Решение задач по дифурам

1)

tgx=

2)x׳ctgt+x=2

X׳ctgt=2-x

ctgt=2-x

ctgtdx=(2-x)dt

ctgtdx=(2-x)dt/:ctgt*(2-x)

=

=+C

=+C

= +C

=+

+=

=

=C(2-x)

3)tdt+xdx+(tdx-xdt)

tdt+xdx+dx-

rcos

dp+

2p

4)(t+2x+1)dt-(2t-3)dx=0

t= x=- —

x=—t=+

(++2—+1)d=(

(+2) = 2z==z‘=z’+z

(z+2)(z’+z)=2z

z’(z+2)+=0

(z+2)dz+=0

dz+=0|:

+)dz+=C

++=C

—+=

—=

z=C =

=

t — = C

C=

5) Теңдеуді шеш: tx-x′=xlnx′

x′=p

tx-p=xlnp

t=

pd()=dx

|:x(z-1)

6)Теңдеуді шеш:

,

8)

-1=Ce

Ce=-1

C=-

9)

12) (y+2)dx-(2x+3y-5)dy=0

y=-2

2x-6-5=0

2x-11=0

2x=11

x=

x=+ = x-

y=-2 = y+2

(-2+2)d-(2+11+3-6-5)d=0

d-(2+3)d=0

-(2+3)=0

=z

d=(2+3)d

z = (2+3z)()

z =2+2 z +3+3= z +3+(2+3z)=z+3+(2+3z)

(z+3) d+ (2+3z)dz=0

(z+3) d+(2+3z)dz=0 :(z+3)

+=C

+=C

+2—=

+2—=

13)(2x+1)=4x+2y

(2x+1)=4x+2y

(2x+1)dy-2ydx=0

(2x+1)dy-2ydx=0 :(2x+1)y

— =0

– =

— =

y=(2x+1)

y=C(x)(2x+1)

=(2x+1)+2C(x)

+2(2x+1)C(x)- 2(2x+1)C(x)=4x

=

C(x)==4= + +

y=( + )(2x+1)+=1+(2x+1) + (2x+1)

14) (xy+)dx-xdy=0

xy+-x=0

xy’=xy+

xy’-xy=

xy’-xy=0

xdy-xydx=0 |xy

— =C

— x =

y=C1

y=C1 (x)

y’=C’(x)+C(x)

xC’(x)+xC(x) — xC(x)=

C’(x)=

C(x)= +C2

C(x)=+C2

y=(+C2)= +C2

15) 2x(x2+y)dx=dy

(2×3+2xy)dx=dy

y’=2×3+2xy

y’-2xy=2×3

y’-2xy=0

dy-2xydx |y, y0

– 2xdx=0

— =C1

— x2 =

y=C1

y’=C1(x)2x+y’(x)

C’(x)+2xC(x)— 2xC(x)=2×3

C1’(x)= dx

C1(x) = — +C2

y=( — +C2)— x2 – 1+

17) ()dx+(+)dy=0

M N

==

= =+

F=

=

+ =

== +C

+ = C

4+= 4C

18)

19)

20) y=xy`-x2 y`3

y`=p

y=xp-x2p3

dy=pdx

d(xp-x2p3)=pdx

xdp+pdx-2xp3dx-3p2x2dp=pdx

(x-3p2x2)dp=2xp3dx

x(1-3p2x)dp=2xp3dx

(1-3p2x)dp=2p3dx

,

x=c(x)*p-3/2

x`=c`(x)p-3/2-3/2p-5/2c(x)

c`(x)p-3/2-3/2c(x)p-5/2+3/2c(x)p-5/2=1/2p3

c`(x)=1/2p3*p3/2

c`(x)=dx/2p3/2

y=xp-x2p3

x=-1/p2+c2

Ж: xp2=c2-1

21) y=xy`+y`2

y=cx+c2

0=x+2c, c=-x/2

y=cx+c2

4y=-x2 , y=-x2/4

22) y=x Лагранж теңдеуі

y=

dy=pdx

d(x)=pdx

(

p(p-1)dx+p(2x-6p)dp=0

(p-1)dx+(2x-6p)dp=0

(p-1)

(p-1)

(p-1)x’+2x=0

(p-1)dx+2xdp=0

ln|x|+2ln|p-1|=ln|C|

x=

x’=C’(p)

C’(p)=

C(p)=

X=

23) y’’-2y’-3y=

-3=0

D=4+4*3=16

r=0

y’=4

y’’=16

16 -3

16A-8A-3A=1

5A=1

A=

y=

24)

a+ib=1+i*0=1, r=1

++Cx+2Bx+C+;

2B+2Bx+C+2Bx+C+2Bx+C=3x

6B=3 B=

y=

25)y

a+bi=1+0=1, r=0

A

2Ax=4x

A=0

B=-A

B=-2

y=

26)

X=

-3y=0

D=4+4*3=16

Y=

X(t=lnx)

Y=

27)

X=

D=25-4*6=1

Y=

28) 2xy’y»=y’2-1

y’=z, y»=z’

2xzz’=z2-1

Dy=

z=

y=

29)yy´´+1=y´2

y´=p, y´´=pp´

y pp´+1=

yp

ypdp=( :

a)C<0

=x+

Cy=

b)C=0

y=x

в)C>0

Cy=

10) (-2xy+y2)dx+ x2dy=0.

(y2-2xy)dx+ x2dy=0

y=zx, dy=zdx+xdz

(z2x2-2zx2)dx+x2(zdx+xdz)=0

z2x2dx-2zx2dx+x2zdx+x3dz=0

z2x2dx-x2zdx+x3dz=0

x2(z2-z)dx+x3dz=0 / : x2

(z2-z)dx+xdz=0

z(z-1)dx+xdz=0/ : x(z2-z), z≠0, z≠1

Cy=

Cy=x(y-x), y=0

11) (2x+3y-5)dx-(x+4y)dy=0

2x+3y-5=0 -8y+3y-5=0

x=-4y -5y=5 x=x1+4 , y=

x=4 y=-1

(2×1+8-3y1-3-5)dx1-(x1+4-4y1-4)dy1

(2×1-3y1) -(x1-4y1)dy1/dx1

y1=zx1 y1’=z’x1+z

(2×1+3zx1)=( x1+4zx1)( z’x1+z)

x1(2+3z)= x1 (1+4z)( z’x1+z)

(2+3z)= z’x1+z +4z’zx1+4z2

z’(x1+4zx1)+4z2+z-2-3z=0

x1z’(1+4z)+4z2-2z-2=0

x1z’(1+4z)+(4z2-2z-2)=0

x1(1+4z)dz+(4z2-2z-2)dx1=0 /:(4z2-2z-2)x1

2az+a+bz-b=4z+1

2az+bz=4z

a-b=1

2a+b=4 => 2+2b+b=4

a-b=1 => a=1+b

3b=2 a=

b=

16)  (2 — 9xy2)x dx + (4y2 — 6×3)y dy = 0

M N

=-18=-18

+

30)x’=x+y, x(0)=0

y’=4y-2x, y(0)=-1

=0

D=1+4*6=25

x(0)=+=0

x(t)=-=

y(0)=-4+=-1

(-1-)

7)

*

x=0

Бернулли теңдеуі

* ; t

dz=dt

z=

y==

y=

+

+

X=