T3. Examples of typical problems

its roots k1 =-1 ,

k2 =3 .

k2 -2k -3 = 0 ,

These are the roots of a given linear independent

partial solutions y

= e-x ,

y

2

= e3x . Then

1

y =C e-x +C

2

e3x

1

is the general solution of this equation.

2. y 6 y 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

2

sin 2x) –

1

k 2 (k 2 4k 4) 0;

k k

2

0,

k

3

k

4

2 .

1

Partial solutions of these equations are:

y 1 ,

y

2

x ,

y e2x , y

4

xe2x .

1

3

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.

y C e x C

2

e x C

3

e2x .

1

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

if:

b) f (x) = x2 +2x +1 ;

c) f (x) = e4x ;

а) f (x) =5 ;

d) f (x) = e2x sin 7x ;

e) f (x) =-cos 2x +3sin 2x ;

f) f (x) = x sin x .

Solution. In all cases the right side has a special form. By formula (3.29),

we have

а) z=0;

b) z=0;

c) z=4;

d) z=2+7і;

e) z=2і;

f) z=і.

8. Find the general solution of right hand member not zeroequation

y 3y 4y x .

Solution. Solve homogeneous equation at first

y 3y 4y 0 .

We have:

= 4; y =C e-x

+C e4x .

k2 -3k -4 = 0; k =-1, k

2

1

1

2

The main

value of

the

right

side of the right hand

member not

zero

z 0 (Without ezx ) which is

not

the root of characteristics equation,

thus,

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

16

16

solution of this equation .

4

4

1 x

3

The general solution of this equation is : y C e x

C

2

e4x

.

1

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 . Then 1, e4x is a fundamental system;

1

y C C

2

e4x

1

to example 2.2 in table. 3.4).

Find the derivatives y * and

y * :

y * 3Ax2 2Bx C,

y * 6Ax 2B .

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

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

y (C cos x C

2

sin x)e2x

1

y* x(B

0

B x)e x (B

0

x B x2 )e x .

1

1

Find derivatives y * and

y * :

after substituting y* , y * and

y *

in the initial equation

e x (2B 2(B

0

2B x) (B

0

x B x2 ))

1

1

1

10B1x 2B1 5B0 x 2.

Comparing the coefficients of the same order x :

x

1

:

10B

1,

1

from this

x0

5B0 2,

: 2B1

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

(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

0

sin 3x .

0

Find derivatives :

y * 3A0 sin 3x 3B0 cos 3x,

y * 9A0 cos 3x 9B0 sin 3x.

229

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

5

, B0

1

.

B0 ) 13

6

6

3(5A0

And now we would consider such constants

C1 and

C2 . Condition

y(0) =1 can be written as follows:

1=C +C + 5 .

1

2

6

Thus,

1

y ¢ = 2C1e2x +3C2 e3x -

(5 sin 3x

+cos 3x) ,

2

and condition y ¢(0) = 0 is equivalent to the equation 0 = 2C +3C

2

-0, 5 .

After solution of this linear system

1

ì

5

ï

ï

+C2

+

=1,

ïC1

6

í

ï

-0.5 = 0,

ï2C +3C

îï

1

2

k1,2 2 4i . Control value of the right side

z 2 4i ( 2,

4) , which

230

y* y* y*

y* , where, y* xex (A cos 2x B sin 2x) is a partial solution of

1

2

3

1

the equation

y 2y 5y

e x cos 2x

(Control number

z =1+2i

is the root

of characteristics equation,

y2*

C cos 2x D sin 2x is a partial solution of the

equation

y 2y 5y sin 2x

(Control

value z = 2i

is not

the root of

characteristics equation) ,

y3* Mx N

is a partial solution of the equation

y 2y 5y x (main value

z = 0

is

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

1

y

5y

6y 1 e2x .

Solution. Characteristic equation

k 2 5k 6 0 has roots

k1 2 and

k2 3 .

So

the general

solution

of

the

homogeneous

equation

y 5y 6y 0 can be written as follows

y C e 2x C

2

e 3x .

1

Function

f (x) 1/(1 e2x )

does not correspond to(3.29). Therefore we

solution if the form of

( 3.31):

y C (x)e 2x C

2

(x)e 3x .

1

As

y1 e 2x , y2

e 3x ,

y1 2e 2x ,

y2 3e 3x , to find the functions

C1 (x)

and C2 (x) , form and solve the systems of the solutions in the form (

3.32):

ì

¢

-2x

¢

-3x

ï

= 0,

ïC1e

+C2 e

ï

í

1

ïC ¢(-2e-2x )

+C

¢(-3e-3x ) =

;

ï

1

2

2x

ï

1+e

îï

231