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Белорусский национальный технический университет

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[НЕСОРТИРОВАННОЕ]

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Математика. Часть 3.pdf

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	∞			2n + 3				∞		1
1)	∑				.		2)	∑			.
1)	∑				.		2)	∑			.
	n=1 n2 +1							n=1	(n	+ 4)!
	∞			(x −1)n								π
5)	∑					.	6)	f (x) = sin 2x, x0 =					.
5)	∑			2n		.	6)	f (x) = sin 2x, x0 =				4	.
	n=1			2n								4
9)	y	′′	− y cos x			= 0, y(0)		′
9)	y		− y cos x			= 0, y(0)	=1, y (0) = 2,k = 5 .

∞

(−1)n

∞

∑

ln(n +1)

n=1

2 dx .

f (x) = x cos 2x .

∫e−x

10)

0, − π ≤ x < 0,

f (x) =

0 ≤ x ≤ π.

x +1,

					Вариант 30
∞	1		∞	3n +1 n
1) ∑		.	2) ∑		.
	+ 4n			4n + 5
n=1 n2			n=1

	∞	(x + 4)n			π
5)	∑		.	6) f (x) = 2 + cos x, x0 =	4	.

	n=1 n(n +1)(n + 2)
9)	y′ = y cos x + 2cos y, y(0) = 0, k = 3 .

	∞	(−1)		n				∞
3)	∑				.		4)	∑xn .
		(2n)!
n=1								n=1
7)	f (x) =		sin x			.	8)	1 sin x		dx .
					x			∫	x
								0
10)				0, − π < x < 0,
	f (x) =					0 ≤ x ≤ π.
				x,

Типовой расчет «Элементы операционного исчисления»

Взадачах № 1, 2 установить принадлежат ли множеству оригиналов данные функции.

Взадаче № 3, пользуясь определением, найти изображение оригинала 1(t) f (t) , где

1(t) = 0, t < 0,1, t ≥ 0.

Взадачах № 4, 5 найти изображение оригинала 1(t) f (t) .

Взадачах № 6 найти свертку данных функций.

Взадаче № 7 найти изображение периодического оригинала 1(t) f (t) .

Взадаче № 8 найти оригинал по данному изображению.

Взадаче № 9 найти частное решение дифференциального уравнения, удовлетворяющее заданным начальным условиям.

Взадаче № 10 найти частное решение системы дифференциальных уравнения, удовлетворяющее заданным начальным условиям.

159

							Вариант 1
	0,	t < 0		0,			t < 0				e−2t sin t
1)	0,	t < 0	2)		1			3)	f (t) = e2t	4) f (t) = sin 4 t 5) f (t) =	e−2t sin t	.
1)	f (t) =		2)	f (t) =	1			3)	f (t) = e2t	4) f (t) = sin 4 t 5) f (t) =		.
	e5t , t ≥ 0						, t ≥ 0				t
	e5t , t ≥ 0					3	, t ≥ 0				t
				t +		3
6)	f1(t) = t, f2 (t) = sh t					7)

8) F ( p)

x′ =

10)y′ = z′ =

=	2
	p( p2 + 4)

−2x −2y −4z, −2x + y −2z, , 5x + 2y +7z,

9) y′′− 2y′ = t2 , y(0) = 0, y′(0) = 0

x(0) =1, y(0) = 2, z(0) = −1

Вариант 2

		0,				t < 0		0,		t < 0				e		3t	, t [0,1]
		0,				t < 0										3t					2
1)	f (t) =					2)					3)								4)	f (t) = ch tsin	2	3t .
1)	f (t) =			t	2	2)		f (t) = sin t		, t ≥ 0.	3)			f (t) =	2,		t [1,	∞]	4)	f (t) = ch tsin		3t .
						, t ≥ 0.

		e							t
									t
5)	f (t) =		cos 2t − cos3t				.	6)	f	(t) = et ,	f	2	(t) = e−t .
5)	f (t) =						.	6)	f	(t) = et ,	f		(t) = e−t .
						t			1
						t
7)

′′

′

8) F ( p) = p2 + 2 p + 7 .

9) y

+ y

= sin t, y(0) = 0,

y (0) = 0 .

10)

x′′−3x′+ 2x + y′− y

= 0,

′

− x′+ x + y′′−5y′+ 4y =

0, t < 0, 1) f (t) = t3, t ≥ 0.

		Вариант 3
	0,	t < 0,
2)			3)	f (t) = et / 2 .	4) f (t) = sin tsin 2t .
2)	f (t) = cost	, t ≥ 0.	3)	f (t) = et / 2 .	4) f (t) = sin tsin 2t .


	t

160

f (t) = 1− e3t .

(t) = cos 2t, f

(t) = e−t .

7) f (t) = e−t , t [0;3],

f (t +3) = f (t) .

te2t

p + 2

′′

− y

= 8t e

′

8) F ( p) =

p( p + 3) .

, y(0) = 0, y (0) = 0 .

′′

′

6 y

= 0,

′

−8x

10)

x(0) =1, x

(0)

= y(0) = y (0) =

0 .

′

+ y

′′

− 6 x

+ 2y = 0,

Вариант 4

t < 0,

f (t) =

t [0, 2],

(t) =

f (t) = 2 −t .

4) f (t) = e2t cos2 t .

, t ≥ 0.

, t > 2.

sin t

f (t) = e2t sin 4t .

(t) = t,

(t) = e3t .

f (t) = e−t , t [0, 2],

f (t + 2) = f (t) .

4 − p

′′

′

8) F ( p) =

p2 + 9 .

− 2y

+ 3y =1, y(0)

= 0, y

(0) = 0 .

x′ = y + z,

10)

x(0) =

3, y(0) = −1, z(0) = 2 .

y′ = z + x, ,

z′ = x + y,

Вариант 5

1)	f (t) =	0,		t < 0,
				, t ≥ 0.
		t +sin 3t
5)	f (t) =		e−αt sin t	.
			t

7)


8) F ( p) =		2 p + 3
			.
	p2 −6 p +12
x′ = 8y,
10) y′ = −2z,
		, x(0)

z′ = 2x +8y −2z,

	0,	t < 0,
2)			3) f (t) = sh 2t .	4) f (t) = cos3 t .
2)	f (t) =		3) f (t) = sh 2t .	4) f (t) = cos3 t .
	t	/ t, t ≥ 0.
	e	/ t, t ≥ 0.
6)	f1(t) = cost, f2 (t) = cost .

9)y′′+ y = t2 , y(0) = 0, y′(0) = 0 .

=2, y(0) = 0, z(0) = −1.

161

	0,		t < 0,
1)		2 t
	f (t) = sin		, t ≥ 0.


	t
5)	f (t) = cos(αt − b) .
7)

Вариант 6

	0,		t < 0,						t
		1				3t		2	t
2)	f (t) =		, t ≥ 0.	3)	f (t) = 2 sh 3t . 4) f (t) = e		sin			.
2)	f (t) =			3)	f (t) = 2 sh 3t . 4) f (t) = e		sin		2	.


	t +1

6) f1(t) = sin 3t, f2 (t) = sin 4t .

8) F ( p) =		5		′′	+ 2y	′	+ y = 2 sin t, y(0) = 0,	′
		p( p2 + 2 p +5) . 9) y		′′		′		′
		p( p2 + 2 p +5) . 9) y						y (0) = 0 .
10)	x′+3x + y = 0,		x(0) = 2, y(0) = 3.
10)		,	x(0) = 2, y(0) = 3.
	y′− x + y = 0,

Вариант 7

1)	f (t) =	0,		t < 0,

			e(2+3t)t , t ≥ 0.

5)	f (t) = cos bt − cos at .
				t
	F ( p) =			2 p + 6
8)					.
			( p − 2)( p + 3)

t < 0,

f (t) = cos3t

, t ≥ 0.

f (t) = 2

− e

. 4) f (t) = cos

4, t

[0, 2),

f1(t) =

, f2

(t) = ch 3t . 7)

f (t) =

[2,3].

f (t +3) =

f (t).

2, t

′′

+ 2y

′

+5y

′

= 5, y(0) = 0, y (0) = 0 .

x′′+ y′′

= 0,

10)

′

x(0) = x (0) = 0, y(0) =

y (0) = −1.

x′+ y =1+et ,

Вариант 8

t < 0,

f (t) =

f (t) = 3 − 2t . 4)

f (t) = ch 3t cos2 t .

f (t) =

t3,

t ≥ 0.

sin t

− e

e−t , t [0, a),

5) f (t) =

6) f

(t) = e−t ,

f (t + 2a) = f (t).

tet

(t) = sin t . 7) f (t) =

e−2t , t [a, 2a],

3p +1

′′

′

F ( p) = p2 ( p2 − 4 p +1)

+ y

= sin t, y(0) = 0, y (0) = 0 .

162

10)	x′′− x′+9x − y′′− y′−3y = 0,					′		′
10)	2x′′+ x′+				x(0) = x (0) =1, y(0) = y (0) = 0
	2x′′+ x′+	7x − y′′+ y′−5y = 0,
						Вариант 9
	0,	t < 0,			0,	t < 0,
1)	0,	t < 0,	2) f (t)				3)	f (t) =1+ e2t . 4) f (t) = sin2 2t cos3t .
1)	f (t) =		2) f (t)	= sin t			3)	f (t) =1+ e2t . 4) f (t) = sin2 2t cos3t .
	e1/ t , t ≥ 0.					, t ≥ 0.
	e1/ t , t ≥ 0.					, t ≥ 0.
					2t

5)	f (t) = e2(t−1) sin(t
8)	F ( p) =	p2	+1	.
		p3	+ 27

10)x′−2y +5x = et , y′− x +6y = e−2t

−1) . 6) f1(t) = t, f2 (t) = cos3t . 7)	2t, t [0, π],	f (t + 2π) = f (t).
−1) . 6) f1(t) = t, f2 (t) = cos3t . 7)	f (t) =	f (t + 2π) = f (t).
	3t, t (π, 2π[,

9) y′′+ y = t et , y(0) = 0, y′(0) = 0 .

, x(0) =1, y(0) = −1. ,

Вариант 10

t < 0,

2, t [0,3),

f (t)

4) f (t) = sin3 t .

f (t) =

, t ≥ 0.

t, t [3,∞).

t −1

f (t) = e−(t−a) cos(t − a) .

(t) = t,

(t) = ch t .

7) f (t) =

3at

, t [0,2π], f (t + 2π) = f (t) .

2π

2 p2 −1

′′

′

F ( p) = p( p2 + 2) .

+ y = e

, y(0)

= 0, y (0) = 0 .

10)

4x′− y′+ 3x = sin t,

x(0) = 2, y(0) = −1.

x′+ y = cost,

Вариант 11

t < 0,

3) f (t) = 2t2 .

f (t) = sh 2tsin 3t .

f (t) =

, t

≥ 0.

t +sin 2t

, t ≥ 0.

−4

f (t) = tsin2 t .

f1(t) = t, f2 (t) = sin 4t . 7)

f (t) =

π, t [0,

π),

f (t + 2π) = f (t).

2π, t [π, 2π],

p2 +1

′′′

′′

′

′′

F ( p) = p2 ( p2 + 9) .

+ y

= cost, y(0)

0 .

= y (0) = y

163

x′′+ y′+ x

= e

′′

′

10)

x(0)

=1, x

′

′′

(0) =

y(0) = −1, y (0) = 2.

+ y

=1,

Вариант 12

t < 0,

f (t)

f (t) = e−2t .

4) f (t) = cos 2t cos3t .

1) f (t) =

= cos3t

, t ≥ 0.

ch 2t, t ≥ 0.

5)		f (t) =	e2t sin t		.
				t

8)	F ( p) =			p2 − p + 2
				p3 − p2 −6 p

10)		x′′+ y	=1,
		y′′+ x		, x(0) =
			= 0,

		0,			t < 0,
1)	f (t) =				t [0,1),
		t,
		et		,	t ≥1

5)	f (t) =		sin 3t cos 2t			.

					t
7)

8) F ( p) =	p
		.
	p2 + 3p + 2

6)	f1	(t) = sin t, f2 (t) = sin 2t . 7)	2t, t [0,1),		f (t + 2) =	f (t).
			f (t) =	t [1,2],
			t,
.		9) 2y′+ 3y = t2 , y(0) = −1.

y(0) = x′(0) = y′(0) = 0.

			Вариант 13
	0,		t < 0,				t
		2					t		4) f (t) = et cos2 t .
2)	f (t) =			, t ≥ 0.		3) f (t) = ch		.
2)	f (t) =					3) f (t) = ch	2	.

			3
	t −		3
	6) f	(t) = cos 2t, f			2	(t) = e2t .
	1				2

9) y′+ ay = b, y(0) = 0 .

10)	x′′+ y′ = sh t −sin t −t,		x(0)	′	y(0) =1, y	′	= 0 .
	y′′+ x′ = ch t −cost,	,		= 0, x (0) = 2,

164

Вариант 14

t < 0,

f (t) =

sin t

, t ≥ 0.

(t) =

t +

, t ≥ 0.

f (t) =

sin 3t

(t) = e3t

(t) =

3)	2, t [0,5),		4)	f (t) = sh t cos 2t .
	f (t) =	∞).
	t, t [5,

cost .

8) F ( p) =

1− p2

9) y

′′

− y

′

−6y = 2,

y(0)

=1,

′

p( p2 +1) .

(0) = 0 .

′′

′

+ y

= e

−t

′

10)

− x

+cost,

y(0)

= 0,

=1.

′

′′

′

x(0) = 2, x (0) =1,

− y

= 2e

+sin t,

Вариант 15

		0,			t < 0,
1)	f (t) =
		sin t			, t ≥ 0.


		t −2
5)	f (t) =		et sin 2t			.
				t

7)

8) F ( p) =						p + 2
					( p2 + 4 p +5)2
	x	′	+3x			−4y = 9e	2t
10)								,
				′				,
	2x + y					−3y = 3e	2t
								,

	0,	t < 0,	3) f (t) = 3sh	t
2)	f (t) =				.	4) f (t) = cos5t sin 3t .

	e(1+i)t , t ≥ 0.		3

6) f1(t) = t, f2 (t) = cos 2t .

.	9) y	′′	− 4y = 2 cos 2t, y(0)	′
				= 0, y (0) = 4 .

x(0) = 2, y(0) = 0.

165

Вариант 16

1)	0, t < 0,
1)	f (t) =
	sh 2t, t ≥ 0.

5) f (t) = 1+ cos 2t . t


	0,			t < 0,
				t [0, 2],
2) f (t) =	t2	,		t [0, 2],

		1
				, t > 2.
				, t > 2.
			4
	t −		4

6) f1(t) = t, f2 (t) = sin 2t .

3)	0,		t [0,4],	4)	f (t) = et sin2 t .
	f (t) =	,	t (4,∞).
	5t
7)	a, t [0,l[,			f (t + 2l) = f (t).
	f (t) =
	−a, t [l, 2l[,

p2 − 2 p −1

′′

′

F ( p) = ( p2 − 2 p + 3)2 .

9) 2y

− 9y =

2 −t, y(0) = 0, y (0) =1.

x′′+ y′

−t,

′

10)

+ y = e

′

′′

−t

x(0) =1, x (0) = 2, y(0) = y

(0) = 0 .

− x + 2y

− y = −e

Вариант 17

t < 0,

f (t) = t2 +1, t [0,1]

f (t) =

f (t) = t2 −1.

4) f (t) = ch 2t cost .

, t ≥ 0.

t >1.

sint,

f (t) = cos3t +t et .

f1(t) = e2t

f2 (t) = sin 3t .

sin t

> 0,

f (t) =

t < 0,

f (t + π) = f (t) .

p2 −3p − 9

′′

′

8) F ( p) =

( p2 + 9)2

0 .

+ y = sin 2t, y(0) = y (0) =

10)

x′′− y′

=1,

′

y′′− x′

x(0) =1, x (0) = y(0), y

(0) =1.

= 0,

Вариант 18

0, t < 0,

t <

t [0, 2),

f (t) =

f (t) = sin 2t cos2 t .

f (t) =

f (t) = 1

, t

≥ 0.

2t, t [2, ∞).

sh it, t ≥ 0.

t −

f (t) =

2 + 3cos 4t

f (t) = t, f

(t) = sh 2t .

166

8) F ( p) =

p2 + 4

9) y

′′′

− y

′

=10e

, y(0)

′

′′

( p2 − 4)2 .

= y (0)

= y (0) = 0 .

10)

x′+3x + y = 0,

x(0) = y(0) =1.

y′− x + y = 0,

Вариант 19

		0,			t < 0,
1)	f (t) =			1

		sin t +			, t ≥ 0.
				t

5)	f (t) =		sin t cos3t			.

			t
7)

2)		0,				t < 0,	3)		f (t) = t2 + 2 . 4) f (t) = ch 2tsin2 t .
2)	f (t) =						3)		f (t) = t2 + 2 . 4) f (t) = ch 2tsin2 t .
		e(1+2i)t , t ≥ 0.

		(t) = e2t					t
6)	f	(t) = e2t	,	f	2	(t) = e	2	.
	1				2

8) F ( p) =		p2 − 2 p −8	9) y	′′	− y	′	= t, y(0)	′
		( p2 − 2 p +10)2 .
								= y (0) = 0 .
10)	x′+ 4x + 4y = 0,			15 .
		, x(0) = 3, y(0) =
	y′+ 2x +6y = 0,

Вариант 20

1)	0, t < 0,
	f (t) =
	sin t, t ≥ 0.
5)	f (t) = tsin 3t cos 4t .
	F ( p) =	2( p −3)
8)			.
		( p2 −6 p + 8)2

	0,			t < 0,			t
		1					t
2)	f (t) =			, t ≥ 0.	3)	f (t) = sh		.	4) f (t) = ch 3tsin 2t .
2)	f (t) =				3)	f (t) = sh	3	.	4) f (t) = ch 3tsin 2t .

			3
	t +		3

6) f	(t) = e5t ,			f	2	(t) = t . 7)	f (t) = t +1, t [0,1], f (t +1) = f (t).
1					2
	9) y	′′	+ y	′			′
	9) y		+ y		=1, y(0) = y (0) = 0 .

x′′+ y′+ x

= e

′

10)

x(0)

y(0) = −1,

′

′′

=1, x (0) =

y (0) = 2.

+ y

=1,

Вариант 21

t < 0,

e2t , t [0,1),

4) f (t) = sh tsin

t .

1) f (t) =

f (t) = 1

, t ≥ 0.

f (t) =

t [1, ∞).

cost, t ≥ 0.

sin t

167

5)	f (t) =	sin t sin 3t			.		6)	f (t) = cos 2t, f						2	(t) = cos3t .		7)	f (t) = 2t, t [0, 3],						f (t +3) = f (t).
5)	f (t) =				.		6)	f (t) = cos 2t, f							(t) = cos3t .		7)	f (t) = 2t, t [0, 3],						f (t +3) = f (t).
			t					1
			t
				p −1							′′					′
8)	F ( p) = ( p2 − 2 p +10)2 .							9)	y		′′		+ y = 3, y(0)			′
8)	F ( p) = ( p2 − 2 p +10)2 .							9)	y				+ y = 3, y(0)			= y	(0) = 0 .
	x′ = −2x −2y −4z,
10) y′ = −2x + y −2z,							x(0) =1, y(0) = 2, z(0) = −1.
10) y′ = −2x + y −2z,						,	x(0) =1, y(0) = 2, z(0) = −1.
	z′ = 5x + 2y			+7z,
	z′ = 5x + 2y			+7z,
													Вариант 22
																				π
								0,					t < 0,			sin t,		t 0,

	0,			t < 0,															2		,				2	t
	0,			t < 0,															2						2	t
1)	f (t) =			t ≥ 0.			2)	f (t) =		2					3)							4)	f (t) = sh 2t cos				.
1)	f (t) =						2)	f (t) =	t	2					3)			π				4)	f (t) = sh 2t cos			2	.
	sin t,								t			,	t ≥ 0.					π								2
								e				,	t ≥ 0.						,∞ .
																0,		t
																0,		t
																		2

5) f (t) = t cos(2t + 3) ;	6) f1(t) = sin 3t, f2 (t) =t.	7)	0,	t [0,1),
				t [1, 4],
			1,

8) F( p) =

. 9)

′′

− 2y

′

+ y = e

, y(0)

′

( p − a)( p −b)

= y (0) = 0 .

10)

′

− x − 2y = t,

y(0) = 4.

x(0) = 2,

y′− 2x − y = t,

Вариант 23

1)	f (t) =	0,		t < 0,
1)	f (t) =
			e2+t , t ≥ 0.

5)	f (t) =		t cost	;
5)	f (t) =		e5t	;
			e5t
7)

8) F( p) =	1		.
		( p − 2)4
x′ = y − z,
10) y′ = z
	−2x, , x(0)

z′ = 2x − y,

	0,			t < 0,		cost,	t [0, π),
2)		1			3)
	f (t) =		,	t ≥ 0.			t [π, ∞].
						0,

		2t +1

6) f1(t) = sin 2t, f2 (t) = e3t .

9) y′+ y = sin t, y(0) = 0 .

=1, y(0) = z(0) = 0.

f (t + 4) = f (t) .

4) f (t) = sin2 t cos 2t.

168

1)	0,		t < 0,
	f (t) =		t ≥ 0.
		3t2,

5)	f (t) = t cos(2t + 3) .
8)	F( p) =	p2	+ 5	.
		( p2	−5)2

			Вариант 24
	0,		t < 0,
2)		1
	f (t) =		, t ≥ 0.


	sin 2t
6)	f1(t) = cos3t, f2 (t) = sin 2t .
9)	y′+ y = t,	y(0) = 0 .

Вариант 25

3)	f (t) = 2 −t.			4)	f (t) = cht sin2 t .
7)	f (t) = 2t, t [0, 1], f (t +1) = f (t);
	x	′′	+ 2y = 0,	x(0)	′	1,
					= 0, x (0) =
10)	y	′′	,	y(0)	′
			−2x = 0,		= y (0) = 0.

1)	f (t) =	0,		t < 0,
				t ≥ 0.
			e2t ,

5)	f (t) = t cos(t + 2) .
			et+2
8)	F( p) =			1	.
			(2 p + 3)3

	0,		t < 0,
2)
	f (t) = 1		, t ≥ 0.

		2
	t −
6)	f1(t) = cos5t, f2 (t) = t .
9)	y′+ y = sin t,		y(0) = 0 .

	x′− x	+ 2y = 0,	x(0)	= 0,			′
10)					1		x (0) = −1,
				= −		,	′
	x′′− 2y′ = 2t − cos 2t, y(0)				2		y (0) = 0.

Вариант 26

3)	f (t) =1−t2.			4) f (t) = sin 2t sin 4t .
7)	t	,	t [0,2],	f (t +3) = f (t) .
	e
	0,		t (2, 3),

1)	f (t) =	0,		t < 0,
1)	f (t) =		, t ≥ 0.
		e57	, t ≥ 0.

5)	f (t) =	e−2t	sin t		.
5)	f (t) =	t			.
		t
7)


8) F ( p) =	p2	−3p − 9		.
8) F ( p) =	( p2		+ 9)2	.
	( p2		+ 9)2

	0,			t < 0,
2)		1			3)	f (t) = e2t .	4) f (t) = sin 4 t .
2)	f (t) =	1		, t ≥ 0.	3)	f (t) = e2t .	4) f (t) = sin 4 t .

			3
	t +		3
6)	f1(t) =t, f2 (t) = sh t .

9) y′′− y′−6y = 2, y(0) =1, y′(0) = 0

10)x′′− y′ =1, , x(0) =1, x′(0) = y(0) = 0, y′(0) =1. y′′− x′ = 0,

Вариант 27

169

		0,	t < 0			0,			t < 0
1)	f (t) =	0,	t < 0		2)		1			3)	f (t) = e3t	4) f (t) = sin4 2t
1)	f (t) =				2)	f (t) =	1			3)	f (t) = e3t	4) f (t) = sin4 2t
		e−6t , t ≥ 0							, t ≥ 0
		e−6t , t ≥ 0							, t ≥ 0
						t −5
5)	f (t) =	e−6t	sin t	.	6) f (t) = t2		, f	2	(t) = sh t
5)	f (t) =			.	6) f (t) = t2		, f		(t) = sh t
		t				1
		t
7)

8) F( p) =

p( p2 + 4)

x′ = −2x −2y −2z, 10) y′ = −2x + y −3z, , z′ = 5x + 2y + 4z,

9) y′′−4y′ = t2, y(0) = 0, y′(0) = 0

x(0) =1, y(0) = 2, z(0) = −1

Вариант 28

1)	f (t) =	0,	t < 0	2)
1)	f (t) =			2)
		e4t , t ≥ 0.

5)	f (t) =	cos3t −cos 2t			.
5)	f (t) =				.
			t
7)

8) F( p) = p2 + 2 p +7 .

0,		t	< 0			e2 t , t [0,1],
				3)		e2 t , t [0,1],	4) f (t) = ch t sin	2	4t .
f (t) = sin 5t				3)		f (t) =	4) f (t) = ch t sin		4t .
			, t ≥ 0.			2, t (1,∞).
		t	, t ≥ 0.			2, t (1,∞).
		t
	6) f		(t) = e3t , f		2	(t) = e−t .
		1			2

9) y′′+ 2y′ = sin t, y(0) = 0, y′(0) = 0 .

10)	x′′−3x′+ 2x + y′−5y = 0,		′	′
10)		, x(0)	= x (0)	= y (0) = 0, y(0) =1.
	− x′+ x + y′′−5y′+ 2y = 0

170

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