Higher_Mathematics_Part_2

1.4. Solve the Bernoulli’s differential equations.

1.4.1. xy′ − 2y − x2 y = 0 . 1.4.2. y ′− y ctg x = y2 .

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1.4.17.(4 + x 2 ) y′ − 2xy = x / y .

1.4.18.3y 2 y′ − y3 / x = x + 1.

1.4.19.3xy′ − y = (x2 + 1) y−2 .

1.4.20.2(1 + x2 ) y′ − 2xy = x / y .

1.4.21.2xy′ + y = x2 y−1 .

1.4.24.y′ + 2y = 2 y .

xcos2 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.(1 − x2 ) y′ − xy = xy2 .

1.4.30.y′ + xy = y 4 (1 − x2 ) .

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

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. y′(4 + x 2 ) = 4 + y 2 , y(0) = π / 2 .

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1.5.19.2yy ′ = ( y2 −1) ctg x , y(π / 2) = 0 .

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

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 + (2x4 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 + xe x )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 .

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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 +(2y 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 +2y)dy = 0 .

1.6.13.(2x sin y +4x3 )dx +(x2 cos y −sin y)dy = 0 .

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

1.6.19.(2xex2 +y2 +3x2 )dx +(2 yex2 +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 .

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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.

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].

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

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

y= ϕ (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

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

Various types of differential equations with appropriate substitution will be considered in the following articles (see table 3.1).

Consider other types of differential equations with appropriate substitution for reduction of order:

1) a differential equation

F(x, y(k) , y(k +1) , ..., y(n) ) = 0,

which does not contain y directly and derivatives of order less than ( k − 1 ) using the substitution y(k) = z(x) . After substituting, we have

F(x, z, z′, ..., z (n−k) ) = 0 ,

where order is ( n − k );

2) a differential equation

F( y, y′, ..., y(n) ) = 0,

which does not contain independent variable x . It can be a lesser order of times one with substituting the new function:

y′ = p(y),

where p(y) is a new function of y. Thereafter

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2.3. Linear differential equations of the order higher than the first

The linear differential equation of the n’th order

the first degree in the form of y and contains its derivatives, but does not include their products.

An equation (3.20) if f (x) ≠ 0 is called right-hand member not zero. If f (x) ≡ 0 as shown,

y = C1 y1 + C2 y2 + ... + Cn yn

is the solution of (3.21).

There are linear dependent and independent systems of the functions.

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In order to determine whether a system of functions y1 , y2 , ..., yn is linearly dependent or it is not, the Vronsky determinant is used.

as well as their derivatives up to the (n–1)-th order, to be independent within a given interval, it is necessary and sufficient for the Vronsky determinant not to be equal to zero in at least one point of this interval.

An arbitrary system with n linearly independent solutions of the homogeneous equation (3.21) is referred to as a fundamental system.

If y1 , y2 , ..., yn is a fundamental system of equation (3.21), the

general solution of the latter has the form:

_

y = C1 y1 + C2 y2 + ... + Cn yn , where C1 , C2 , ..., Cn are arbitrary constants.

The general solution of the non-homogeneous equation (3.20) can be written in the form:

y = y + y ,

where y is the general solution of the corresponding homogeneous equation (3.24),

y is a partial solution of the non-homogeneous equation (3.20)

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Solution. An equation does not contain x , directly, the substitution

y′ = p( y) , y′′ = p dpdy

will give a new differential equation in p and y of order lesser than the original equation

thereafter, the following two cases: 1) p = 0 , dydx = 0 , y =C ;

is general solution of the original equation (3.23).

Particular, that solution y =C is proceeded from the general solution, if C1 = 0 .

4. Solve

y′′′−(y′′)2 =0 .

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Solution. An equation is of order 3 and does not contain x and y, directly,

.Eventually we would obtain the resultant equation in the form of First order, as follows; z ′−z2 = 0 .

Then we have:

y = ∫ (−ln x +C1 +C2 )dx =−(x +C1 ) ln x +C1 +C2 x +C3 –

Is the general solution of the given equation ( C1 , C2 , C3 –constants)

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