Математика / 3 семестр / 1 МОДУЛЬ Дифференциальные уравнения / Терехина Фикс - Дифференциальные уравнения

1: y(4) = sin x ; 2x:

pOSLEDOWATELXNYM INTEGRIROWANIEM NAHODIM

y000

= Z y(4) dx = Z (sin x ;

2x) dx = ; cos x ; x2 + C1:

x3

y00

= Z y000 dx = Z (; cos x ; x2 + C1) dx = ; sin x ; 3 + C1x + C2:

y0 = Z y00 dx = Z (; sin x ;

x3

3 + C1x + C2) dx =

x4

x2

x4

x2

= cos x ;

12 + C1

2 + C2x + C3:

y = Z y0 dx = Z (cos x ; 12

+ C1 2

+ C2x + C3) dx =

x2

x4

x3

= sin x ; 60 + C1

6 + C2

2 + C3x + C4:

oB]EE RE[ENIE URAWNENIQ

x4

x3

x2

y = sin x ; 60 + C1 6 + C2 2 + C3x + C4:

rE[AQ \TO URAWNENIE, NAHODIM FUNKCI@ z(x C1) =

y0 A ZATEM

x

ISKOMU@ FUNKCI@

Z

y(x C1 C2)

x

Z

y =

y0 dx =

z(x C1) dx + C2

:

|TO URAWNENIE DOPUSKAET RAZDELENIE PEREMENNYH

dz

dz

dx

dz

Z

dx

dx x ln x = z

z =

=

x ln x

z

x ln x

ln jzj = ln j ln xj + ln C1

z = C1 Zln x:

nAHODIM FUNKCI@ y(x): tAK KAK

y0 = z

(x) = C1 ln x TO

Z

x

Z

x

;

y =

y0 dx =

C1 ln x dx = C1 x(ln x

1) + C2:

oB]EE RE[ENIE URAWNENIQ

y = C1 x(ln x ; 1) + C2:

3: x3y00 + x2y0

= 1 y(1) = 1

y0(1) = 0:

x3y00 + x2y0 = 1 )

NET QWNO y

) x3 z0 + x2 z = 1:

y0 = z(x) y00 = z0(x)

3

z

1

z0 + x

= x3 :

rAZDELIM NA x

POLU^AEM LINEJNOE URAWNENIE

z(x) = U(x)V (x) z0 = U0V + UV 0 U0V + UV 0 +

UV =

1

3

V 0 + Vx ! =

1

x

x

U0V + U

:

x3

1

V 0 + V = 0

U0 V =

:

x3

x

U0 x1 =

1

:

dV

V

dV

dx

x3

dU =

1

dU = dx2

dx

= ; x

V

= ; x

2

ln V = ; ln x V = x1 :

dx

x

x

U =

dxx2 = ;x1

+ C1:

1Z

1

tAKIM OBRAZOM

z(x) = U V = C1 ; x! x:

sLEDOWATELXNO

y0 = z(x) = C1

1 1 y

=

Z1

z(x) dx =

Z

C1

1

1 dx =

= Z

C1

1

!

; x

! x

; x!

x

;

dx = C1 ln jxj + x

+ C2:

x

x2

1

+ C2:

oB]EE RE[ENIE

y = C1 ln jxj + x

nAJDEM ^ASTNOE1RE[ENIE URAWNENIQ.

>

y(1) = 1

1 = C

ln 1 + (1=1) + C

+ C2

8y

0

(1) = 0

1

j 1j=1)

(1=1) 2

8y = C1 ln jxj + x

80 = (C1

<y0 = (C1

;

1=x)

(1=x)

<

<

;

>

:

:

:

8

1 = 1 + C2

8

C2 = 0

:

<

0 = C1 ; 1

<

C1 = 1:

:

:

y = ln jxj + x1 :

~ASTNOE RE[ENIE:

4: y000tg x = y00

+ 1:

dANNOE URAWNENIE { URAWNENIE TRETXEGO PORQDKA, KOTOROE NE SO-

DERVIT W QWNOM WIDE FUNKCI@ y.

dELAEM ZAMENU

y00

= z(x)

TOGDA

y000 = z0(x):

+ 1:

uRAWNENIE PRIMET WID

z0 tg x = z

|TO URAWNENIE S RAZDELQ@]IMISQ PEREMENNYMI

dz

=

dx

dz

= ctg x dx

z + 1

tg x

z + 1

ln jz + 1j = ln j sin xj

+ ln C1

z + 1 = C1 sin x:

nAHODIM

z(x) = C1 sin x ; 1:

) y0 =

dALEE

:

y00 = z(x) = C1

sin x ; 1

z(x) dx

; x + C2:R

y0 =

(C1

sin x ; 1) dx = ;C1 cos x

i, NAKONECR

, POLU^AEM OB]EE RE[ENIE ISHODNOGO URAWNENIQ

y =

R

y0 dx =

R

(;C1 cos x;x+C2)dx = ;C1 sin x;x2=2+C2 x+C3:

2

y = ;C1 sin x ; x =2 + C2 x + C3:

MOVNO PONIZITX DO PERWOGO PODSTANOWKOJ

p = dp

y0 = p(y)

TOGDA y00

= p0

y0

= p0

p:

x

xx

y

x

y

dy

5: y00 ; y0ey = 0:

y00

;

y0ey = 0

)

nET QWNO x

)

p0

p

; pey = 0

)

y0 = p(y) y0 = p0 p

p (p0

;

ey) = 0:

x

x

y

1) p = 0

y0

= 0

y = const:

dp

x

Z dp = Z

2)

y

y

y

+ C1

dy

= e )

e dy ) p = e

tOGDA

dy

) Z

dy

= Z

y0

= ey + C1

) dx = ey + C1

dx:

ey + C1

1

y

,

POLU^AEM

C1

(y ; ln(C1 + e )) = x + C2

iNTEGRIRUQ

1

6:

rE[ITX ZADA^U kO[I

y00 = p

y(0) = y0(0) = 0:

y

y00 =

1

NET QWNO x

) p0

1

dp

1

)

y0 = p(y) y0

= p0

p

p =

) dy p =

p

dy

x

x

y

p

p

p2

C1

Z p dp = Z p

)

2 = 2py +

2

) p = q4py + C1:

y

tAK KAK y0(0) = p(0) = 0

I y(0) = 0

TO MOVNO SRAZU NAJTI C1.

q

4p

+ C1

)

0 =

0

C1 = 0:

iTAK,

IMEEM

y0 = 4p

+ 0 = 2p

)

y

y

qdy

4

dy

2

4

2py

=

)

=

dx

)

3q

y3

=

x

+ C2:

dx

2py

4

4

4

tAK KAK

y(0) = 0, TO

32p03 = 0 + C2

) C2 = 0:

2

4

3

4=3

oKON^ATELXNO, ^ASTNOE RE[ENIE

x = 3qy3: ILI

y =

2 x!

:

7:

rE[ITX ZADA^U kO[I

y00

= 32 sin3 y cos y

y(1) =

2

y0(1) = 4:

y0 = p(y) y0

= p0 p:

x

x

y

uRAWNENIE PRIMET WID

p0 p = 32 sin3 y cos y

)

dydp p = 32 sin3 y cos y

Z

p dp = 32 Z sin3 y cos y dy

)

p2=2 = 8 sin4 y + C1:

tAK KAK y(1) = =2

I y0(1) = p = 4 TO MOVNO SRAZU NAJTI C1.

16

= 8 sin4( =2) + C1

8 = 8 + C1

C1 = 0:

2

iTAK, IMEEM

p2

= 4 sin2 y:

2 = 8 sin4 y

p2 = 16 sin4 y

p = y0

dy

dy

dx = 4 sin2 y

)

= 4 dx

) ;ctg y = 4x + C2:

sin2 y

tAK KAK

y(1) = =2,

TO

;ctg ( =2) = 4 + C2

0 = 4 + C2

C2 = ;4:

oKON^ATELXNO, ^ASTNOE RE[ENIE

4x = 4 ; ctg y:

y1(x) = y2(x) ILI

y1(x)

= = const:

y2(x)

W [y1

y2] =

y1

y2

= y1 y0

;

y2 y0 :

y0

y0

2

1

1

2

CFORMULIRUEM KRITERIJ LINEJNOJ ZAWISIMOSTI ^EREZ OPREDE-

LITELX wRONSKOGO.

t E O R E M A

1.

eSLI OPREDELITELX wRONSKOGO

W[y1 y2] = 0

NI PRI

KAKIH x TO SISTEMA FUNKCIJ y1

6

y2 QWLQETSQ LINEJNO NEZAWISIMOJ.

t E O R E M A 2.

eSLI OPREDELITELX wRONSKOGO W [y1 y2] TOV-

DESTWENNO RAWEN NUL@, TO SISTEMA FUNKCIJ y1

y2 QWLQETSQ LINEJNO