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Answers

1. y =								x 2			ln x −			x 2			+ C									x 2			+ C			2		x + C	3	.					2. y = e3x + x6 + C + C																			2		x + C		3		x2 .
1. y =											ln x −						+ C										2		+ C					x + C		.					2. y = e3x + x6 + C + C																					x + C				x2 .
						2									4							1					2																													1
3. y				=			1			e5x + sin x								+ C							x2			+ C			2	x		+ C .							4.			y =	1 cos2x − sin x + C																x2		+ C		2	x + C .
3. y				=						e5x + sin x								+ C						2				+ C				x		+ C .							4.			y =	1 cos2x − sin x + C															2			+ C			x + C .
						125																1		2										3											8														1	2						3
5. y = ex +											x4	+ C			x2			+ C					2	x + C						. 6. y =					x5			−			1			sin 3x + C						x2				+ C	2	x + C .							7. y =				x2		ln x −
										24			1		2								2						3						60					27									1		2				2				3							2
−	x2					+			x3		+ C x + C									.		8.					y =						5 x		+ x3 + C x + C														.		9.			y =				3x					− sin x + C								x2	+
−						+					+ C x + C									.		8.					y =			(ln 5)2					+ x3 + C x + C														.		9.			y =			(ln 3)3						− sin x + C							2		+
4						6						1						2												(ln 5)2														1			2										(ln 3)3									1				2
+C x + C .													10. y =											4x					−		1			sin3x +				1				cos5x + C							x2			+ C x + C .											11.			y =		ln2 x				+
2						3																(ln 4)3									27						125										1		2					2				3												2
+C1 ln x + C2 .																			12.							y = C1 (x − e− x )+ C2 .																					13.					y = arcsin2 x +C1arcsin x + C2 .
14.						y = C e5x							+ C				2		.					15.					ln				C ( y + 1) − 1										= C (x + C							2		), C ≠ 0;							y = C ;							y = − x + C .
14.						y = C e5x							+ C						.					15.					ln				C ( y + 1) − 1										= C (x + C									), C ≠ 0;							y = C ;							y = − x + C .
										1																								1										1										1									3
16.				e y + C								= (x + C							2		)2			.			17.						1+ C y 2					= C x + C								2		.	18.					y = C				2	eC1x . 19. x + C													=
										1									2															1							1					2												2												2
		2																																										1							20e y + C							1	−				C
=		2	(					y − 2C )								y + C											).			20. x + C									2		=			1		ln												1					1 .			21.Linear
=			(					y − 2C )								y + C											).			20. x + C											=					ln																	1 .			21.Linear
3												1											1																					C1							20e y + C								+				C
3																																												C1							20e y + C							1	+				C
dependency.												22. Linear dependency 23.																										Linear independency .																				1					1
dependency.												22. Linear dependency 23.																										Linear independency .
																		Individual tasks
								Т.2										Individual tasks

						2.1. Integrate the given equations of the second order
					2.1.1. y′′ =											sin x										.																		2.1.2. y′′(4 + x2 ) = 2 .
					2.1.1. y′′ =																					.																		2.1.2. y′′(4 + x2 ) = 2 .
																cos2 x
					2.1.3. y′′ = 4cos2 x .																																							2.1.4. y′′ = 2x arctg x .
					2.1.5. y′′								1 − x2									+ x = 0 .																						2.1.6.							y′′ = arctg x
				2.1.7.								y′′ = xex .																																2.1.8. y′′										1 − x2					= 1.
				2.1.9. y′′ = x ln x .																																								2.1.10. y′′ = x sin 2 x .
				2.1.11. y′′(1 − x2 ) = x3 .																																								2.1.12. y′′ = 3x2 + ln x .
				2.1.13. y′′												x2 − 1 = x .																												2.1.14. y′′ = x sin 2x .
				2.1.15. y′′ = 6x arctg x .																																								2.1.16.									y′′ = sin3 x .
				2.1.17. y′′ = cos3 x .																																								2.1.18. y′′ = xe2x .
				2.1.19. y′′ = 4 sin 2 x .																																								2.1.20. xy′′ = 1 + x2 .
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2.1.21.	y′′ 1+ x = x .	2.1.22.	y′′ = x x − 1 .
2.1.23.	y′′ cos3 x = sin x .	2.1.24.	y′′ = ex (x + 1) .
2.1.25.	y′′ = x + ln x .	2.1.26.	y′′ = cos4 x .
2.1.27.	x2 y′′ = ln x .	2.1.28.	y′′ = (x + 3)e x .

2.1.29. (x −1)2 y′′ = x2 − 2x .	2.1.30. 2 x ( x + 1)2 y′′ = 1 .
2.2. Integrate the following equations using substitution y′ = z(x)
2.2.1. (9 + x2 )y′′ + 2xy′ = 0 .	2.2.2. xy′′ = y′ + x .
2.2.3. y′′ − 2y′ ctg x = cos x .	2.2.4. y′′(x2 + x) = (4x + 2) y′ .
2.2.5. y′′ − 2 ctg x y′ = sin3 x .	2.2.6. (2 + x2 )y′′ + 2xy′ = x2
2.2.7. xy′′ + x( y′)2 − 2y′ = 0 .	2.2.8. y′′ sin x − y′ cos x = sin x .
2.2.9. xy′′ − y′ = e x x2 .	2.2.10. y′′ + 4(tg x)y′ = cos2 x .
2.2.11. xy′′ + y′ = ( y′)2 .	2.2.12. xy′′ = y′ + x3 .
2.2.13. y′′x ln x = y′ .	2.2.14. y′′ + 2(tg x)y′ = cos3 x .
2.2.15. y′′ − 2xy′ = 4x .	2.2.16. xy′′ − y′ = x2 cos x .
2.2.17. y ′′−2 y ′ctg x = 0 .	2.2.18. xy′′ = y′ + xex .
2.2.19. (x2 + 1)y′′ = 2x( y′ + 1) .	2.2.20. xy′′ = y′ + x2 sin x .
2.2.21. (x2 + 1) y′′ = 4x( y′ −1) .	2.2.22. x(ln x + 2) y′′ = y′ .
2.2.23. xy′′ = y′ ln(y′ / x) .	2.2.24. (1− x2 ) y′′ = 2xy′ .
2.2.25. xy′′ = y′ + x2 .	2.2.26. (1 + x2 ) y′′ = 2xy′ .
2.2.27. xy′′ + y′ = x .	2.2.28. y′′ − 2(tg x) y′ = cos x .
2.2.29. x2 y′′ + xy′ = 1.	2.2.30. y′′ − y′ /(x − 1) = x2 − x .

2.3. Integrate the following equations using substitution y′ = p( y) .

2.3.1.	yy′′ = ( y′)2 .	2.3.2. 2yy′′ + ( y′)2 = 0 .
2.3.3. yy′′ = 2y′ + ( y′)2 .		2.3.4. yy′′ + 2( y′)2 = 0 .
2.3.5. yy′′ − 2( y′)2 = 2y3 y′ .		2.3.6. y3 y′′ + 2y′ = 0
2.3.7. yy′′ = 3(y′)2 .		2.3.8. y′′ = ( y′)2 + 2y′ .
2.3.9.	y′′y3 = 2y′ .	2.3.10. 33 y2 y′′ = y′ .
2.3.11. y′′ = 2yy′ .		2.3.12. 2y 2 y′′ = y′ .
2.3.13. y3 y′′ = 6 .		2.3.14. y2 y′′ = y′ .
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2.3.15. y′′ − 2yy′ = 0 .	2.3.16.	y′′ = y′ + (y′)3 .
2.3.17. 4 y y′′ = 1 .	2.3.18.	2yy′′ − ( y′)2 = y′ .
2.3.19. yy′′ + (y′)2 + 1 = 0 .	2.3.20. 2yy′′ + ( y′)2 + y′ = 0 .
2.3.21. yy′′ = 2(y′)2 − ( y′)3 .	2.3.22. y′′ = 2y′( y + 1) .
2.3.23. yy′′ − ( y′)2 = y 2 y′ .	2.3.24. yy′′ − 2yy′ ln y = ( y′)2 .
2.3.25. 3y′′ = y −5 / 3 .	2.3.26. yy′′ = 2(y′)2 + y′
2.3.27. y′′ = ( y′)2 + y′ .	2.3.28. y′′y3 + 9 = 0 .
2.3.29. y′′ + (y′)2 = 2e− y .	2.3.30. y′′( y + 1) = ( y′)2 + y′ .

Тopic 3. Linear differential equation with constant coefficients

Linear differential equation with constant coefficient Euler’s method. Non-homogeneous linear equations with constant coefficients. Method of independent coefficients and method of variation of arbitrary constants (Lagrange’s method).

Literature: [2, chapter 3, sec. 3.3], [3, chapter. 8, §4], [4, chapter 8, § 26], [6, chapter 11, sec.11.3, 11.4], [8, chapter 13, §§21—28], [10, §§4—5].

Т.3	Main theoretical information

3.1. Homogeneous linear equations

Homogeneous linear equation of the n’th order with constant coefficients is written below

	a	y(n) +a y(n−1)	+...+a		y′	+a y = 0,	(3.24)
	0	1	n−1			n
where a0 , a1 , …, an		— any real numbers, a0		≠ 0 .
General solution of homogeneous linear equation:
		y =C1 y1 +C2 y2 +...+Cn yn ,
where y1 , y2 , ..., yn — linearly			independent			particular	solutions of the

homogeneous equation (3.24), C1 , C2 , ..., Cn — any constants.

Euler’s method. Particular solution of the equation (3.24) is evaluated in the form y = ekx , where k — unknown constant. Using the substitution, we will have:

ekx (a0 kn +a1kn−1 +...+an−1k +an ) = 0 , 213

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Therefore

a kn +a kn−1		+...+a	n−1	k +a	n	= 0.	(3.25 )
0	1		n−1		n

Equation (3.25) is called Auxiliary equation. This equation is obtained from (3.24) with the help of substitution of the derivatives of the evaluating function by using the powers of k and y equals to one. It means:

y(n) → kn , y(n−1) → kn−1 , ..., y′ → k, y →1 .

Auxiliary equation (3.25) is algebraic one of the nth order and has n number of roots (real or complex). A particular solution of the equation (3.24) depends on the roots of the auxiliary equation (3.25). Correspondence between roots of the auxiliary equation and the particular solutions are given in table 3.2.

									Table 3.2

№		Characteristic of the root	The number of the					Particular solutions which
№		Characteristic of the root	linear independent					Particular solutions which
		in the Auxiliary equation	solutions which					corresponds to the given root
			corresponds to the
			given root
1		k – real simple root		one				ekx
		divisible by 1		one				ekx

2		k – real simple root		m				ekx , xekx ,	..., xm−1ekx
2		divisible by m		m				ekx , xekx ,	..., xm−1ekx
3		k =α±βi – pair of		two				eαx cos βx, eαx sin βx
3		complex compared		two				eαx cos βx, eαx sin βx
		simple roots

		k =α±βi – Pair of						eαx cos βx, eαx sin βx,
4		complex compared		2m				…,
4		simple roots divisible		2m				…,
		by m				xm−1eαx cosβx,			xm−1eαx sin βx

	Let’s consider the equation of the second order
			a0 y′′ + a1 y′ + a2 y = 0 .						(3.26)
	Its auxiliary equation has the following structure
			a k2	+a k +a	2	= 0 .			(3.27)
			0	1	2
	Dependent on the denominator D = a2 − 4a a						2	we have 3 such conditions,
				1		0	2

according to table 3.3.

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									Table 3.3
	Roots of the auxiliary equation ( 27 )	General solution of the equation
k1	and k2 – real and different	y = C ek1x	+ C		2	ek2x
		1			2
k1	and k2 – real and equal	y = ek1x (C	1	+ C		2	x)
			1			2
k1 = α + βi , k2 = α − βi – complex roots		y = eαx (C		cos βx + C				2	sin βx)
		1						2

3.2. Nonhomogeneous linear equation

The following equation:

	a	y(n) +a y(n−1)	+...+a	y′+a y = f (x),	(3.28)
	0	1	n−1	n
where a0 , a1 , ..., an		— constants,	f (x) — continues on (a,b)		function is a

linear nonhomogeneous differential equation with constant coefficients.

General solution of non homogeneous equation (3.28) looks like the following

y = y + y ,

where y is the general solution of the corresponding homogeneous equation

(3.24), y is a particular solution of the nonhomogeneous equation (3.28). General solution of y equation (3.24) is considered before.

To find particular solution of the equation (3.28) such methods are used:

1)independent coefficients;

2)variation of any constants (Lagrange method).

3.2.1.Method of independent coefficients

This method is used for evaluating linear equations with constant coefficients and right part of the special form

f (x) = eαx (P (x) cos βx +Q (x) sin βx)		,	(3.29)
n	m

or which is the sum of functions of the same type. Here α and β are constants,

Pn (x) and Qm (x)	are given binomial	of the	variable x		powered
correspondingly by n	and m . Let’s name	number		z =α+βi	auxiliary

(control) number of the right part of the equation.

Particular solution of the equation (3.28) with right part (3.29) is evaluated as follows:

		y* = xr eαx (P (x) cos βx +Q (x) sin βx)					,
		l		l
where P (x) ≡ A xl + A xl−1 + ... + A ,				Q (x) ≡ B	0	xl + B xl−1		+ ... + B		are
l	0	1	l	l	0	1			l
binomial of the order		l = max(n, m)	with independent coefficients A0 ,						A1 , ...,	Al ,

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B0 , B1 , ..., Bl ; r is divisible by roots z = α + βi in the auxiliary equations (3.25). If z is not the root of the auxiliary equation, then r = 0 .

For some graphs of functions f (x) particular solutions are evaluated in such a form (tаble 3.4):

Table 3.4

№

View of the right side

Control number of the right

Structure of the particular

( f (x) )

side ( z )

solution ( y* )

1.1

αx

z = α – is not the root

αx

of the auxiliary equation

1.2

αx

z = α – the root of the

αx

auxiliary equation r

2.1

Pn (x)

z = 0 – is not the root

Pn (x) ≡ A0 x

of the auxiliary equation

+ A xn−1

+ ... + A

2.2

Pn (x)

z = 0 – the root of the

x r P (x)

auxiliary equation

3.1

P (x)e αx

divisible by r

P (x)eαx

z = α – is not the root of

3.2

the auxiliary equation

P (x)e αx

z = α – is the root of

xr P (x)eαx

the auxiliary equation

divisible by r

4.1

A cos βx + B sin βx

z = β i – is not the root

A1 cos β x + B1 sin β x

of the auxiliary equation

4.2

A cos βx + B sin βx

z = β i – root of the

xr ( A cos β x + B sin β x)

auxiliary equation

divisible by r

5.1

α x

cos β xPn ( x)

z = α + β i – is not the

α x

root of the auxiliary

(Pn ( x) cos β x +

equation

+ Q n ( x) sin

β x)

5.2

α x

cos β xPn ( x)

z = α + β i – root of the

α x

( x) cos β x +

auxiliary equation

(Pn

divisible by r

+ Q n ( x) sin

β x)

3.2.2. Lagrange’s Method (variations of any constants):

This method is used for evaluating the linear differential equation both with variable and constant coefficient when the general solution of the homogeneous equation is known.

Let us consider the Lagrange’s method for the example of the equation of the second order

y′′ + a1 (x) y′ + a2 (x) y = f (x).

(3.30)

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Let y = C1 y1 + C2 y2 – be the general solution of equation y′′ + a1 (x) y′ + a2 (x) y = 0 . The General solution of not zero equation (3.30) is evaluated in the form

		y = C1 (x) y1 + C2 (x) y2 ,
where the functions C1 (x)				і C2 (x)	depend on the system
			′	(x) y +C	′	(x) y		= 0,
	C				2		2
		1		1
			′(x) y′+C		′	(x) y	′	= f (x).
	C
Using Cramer’s		1		1	2		2
	formula can be evaluated

the homogeneous right hand member

(3.31)

(3.32)

;

1 =

;

2 =

;

y′

f (x)

y′

f (x)

C′ (x) =

1 ; C

(x) =

2 ;

C (x) =

′ (x)dx + C

; C

(x) =

∫

C′ (x)dx + C

∫ 1

Substituting meaning C1 (x) і

C 2 (x)

into the formula (3.31) we will obtain

the general solution of the right-hand member not zero equation (3.30).

Т.3

Examples of typical problems

Find the general solution of linear homogeneous equation

1. y′′−2y′−3y = 0 .

Solution, Form the equation

its roots k1 =−1 ,

k2 =3 .

k2 −2k −3 = 0 ,

These are the roots of a given linear independent

partial solutions y

= e−x ,

= e3x . Then

y = C e−x +C

e3x

is the general solution of this equation.

2. y′′ − 6y′ + 13y = 0 .

Solution, The equation

k2 −6k +13 = 0

We have two complex roots k1 = 3 + 2i , k2 = 3 − 2i . Then

y = e3x (C cos 2x + C

sin 2x)

is the general solution of this equation.

3. y′′ + 8y′ + 16y = 0 .

Solution: Simplifying the equation

k2 +8k +16 = 0 217

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And its roots are k1 = −4 , k2 = −4 , then characteristics equation has roots

k = −4 , each one is divisible by two. General solution of this equation can be written in this form,

y = e−4x (C1 +C2 x) .

4. y (4) − 4y(3) + 4y′′ = 0 .

Solution: Making the equation

k4 −4k3 +4k2 = 0

the solution is:

k 2 (k 2 − 4k + 4) = 0; k = k					2	= 0, k			3	= k	4	= 2 .
				1	2				3		4
Partial solutions of these equations are:
y = 1 ,	y	2	= x ,	y = e2x , y			4	= xe2x .
1		2		3			4

The general solution of the initial form of these equations are:

y=C1 +C2 x +e2x (C3 +C4 x) .

5.y′′′ − 2y′′ − y′ + 2y = 0 .

Solution, Characteristics equation is:

k3 −2k2 −k +2 = 0 .

Simplify the left side of this equation on multipliers so that

k 3 − k − 2(k 2 − 1) = 0;	(k 2 − 1)(k − 2) = 0 ;
(k − 1)(k + 1)(k − 2) = 0 ;	k1 = 1, k2 = −1, k3 = 2.

The general solution can be written as follows:

y = C e x + C	2	e− x + C	3	e2x .
1	2		3
6. y (4) − y = 0 .
Solution, Make and solve the characteristics equation
k 4 − 1 = 0; (k 2 + 1)(k − 1)(k + 1) = 0 ; k			= ±i, k		3	= 1 , k	4	= −1 .
		1,2			3		4

General solution is

y= C1 sin x + C2 cos x + C3e x + C4e− x .

7.Find the main value “Z” of the right side of this equation

a	0	y(n) + a y(n−1) + ... + a		n−1	y′ + a	n	y = f (x),
if:	0		1	n−1		n
if:			b) f (x) = x2 +2x +1 ;				c) f (x) = e4x ;
а) f (x) =5 ;			b) f (x) = x2 +2x +1 ;				c) f (x) = e4x ;
d) f (x) = e2x sin 7x ;			e) f (x) =−cos 2x +3sin 2x ; f) f (x) = x sin x .
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Solution. In all cases the right side has a special form. By formula (3.29), we

have	b) z = 0;	c) z = 4;			d) z = 2 + 7і; e) z = 2і;				f) z = і.
а) z = 0;
8. Find the general solution of right hand member not zeroequation
		y′′ − 3y′ − 4y = x .
Solution. Solve homogeneous equation at first
We have:		y′′ − 3y′ − 4y = 0 .
					= 4; y = C e−x +C			e4x .
k2 −3k −4 = 0; k =−1, k				2			2
The main	value of	1	side		1				not zero
		the right			of the right hand	member

z = 0 (Without ezx ) which is not the root of characteristics equation, thus, partial solution of the right hand member not zero equation can be written as follows:

y* = Ax + B .

When Ax + B is a linear equation which is the general form of the right side of this equation, А and В are unknown constants which can be calculated.

Find the derivatives y′* and y′′* :

y′* = A, y′′* = 0 .

Now we would substitute these values for y* , y′* and y′′* in the initial equation:

−3A − 4(Ax + B) = x , −4Ax − 3A − 4B = x .

After relating the coefficients which have the same power. x , we would obtain the system of two equations:

x1 : −4A = 1 , x0 : −3A − 4B = 0 ,
The solutions of these are	A = −	1	,	B =	3	.	so, y* = −		1		x +			3		–		is a
partial solution of this equation.	4			16					4			16
partial solution of this equation.								1					3
The general solution of this equation is : y = C e− x + C e4x −								1		x +			3		.
										x +					.
				1			2	4				16
9. Find the general solution of the right hand member not zero equation:
y′′ − 4y′ = 48x2 − 2 .
Solution. Characteristic equation k2 − 4k = 0 has two roots									k			= 0, k					2	= 4 .
									1								2

Then 1, e4x is a fundamental system

y = C1 + C2 e4x

is the general solution of the homogeneous equation y′′ − 4y′ = 0 .

Now we are supposed to find the partial solution of this equation similar to the previous example. The general value of right hand member not zero equation

219

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z = 0 (without multiplier ezx ), but this value is the root of characteristic

equation and can be divided by one, then the partial solution of the right hand member not zero equation we would calculate in such a form

y* = x(Ax2 + Bx + C) = Ax3 + Bx2 + Cx ,
where А, В and С are unknown	constants, so Ax2 + Bx + C ,	which is the
simplified form of the right side x2 − 2 , should be multiplied by		x1 (according
to example 2.2 in table 3.4).
Find the derivatives y′* and	y′′* :

y′* = 3Ax2 + 2Bx + C, y′′* = 6Ax + 2B .

After putting the values y′* and y′′* in the initial equation we will obtain : 6Ax + 2B − 4(3Ax2 + 2Bx + C) = 48x2 − 2 ,

−12Ax2 + (6A − 8B)x + 2B − 4C = 48x2 − 2 .

Comparing the coefficients which has the same order x , we will obtain a system of three equations:

x1 : −12A = 48 , x1 : 6A − 8B = 0 , x0 : 2B − 4C = −2 ,

the solutions of which are A = −4, B = −3 , С= –1. so, y* = −4x3 − 3x2 − x is

the partial solution of the given equation.

The general solution of the initial equation:

y= C1 + C2 e4x − 4x3 − 3x2 − x .

10.Find the general solution of the right hand member not zero equation y′′ − 4y′ + 5y = e2x .

	Solution. Characteristic equations k2 − 4k + 5 = 0 has two complex roots
k	= 2 + i , k	2	= 2 − i . Тhen e2x cos x , e2x sin x is a fundamental system;
1		2
			y = (C cos x + C	2	sin x)e2x
			1	2

is the general solution of the homogeneous equation y′′ − 4y′ + 5y = 0 .

Initial value of the right side of the right member not zero equation z = 2 is not a root of the characteristic equation. Partial solution of the right hand member not zero equation in this form

y* = Ae2x .
After putting the value in the initial equation	y* = Ae2x , y′* = 2Ae2x and
y′′* = 4Ae2x , we would obtain:
4Ae2x − 4(2Ae2x ) + 5(Ae2x ) = e2x ,	Ae2x = e2x , А = 1.
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