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Национальный авиационный университет

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DDR 4 p.190-249.pdf

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1.4.23.	y		2y			y 0 .			1.4.24.	y	2 y			2 y
					2										.
		x 1									x		cos2 x
1.4.25. xy y y 2 ln x .									1.4.26. 3x2 y xy y 2						0 .
1.4.27. xy 2y 2x3						y .			1.4.28.	2y y tg x y3 tg x 0 .
1.4.29.			2	) y xy xy			2		1.4.30.	y	y		4		2
	(1 x							.				y		(1 x		) .
											x

1.5. Find the general solution and also the particular solution through the point written opposite the equation.

1.5.1. y sin x y ln 2 y , y( / 2) e .

1.5.2.	y ¢-2y tg x =sec x , y(0) 1 .
1.5.3.	2( y x) (x 2y) y , y(1) 0 .

1.5.4. 2y 2 x 2		x2 y 0 , y(1) 0 .
1.5.5.		2	) 4 y	2	,	y(0) / 2 .
	y (4 x
1.5.6.	(x2 4) y 2xy x ,					y(0) 1 .

1.5.7.(xy¢- y) arctg xy = x ln x , y(e) 0 .

1.5.8.xy y(1 ln 2 xy ) , y(1) e .


1.5.9. 2(1 e x ) yy e x y2			,	y(0) 0 .
1.5.10. y ¢+ y tg x = cos2 x ,				y( / 4) 0.5 .
1.5.11.	y ¢- y ctg x =sin 2x cos x ,					y( / 2) 0 .
1.5.12. y ¢+ y tg x = ex cos x , y(0) 1 .
1.5.13. sin x sin yy cos x cos2					y ,	y( / 2) 0 .
1.5.14. y		x2 y 2 xy 0 ,			y(1) 1 .
1.5.15. y 4 y / x ( y / x)2 , y(1)						2 .
1.5.16. y sin2 x y 1,			y( / 4) 1.
1.5.17.	(x2 y2 )dy 2xydx 0 , y(1) = 2 .
1.5.18.	y cos2 x y , y( / 4) e .
1.5.19.	2yy ¢ = ( y2 -1) ctg x , y( / 2) 0 .

209

1.5.20. tg ydx -x ln xdy = 0 , x( / 2) e .

1.5.21. xy y / x 1/ x ,

y(1) 0 .

1.5.22.

y x

y(1)

0 .

y x

1.5.23.

y(1) 1 .

1 x2

1.5.24.

2 y 2

, y(1) 1 .

y 2 x2

1.5.25.

y 3x 2 y 3x5 ,

y(0) 1 .

1.5.26.(x y) y y 2x , y(1) 0 .

1.5.27.xy y(1 ln xy ) , y(1) ee .

1.5.28.x2 y y 2 1 , y(1) 0 .

1.5.29.	(x	2			,	y(0) / 6 .
			1) y x tg y
1.5.30.	yy xey2			0 ,		y(1) 0 .

1.6. Solve the exact differential equations.

1.6.1.(4x3 y3 3x2 y 2 2xy)dx (3x4 y 2 2x3 y x2 )dy 0 .

1.6.2.(4x3 y 2 3x2 y 2x)dx (2x 4 y x3 2y)dy 0 .

1.6.3.(ln x 2xy 2 )dx (2x2 y ln y)dy 0 .

1.6.4.(cos x sin y xex )dx (sin x cos y ye y )dy 0 .

1.6.5.(ln x y)dx (ln y x)dy 0 .

1.6.6.(arctg x +ln y)dx +( y /(1+ y2 ) +x / y)dy = 0 .

1.6.7.(2x sin y 3x2 )dx (x2 cos y 1/ y)dy 0 .

1.6.8. (3x2e y	x	)dx (x3e y y3 )dy 0 .

	1 x2

1.6.9.( y sin x cos2 x)dx (x cos y sin3 y)dy 0 .

1.6.10.(2x cos(x2 + y2 ) +x2 )dx +(2 y cos(x2 + y2 ) + y)dy = 0 .

1.6.11.(x( y +2)exy +2x)dx +(2x + x2 exy )dy = 0 .

1.6.12.( y cos x +cos y)dx +(sin x -x sin y +2 y)dy = 0 .

210

1.6.13.	(2x sin y +4x3 )dx +(x2 cos y -sin y)dy = 0 .
1.6.14. (y -x		1+x2 )dx +(y			y2 -1 +x)dy = 0 .
1.6.15.	(2xe x2 y cos x)dx (e x2 y						sin y)dy 0 .
	æ2x -1	ö	æ	cos	y				ö
	æ2x -1	ö	ç	cos	y			2	÷
	ç	÷	ç					2	÷
1.6.16.		+2xy÷dx +ç				+x			÷dy = 0 .
1.6.16.		+2xy÷dx +ç				+x			÷dy = 0 .
	ç	÷	ç						÷
	ç	÷	ç		y
	è 1+x	ø	ç		y				÷
			è						ø
1.6.17.	cos(ln x) ln y dx ln x sin(ln y) dy 0 .
		x				y

1.6.18.(ex-y + y2 +3x2 )dx +(2xy -ex-y )dy = 0 .

1.6.19.(2xex2 +y2 +3x2 )dx +(2yex2 +y2 -3y2 )dy = 0 .

1.6.20.(4x3 y2 +2xy3 )dx +(2yx4 +3x2 y2 +4y3 )dy = 0 .

1.6.21.(3x2 y + y2 +2x)dx +(x3 +2xy)dy = 0 .

1.6.22.(sin x y)dx ( y cos y2 x)dy 0 .

1.6.23.ln x ex y dx ex ey ey dy 0 .

1.6.24.(2xey + y3ex +2)dx +(x2 ey +3y2 ex )dy = 0 .

1.6.25.(x-1 +2xy2 )dx +( y-1 +2x2 y)dy = 0 .

1.6.26.4x3 sin y 2x cos y dx x4 cos y x2 sin y dy 0 .

1.6.27.y 2 3x 2 y 4 2x dx 2xy 4x3 y3 3y 2 dy 0 .

1.6.28.

+ln y +2x÷dx +çln x +

+2y÷dy =

0 .

èx

1.6.29.

+ y÷dx +çx -

÷dy = 0 .

çè 1-x

1- y

÷ø

èç

÷ø

1.6.30.

(sin2 x +2xy2 )dx +(2x2 y -cos2 y)dy = 0 .

Topic 2. Differential equations of order higher than the first

Main concepts and the definitions. Reduction of order by substitution. Absence of the dependent variable. Absence of the independent variable. Homogeneous linear equation with constant coefficients.

211

Literature: [2, ch.3], [3, ch. 8, §2], [4, section 8, §26], [6, section 11], [7, ch.11, §11.2, 11,3], [8, section 13, §§16–18], [10, §3].

T 2. Main concepts

2.1. Main concepts and definitions

The general form of a n – order differential equation is

F(x, y, y , y ,..., y(n) ) 0 .

When a differential equation is solvable for y(n), it may be written in the

form

y	(n)		(n 1)	)	(3.18)
		f (x, y, y ,..., y

There are n constants C1 ,C2 ,...,Cn in the most general solution of the n – order differential equation:

y(x, C1 ,C2 ,..., Cn ) or Ф(x, y, C1 ,C2 ,..., Cn ) =0.

2.2.Reduction of Order by Substitution

Certain types are readily solvable. One method of attack is to make such a substitution as to reduce the order and then try to solve the result. For example, consider the equation

y(n) f (x),

where f (x) is a given

continuous

function. To

solve this, substitute p for

d n 1 y

and obtain

dxn 1

(n)

n 1

= f(x).

dxn

dxn 1

Therefore

n 1

p = f(x)dx + C1, or

f(x)dx + C1.

( 3.19 )

n 1

Clearly we can treat equation ( 3.19 ) in a similar manner and obtain

n 2

= ( f(x)dx)dx + C1x + C2,

n 2

And it appears that we may continue this process until we find y in terms of x by n successive integrations:

y		f (x)dx dx C		x n 1	C	n 1	x C	.
y		f (x)dx dx C			C		x C	.
			1	(n 1)!			n
		n		(n 1)!
	n	n
	n

212

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