a11 k	a12		0 ,	(3.38)
	a21	a22 k

or

k2 (a11 a22 )k a11a22 a12 a21 0.

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		1	we		consider			its	own		vector ( p(1)					, p(1) )			and partial
																		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
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			1				1	2			2

In matrix form it can be written as follows:

	y		C1	p(1)			C2	p(2)
	1				1	ek1t			1	ek2t .
y2				p(1)				p(2)
					2				2

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

T 4. Examples of typical problems solving

1. Solve simultaneous equations

dy 5y 4z,

dt

dz 2y 3z t.dt

Solution. We can solve this system by the eliminating method. Differentiating the first equation of the given system:

d 2 y 5 dy 4 dz . dt2 dt dt

After substituting into obtained system instead of dzdt its value from the second equation of the system. We will obtain

241

d 2 y

5

dy

8y 12z

4t.

(3.39)

dt2

dt

From the first equation of the system we can find

z

1 ( dy

5y)

(3.40)

4

dt

And substituting into equation (3.39) instead of z

the value 1

( dy

5y) we

would obtain linear equation of the second order

4

dt

d 2 y

8

dy

7 y 4t.

(3.41)

dt2

dt

Make and solve characteristic equation:

k 2 8k 7 0;

k 1, k

2

7.

1

So the general solution of the corresponding homogeneous equation is

y C et

C

2

e7t .

1

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

8A 7B 0,

or

A

, B 49 .

7

In such case general solution of the equation (4.41) has such a form

y C et C e7t

4

t 32 .

1

2

7

49

From formula ( 3.40 ) we can find

z C1et 12 C2 e7t 75 t 4943 .

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

y	C		1 et C	2		1 e7t			4t / 7	32 / 49 .
	1					0.5			5t / 7 43 / 49
z			1

2. Find general solution of the system

242

dx x2 xy,

dt

dy xy y2 .dt

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

dxdt dydt x2 2xy y2 ,

or

d(x y) (x y)2 , dt

from where

d(x y)	dt ,		1	t C .
(x y)2			x y	1

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

dx

x2 xy

,

dx

x

,

dx

dy

, y C x .

dy

xy y2

dy

y

x

y

2

From equations

1

t C ,

y C x

we would obtain the general

x y

1

2

solution of the given system:

x		1			,	y		C2			.
	(C	t)(1 C	2	)			(C	t)(1	C	)
1						1			2

3. Find the general solution of the system

dy1 5y 2y ,

dt 1 2

dy2 4y1 3y2 .dt

Solution. We can use generalized Euler’s method. Make characteristic equation of the system

	5 k	2		0 ,	або k 2 8k 7 0 .
	4	3 k

Roots are k1 1 , k2 7 .

When k 1 the system (3.37) is equivalent to one equation

243

	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.
1	2
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

2

e7t ,

y

2

2C et

C

2

e7t ,

1

1

1

or

y1

C

1

et

C

1 e7t .

1

2

2

y2

1

4. Find general solution of the system

dy1

y

y

,

2

1

dt

dy2

2y1 3y2 .

dt

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

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
			1		1
	Y		e(2 i)t		e2t (cos t i sin t)
		1	i	1	i

			cos t				sin t
			e2t		i				e2t .
Then		cos t sin t					cos t sin t
y			cos t			2t			sin t	2t
1		C1		e			C2		e
y2			cos t sin t					cos t sin t

244

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

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 ,
	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 ( t)e2t , y ( t)e2t .

After substitution of this expression into the given system we obtain

2( t) 3( t) t ,2( t) t t .

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

So we can write linear independent solutions of the system as: x1 (t) e2t , x2 (t) (1 t)e2t ;

		y (t) e2t ,			y	2	(t) te2t .
		1
General solution of the given system is
x C e2t C			2	(1 t)e2t ,			y C e2t	C	2	te2t .
	1						1
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.

245