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Национальный авиационный университет

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where y1 (t), y2 (t), …, yn (t) are the unknown functions, t is independent

variable containing system in a normal form or system, which can be obtained concerning derivatives of unknown functions yi (t) , i = 1, 2, …, n .

The solution of the system (3.33) on interval ( a, b ) is the set of n of continuously differentiated functions

y1 = ϕ1 (t), y2 = ϕ2 (t), ..., yn = ϕn (t),

which transform each equation of this system into equality.

Cauchy’s test for system (3.33) is to find such a solution, which depends on the initial condition:

y1 (t0 ) = a1 , y2 (t0 ) = a2 , ..., yn (t0 ) = an , where a1 , a2 , ..., an — are the real values.

For solving simultaneous differential equations in normal form we use the following methods:

1)method of eliminating;

2)method of integrated combinations

General form of the method of eliminating. After simultaneous equations differentiating and eliminating all unknown functions yi (x) , except one, we

obtain a differential equation of n order concerning the one function (for example, y1 ). After integration of this equation it is possible to find other

unknown functions.

The meaning of the method of integrated combinations is to make so-called integrated combinations from the equation of the given system with the help of arithmetical operations, that is that the equation concerning some new function

u= u(t, y1 , y2 , .., yn ), which can be easily integrated.

4.2.Euler’s method for solving simultaneous differential equations with constant coefficients

The system of normal linear differential equations with constant coefficients we can name such system

dy1

= a

+ a

+ ... + a

+ f (t),

dy2

= a21 y1 + a22 y2

+ ... + a2n yn

+ f2

(t),

(3.34 )

dyn

= a

+ a

+ ... + a

+ f

(t),

n1 1

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where

y1 ,

y2 ,

...,

are

unknown

functions

independent

variable

t ; f1 ,

f 2 , ..., fn

are given and continuously on interval (a,b) functions;

aij

are constant

values

(i, j =

) . If fi (x) ≡ 0

(i =

) ,

then system

(3.34)

1, n

called homogeneous, in other case — right hand member not zero.

Let’s consider algebraic method of solving linear homogeneous

simultaneous differential equation (generalized Euler’s method).

For example;

dy1

= a

+ a

(3.35)

dy2

= a21 y1 + a22 y2 .

This system can be written as a single matrix equation as follows;

= AY .

Here

dy1

A = 11

, Y

a22

dy2

General solution of the system (3.35) can be written as

= C

(1)

+ C

(2)

(1)

(2)

where y(1) , y

(1)

and

y(2) , y

(2)

are linear independent partial solutions of the

given system.

Partial solution of the system can be found from

= p ekt ,

= p

ekt ,

(3.36)

where p1 , p2 , k are unknown constants. After substituting of the formula (3.36) in system (3.35) we obtain homogeneous system of linear algebraic equations

concerning the unknown p1		and	p2	of the equations
(a11 − k) p1 + a12 p2 = 0,								(3.37)
a		p	+ (a	22	− k) p	2	= 0.
	21 1

The obtained system should be not equal to zero. So,

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		a11 − k		a12		= 0	,	(3.38)
		a11 − k		a12
or			a21	a22 − k
or

	k2 − (a + a )k + a a − a a = 0.
	11		22	11	22		12	21

This equation is called characteristic equation of the system (3.35).

Let k1 , k2		be different real roots of characteristic equation.
Then for the root k				we consider its own vector							( p(1) , p(1) )						and partial
			1									1			2
solution y(1)	= p(1) ek1t , y(1)			= p(1) ek1t . Similarly for the root									k	2	we consider its
1	1		2	2										2
own vector	( p(2)	, p	(2) ) and partial solution y(2)							= p(2) ek2t ,			y(2)			= p(2) ek2t .
	1		2						1	1				2			2
General solutions of the system are:
		y	= C y(1) + C		2	y(2) ,	y	2	= C y(1) + C		2	y(2)	,
		1	1	1	2	1		2	1	2	2	2

In matrix form it can be written as follows:

= C

p1(1)

k t

+ C

p1(2)

e 1

e 2

(1)

(2)

In case when the characteristic equation (3.38 ) has multiple roots, system (3.35) is simpler to solve by eliminated method.

Т.4 Examples of typical problems solving

1. Solve simultaneous equations

= 5y + 4z,

= 2y + 3z

+ t.

Solution. We can solve

this system by the

Differentiating the first equation of the given system:

d 2 y

= 5

+ 4

dt2

After substituting into obtained system instead of

second equation of the system. We will obtain

d 2 y

+ 8y + 12z + 4t.

dt2

233

eliminating method.

dzdt its value from the

(3.39)

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From the first equation of the system we can find

	z =		1		(		dy			− 5y)															(3.40)
	z =	4			(		dt			− 5y)															(3.40)
		4					dt
And substituting into equation (3.39) instead of																			z the value					1	(	dy	− 5y)	we
And substituting into equation (3.39) instead of																			z the value						(		− 5y)	we
would obtain linear equation of the second order																							4			dt
would obtain linear equation of the second order
	d 2 y			− 8					dy		+ 7 y			= 4t.											(3.41)
	dt2			− 8					dt		+ 7 y			= 4t.											(3.41)
	dt2								dt
Make and solve characteristic equation:
k 2 − 8k + 7 = 0; k = 1, k																2	= 7.
											1					2
So the general solution of the corresponding homogeneous equation is
	y = C et									+ C		2	e7t .
					1							2
Partial solution of right hand member not zero equation (3.41) is
		y* = At + B .
After substituting y* into equation (4.41), we would obtain a parity
−8A + 7(At + B) = 4t,
which can be made when t is initial and in condition of
7A = 4,																	4					32
							or							A =			4			, B =		32	.

																	7					49
− 8A + 7B = 0,																	7					49
In such case general solution of the equation (4.41) has such a form
y = C et + C e7t +													4	t +		32			.
y = C et + C e7t +														t +					.
1								2					7			49
From formula ( 3.40 ) we can find													7			49
From formula ( 3.40 ) we can find
z = −C et				+		1		C			e7t		−	5	t −			43			.
z = −C et				+	2			C			e7t		−	7	t −		49				.
1					2					2				7			49

In matrix form the solution of initial system can be written as follows:

= C1

+ C2

4t / 7 + 32 / 49

0.5

−5t / 7 − 43 / 49

−1

2. Find general solution of the system

dx = x2 + xy,

dy = xy + y2 .

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Solution. Solve this system by the method of integrated combinations. After making the equation we obtain the first integrated combination

= x2

+ 2xy + y2 ,

d(x + y)

(x + y)2 ,

from where

d(x + y)

= dt , −

= t + C .

(x + y)2

x + y

After dividing the first equation by the second one we obtain one more integrated combination:

x2 + xy

, y = C x .

xy + y2

From equations

−

= t + C

y = C x

we would obtain the general

x + y

solution of the given system:

x =

, y =

(C − t)(1+ C

)

(C − t)(1+ C

)

3. Find the general solution of the system

dy1

= 5y

+ 2y

dy2

= 4y1 + 3y2 .

Solution. We can use generalized Euler’s method. Make characteristic

equation of the system

5 − k

= 0 ,

або k 2 − 8k + 7 = 0 .

3 − k

Roots are k1

= 1 , k2 = 7 .

When k

=1 the system

(3.37) is equivalent to one equation

4 p1 + 2 p2 = 0 .

Take p1 = 1, then

p2 = −2 . So for root

k = 1 its own vector can be written

(1; − 2) . Then y(1) = et

y(1)

= −2et

–

are partial solutions of the given system.

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When k = 7 from the system (3.37) we can obtain the equation

−2 p1 + 2 p2 = 0 , or p1 = p2 ,
which designates its own vector	(1; 1) . Then y(2)	= e7t ,	y(2)	= e7t	are also
	1		2

partial solutions of the given system.

General solution of the given system can be written as follows

y = C et

+ C

e7t , y

= −2C et

+ C

e7t ,

= C1

−2

4. Find general solution of the system

dy1

= y

+ y

dy2

= −2y1 + 3y2 .

Solution. Like in previous example we can use generalized Euler’s method.

Characteristic equation is	1 − k	1	= 0 , it means that k2 − 4k + 5 = 0
Characteristic equation is	1 − k	1	= 0 , it means that k2 − 4k + 5 = 0
	− 2	3 − k
has two complex connected roots	k1,2	= 2 ±	i . In this case to make the solution

of the given system we need to know just the solution depending on the value k = 2 + i .

When k = 2 + i the

system (3.37)

can

be transformed

into the equation

(−1 − i) p1 + p2 = 0 . If p1 = 1, then

p2 = 1 + i . So for root 2 + i its own vector

can be written (1;

1+ i)

and partial solution

(2+i)t

(cos t + i sin t)

Y =

+ i

e2t .

cos t

e2t + i

sin t

Then

cos t − sin t

cos t + sin t

cos t

sin t

= C

2 cos t + sin t

1 cos t − sin t

is the general solution of the system 5. Solve Coushy’s test

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dx		= 3x + y,


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

dy		= − x + y,
	dt

Solution. Characteristic equation of the given system can be written as follows

3 − k	1	= 0 ,
3 − k	1
− 1	1 − k

from where

(3 − k)(1− k) + 1 = 0 , k2 − 4k + 4 = 0 , k1 = k2 = 2 .

The own numbers are identical than the solutions of the given system can be found from

x = (α + γt)e2t , y = (β + δt)e2t .

After substitution of this expression into the given system we obtain

γ + 2(α + γt) = 3(α + γt) + β + δt , δ + 2(β + δt) = −α − γt + β + δt .

This equalities can be obtained when t	is any number and when equalities
α − γ + β = 0 , γ + δ = 0 are possible. From here		we can obtain two linear
independent solutions, for example α = 1 ,	β = −1,	γ = δ = 0 і α = 1 , β = 0 ,
γ = 1 , δ = −1.

So we can write linear independent solutions of the system as:

	x (t) = e2t ,			x	2		(t) = (1 + t)e2t ;
	1				2
	y (t) = −e2t ,			y		2	(t) = −te2t .
	1					2
General solution of the given system is
x = C e2t + C		2	(1 + t)e2t ,				y = −C e2t	− C	2	te2t .
	1	2					1		2
Determining the	constants C1			and C2 . Taking						into account initial
conditions x(0) = 1,	y(0) = 0 , we obtain					the system of the equations

C1 + C2 = 1, 0 = −C1 ,

the solutions of which are C1 = 0 , C2 = 1 . So,

x = (1 + t)e2t , y = −te2t – is the solution of Cauchy’s test.

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	Tests for general and self-studying
Т.4	Tests for general and self-studying

Solve the simultaneous differential equations by the method of integrated combinations

= ,

−

x(0)

= 2,

y(0) = 4.

;

=−

(x − y)

x + y

Solve the simultaneous differential equations by the method of eliminating or generalized Euler’s method

= −3x + y,

= −

= −20x + 6y.

= 2x.

= x + 4y,

= 2x + y,

= y − 3x.

= 3x + 4y.

= 4x + 6y,

= y + 2e

= 2x + 3y + t.

= x + t 2 .

= 2x − y,

= 2x + y + cos t,

10. dt

11.

= 2y − x − 5et sin t.

= − x + 2 sin t.

Answers

1.	x = C			eC1t2 ,		y =	1		e−C1t2 .	2. x2 − y2 = C ,	2x +(x − y)2 = C	2	.

			2				2C1C2			1
		t			2t
3.	x =		+ 2, y =			+ 4 .	4.	x = C1e2t + C2et , y = 5C1e2t + 4C2et . 5.			x = C1 cost + C2 sin t,
		3			3

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y = 2C sin t − 2C								2	cost.			6.	x = et (C cos 2					3t + C			2		sin 2	3t), y =		3	et (C	2	cos2 3t −
		1						2							1						2				2			2
																									2				3
−C sin 2				3t). 7.					x = C et			+ C		e5t , y = −C et + 3C						e5t .				8. x = C	+ C		e7t −		3	(7t + 2)t,
−C sin 2				3t). 7.					x = C et			+ C		e5t , y = −C et + 3C						e5t .				8. x = C	+ C		e7t −			(7t + 2)t,
	1								1				2				1		2					1		2		49
		2			1				1																			49
		2			1				1														9. x = C1et +C2e−t +tet
y = −			C1	+			C2e7t +				(14t		2 − 3t − 1).																	−t2 −2,
y = −		3	C1	+		2	C2e7t +			49	(14t		2 − 3t − 1).																	−t2 −2,
y = C et − C					e−t + (t − 1)et − 2t . 10.										x = C et +C				e3t +et (2cost −sin t),
	1			2											1			2
C	e3t	+et		(3cost +sin t) . 11.									x =		1	cost − (C + (1 + t)C						2	)et ,	y = C et + C		tet − 2 cost −					sin t	.
2															2		1					2		1	2					2
															2															2
									Individual tests
		Т.4							Individual tests

4.1. Solve simultaneous differential equations by eliminating method. In

every system x = x(t), y = y(t) ,

x = 4x + 3y + t,

y = −2x − y.

4.1.3.x = x − y + cos 3t,

y = x − 3y.

x = x − 3y + sin t,

y = x + 5y − cos t.

x = −6x − 4y + t,

y = 3x + 2y − 5.

x = 2x + 2y − sin 2t,

y = x + 3y.

4.1.11.x = 4x − 5y + e2t ,y = 2x − 2y.

4.1.13.x = 2x + 5y,

y = 2y + sin t.

		t	,
4.1.15.	x = 4x − y + e		,
4.1.15.		+ 2y + 2et .
	y = x	+ 2y + 2et .

x = dxdt , y = dydt .

4.1.2.x = x − 2y + et ,y = x + 4y + 1.

4.1.4.x = x − y + t2 − 1,

y = 5x + 5y − 2.

x = −3x − y + 2t,

y = − x + 4y − t.x = 2x + y + t − 2,

y = 3x + 4y + 3.

4.1.10.			+ e		t	,
4.1.10.	x = −2x − 5y		+ e			,
		y = x + 4y + 2.

			−t	,
4.1.12.	x = x − y − e			,
4.1.12.
	y = 2x + 3y + 2e−t .

4.1.14.		x = − x + 4y,
4.1.14.
	y = x + 2y + et .

4.1.16.x = −2x + 5y + 3,y = − x + 2y − 4.

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x = −2x − 2y − cos t,

y = 4x + 2y + 2.

4.1.19.x = 2x − 4y + e−2t ,

y = 5x − 2y + 4.

4.1.21.x = x − y − t − 2,y = 2x + 3y + 2t.

4.1.23.x = 2x + 3y + sin t,y = −3x + 2y + cos t.

4.1.25.x = −5x + 5y + t2 ,

y = −5x + y + t − 4.

4.1.27.x = − x − 2y + et ,y = x − 3y + t.

4.1.29. x = −3x − y − e−2t ,

y = x − y + 1.

x = x + 2y − 2t,

y = −5x − y + 3t.

4.1.20.x = 6x + 8y + e2t ,y = −2x − 2y − 3e2t .

4.1.22.x = −3x + 4y + et ,

= − + + t

y 5x 5y 3e .

4.1.24. x = −6x − y + e−t ,

= + − −t

y 17x 2y e .

4.1.26.x = 2x + 2y + t2 ,y = −2x − 3y.

4.1.28.x = −3x − 2y + e−t ,y = 2x + y.

4.1.30.x = 3x + 17 y + t2 ,y = −2x − 3y − t.

4.2. Solve the system of linear			homogeneous	differential equations by
dx		= a11 x + a12 y,

dt
dt		if:
Euler’s method	dy	if:
	dy	= a21x + a22 y,
		= a21x + a22 y,
dt
4.2.1. a11 = 3,		a12 = 1,	a21 = 2,	a22 = 4 .
4.2.2. a11 = 4,		a12 = 1,	a21 = −1,	a22 = 2 .
4.2.3. a11 = 1,		a12 = −1,	a21 = 2,	a22 = 3 .
4.2.4. a11 = 3,		a12 = 8,	a21 = 1,	a22 = 1 .
4.2.5. a11 = 7 ,		a12 = 4,	a21 = −1,	a22 = 3 .
4.2.6. a11 = 2,		a12 = 1,	a21 = −5,	a22 = 4 .
4.2.7. a11 = −2,		a12 = 3,	a21 = 3,	a22 = −2 .
4.2.8. a11 = −5,		a12 = −3,	a21 = 3,	a22 = 1 .
4.2.9. a11 = 2,		a12 = −4,	a21 = 1,	a22 = 2 .
4.2.10. a11 = 2,		a12 = 1,	a21 = 3,	a22 = 4 .
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