3952

sin3

x

1

sin 2x

Итак,

y = -

+

x -

-

частное решение.

3

2

2

полагая y′ = P( y) . Тогда y¢¢ =

dP

×

dy

= P

dP

= P × P¢ . Теперь данное уравнение

dy dx

dy

первого порядка

с разделяющимися переменными

P

dP

× y3 = -1.

dy

P2

1

Разделяя переменные и интегрируя, получим:

=

+ С ,

2

2 y2

1

Так как P = y′, то

dy

= ±

1

откуда P = ±

1

+ 2С .

+ 2С .

y2

y 2

1

dx

1

−

1 − y 2

= ± x

или x2 + y2 =1.

Частным

решением,

удовлетворяющим заданным начальным условиям

y(0) =1 и

y′(0) = 0 будет функция

y = +

.

1 − x2

6.01

y′′′ = sin x ,

y(0) = y′(0) = y′′(0) = 0

6.02

y ′′′ =

1

,

y(1) =

3

;

y′(1) = y′′(1) = 0

x

4

6.03

y′′′ = cos x ,

y(0) = y′(0) = 0;

y′′(0) = 1

6.04

y′′′ = sin2 x ,

y(0) = 0;

y′(0) =

1

; y′′(0) = 0

8

6.05

y′′′ = ln x ,

y(1) = −

7

; y′(1) =

1

;

y′′(1) = 0

36

4

6.06

y′′′ =

1

,

y(1) = 0;

y′(1) = 1;

y′′(1) = 0

x2

6.07

y′′′ = cos2 x ,

y(0) = 0;

y′(0) =

1

;

y′′(0) = 0

8

6.08

y′′′ = x sin x ,

y(0) = 2;

y′(0) = y′′(0) = 0

6.09

y′′′ = x cos x ,

y(0) = y′(0) = 0

y′′(0) = 1

6.10

y′′′ = sin2 2x ,

y(0) = 0;

y′(0) =

1

;

y′′(0) = 0

32

6.11

y′′′ = e2x ,

y(0) =

1

;

y′(0) =

1

; y′′(0) =

1

8

4

2

6.12

y′′′ = cos2 2x ,

y(0) = 0;

y′(0) = −

1

; y′′(0) = 0

32

6.13

y′′′ =

1

,

y(1) = 0; y′(1) =

1

; y′′(1) = −

1

x3

2

2

6.14

y′′′ = sin2 3x ,

y(0) = 0;

y′(0) =

1

;

y′′(0) = 0

72

31

7.01

xy′′ + xy′ = y′ ,

y(1) = 0;

y′(1) = 1

7.02

x2 y′′ = (y′)2 ,

y(1) = 2;

y′(1) = 2

7.03

x3 y′′ + x2 y′ = 1,

y(1) = y′(1) = 0

7.04

y′′ + y′ tg x = sin 2x

y(0) = 1;

y′(0) = 0

7.05

xy′′ − y′ = x2ex ,

y(1) = y′(1) = 0

7.06

y′′ + 2x(y′)2 = 0 ,

y(0) = 2;

y′(0) = 9

7.07

y′′x ln x = y′ ,

y(e) = 2;

y′(e) = 4

7.08

y′′ − y′ctg x = sin 2x ,

π

=

π

;

π

= 0

y

2

y′

2

2

7.09

y′′(ex + 1)+ y′ = 0 ,

y(0) = 3;

y′(0) = 2

7.10

y′′ + y′ = xy′′ ,

y(0) = 2;

y′(0) = −1

7.11

xy′′ − 2 y′ = −

2

,

y(1) = 0;

y′(1) = 0

x2

7.12

xy′′ = y′ + x sin

y′

,

y(1) = π − 1;

y′(1) = π

x

2

2

7.13

x2 y′′ + xy′ =1,

y(1) = 0;

y′(1) = 1

33

7.14

(1 − x2 )y′′ + xy′ = 2 ,

y(0) = 0;

y′(0) = 2

7.15

xy′′ − y′ = x ,

y(1) = 0;

y′(1) = 0

7.16

xy′′ − y′ = x3 ,

y(1) = 0;

y′(1) = 0

7.17

(x − 3)y′′ + y′ = 0 ,

y(4) = 0;

y′(4) = 1

7.18

(1 + x2 )y′′ + xy′ = 0 ,

y(0) = y′(0) = 1

y′

2 x

1

1

7.19

y′′ −

= xe

,

y

= y′

= 0

x

2

2

7.20

x4 y′′ + x3 y′ = 4 ,

y(1) = 1; y′(1) = 0

1

π

π

1

7.21

tg xy′′

− y′ +

= 0 ,

y

= 0;

y′

=

sin x

2

2

2

7.22

(1 + x2 )y′′ + 2xy′ = x3 ,

y(0) = y′(0) = 0

7.23

(1 + x2 )y′′ + 2xy′ = 12x3 ,

y(0) = 0;

y′(0) = 1

7.24

y′′ +

2x

y′

= 2x ,

y(0) = 0;

y′(0) =

1

1

+ x2

2

1

′

7.25

xy′′ +

y′ =

,

y(1) = 4;

= 0

x

y (1)

7.26

(x + 1)y′′ + y′ = x + 1,

y(0) = 0;

y′(0) =

1

2

π

π

7.27

y′′tg 5x = 5y′,

y

= 0;

y′

=1

10

10

7.28

(1 + sin x)y′′ = y′cos x ,

y(0) = 0;

y′(0) = 1

7.29

y′′ tg x = y′ + 1,

π

= 0;

π

= 0

y

y′

2

2

34

7.30 tg x × y′′ = 2 y′ ,

π

=

π

π

= 1

y

2

4

; y′

2

8.01

y′′ = 128 y3 ,

y(0) = 1;

y′(0) = 8

8.02

y′′y3 = −64,

y(0) = 4;

y′(0) = 2

8.03

y′′ + 8sin y cos3 y = 0 ,

y(0) = 0;

y′(0) = 2

8.04

yy′′ − (y′)2 = y′y2 ,

y(0) = 1;

y′(0) = 2

8.05

y(y − 1)y′′ = (y′)2 ,

y(0) = y′(0) = 2

8.06

(y + 2)(y + 3)y′′ = (y′)2 ,

y(0) = −1;

y′(0) = 1

8.07

(y − 4)(y − 5)y′′ = (y′)2 ,

y(6) = 6;

y′(6) = 1

8.08

2(y′)2 = y′′y ,

y(0) = 1;

y′(0) = 1

8.09

y′′ = (y′)2 tg y ,

y(0) = π ;

y′(0) = 1

4

8.10

y′′sin y = (y′)2 ,

y(0) = π ;

y′(0) = 1

3

8.11

yy′′ + 1 = (y′)2 ,

y(0) = 1;

y′(0) =

10

8.12

(1 + y)y′′ = (y′)2 ,

y(0) = 1;

y′(0) = 2

35

8.13

y3 y′′ + 9 = 0 ,

y(0) = 1;

y′(0) = 3

8.14

y′′ + 2 y(y′)3 = 0 ,

y(0) = 0;

y′(0) = −3

8.15

y′′ tg y = 2(y′)2 ,

y(0) = π ;

y′(0) = 1

2

8.16

(y′)2 + 2yy′′ = 0 ,

y(0) = y′(0) = 1

8.17

y2 y′′ = (y′)3 ,

y(0) = 1;

y′(0) = 2

8.18

2(y′)2 = ( y − 1) y′′ ,

y(0) = 2;

y′(0) = 1

8.19

yy′′ + (y′)2 = 0 ,

y(0) = y′(0) = 1

8.20

y′′sin y = 2(y′)2 ,

y(0) = π ;

y′(0) = 1

2

8.21

y3 y′′ = y4 − 16 ,

y(0) = 2

2; y′(0) = 2

8.22

yy′′ − (y′)2 = y 2 ln y ,

y(0) = y′(0) = 1

8.23

y′′ + 50sin y cos3 y = 0 ,

y(0) = 0;

y′(0) = 5

8.24

y′′ + y = b ,

y(0) = 26;

y′(0) = 0

8.25

y′′ctg y = 2(y′)2 ,

y(0) = 0;

y′(0) = 1

8.26

yy′′ + y = (y′)2 ,

y(0) = 1;

y′(0) = 2

8.27

y′′ = 72 y3 ,

y(2) = 1;

y′(2) = 6

8.28

2 yy′′ = y 2 + (y′)2 ,

y(0) = 1;

y′(0) = 0

8.29

y′′ = 50sin3 y cos y ,

y(1) = π ;

y′(1) = 5

2

8.30 y4 − y3 y′′ =1,

y(0) =

2

2; y′(0) =

2

Оглавление

§1. Основные понятия...............................................................................................

3

§2. Задачи на составление дифференциальных уравнений ..................................

4

Задание №1 ..................................................................................................................

8

§3. Дифференциальные уравнения первого порядка. Типы уравнений и

методы их решений...................................................................................................

10

3.1. Уравнения с разделяющимися переменными.............................................

10

Задание №2 ................................................................................................................

12

3.2. Однородные дифференциальные уравнения ..............................................

14

Задание №3 ................................................................................................................

17

3.3. Линейные уравнения первого порядка........................................................

19

Задание №4 ................................................................................................................

21

3.4. Уравнение Бернулли......................................................................................

23

Задание №5 ................................................................................................................

25

§4. Дифференциальные уравнения второго порядка, допускающие

понижение порядка...................................................................................................

27

Задание №6 ................................................................................................................

31

Задание №7 ................................................................................................................

33

Задание №8 ................................................................................................................

35

Литература .................................................................................................................

37