Higher_Mathematics_Part_2

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y = C (x)e−2x + C

= e−3x , y′

= −2e−2x ,

= −3e−3x , to find the functions

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The general solution of the initial equation is:

y= (C1 cos x + C2 sin x)e2x + e2x .

11.Find the general solution if the right hand member not zero equation y′′ − 7 y′ + 6 y = (x − 2)e x .

Solution . Evaluate the homogeneous equation at first y′′ − 7 y′ + 6y = 0 .

We would obtain:

k2 −7k +6 = 0; k1 =1; k2 = 6; y = C1ex +C2 e6x .

Control number of the right side of right hand member not zero equation z = 1 is simple root of the characteristic equation (divisible by r = 1).

Partial solution of right side member not zero equation according to the 3.2 table 3.4 could by found in a form

			y* = x(B				0		+ B x)e x = (B				0	x + B x2 )e x .
							0			1			0				1
Find derivatives y′*					and					y′′* :
y′* = (B			0	+ 2B x)e x + (B							0	x + B x2 )e x ,
			0		1						0	1
y′′* = 2B e x					+ 2(B				0	+ 2B x)e x + (B						0	x + B x2 )e x ,
			1						0	1						0	1
after substituting		y* , y′*			and				y′′* in the initial equation
		e x (2B + 2(B							0	+ 2B x) + (B			0	x + B x2 )) −
				1					0	1			0				1
− 7e x (B	0	+ 2B x + (B				0		x + B x2 )) + 6(B							0	x + B x2 )e x = (x − 2)e x .
	0			1		0				1					0		1

After simplifying we would obtain the equation: −10B1x + 2B1 − 5B0 = x − 2.

Comparing the coefficients of the same order x :

x1 :	− 10B1 = 1,			from this
		− 5B0
x0 : 2B1			= −2,

The general solution of initial equation is:

			9		x	2
y = C e x + C	2	e6x +		x −			e x .

1			25	10

12. Solve the Cauchy’s test

y′′ − 5y′ + 6y = 13sin 3x,	y(0) = 1,		y′(0) = 0 .
Solution. Characteristics equation k 2 − 5k + 6 = 0			has roots e2x , e3x –
fundamental system
The general solution of homogeneous equation y′′ − 5y′ + 6y = 0 . Then
y = C e2x + C	2	e3x .
1	2
221

Right side of these equation can be written in such a form : 13sin 3x = e0 x (0 cos 3x + 13sin 3x),

then control value of right

side z = 0 +3i =3i, . So for

this case in formula

(3.29) Pn (x) = 0, Qm (x) =13.

Because z = 3i

is not a root of characteristics

equation then we would obtain partial solution in such a form

y* = A cos 3x + B

sin 3x .

Find derivatives :

y′* = −3A sin 3x + 3B

cos 3x,

y′′* = −9A cos 3x − 9B

sin 3x.

After putting the values

y* , y′* and y′′* and simplifying in initial equation

we would obtain.

−3(A0 + 5B0 ) cos 3x + 3(5A0 − B0 ) sin 3x = 13sin 3x.

Compare the coefficients

when sin 3x and cos 3x :

− 3(A0 + 5B0 ) = 0

A0 =

, B0 = −

3(5A0 − B0 ) = 13

Therefore,

y* =

cos 3x −

sin 3x

is a partial solution of right hand member not zero equation, and general solution of those equation can be written as follows:

y = y + y* = C1e2x + C2e3x + 16 (5cos 3x − sin 3x).

And now we would consider such constants C1 and C2 . Condition y(0) =1 can be written as follows:

	1=C +C +									5	.
	1=C +C +										.
					1		2			6
Thus,										6
Thus,
y′ = 2C e2x +			3C		e3x	−		1	(5sin 3x +cos 3x) ,
y′ = 2C e2x +			3C		e3x	−			(5sin 3x +cos 3x) ,
	1		2					2
								2
and condition y′(0) = 0 is equivalent to the equation 0 = 2C +3C												2	−0, 5 .
After solution of this linear system										1		2
After solution of this linear system
							5
							5
			+C2 +
	C1		+C2 +				6	=1,
							6
							−0.5 = 0,
	2C +3C					2	−0.5 = 0,
			1			2
we would obtain C = 0,	C =	1		.
we would obtain C = 0,	C =			.
1	2

In this case the solution of Cauchy’s test will be obtained: 222

y= 16 e3x + 16 (5cos 3x − sin 3x).

13.General form of partial equation can be written as:

y′′ − 4y′ + 20y = xe2x sin 4x .
Solution. Characteristics equation		k2 − 4k + 20 = 0		has two complex roots
k1,2 = 2 ± 4i . Control value of the right side z = 2 + 4i				(α = 2,	β = 4) , which is
equal to the root of characteristic equation. Then			r = 1 and		partial solution
should be found in such a form
y* = xe2x ((A x + B ) cos 4x + (A x + B ) sin 4x) .
1	1	2	2
14. Write the general form of partial solution
y′′ − 2y′ + 5y = e x cos 2x + sin 2x + x .
Solution. Characteristics equation		k2 −2k +5 = 0 has two complex roots

k1,2 =1±2i . Partial solution of this equation should be found in such a form:

y* = y* + y* + y* , where, y* = xex (A cos 2x + B sin 2x)

a partial solution of

the equation y′′ − 2 y′ + 5y = e x cos 2x (Control number

z =1+2i

is the root of

characteristics

equation,

y2* = C cos 2x + D sin 2x

partial solution of

the

equation

y′′ − 2y′ + 5y = sin 2x

(Control

value

z = 2i

is not

the

root

characteristics

equation)

, y3* = Mx + N

is a

partial

solution of

the

equation

y′′ − 2y′ + 5y = x (main

value

z = 0

not

the

root

of the

characteristic

equation), where A,

B, C, D, M ,

N – are unknown constants.

15. Find the general solution of the equation

y′′ + 5y′ + 6y =

1+ e2x

Solution. Characteristic equation

k 2 + 5k + 6 = 0

has roots

k1 = −2

and

k2 = −3 .

the

general

solution

the

homogeneous

equation

y′′ + 5y′ + 6y = 0 can be written as follows

y = C e−2x

+ C

e−3x .

Function

f (x) = 1/(1 + e2x )

does not correspond to (3.29). Therefore we

can use the Lagrange’s method in order to solve these problems to obtain the solution if the form of (3.31):

As y1 = e−2x

C1 (x) and C2 (x)

, form and solve the systems of the solutions in the form (3.32):

223

−2x

−3x

= 0,

C ′e

+C ′e

C ′(−2e−2x ) +C

′(−3e−3x ) =

;

1+e

e−2x

e−3x

= −e−5x ,

1 =

e−3x

= −

e−3x

−3e−3x

−2e−2x −3e−3x

+e2x

1+e2x

e−2x

−2x

− 2e

−

;

C′

′

= −

;

1+ e

+ e

1+ e

1+ e2x

C =

e2x

dx =

d(e2x

+ 1)

ln(1+ e2x ) + C

;

∫ 1+ e2x

2 ∫ e2x

+ 1

C =−

e3x dx

ex = t

=−

t2 dt

=−(t −arctg t)

∫ 1

+e2x

ex dx = dt

∫ 1+t2

So,

=−(ex −arctg ex ) +C4 .

−2x

−3x

y = e

ln(1+e

) +C3

−e

−arctg e

+C4

–

is the general solution of the given equation where,

C3 ,C4

are

arbitrary

constants.

Т.3

Class and self assignments

Find the general solutions of the given homogeneous equations.

1.	y′′ − 3y′ + 2y = 0 .		2.	y′′ − 6y′ + 8y = 0 .
3.	y′′ − 2y′ + y = 0 .		4.	y′′ + 4y′ + 4y = 0 .
5.	y′′ − 2y′ + 2y = 0 .		6.	y′′′ − y′′ − y′ + y = 0 .
7.	y′′ + y′ + y = 0 .		8.	y′′′ − 3y′′ + 3y′ − y = 0 .
9.	y′′′ + y = 0 .		10.	y (4)	+ y = 0 .
11.	y (4)	+ 5y′′ + 4y = 0 .	12.	y (5)	+ 8y′′′ + 16y′ = 0 .
13.	y (5)	− 5y (4) + 12 y′′′ − 16y′′ + 12 y′ − 4y = 0 .

Find the general solutions of the right hand member not zero equations.

14. y′′ − 7 y′ + 12y = 5 .	15. y′′ − 7 y′ + 6y = sin x .
16. y′′ + y′ + y = 3e2x .	17. y′′ + 2y′ − 3y = 4e− x .
	224

18.y′′ − 8y′ + 7 y = 3x2 + 7x + 8 .

19.y′′ − 2y′ + 4y = (x + 2) e3x .

20.y′′ − 2y′ = x3 + 2x − 1 .

21.y′′ − 6y′ + 25y = 2 sin x + 3 cos x .

22.y′′ + y = (3x + 2) sin 2x + (x 2 + x + 2) cos 2x .

23.y′′ + 4y = sin 2x .

24.y′′′ − y′′ = −3x + 1 .

25.y′′ − 2y′ + y = 4e x .

26.y′′ + y = xe x + 2e− x .

Find the general solutions of the right hand member not zero equations by the method of arbitrary constants variation.

27.y′′ + y = sin1 x .

28.y′′ − y′ = 2 − x e x .

Solve using Cauchy’s method.

29.	y ′′+4y ′+4 y = 3e−x , y(0) = y′(0) = 0 .
30.	y′′+4y =sin 2x , y(0) = y′(0) = 0 .

Answers

1.	y = C1e x + C2e2x .										2. y = C1e2x + C2e4x .									3. y = C1e x + C2 xe x .
4.	y = (C1 + C2 x)e−2x .								5. y = (C1 cos x + C2 sin x) e x . 6.										y = C1e x + C2 xe x + C3e− x .
												1			x .
	y = (C1 cos			3	x + C						3 x) e−								y = C1e x + C2 xe x + C3 x 2e x .
7.	y = (C1 cos			3	x + C			2 sin			3 x) e−			2				8.	y = C1e x + C2 xe x + C3 x 2e x .
				2						2
	y = C1e− x + (C2 cos						3					3				x			2	x (C1 cos	2		2
9.							3		x + C3 sin			3	x)e 2 .				10.	y = e	2		2	x + C2 sin	2	x) +
9.							2		x + C3 sin			2	x)e 2 .				10.	y = e	2			x + C2 sin		x) +
							2					2							2			2
	−	2	x		2						2							y = C1 sin x + C2 cos x + C3 sin 2x +
+e		2	(C3 cos		2	x	+C4 sin				2	x) .					11.
+e		2	(C3 cos		2	x	+C4 sin				2	x) .					11.
					2						2											13. y = C1ex +
+C4 cos 2x .					12.		y = C1 + (C2 + C3 x) cos 2x + (C4 + C5 x) sin 2x .															13. y = C1ex +
+C2ex cos x + C3ex sin x + C4xex cos x + C5xex sin x .																				14. y = C1e3x + C2e4x +					5	.
+C2ex cos x + C3ex sin x + C4xex cos x + C5xex sin x .																				14. y = C1e3x + C2e4x +						.
																						12

225

																																1		x (C
15.			y = C ex + C								e6x +			7						5		sin x .						16.			y = e−					cos 3 x + C									3 x) + 3 e2x .
15.			y = C ex + C								e6x +			7			cos x +			5		sin x .						16.			y = e−		2			cos 3 x + C					2	sin			3 x) + 3 e2x .
				1				2					74							74												1				2					2			2						7
													74							74																2								2						7
17.			y = C e x					+ C				2	e−3x			− e− x .															18.	y = C ex + C							e7x +	3		x2		+	97		x +				1126				.
				1								2																				1				2				7					49					343
																																								7					49					343
19.																				x				3x 1						10			20. y			= C1 + C2e				2x			−	1	x	4		−		1		x	3	−
			y = (C1 cos							3x + C2 sin									3x)e		+ e						x +				.
			y = (C1 cos							3x + C2 sin									3x)e		+ e					7	x +			49	.													8						4
																										7				49														8						4
−		7	x2−		3	x .21. y = e3x (C												cos 4x + C					2	sin 4x) +					14cos x + 5sin x						.			22. y = C cos x + C												sin x −
8				8												1							2								102											1							2
8				8																											102
−	9x2 + 9x + 28								cos 2x −						x + 2				sin 2x .					23.		y = C cos 2x + C sin 2x −											1	x cos 2x .					24. y = C ex											+
				27												9														1		2				4														1
				27			1									9																				4
+C			2	+ C x +			1			x3 + x2 .								25. y = (C							+ C		2	x)e x			+ 2x 2e x .					26. y = C cos x + C													2	sin x +
+C				+ C x +						x3 + x2 .								25. y = (C							+ C			x)e x			+ 2x 2e x .					26. y = C cos x + C														sin x +
				3			2																	1															1
							2

+e− x +

(x − 1)ex . 27. y = (C

+ ln

sin x

) sin x + (C

− x) cos x .

28.

y =

+ C +

+C2ex

. 29. y =1,5x2e−2x . 30.

y = −

cos 2x +

sin 2x .

Т.3		Individual tasks
3.1. Find the general solutions of			linear homogeneous equations
3.1.1. а)		y′′ + y′ − 2y = 0 ;	б)	y′′ − 2y′ + 5y = 0 .
3.1.2. а)		y′′ − 4y′ = 0 ;	б)	y′′ + 4y′ + 4y = 0 .
3.1.3. а)		2y′′ − y′ − y = 0 ;	б)	y′′ − 2y′ + 10y = 0 .
3.1.4. а)		y′′ + 6y′ + 13y = 0 ;	б)	y′′ − 4y′ + 4y = 0 .
3.1.5. а)		2y′′ − 3y′ − 5y = 0 ;	б)	y′′ + 2y′ + 10y = 0 .
3.1.6.	а)	y′′ − 9y = 0 ;	б)	3y′′ + 12y′ + 15y = 0 .
3.1.7.	а)	y′′ − 2y′ − 3y = 0 ;	б)	3y′′ − 4y′ + 4y = 0 .
3.1.8.	а)	4y′′ + 4y′ + 5y = 0 ;	б)	y′′ + 8y′ + 16y = 0 .
3.1.9.	а)	4y′′ − 8y′ + 5y = 0 ;	б)	y′′ − 8y′ + 20y = 0 .
3.1.10. а)		y′′ − 4y′ + 5y = 0 ;	б)	y′′ − 6y′ + 8y = 0 .
3.1.11. а)		y′′ − 4y′ + 29y = 0 ;	б)	y′′ − 7 y′ + 10y = 0 .
3.1.12. а)		4y′′ − 4y′ + y = 0 ;	б)	y′′ + 2y′ + 5y = 0 .
3.1.13. а)		y′′ + 7 y′ + 12y = 0 ;	б)	4y′′ + 4y′ + y = 0 .
3.1.14. а)		y′′ + 2y′ − 8y = 0 ;	б)	9y′′ − 12y′ + 4y = 0 .
3.1.15. а)		4y′′ − 12y′ + 9y = 0 ;	б)	5y′′ − 6y′ + 5y = 0 .
3.1.16. а)		y′′ + y′ + y = 0 ;	б)	y′′ − 4y′ = 0 .
			226

3.1.17.а) y′′ + 3y′ + 2y = 0 ;

3.1.18.а) y′′ + 5y′ + 4y = 0 ;

3.1.19.а) y′′ − 7 y′ + 6y = 0 ;

3.1.20.а) y′′ + 9y′ + 8y = 0 ;

3.1.21.а) y′′ − 5y′ + 4y = 0 ;

3.1.22.а) y′′ + 4y′ + 13y = 0 ;

3.1.23.а) y′′ − 2y′ + 10y = 0 ;

3.1.24.а) y′′ − 10y′ + 9y = 0 ;

3.1.25.а) y′′ − 12y′ + 11y = 0 ;

3.1.26. а) 2y′′ − 3y′ + y = 0 ;

3.1.27.а) y′′ − 6y′ + 5y = 0 ;

3.1.28.а) y′′ + 7 y′ − 8y = 0 ;

3.1.29. а)	5y′′ + 2y′ − 7 y = 0 ;
3.1.30. а)	3y′′ − 5y′ − 8y = 0 ;

б) y′′ − 4y′ + 4y = 0 .

б) 9y − 12y′ + 4y = 0 .

б) y′′ − 6y′ + 9y = 0 . б) y′′ − 10y′ + 25y = 0 . б) y′′ + 4y = 0 .

б) y′′ − 6y′ + 9y = 0 .

б) y′′ − 12y′ + 36y = 0 . б) y′′ + 4y′ + 13y = 0 . б) y′′ − 2y′ + y = 0 .

б) 25y′′ − 10y′ + y = 0 . б) 4y′′ − 12y′ + 9y = 0 . б) 36y′′ − 12y′ + y = 0 . б) 16y′′ − 8y′ + y = 0 .

б) 8y′′ − 4y′ + y = 0 .

3.2. Find the general solutions of linear homogeneous equations.

3.2.1. y′′′ − y′′ + 4y′ − 4y = 0 .			3.2.2.	y(4)	+ 16y = 0 .
3.2.3.	y′′′ − 2y′′ + y′ − 2y = 0 .		3.2.4.	y(4)	+ 5y′′ + 4y = 0 .
3.2.5. y′′′ + 2y′′ + 4y′ + 8y = 0 .			3.2.6.	y(4)	+ 8y′′ − 9y = 0 .
3.2.7. y′′′ − 4y′′ + 5y′ − 2y = 0 .			3.2.8.	y(4)	+ 4y′′ + 4y = 0 .
3.2.9.	y′′′ − 3y′′ − y′ + 3y = 0 .		3.2.10.	y′′′ + 5y′′ + 7 y′ + 3y = 0 .
3.2.11.	y′′′ − y′′ − 4y′ + 4y = 0 .		3.2.12.	y(4)		− 13y′′ + 36 y = 0 .
3.2.13.	y′′′ − 4y′′ + y′ + 6y = 0 .		3.2.14.	y(4)		+ 2y′′′ − 2y′ − y = 0 .
3.2.15.	y′′′ + 3y′′ − 9y′ − 27 y = 0 .		3.2.16. y′′′ − 3y′′ + 3y′ − y = 0 .
3.2.17.	y′′′ + 2y′′ − 4y′ − 8y = 0 .		3.2.18. y(4)		+ 3y′′′ + 3y′′ + y′ = 0 .
3.2.19.	y′′′ − 6y′′ + 12y′ − 8y = 0 .		3.2.20. y(4)		− 3y′′′ − y′′ + 3y′ = 0 .
3.2.21.	y (4)	− 7 y′′′ + 6y′′ = 0 .	3.2.22. y′′′ + 3y′′ − 4y′ − 12y = 0 .
3.2.23.	y (4)	− 2y′′′ + 2y′′ = 0 .	3.2.24. y(4)			− 3y′′ − 4y = 0 .
3.2. 25.	y(4)	+ 2y′′′ + 2y′′ + 2y′ + y = 0 .

3.2.26.y(4) − y′′′ − 3y′′ + y′ + 2y = 0 .

3.2.27.y′′′ − 2y′′ + 9y′ − 18y = 0 .

3.2.28.y (4) − 2 y′′′ + 2y′′ − 2y′ + y = 0 . 3.2.29. y′′′ + 9y′′ + 27 y′ + 27 y = 0 .

3.2.30.y(4) − 3y′′′ + 3y′′ − 3y′ + 2y = 0 .

227

3.3. Find general the solutions of linear homogeneous equations with right part of special form.

3.3.1.	y′′ − 2 y′ + y = xe x .	3.3.2.	y′′ + y = x sin x .
3.3.3.	y′′ − 6 y′ + 9 y = e x sin x .	3.3.4.	y ′′+2 y′+5y = ex sin 2x .
3.3.5.	y′′ − 6 y′ + 9 y = 2x2 − 2x + 3 .	3.3.6.	y′′ − 2y′ + 2 y = e x cos x .
3.3.7.	y′′ + 4 y′ + 5y = x2 + 3 .	3.3.8.	y′′ − 3y′ + 2y = 2xe x .
3.3.9.	y′′ + 2 y′ = 2 + x − x2 .	3.3.10.	4y′′ − 16y′ + 15y = e1,5x .
3.3.11.	y′′ + 4y′ + 4y = e−2x + e x .	3.3.12. y′′ − y′ = 2x2 .
3.3.13.	y′′ + y′ = e− x + x + 1 .	3.3.14. y′′ + 2y′ + 10y = xe− x .
3.3.15.	y′′ + y′ − 2y = e x cos x .	3.3.16. y′′ + 2y′ + y = e− x + sin x .
3.3.17.	y′′+4y = cos 2x .	3.3.18. y′′ + 9y = xe3x .
3.3.19.	y′′ − 2y′ + 2y = e x sin 2x .	3.3.20.	y′′ + 9y = e3x .
3.3.21.	y′′ − 2y′ + y = e x .	3.3.22. y′′ − 3y′ = 1 − 2x − x2 .
3.3.23.	y′′ + 4y′ − 5y = xe x .	3.3.24. y′′ − 4y′ + 4y = e2x .
3.3.25.	y′′ + 2y′ + 2y = (x + 1)e x .	3.3.26. y′′ + 2y′ + 5y = cos x .
3.3.27. y′′ + 4y′ + 3y = e− x + x2 .		3.3.28. y′′ + 4y′ + 8y = cos 2x .
3.3.29. y′′ − 3y′ + 2y = e2x .		3.3.30. y′′ − 8y′ = sin 4x + x .
3.4. Solve Cauchy’s test for equations of the second order.
3.4.1. y′′ + 4y′ + 8y = sin 4x ,		y(0) = 0 ,		y′(0) = 1.
3.4.2. y′′ − 3y′ + 2y = e x ,		y(0) = 2 ,		y′(0) = 1.
3.4.3. y′′ + y′ + y = cos 2x ,		y(0) = −1,		y′(0) = 3 .
3.4.4. y′′ − 4y′ + 3y = x2 − 3x ,		y(0) = 2 ,		y′(0) = 4 .
3.4.5. y′′ + 2y′ + 5y = e2x ,		y(0) = 0 ,		y′(0) = 0 .
3.4.6. y′′ − 10 y′ + 9 y = xe x ,		y(0) = 1 ,		y′(0) = 0 .
3.4.7.	y′′ − 4y′ + 4y = e2x ,	y(0) = 1 ,		y′(0) = 1.
3.4.8.	y′′ − 2y′ + 2y = 3x − 2 ,	y(0) = −2 ,		y′(0) = 2 .
3.4.9.	y′′ + 5y′ − 6y = e4x ,	y(0) = 3 ,		y′(0) = 2 .
3.4.10. y′′ + 2y′ + 10y = x2 − 4 ,		y(0) = 0 ,		y′(0) = 4 .
3.4.11. y′′ − 4y′ = x2 − 5x + 2 ,		y(0) = 0 ,		y′(0) = −1 .
3.4.12. y′′ − 2y′ + y = e x ,		y(0) = 3 ,		y′(0) = 5 .
3.4.13. y′′ + 9y = sin 3x ,		y(0) = 2 ,		y′(0) = −1 .
3.4.14. y′′ + 2 y′ + 2 y = e x sin 3x ,		y(0) = 1 ,		y′(0) = 3 .
	228

3.4.15.y′′ + 2y′ + y = e− x ,

3.4.16.y′′ + y′ − 2y = e2x sin x ,

3.4.17.y′′ + 6y′ + 9 y = e−3x ,

3.4.18.y′′ + 4y′ + 5y = e2x ,

3.4.19.4y′′ − 16 y′ + 15y = x2 − 1,

3.4.20.4y′′ + 4 y′ + 5y = xe x ,

3.4.21.y′′ − 3y′ + 2y = e2x ,

3.4.22.y′′ + 4y′ + 5y = x2 + 2x ,

3.4.23.y′′ − 2y′ + 2 y = e x cos x ,

3.4.24.y′′ − 6y′ + 9y = 2x2 + 5 ,

3.4.25.2y′′ − y′ − y = ex + x ,

3.4.26.y′′ − 2y′ + 10y = cos x ,

3.4.27.4y′′ − 8y′ + 5y = xex ,

3.4.28.3y′′ − 12y′ + 4y = e x sin 2x ,

3.4.29.y′′ − 4y′ = 2x2 + 3x − 1 ,

3.4.30.y′′ − 8y′ + 16y = e4x ,

y(0) = 4 ,	y′(0) = 0 .
y(0) = −5 ,	y′(0) = 1.
y(0) = −3 ,	y′(0) = 2 .
y(0) = 2 ,	y′(0) = 6 .
y(0) = 3 ,	y′(0) = −1 .
y(0) = 4 ,	y′(0) = −1 .
y(0) = 1 ,	y′(0) = 0 .
y(0) = 1 ,	y′(0) = 4 .
y(0) = 2 ,	y′(0) = −5 .
y(0) = 0 ,	y′(0) = 3 .
y(0) = 0 ,	y′(0) = 0 .
y(0) = −1,	y′(0) = −3 .
y(0) = 2 ,	y′(0) = −4 .
y(0) = 1 ,	y′(0) = 5 .
y(0) = 6 ,	y′(0) = −2 .
y(0) = 3 ,	y′(0) = 8 .

3.5. Solve the equations using the Lagrange’s method.

3.5.1.

y′′ − 2y′ + y =

3.5.2. y′′ − 2y′ + y =

2x + 1

x2 + 4

3.5.3.

y′′ − 4 y′ + 3y = ln(1+ e− x ) .

3.5.4. y′′ + 4y = tg2 2x .

3.5.5.

y′′ − 3y′ + 2y =

e2x

3.5.6. y′′ + y =

(1 + e x )2

cos3 x

3.5.7. y′′ − 4y′ + 4y = e2x ln(x2 + 1) .

3.5.8. y′′ + 2y′ + y = e− x ln(x2 + 4) .

3.5.9.

y′′ + 4y =

3.5.10. y′′ + y =

3 + cos2

sin3

3.5.11.

y′′ + y =

3.5.12. y′′ − y =

e3x

sin2 x + 2

1− e2x

3.5.13. y′′ − 2y′ + 2y = ex tg x .

3.5.14. y′′ − 2y′ =

e3x .

e x

1 − e2x

3.5.15. y′′ − 2y′ + 2y

3.5.16. y′′ − 3y′ + 2y = sin(e− x ) .

sin x

229

3.5.17. y′′ − 4y′ + 5y =

e2x sin x

3.5.18.

y′′ − 4y′ + 4y =

e2x .

sin2 x + 1

+ 1

3.5.19. y′′ − 4y′ + 5y =

e2x

3.5.20. y′′ + 2y′ + y = e− x arctg x .

1 + sin 2 x

3.5.21. y′′ + 2y′ + y =

− x

3.5.22.

y′′ − 3y′ + 2y =

e x

e x +

− 1

3.5.23. y′′ + 3y′ + 2y = cos(e x ) .

3.5. 24.

y′′ + 4 y′ + 3y = arctg(ex ) .

3.5.25. y′′ + y =

sin x

3.5.26. y′′ + 2y′ + y = e− x ln x .

cos2 x −

cos 3x

3.5.27. y′′ + 2y′ + y = 3

xe− x .

3.5.28.

y′′ + 9y =

sin2 3x + 1

3.5.29. y′′ − 2y′ + y = ex

1− x .

3.5.30. y′′ + 5y′ + 6 y = e− x ln(1 + e x ) .

Topic 4. Simultaneous differential equations

Normal simultaneous differential equations. Methods of eliminating and integrated combinations of solving simultaneous differential equations in normal form. Simultaneous differential equations with constant coefficients. Generalized Euler’s method.

Literature: [2, 3, chapter 3.3], [3, chapter 8, §6], [4, chapter 8, § 26], [6, chapter 11, item 11.5], [7, chapter 13, §§29—30], [8, 2 chapter, §§6].

Т.4	General theoretical information

4.1. Normal simultaneous differential equations

Simultaneous differential equations of 1st kind (or first order):

dy1

(t, y

, .., y

dy2

= f2 (t, y1

, y2

, .., yn ),

(3.33)

dyn

(t, y

, y

, .., y

230

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