Добавил:

Upload Опубликованный материал нарушает ваши авторские права? Сообщите нам.

Вуз:

Национальный авиационный университет

Предмет:

[НЕСОРТИРОВАННОЕ]

Файл:

Higher_Mathematics_Part_2

.pdf

Скачиваний:

Добавлен:

19.02.2016

Размер:

9.73 Mб

Скачать

☆

<<< < Предыдущая 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1920 / 2520 21 22 23 24 25 > Следующая >>>

Newton’s law of motion, we have F = ma, where a = dvdt and v represents

velocity.

The differential equations are used to solve other rate problems.

Т.1

Typical problems

1. Solve the separable differential equation

xydx + (1− x 2 ) ln ydy = 0 .

Solution. Division by

y(1 − x2 ) ≠ 0 gives variables separable:

dx +

ln y

dy = 0,

1 − x2

integrating this, we get

−

1 − x2

ln2

y = C .

The given equation has the other singular solution x = ±1 .

2. Find the general solution and also the particular solution of y′ =

1 − x2

through the point

y(0) = e .

Solution. Separating the variables, we obtain

, ∫

= ∫

= arcsin x + C .

1−x2

Substituting of (0, e) in the general solution gives ln e = arcsin 0 + C C = 1 .

Hence, ln y = arcsin x + 1 is a particular solution for (0, e).

3. Find the general integral of y ′ =									x(1+ y						2 )		.
3. Find the general integral of y ′ =									y(1+ x2 )								.
									y(1+ x2 )
Solution. Separating the variables, we obtain
		y(1+ x	2	)		dy	= x(1+ y				2	) ,			ydy			=		xdx	,
		y(1+ x		)		dx	= x(1+ y					) ,			+ y2			=	1+ x2		,
						dx						1			+ y2				1+ x2
	1	ln(1+ y2 ) =			1	ln(1+ x2 ) +					1		ln	C		, 1+ y2 = C(1+ x2 )
	1				1						1
2				2							2
2				2			x(1+ y2 )				2
is the general integral of y′ =							x(1+ y2 )			.
is the general integral of y′ =							y(1+ x2 )			.
							y(1+ x2 )
4. Solve the equation				y′ =			y2
								.
							xy − x2	.
							191

http://vk.com/studentu_tk, http://studentu.tk/

Solution. Since the equation is homogeneous, write

y = xu(x) ,

= x

+ u .

Substituting these values for y and y′ in the equation, we get

+u =

u2 x2

+ u

u 2

, x

x2u −x2

− 1

u −1

∫

u − 1

du = ∫

u − ln

= ln

+ ln

u = ln

Cxu

Replacing u by its equal

, we have

y = x ln

5. Find the general solution of

′ =

x + 2y − 3

2x − 2

Solution.

Finding

the

determinant

= −4 ≠ 0 . Since

≠ 0 ,

transforming this to the homogeneous equation substituting

x = t + α,

y = v + β.

By (3.4), we get α + 2β − 3 = 0,

then α = 1,

β = 1.

Thus

2α − 2 = 0,

d(v + 1)

x = t + 1,

y = v + 1,

d(t + 1)

Substituting the values of t and v, we obtain

t + 1+ 2v + 2 − 3

, or

t + 2v

2t + 2 − 2

Since this equation is homogeneous, write

Then

v = ut,

u = u(t),

v′ = u′t + u.

t + 2ut

1+ 2u

u′t + u =

; u′t + u =

; u

′t =

; 2

; 2du =

Hence

u =

ln | Ct | , or v =

t ln | Ct |

(3.15)

Replacing vine (3.15) t and v by x – 1 and y – 1, we have

y − 1 =

(x − 1) ln | C(x − 1) | ,

або

y = 1+

(x − 1) ln | C(x −1) | .

6. Solve the equation

(2x + 4y + 3)dy − (x + 2y + 1)dx = 0 . 192

http://vk.com/studentu_tk, http://studentu.tk/

Solution. Transform this to the form

′ =

x + 2y + 1

2x + 4y + 3

The determinant =

= 0 . That is why substitute z = x + 2 y . Hence

=1+

−1 ,

2 dx

z +1

4z +5

2z +3

1 dz

dz = dx ,

−1

2z +3 4z +5

2 dx

2z +3

∫ (1+

)dz = 2∫ dx , z +

4z + 5

= 2x + C / 4 ,

4z +5

4z + ln 4z + 5 = 8x + C ,

ln 4x + 8y + 5 = 4x − 8y + C is the general integral of the given equation. 7. Find the general solution of y′ + xy = sinx x .

Solution. This equation has the form (3.5) and we can use the Bernoulli’s method:

y =u(x)v(x) ,

v +u

sin x

v +u

= −

u =

sin x

;

= sin x;

= − cos x + C.

Namely, the general solution is

y =

(C − cos x).

8. Solve

y′ +

y + y 2 = 0 .

x2 + 1

Solution. This is the Bernoulli’s equation. To solve it, we first divide it by y2

and obtain

y′

= −1.

x2 + 1 y

193

http://vk.com/studentu_tk, http://studentu.tk/

This form suggests the substitution

z = y−1 . Then

y =

=−

Substituting, we get

−

dz +

= 0 or

−

z = 1 .

(3.16)

x2 + 1 z

z 2

x2 +

z 2 dx

The last equation has the form (3.5) and we can use the Lagrange’s method.

First find the general solution of the homogeneous equation:

−

z =

0 ,

dx ,

∫

= ∫

d(x2 + 1)

+ 1

ln | z | = ln | C | + ln(x2 + 1) ,

z = C(x2 + 1).

Since С is an unknown function of x , namely C =C(x) , then

z = C(x)(x2 +1) ,

(x2 + 1) + 2xC(x) .

Substituting this in (3.16), we get

(x2 + 1) + C(x)2x −

C(x)(x2 + 1) = 1,

2 +

therefore

Then

x2 + 1

C(x) =

= arctg x + C

( C

= const ).

∫ x2 + 1

Thus,

1/ y = z = (arctg x + C )(x2 + 1) ,

y =

(arctg x + C )(x2 + 1)

is the general solution of the given equation.

y = 0.

The given equation has another singular solution

9. Solve the equation

yex dx +( y +ex )dy = 0.

Solution. From §1.5, it appears that an equation

is exact (3.11) if

Mdx + Ndy = 0

∂M

= ∂N .

∂y

∂x

194

http://vk.com/studentu_tk, http://studentu.tk/

In our case, we have

M (x, y) = ye x , N (x, y) = y + e x ;

дM

= e x ,

дN

= e x .

дx

дy

Therefore

∂M

∂N

, and the equation is exact. Consequently,

∂y

∂x

дu

= M (x, y) = ye

дx

дu

= N(x, y) = y + e

дy

Integrating of the first equation with respect to x gives

u = ∫ yex dx = yex

+ ϕ( y).

( 3.17)

Differentiating with respect to y gives

y :

дu

= e x

+ ϕ′( y) , or

дy

e x + ϕ′( y) = y + e x ,

ϕ′( y) = y , ϕ =

y 2

+ C .

Thus,

y 2

u = ye x

+ C .

Finally,

yex +

=C –

is the general solution of the given equation (C = const).

10. A curve is passing through М (1; 2). Any its tangent intercepts with straight line у=1 at the point where the abscissa is half of the abscissa at the tangent point. Find the equation of its curve.

Solution. Let (х, у) be any point lying on the curve (fig. 3.1). The equation of the tangent at Y (х, у) is

Y − y = y′(X − x) ,

where Х, Y are

the set of points in the

coordinate plane of the tangent. As the given

tangent intercepts the line Y = 1 at point 2х,

therefore

1− y = y′(2x −x) , або x

=1− y .

х 2х

Fig. 3.1

This is the variables separable differential

equation. Thereafter

=−

, ln | y −1|=−ln | x | +ln C ,

y −1=

y −1

195

http://vk.com/studentu_tk, http://studentu.tk/

If the curve which we find is passing through М(1; 2), we obtain С = 1. Hence, y = 1x +1 is the found curve (hyperbola).

11. A particle of 1 кg mass moves along a straight line under the influence of several applied forces whose resultant is proportional to the time from the moment t = 0 , and disproportional of the velocity of it. There is the velocity

50 м/seс and the force 4 N at the moment of time t = 10 seс. Find the speed of the particle after 2 min from rest.

Solution. In accordance with Newton’s law of motion, we have F = ma ,

where acceleration a =

. In our case

F =

. Simultaneous by the equation

F = k

occurs, where

from 4 = k

is k = 20. Comparing the right parts of

F =

and F = 20

, we then write our differential equation of motion in the

form

= 20

This is the variables separable differential equation. Thereafter v = 20t 2 + C .

The constant С is defined from the initial condition: 50 = 20 100 + C , thus С=500.

Consequently, v = 20t 2 + 500 . Finally, the speed of the particle after 2 min from rest is

v = 20 1202 +500 = 288500 = 10 2885 (m/seс).


Т.1		Self-test and class assignments
Solve the separable differential equations.
1.	xydx + (x + 1)dy = 0 .		2.	xydx + (1+ y 2 ) 1+ x2 dy = 0 .
3.	y′ = 5 y .		4.	y′ + y 2 = 1 .
5.	x 2 ( y 3 + 5)dx + (x3 + 5) y 2 dy = 0 .		6. y′ = 33x+2 y .
7.	4yy′ = − tg x(4 + y 4 ) .		8. 1− y 2 dx + 1− x2 dy = 0
9*. y′ = 3x + 4y .			10*. y′( y + x) = 1 .

Find the general solution and also the particular solution through the point written opposite it.

196

http://vk.com/studentu_tk, http://studentu.tk/

11.

yy′

+ e y = 0 , y(1) = 0 .

12.

tg ydx − x ln xdy = 0 , x

= e .

13.

cos2 x sin 2

ydy + sin x cos2

ydx = 0 , x(0) = 0 .

Solve the homogeneous differential equations.

x2 + y2

−

14.

y′ +

= 0 .

15.

xydy − y

= (x + y)

x dx .

16.

(x − 2y)dx − xdy = 0 .

17.

xy′− y = x tg

18.

(x + y)dx + ( y − x)dy = 0 .

19. (x2 − xy)dy + y2dx = 0 .

20.

xy′ = x sin

+ y .

21.

y′ = e

+ 1 .

Transform the given equations into homogeneous ones and solve them.

22. y′ = −	2x + y − 1	.	23. y′ =	x − 2 y + 3	.

	x − 2 y + 3			2x − 4y − 1

24.(x − y)dx + (2y − x + 1)dy = 0 .

25.(x + y + 1)dx + (2 y + 2x − 1)dy = 0 .

Solve the linear differential equations of the first order.

26.	y′ + y = e x .						27.	y′ − 2xy = 1− 2x 2 .
28.	xy′ + y = x2 + 3x + 2 .						29. tdx + (x − t sin t)dt = 0 .
30.	2y	dx		+ x = 2y3 .			31.	y′+ y cos x =sin x cos x .
		dy

32.	y′ − 4y = cos x .						33.	y′+ y tg x = cos2 x .
34.	y′ −		2	y =	ex (x − 2)	.	35.	y′− y ctg x = 2x −x2 ctg x .
			x
					x
Solve the Bernoulli’s differential equations.
36.	xy 2 y′ = x 2 + y 3 .						37. y ′ = y4 cos x + y tg x .
38.	xy′ + 2y = 2x y .						39. (1+ x2 ) y′−2xy = 4 (1+ x2 ) y arctg x .
					cos2 x
Solve the exact differential equations.
40.	2xydx + (x 2 − y 2 )dy = 0 .
							197

http://vk.com/studentu_tk, http://studentu.tk/

41.(2xy + 3y 2 )dx + (x 2 + 6xy − 3y 2 )dy = 0 .

42.(x + sin y)dx + (x cos y + sin y)dy = 0 .

43.(x2 + y2 + y)dx + (2xy + x + e y )dy = 0 .

44.(3x 2 y + sin x)dx + (x3 − cos y)dy = 0 .

				x	3
45.	3x2	(1+ ln y)dx =	2y −	x		dy .
45.	3x2	(1+ ln y)dx =	2y −			dy .
				y
				y

Compose and solve the differential equations.

46.Radium decomposes at a rate proportional to the amount present. If of 100 mg. set aside now there will be left 98,3 mg. 50 years hence, find how much will be left t centuries from the time when the radium was set aside and also how long a time will elapse before one tenth of the radium has disappeared.

47.Find the equation of all curves which cut the circles x2 + y2 = R2 16 at an angle of 450.

48.Find the equation of the curve such that each of its tangent lines together with the coordinate axes encloses a constant area equal to 9.

49.The force exerted by a spring is proportional to the amount the spring is stretched and is 10 kg when the spring is stretched 2 cm. A 10 kg weight is hung on the spring and is drawn down slowly until the spring is stretched 4 cm and then released. Find the equation, the period, the amplitude, and the frequency of the resulting motion.

Answers

1. y = C(x + 1)e− x , x = −1 .

2. 1+ x 2 + ln y + y 2 = C .

y = x + C .

1+ y

= x + C .

(y 3 + 5)(x3 + 5) = C .

3 3−2 y + 2 33x

= C .

arctg y2 =

1− y

= ln

cos x

+C .

arcsin x +arcsin y = C .

9*.

y = Ce4x −

x −

Hint. Use the

substitution 4y + 3x = z(x) .

10*. y − ln

x + y + 1

= C .

Hint. Use the

substitution

y + x = z(x) . 11.

x2 + 1 = 2e− y ( y + 1) . 12. x = esin y . 13. tg y − y +

= 1 .

cos x

x2 (x2 + 2y2 ) = C .

14.

15.

(x + y) ln Cx = xe x .

16.

y = (Cx −1)x, x = 0 .

= ln(C x2 + y2 ) .

17. y = x arcsin(Cx) .

18. arctg

19. y = Ce x . 20. y = 2xarctgCx .

198

http://vk.com/studentu_tk, http://studentu.tk/

22. x2 + xy + 3y − x − y2

23. x2 + 4y2 − 4xy + 6x + 2y = C .

21. e x

= C .

− Cx

24.

x 2

+ y

− xy + y = C .

25. x + 2y + 3ln

+ y − 2

= C .

26. y = Ce

− x

x 2

27. y = Ce x2 + x .

28. y =

+ 2 .

29. x =

sin t

− cos t . 30. x =

y3 .

31. y = Ce− sin x + sin x − 1 .

32. y = Ce4x +

(sin x − 4 cos x) .

33.

34. y = Cx2 + e x .

y = C cos x + sin x cos x .

35. y = C sin x + x 2 . 36. y3 = Cx3 − 3x 2 .

37. y−3 = (C −3tg x)cos3 x,

y = 0 .

38.

y = tg x +

C +ln

cos x

39. y = (1+ x2 )(arctg2 x +C)2 .

40. 3x 2 y − y3 = C .

41. x 2 y + 3y 2 x − y3 = C .

42.

x 2

+ x sin y − cos y = C .

43.

+ y 2 x + yx + e y

= C .

44. x3 y −cos x −sin y = C .

x2 + y2 ± arctg

45. x3 (1+ ln y) − y 2 = C . 46.100e-0,0344t; 310 years. 47. ln C

= 0 .

1 cm, 2,21.

48. 2xy = 9. 49. x = –

cos

6gt

, 0,452 sec,

Т.1

Individual tasks

1.1. Solve the separable differential equations.

1.1.1. 3(x2 y 2 + x2 )dx + (2y − x3 y)dy = 0 .	1.1.2.	xey y′ = e2 y
1.1.3. y(4 + x2 )dy + 1 − y2 dx = 0 .	1.1.4.	y′ = x4 x+ y .
1.1.5. (x2 − y2 x2 )dx +( y2 −x2 y2 )dy = 0.	1.1.6.	1 − x2 dy −

+1.

ydx = 0 .

1.1.7. x(1 − y 2 )dx −	1 − x2 dy = 0 .	1.1.8. yy′ + 4 + y 2 = 0 .
1.1.9. cos yex dx +(1+e2x ) sin ydy = 0 .			4 − x2
		1.1.10. xy′−4 = y2 .
1.1.11. y′ − x cos2 y sin 2 x = 0 .		1.1.12. yy′(1+ x2 ) = 1+ y 2 .
1.1.13. (1 + y 4 )dx −	x ydy = 0 .	1.1.14. (x − 1) yy′ = 1 + y 2 .
1.1.15. y(1+ 2x) y′ = (1− 2x) .		1.1.16. x2 yy ′+3 = y2 .
1.1.17. y(1 + x2 ) y′ + x(1 + y 2 ) = 0 .		1.1.18. y′ =	y3 −1	.

			y2 x2
	199

http://vk.com/studentu_tk, http://studentu.tk/

1.1.19.	ydy −x sin x 9− y2 dx = 0 .	1.1.20. (1+ x3 ) y′= x2 (1+ y) .
1.1.21.	1+ x2 dy = tg ydx .	1.1.22. x − xy2 = y′(4 + x2 ) .
1.1.23. (1+ x2 )dy = 1− y−2 dx .		1.1.24. (1 + x) yy′ = e− y2 .
1.1.25. tg xdy + 1− ydx = 0 .		1.1.26. y′ = 2y ctg x .
1.1.27. xx′ = y cos( y2 ) 1+ x2 .		1.1.28. e1/ y y′ = y 2 x2 ln x .
1.1.29. e−sin y dx = x cos y ln xdy .		1.1.30. y 2 dx + x ln xdy = 0 .
1.2. Solve the homogeneous differential equations.
1.2.1. 2y ′ = ey / x +2y / x .		1.2.2. y ′ =	x2	+2	y	.
			y2
					x

1.2.3.y′ = x + 2y − 1.

x− y

1.2.5. (x + y)dx + ( y − 2x)dy = 0 . 1.2.7. ( y − 2x)dx + ( y + 2x)dy = 0 .

1.2.9. (3y 2 + 2x2 )dx = ( y 2 − x2 )dy .

1.2.11. (2y −x)dx = (3x + y)dy .

1.2.13. (2x + 3y)dx = ( y + 2x)dy .

1.2.15.	xy′ = x sin	y	+ y .

		x
1.2.17.	4xy + y 2 = x(x + y) y′ .
1.2.19.	x2 y′ = y2 −2x2 .

1.2.21. (x + 2y)dy + (2x + y)dx = 0 .

1.2.23. (x2 − y 2 ) y′ = 2( y 2 + xy) . 1.2.25. (2x2 + y 2 ) y′ = y 2 + 2xy .

1.2.27. xy′ = y(ln2 ( y / x) + 1) . 1.2.29 xy′ sin( y / x) + x = y sin( y / x) .

1.3. Solve the linear differential equations.

1.3.1. (1− x2 ) y′ + 2xy = x.

1.3.3. (4 + x2 ) y′ − 2xy = (4 + x2 )2 .

1.2.4. 2y′ =	y 2	+ 2	y	− 4 .
	x2		x

1.2.6. (x2 + y 2 )dx + x2 dy = 0 . 1.2.8. 2xydy = (x2 − y2 )dx .

1.2.10. xy′− y = x ctg xy .

1.2.12.y′ = x + 2y .

x− 4y

1.2.14.2yxy ′ = x2 + y2 .

1.2.16.	xy′ = xe2 y / x + y .
1.2.18.	xy′ = y + x sec(y / x) .
1.2.20.	xy′ = y − xe y / x .
1.2.22.	x2 y′ + y 2 = xyy′ .
1.2.24.	2x2 + y 2 = 2x2 y′ .

1.2.26. (x − y) y′ = x +2y .

1.2.28. xy′ − y = y 2 + x2 . 1.2.30. xy + y 2 = (3x2 + xy) y′ .

1.3.2. y′ − y / x = x cos x .

200

http://vk.com/studentu_tk, http://studentu.tk/

<<< < Предыдущая 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1920 / 2520 21 22 23 24 25 > Следующая >>>

Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]

#
15.08.2019637.53 Кб4Gulenko.docx
#
12.11.2018765.44 Кб6Hagakure_-_Sokrytoe_v_listve.doc
#
19.02.20166.95 Mб201Higher Mathematics. Part 3.pdf
#
19.02.20169 Mб95Higher_Mathematics_Part_1.pdf
#
19.02.20166.63 Mб1052Higher_Mathematics_Part_1.pdf
#
19.02.20169.73 Mб45Higher_Mathematics_Part_2.pdf
#
19.02.20167.48 Mб408Higher_Mathematics_Part_2.pdf
#
19.02.20167.31 Mб123Higher_Mathematics_Part_2.pdf
#
19.02.20169.11 Mб304Higher_Mathematics_Part_3.pdf
#
19.02.2016694.82 Кб11HiraganaKatakanaWorkSheet (1).pdf
#
11.09.2019192.23 Кб1hjuj;f.docx