Phng trnh vi phn cp 1 PT c bin phn ly PT ng cp cp 1 PT vi phn tuyn tnh cp 1 2. Phng trnh vi phn cp 2 Phng trnh vi phn cp 2 gim cp c Phng trnh vi phn cp 2 tuyn tnh3. ng dng phn gii mch in2/5/20152Phng Trnh Vi Phn Phng trnh vi phn tng qut, , (1), (2), , () Phng trnh vi phn tuyn tnh bc n tng qut =0

+1

1

1 +1

+

Phng Trnh Vi Phn Cp 1 Dng tng qut:

= , hoc

= , Nghim: Nghim tng qut: = + Nghim ring: = () + 02/5/20153Cc Dng c Bit PTVP c bin phn ly

= , = () PTVP ng cp

=

PTVP tuyn tnh

+ = ()PTVP Cp 1: C Bin Phn Ly

= () B1: Kim tra? B2: Tm n0 hng s: () = 0 (Lu : Ch nhn n0 thc) B3: Tm n0 tng qut:1() = 1() = = + B4: Tm n0ring, t kb 0 = +02/5/20154

= 1 +2 + 2 + 2

2; 0 = 1 B1: Kim tra = = 1 +21 +2 B2: = 1 +2= 0 2 = 1 k0 nhn B3:

=

1 +2 = 1 +2

1 +2 = 1 +2

= tan +

33+0

= 1 +2 + 2 + 2

2; 0 = 1 B4: T iu kin ban u 0 =tan +

33+0=0 tan0= 1 0 = 4 = tan +

33+

42/5/20155

2

2 1 + 2 = 0; 0 = 1 B1: Kim tra

= =

21+2

2 B2: = 2 = 0 = 0 nhn B3:

2=

21 +2

1

= arctan +0 =1arctan 0

2

2 1 + 2 = 0; 0 = 1 B4: T iu kin ban u 0 =1arctan 0=0 0 = 1 =1arctan +12/5/20156PTVP Cp 1: ng Cp =

B1: Kim tra? B2: t z =

= = +. B3: Tm nghim z Gii nh PT c Bin Phn Ly B4: Tm nghim y: =

= 2 +52 + B1: Kim tra? B2: t z =

= = +. B3: Th vo pt: + = 2 +52 + + = 2 +52 + = 1

2 + 52 + = ()PT c bin phn ly2/5/20157

=

= 1

2 +52 + B41: Tm n0hs:2 +52 + = 0 = 1 = 2 B42: Tm n0 tng qut

2 +52 += 1

2 +52 += 1

=

= 1

2 +52 + 1

= ln +0

2+52+= 2

23+2

Xt 2 3 +2 = 0 c 2 n0 = 1 v = 22

23+2 =

1+

2 2 + 1 = 2 = 3; = 4 31 42 = 3ln 14ln 2 3ln 14ln 2 = ln +0 Ta c: =

3ln 4ln 2 = 02/5/20158PTVP Cp 1: Tuyn Tnh

+ = () Dng tng qut: + = () Thun nht: = 0 + = 0

=

= PTVP Cp 1: Tuyn Tnh

+ = () Dng tng qut: + = () Khng thun nht: 0 + = () B1: Kim tra B2: = B3: B4: = +0()2/5/20159

+tan =cos2; (0) =2 B1: = cos2 0 B2: = = = ln1cos=1cos B3: =1coscos2 = sin B4: = +() = sin + cos = sin +0cos |=0 = 2 0 = 2 = sin +2 coscos += sin() ; (0) =2Vit li phng trnh

+1cos= tan() B1: =1cos; = tan 0 B2: = =

1cos

= ln1cos+tan =1cos+tan = 1 +sincos2/5/201510cos += sin() ; (0) =2 B3: = 1 +sincostan = 1costan +tan2 =1cos+tan + = 1 +sincos +cos += sin() ; (0) =2 B4: = +()=1 +sincos +1 +sincos = 1 cos1 +sin+ cos1 +sin = 1 cos1 + sin+ cos1 +sin| =0 = 2 = 1; = 1 1 cos1 + sin2/5/201511Phng trnh vi phn cp 2 Dng tng qut:, , , = 0 Nu gii c pt trn theo y, n c dng

= , , Phng trnh vi phn cp 2 Phng trnh gim cp c Khng cha y: , , = 0 = , Khng cha x: y, , = 0 = , Phng trnh tuyn tnh Thun nht: + xy + qxy = 0 Khng thun nht: + xy + qxy = ()2/5/201512PTVP Cp 2: Gim Cp c Khng Cha

= (, ) B1: Kim tra B2: = = B3: = , PTVP cp 1 theo bin B4: Tm n0y =

B1: y l PTVP cp 2 khng cha y B2: t bin ph = = = 2 B3: Tm n0 : =1

1 B4: Tm n0 : = 1

1 = 1 +22/5/201513PTVP Cp 2: Gim Cp c Khng Cha

= (, ) B1: Kim tra B2: t bin ph = =

=

= (, )y l ptvp cp 1 (theo ) v l PT c bin phn ly B3: Gii PT c bin phn ly B4: Tm n0yPTVP Cp 2: Tuyn Tnh Thun Nht

+ + = 0 B1: Kim tra B2: ptt 2 + + = 0 1,2 = 2 422/5/201514PTVP Cp 2: Tuyn Tnh Thun Nht

+ + = 0 B3: Tm n0 tng quto 1 2 = 1

1

+ 2

2

o 1 = 2 = 1 + 2

2

o 0 phc: 1,2 = =

1cos + 2sin B4: Tm n0ringy + 2y + 2y = 0; y(/4) = 2; y(/4) = -2 B1: ptvp tt hsh thun nht B2: ptt 2 +2 +2 = 0< 0: nghim phc 1,2 = 1 = 1; = 1 B3:Tm 0 tng qut =

1cos + 2sin 2/5/201515y + 2y + 2y = 0; y(/4) = 2; y(/4) = -2 B4: Tm n0ring =

1cos +2sin =1cos sin +2sin +cos 4 = 4

1 22 +2 22 = 2

4 = 122 22 4 = 12 4 = 2 () & 1 = 2 = 2 4 = cos +sin 2 4PTVP Cp 2: Tuyn Tnh Khng Thun Nht Nghim tng qut cho bi phng trnh = +

Trong yh- Nghim tng qut yp- Nghim ring2/5/201516PTVP Cp 2: Tuyn Tnh Khng Thun Nht H S HngGii ptr bng phng php Nhm Nghim

+ + = ; , , iu kin 1: H s phng trnh thun nht phi lhng s.iu kin 2: Biu thc g(x) phi c dng c bit. =

cos +

sinPhng Php(1) Kim tra biu thc phi tha 02 iu kin trn(2) Nu () c dng

+ + = =1

Gii tng hm

ring l c

&

(3) Vit a thc c trng v tm nghim 1 & 2(4) Vit ra dng nghim ring

=

cos +

sin2/5/201517Phng Php (tt)(5) So snh nghim

v 1 & 2 Nu

1 & 2, th = 0 Nu

= 1 || 2, th = 1 Nu

= 1 & 2, th = 2(6) Tm dng biu thc

&

theo

&

(7) Nghim ring ca phng trnh

=1

3 4=32 +2 sin()(1) Kim tra32 = 23cos0 +

sin(0) 1 = 2; 1 = 02sin = 00cos +2sin 2 = 0; 2 = 1(2) Gii tng phng trnh 3 4=32(1) 3 4= 2sin (2)(3) Gii ptt: 2 3 4 = 0 1 = 1 & 2 = 42/5/201518 3 4=32(1)(4) Tm nghim ring ca pt (1), c 1 =2; 1 =0(5) Do 1 +1 = 2 1 = 1 & 2 = 4 = 0(6) Nghim ring cho bi 1=2 Th vo pt (1), ta c42 62 42 = 32 Suy ra = 12 Vy nghim 1= 12

2 3 4=2 sin (2)(4) Tm nghim ring ca pt (2), c 2 =0; 2 =1(5) Do 2 +2 = 1 = 1 & 2 = 4 = 0(6) Nghim ring cho bi 2= cos() +sin() Th vo pt (2), ta ccos sin 3 sin +cos 4 cos +sin =2sin Suy ra 5 3 = 03 5 = 2 =317; = 517 Vy nghim 2=317cos 517sin2/5/201519 3 4=32 +2 sin()(7) Vy nghim ring: =1 +2 =12

2 +317cos 517sin