differential equations - pick my...

MATHEMATICS 449

Notes

MODULE - VCalculus

Differential Equations

28

DIFFERENTIAL EQUATIONS

Having studied the concept of differentiation and integration, we are now faced with the questionwhere do they find an application.In fact these are the tools which help us to determine the exact takeoff speed, angle of launch,amount of thrust to be provided and other related technicalities in space launches.Not only this but also in some problems in Physics and Bio-Sciences, we come across relationswhich involve derivatives.

One such relation could be 2ds 4.9 tdt

= where s is distance and t is time. Therefore, dsdt

represents velocity (rate of change of distance) at time t.Equations which involve derivatives as their terms are called differential equations. In this lesson,we are going to learn how to find the solutions and applications of such equations.

OBJECTIVESAfter studying this lesson, you will be able to :• define a differential equation, its order and degree;• determine the order and degree of a differential equation;• form differential equation from a given situation;• illustrate the terms "general solution" and "particular solution" of a differential equation

through examples;• solve differential equations of the following types :

(i)dy f ( )dx

= x (ii)dy f ( )g ( )dx

= x y

(iii) ( )dy f ( )dx g

=xy (iv) ( ) ( )dy P y Q

dx+ =x x (v)

2

2d y

f( )dx

= x

• find the particular solution of a given differential equation for given conditions.

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In fact these are the tools which help us to determine the exact takeoff speed, angle of launch,

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In fact these are the tools which help us to determine the exact takeoff speed, angle of launch,amount of thrust to be provided and other related technicalities in space launches.

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amount of thrust to be provided and other related technicalities in space launches.Not only this but also in some problems in Physics and Bio-Sciences, we come across relations

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Not only this but also in some problems in Physics and Bio-Sciences, we come across relations

is distance and

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is distance and

represents velocity (rate of change of distance) at time

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represents velocity (rate of change of distance) at time t

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t.

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.Equations which involve derivatives as their terms are called differential equations. In this lesson,

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Equations which involve derivatives as their terms are called differential equations. In this lesson,we are going to learn how to find the solutions and applications of such equations.

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we are going to learn how to find the solutions and applications of such equations.

After studying this lesson, you will be able to :

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After studying this lesson, you will be able to :define a differential equation, its order and degree;

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define a differential equation, its order and degree;determine the order and degree of a differential equation;

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determine the order and degree of a differential equation;form differential equation from a given situation;

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form differential equation from a given situation;illustrate the terms "general solution" and "particular solution" of a differential equationwww.pi

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illustrate the terms "general solution" and "particular solution" of a differential equationthrough examples;www.pi

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through examples;solve differential equations of the following types :

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solve differential equations of the following types :

MATHEMATICS

Notes

MODULE - VCalculus

450


EXPECTED BACKGROUND KNOWLEDGE

• Integration of algebraic functions, rational functions and trigonometric functions

28.1 DIFFERENTIAL EQUATIONSAs stated in the introduction, many important problems in Physics, Biology and Social Sciences,when formulated in mathematical terms, lead to equations that involve derivatives. Equations

which involve one or more differential coefficients such as 2

2dy d y

,dx dx

(or differentials) etc. and

independent and dependent variables are called differential equations.

For example,

(i)dy c os xdx

= (ii)2

2d y

y 0dx

+ = (iii) xdx ydy 0+ =

(iv)2 32

22

d y dyx 0dxdx

+ =

(vi)2dy dyy 1

dx dx = + +

28.2 ORDER AND DEGREE OF A DIFFERENTIAL EQUATIONOrder : It is the order of the highest derivative occurring in the differential equation.Degree : It is the degree of the highest order derivative in the differential equation after theequation is free from negative and fractional powers of the derivatives. For example,

Differential Equation Order Degree

(i)dy s i n xdx

= One One

(ii)2

2dy 3y 5xdx

+ =

One Two

(iii)2 42

22

d s dst 0dtdt

+ =

Two Two

(iv)3

3d v 2 dv

0r drdr

+ = Three One

(v)2 54 3

34 3

d y d yx s i n xdx dx

+ =

Four Two



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d

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dy

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yy 1

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y 1dy 1d

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dy 1dyy 1y

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yy 1yd

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dy 1

dy 1

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y 1d

y 1x

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xy 1= + +y 1

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y 1= + +y 1y 1= + +y 1

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y 1= + +y 1yy 1y= + +

yy 1y

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yy 1y= + +

yy 1y

ORDER AND DEGREE OF A DIFFERENTIAL EQUATION

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ORDER AND DEGREE OF A DIFFERENTIAL EQUATION It is the order of the highest derivative occurring in the differential equation.

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It is the order of the highest derivative occurring in the differential equation. It is the degree of the highest order derivative in the differential equation after the

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It is the degree of the highest order derivative in the differential equation after theequation is free from negative and fractional powers of the derivatives. For example,

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equation is free from negative and fractional powers of the derivatives. For example,

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Differential Equation

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Differential Equation

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s i n x

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s i n x

2

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23

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3

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dy dy

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dy dy+ =

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+ =3+ =3

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3+ =3y+ =y

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y+ =y

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dy dy

dy dy

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dy dy

dy dy

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dx dx

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dx dx

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dx dx dx dx

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dx dx dx dx

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2 2

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2 2d d

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d d s s

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s s

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d d

d d

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d d

d d

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dt dtwww.pick

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dt dt

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dt

dt dt

dtwww.pick

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dt

dt dt

dtwww.pick

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(iii)

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(iii)

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MATHEMATICS 451

Notes

MODULE - VCalculus


Example 28.1 Find the order and degree of the differential equation :

322 2

2d y dy1 0

dxdx

+ + =

Solution : The given differential equation is3

22 2

2d y dy1 0

dxdx

+ + =

or

322 2

2d y dy

1dxdx

= − +

The term

32 2dy1

dx

+

has fractional index. Therefore, we first square both sides to remove

fractional index.

Squaring both sides, we have32 22

2d y dy1

dxdx

= +

Hence each of the order of the diferential equation is 2 and the degree of the differential equationis also 2.

Note : Before finding the degree of a differential equation, it should be free from radicals andfractions as far as derivatives are concerned.

28.3 LINEAR AND NON-LINEAR DIFFERENTIAL EQUATIONSA differential equation in which the dependent variable and all of its derivatives occur only in thefirst degree and are not multiplied together is called a linear differential equation. A differentialequation which is not linear is called non-linear differential equation . For example, the differentialequations

2

2d y

y 0dx

+ = and3

2 33

d y dycos x x y 0

dxdx+ + = are linear..

The differential equation 2dy y log x

dx x + =

is non-linear as degree of dydx

is two.

Further the differential equation 2

2d y

y 4dx

− = x is non-linear because the dependent variable

y and its derivative 2

2d ydx

are multiplied together..



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Hence each of the order of the diferential equation is 2 and the degree of the differential equation

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Hence each of the order of the diferential equation is 2 and the degree of the differential equation

Before finding the degree of a differential equation, it should be free from radicals and

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Before finding the degree of a differential equation, it should be free from radicals andfractions as far as derivatives are concerned.

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fractions as far as derivatives are concerned.

LINEAR AND NON-LINEAR DIFFERENTIAL EQUATIONS

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LINEAR AND NON-LINEAR DIFFERENTIAL EQUATIONSA differential equation in which the dependent variable and all of its derivatives occur only in the

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A differential equation in which the dependent variable and all of its derivatives occur only in thefirst degree and are not multiplied together is called a

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first degree and are not multiplied together is called a equation which is not linear is called non-linear differential equation . For example, the differential

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equation which is not linear is called non-linear differential equation . For example, the differential

y 0

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y 0+ =

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+ =y 0+ =y 0

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y 0+ =y 0 and

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and

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MATHEMATICS

Notes

MODULE - VCalculus

452


28.4 FORMATION OF A DIFFERENTIAL EQUATIONConsider the family of all straight lines passing through the origin (see Fig. 28.1).

This family of lines can be represented byy = mx .....(1)

Differentiating both sides, we getdydx

= m .....(2)

From (1) and (2), we getdyydx

= x .....(3)

Sodyy x anddx

= =m y x represent the same family..

Clearly equation (3) is a differential equation.

Working Rule : To form the differential equation corresponding to an equation involvingtwo variables, say x and y and some arbitrary constants, say, a, b, c, etc.(i) Differentiate the equation as many times as the number of arbitrary constants in the

equation.(ii) Eliminate the arbitrary constants from these equations.

Remark

If an equation contains n arbitrary constants then we will obtain a differential equation of nth

order.

Example 28.2 Form the differential equation representing the family of curves.2y ax bx.= + ......(1)

Differentiating both sides, we getdy 2 ax bdx

= + ......(2)

Differentiating again, we get2

2d y

2 adx

= ......(3)

⇒2

21 d y

a2 dx

= ......(4)

(The equation (1) contains two arbitrary constants. Therefore, we differentiate this equationtwo times and eliminate 'a' and 'b').On putting the value of 'a' in equation (2), we get

2

2dy d yx bdx dx

= +

Fig. 28.1



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To form the differential equation corresponding to an equation involving

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To form the differential equation corresponding to an equation involvingand some arbitrary constants, say,

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and some arbitrary constants, say, Differentiate the equation as many times as the number of arbitrary constants in the

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Differentiate the equation as many times as the number of arbitrary constants in the

Eliminate the arbitrary constants from these equations.

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Eliminate the arbitrary constants from these equations.

If an equation contains n arbitrary constants then we will obtain a differential equation of n

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If an equation contains n arbitrary constants then we will obtain a differential equation of n

Form the differential equation representing the family of curves.

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Form the differential equation representing the family of curves.

y

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y = +

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= +a= +a

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a= +a

Differentiating both sides, we get

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Differentiating both sides, we get

Differentiating again, we get

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MATHEMATICS 453

Notes

MODULE - VCalculus


⇒2

2dy d y

b xdx dx

= − ......(5)

Substituting the values of 'a' and 'b' given in (4) and (5) above in equation (1), we get2 2

22 2

1 d y dy d yy x x x

2 dxdx dx

= + −

or2 2 2

22 2

x d y dy d yy x x

2 dxdx dx= + −

or2 2

2dy x d y

y xdx 2 dx

= −

or2 2

2x d y dy

x y 02 dxdx

− + =

which is the required differential equation.

Example 28.3 Form the differential equation representing the family of curves

y = a cos (x + b).

Solution : y = a cos (x +b) .....(1)Differentiating both sides, we get

dy a sin(x b)dx

= − + ....(2)


( )2

2d y

acos x bdx

= − + .....(3)

From (1) and (3), we get2

2d y

ydx

= − or2

2d y

y 0dx

+ =


Example 28.4 Find the differential equation of all circles which pass through the origin andwhose centres are on the x-axis.Solution : As the centre lies on the x-axis, its coordinates will be (a, 0).Since each circle passes through the origin, its radius is a.Then the equation of any circle will be

( )2 2 2x a y a− + = (1)

To find the corresponding differential equation, we differentiate equation (1) and get



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Form the differential equation representing the family of curves

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Form the differential equation representing the family of curves

( )

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( )( )x b( )

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( )x b( )( )− +( )

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( )− +( )( )x b( )− +( )x b( )

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( )x b( )− +( )x b( )

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2

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2d y

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d y

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y

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ydx

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dx= −

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= −

which is the required differential equation.www.pick

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Find the differential equation of all circles which pass through the origin andwww.pick

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Find the differential equation of all circles which pass through the origin and

MATHEMATICS

Notes

MODULE - VCalculus

454


( ) dy2 x a 2y 0dx

− + =

ordyx a y 0dx

− + =

ordya y xdx

= +

Substituting the value of 'a' in equation (1), we get2 2

2dy dyx y x y y xdx dx

− − + = +

or2 2

2 2dy dy dyy y x y 2xydx dx dx

+ = + +

or 2 2 dyy x 2xydx

= +


RemarkIf an equation contains one arbitrary constant then the corresponding differential equation isof the first order and if an equation contains two arbitrary constants then the correspondingdifferential equation is of the second order and so on.

Example 28.5 Assuming that a spherical rain drop evaporates at a rate proportional to itssurface area, form a differential equation involving the rate of change of the radius of the raindrop.Solution : Let r(t) denote the radius (in mm) of the rain drop after t minutes. Since the radius isdecreasing as t increases, the rate of change of r must be negative.If V denotes the volume of the rain drop and S its surface area, we have

34V r3

= π .....(1)

and 2S 4 r= π .....(2)It is also given that

dV Sdt

∝

ordV Sdt

= −k

ordV dr. Sdr dt

= −k .....(3)

Using (1), (2) and (3) we have

2 2dr4 r . 4 rdt

π = − πk



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2xy

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2xydx

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+ +

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+ +

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2xydx

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dx

If an equation contains one arbitrary constant then the corresponding differential equation is

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If an equation contains one arbitrary constant then the corresponding differential equation isof the first order and if an equation contains two arbitrary constants then the corresponding

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of the first order and if an equation contains two arbitrary constants then the correspondingdifferential equation is of the second order and so on.

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differential equation is of the second order and so on.

Assuming that a spherical rain drop evaporates at a rate proportional to its

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Assuming that a spherical rain drop evaporates at a rate proportional to its

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surface area, form a differential equation involving the rate of change of the radius of the rain

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surface area, form a differential equation involving the rate of change of the radius of the rain

Let r(t) denote the radius (in mm) of the rain drop after t minutes. Since the radius is

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Let r(t) denote the radius (in mm) of the rain drop after t minutes. Since the radius isdecreasing as t increases, the rate of change of r must be negative.

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decreasing as t increases, the rate of change of r must be negative.If V denotes the volume of the rain drop and S its surface area, we have

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If V denotes the volume of the rain drop and S its surface area, we have

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It is also given that

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It is also given that

MATHEMATICS 455

Notes

MODULE - VCalculus


ordrdt

= k


CHECK YOUR PROGRESS 28.11. Find the order and degree of the differential equation

dy 1y x dydxdx

= +

2. Write the order and degree of each of the following differential equations.

(a)4 2

2ds d s3s 0dt dt

+ =

(b)2dy dyy 2x x 1

dx dx = + +

(c) 2 21 x dx 1 y dy 0− + − = (d)2 32

2d s ds3 4 0

dtdt + + =

3. State whether the following differential equations are linear or non-linear.

(a) ( ) ( )2 2xy x dx y x y dy 0− + − = (b) dx dy 0+ =

(c)dy c os xdx

= (d) 2dy sin y 0dx

+ =

4. Form the differential equation corresponding to

( ) ( )2 2 2x a y b r− + − = by eliminating 'a' and 'b'.5. (a) Form the differential equation corresponding to

( )2 2 2y m a x= −

(b) Form the differential equation corresponding to2 2 2y 2ay x a− + = , where a is an arbitrary constant.

(c) Find the differential equation of the family of curves 2x 3xy Ae Be−= + where Aand B are arbitrary constants.

(d) Find the differential equation of all straight lines passing through the point (3,2).(e) Find the differential equation of all the circles which pass through origin and whose

centres lie on y-axis.

28.5 GENERAL AND PARTICULAR SOLUTIONSFinding solution of a differential equation is a reverse process. Here we try to find anequation which gives rise to the given differential equation through the process ofdifferentiations and elimination of constants. The equation so found is called the primitive orthe solution of the differential equation.



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Write the order and degree of each of the following differential equations.

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Write the order and degree of each of the following differential equations.

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.com2

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x 1

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.com dy dy

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+ +

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.comx 1+ +x 1x 1+ +x 1

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.comx 1+ +x 1x 1+ +x 1

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dy

dy

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dy

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dy dy

dy dy

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dy dy

+ + + +

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+ +

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+ +

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dx dx

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dx dx

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dx dx dx dx

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dx dx dx dx+ + + +

+ + + +

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+ + + +

3

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33

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3

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ds ds

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ds ds+ =

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+ =

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ds ds

ds ds

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ds ds

ds ds

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dt dt

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dt dt

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dt dt dt dt

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dt dt dt dt

State whether the following differential equations are linear or non-linear.

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State whether the following differential equations are linear or non-linear.

(b)

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(b) d

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d

2

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2 y 0

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y 0+ =

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+ =2+ =2

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2+ =2n+ =n

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n+ =n y 0+ =y 0

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y 0+ =y 0

Form the differential equation corresponding to

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Form the differential equation corresponding to2

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2b r

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b rb r− =b r

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b r− =b r by eliminating 'a' and 'b'.

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by eliminating 'a' and 'b'.Form the differential equation corresponding to

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)

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)

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2 2

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2 2a x

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a x


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Form the differential equation corresponding to2 2

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2 2y

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y x a

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x a2 2x a2 2

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2 2x a2 2

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+ =

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+ =2 2+ =2 2

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2 2+ =2 2x a+ =x a

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x a+ =x a2 2x a2 2+ =2 2x a2 2

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2 2x a2 2+ =2 2x a2 2

Find the differential equation of the family of curves

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Find the differential equation of the family of curves and B are arbitrary constants.www.pi

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and B are arbitrary constants.Find the differential equation of all straight lines passing through the point (3,2).www.pi

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Find the differential equation of all straight lines passing through the point (3,2).

MATHEMATICS

Notes

MODULE - VCalculus

456

Differential Equations Remarks

(1) If we differentiate the primitive, we get the differential equation and if we integrate thedifferential equation, we get the primitive.

(2) Solution of a differential equation is one which satisfies the differential equation.

Example 28.6 Show that 1 2y C sin x C cos x,= + where 1 2C and C are arbitrary

constants, is a solution of the differential equation :2

2d y

y 0dx

+ =

Solution : We are given that

1 2y C sinx C cosx= + .....(1)Differentiating both sides of (1), we get

1 2dy C cosx C sin xdx

= − .....(2)

Differentiating again, we get2

1 22d y

C sinx C cosxdx

= − −

Substituting the values of 2

2d ydx

and y in the given differential equation, we get

21 2 1 22

d yy C sin x C cosx ( C sinx C cosx)

dx+ = + + − −

or2

2d y

y 0dx

+ =

In integration, the arbitrary constants play important role. For different values of the constantswe get the different solutions of the differential equation.

A solution which contains as many as arbitrary constants as the order of the differential equationis called the General Solution or complete primitive.

If we give the particular values to the arbitrary constants in the general solution of differentialequation, the resulting solution is called a Particular Solution.

Remark

General Solution contains as many arbitrary constants as is the order of the differential equation.

Example 28.7 Show that ay cxc

= + (where c is a constant) is a solution of the differential

equation.



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1 2

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1 2C

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C1 2C1 2

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1 2C1 2 si

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sin x

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n x

1 2

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1 2C

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C sin

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sin1 2sin1 2

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1 2sin1 2x

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x1 2x1 2

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1 2x1 2− −

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− −1 2− −1 2

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1 2− −1 2C− −C

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C− −C1 2sin1 2− −1 2sin1 2

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1 2sin1 2− −1 2sin1 21 2x1 2− −1 2x1 2

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1 2x1 2− −1 2x1 2


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1

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1y

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y C

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C si

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si= +

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= +1= +1

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1= +1C= +C

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C= +C si= +si

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si= +sin= +n

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n= +n x= +x

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x= +x

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2

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2d y

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d yy 0

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y 0dx

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dx

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+ =

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+ =y 0+ =y 0

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y 0+ =y 0

In integration, the arbitrary constants play important role. For different values of the constants

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In integration, the arbitrary constants play important role. For different values of the constantswe get the different solutions of the differential equation.

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we get the different solutions of the differential equation.

A solution which contains as many as arbitrary constants as the order of the differential equation

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A solution which contains as many as arbitrary constants as the order of the differential equationis called the www.pi

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is called the General Solutionwww.pick

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General Solution

If we give the particular values to the arbitrary constants in the general solution of differentialwww.pick

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If we give the particular values to the arbitrary constants in the general solution of differentialequation, the resulting solution is called a

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equation, the resulting solution is called a

MATHEMATICS 457

Notes

MODULE - VCalculus


dy dxy x adx dy

= +

Solution : We have ay cxc

= + .....(1)

Differentiating (1), we get

dy cdx

= ⇒dx 1dy c

=

On substituting the values ofdydx

and dxdy in R.H.S of the differential equation, we have

( ) 1 ax c a cx y

c c + = + =

⇒ R.H.S. = L.H.S.

Henceay cxc

= + is a solution of the given differential equation.

Example 28.8 If 2y 3x C= + is the general solution of the differential equation

dy 6x 0dx

− = , then find the particular solution when y = 3, x = 2.

Solution : The general solution of the given differential equation is given as

2y 3x C= + ....(1)

Now on substituting y = 3, x = 2 in the above equation , we get3 = 12 + C or C 9= −

By substituting the value of C in the general solution (1), we get

2y 3x 9= −

which is the required particular solution.

28.6 TECHNIQUES OF SOLVING A DIFFERENTIAL EQUATION

28.6.1 When Variables are Separable

(i) Differential equation of the type dy f(x)dx

=

Consider the differential equation of the type dy f(x)dx

=

or dy = f (x) dxOn integrating both sides, we get

( )dy f x dx=∫ ∫



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is the general solution of the differential equation

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is the general solution of the differential equation

, then find the particular solution when y = 3, x = 2.

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, then find the particular solution when y = 3, x = 2.

The general solution of the given differential equation is given as

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The general solution of the given differential equation is given as

Now on substituting y = 3, x = 2 in the above equation , we get

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Now on substituting y = 3, x = 2 in the above equation , we getor

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or C 9

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C 9C 9= −C 9

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C 9= −C 9By substituting the value of C in the general solution (1), we get

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By substituting the value of C in the general solution (1), we get

x 9

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x 9x 9= −x 9

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x 9= −x 9

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QUES OF SOLV

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QUES OF SOLV

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28.6.1 When Variables are Separablewww.pick

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28.6.1 When Variables are Separable

MATHEMATICS

Notes

MODULE - VCalculus

458


( )y f x dx c= +∫where c is an arbitrary constant. This is the general solution.

Note : It is necessary to write c in the general solution, otherwise it will become a particularsolution.

Example 28.9 Solve

( ) 2dyx 2 x 4x 5dx

+ = + −

Solution : The given differential equation is ( ) 2dyx 2 x 4x 5dx

+ = + −

or2dy x 4x 5

dx x 2+ −

=+

or2dy x 4x 4 4 5

dx x 2+ + − −

=+

or ( )2x 2dy 9dx x 2 x 2

+= −

+ +or

dy 9x 2dx x 2

= + −+

or9

dy x 2 dxx 2

= + − + .....(1)

On integrating both sides of (1), we have

9dy x 2 dx

x 2 = + − + ∫ ∫ or

2xy 2x 9 log | x 2 | c2

= + − + + ,

where c is an arbitrary constant, is the required general solution.

Example 28.10 Solve

3dy 2x xdx

= −

given that y = 1 when x = 0

Solution : The given differential equation is 3dy 2x xdx

= −

or ( )3dy 2x x dx= − ...(1)

On integrating both sides of (1), we get

( )3dy 2x x dx= −∫ ∫ or4 2x xy 2. C

4 2= − +

or4 2x xy C

2 2= − + .....(2)

where C is an arbitrary constant.Since y = 1 when x = 0, therefore, if we substitute these values in (2) we will get

1 0 0 C= − + ⇒ C = 1



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y x

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y xd

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=

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d

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dy 9

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y 9d

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dx

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x= + −

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= + −y 9= + −

y 9

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y 9= + −

y 9

dx

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dx


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9 9

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9 9

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9 9

9 9

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9 9

9 9+ − + −

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+ − + −+ − + −

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+ − + − + − + −

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+ − + −

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x 2 x 2

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x 2 x 2

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x 2 x 2 x 2 x 2

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x 2 x 2 x 2 x 2+ − + −

+ − + −

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+ − + −

+ − + −x 2+x 2 x 2+x 2

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x 2+x 2 x 2+x 2x 2 x 2+x 2 x 2 x 2 x 2+x 2 x 2

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x 2 x 2+x 2 x 2 x 2 x 2+x 2 x 2where c is an arbitrary constant, is the required general solution.

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where c is an arbitrary constant, is the required general solution.

Solve

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Solve

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dy

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dydx

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dx= −

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= −

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Solution :

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Solution : The given differential equation is

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The given differential equation is

On integrating both sides of (1), we getwww.pick

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MATHEMATICS 459

Notes

MODULE - VCalculus

Differential EquationsNow, on putting the value of C in (2), we get

( )4 21y x x 12

= − + or ( )2 21y x x 1 12

= − +


(ii) Differential equations of the typedy f(x) g(y)dx

= ⋅

Consider the differential equation of the typedy f (x) g(y)dx

= ⋅

ordy f (x) dx

g(y)= .....(1)

In equation (1), x's and y's have been separated from one another. Therefore, this equation isalso known differential equation with variables separable.To solve such differential equations, we integrate both sides and add an arbitrary constant onone side.

To illustrate this method, let us take few examples.

Example 28.11 Solve

( ) ( )2 21 x dy 1 y dx+ = +

Solution : The given differential equation

( ) ( )2 21 x dy 1 y dx+ = +

can be written as 2 2dy dx

1 y 1 x=

+ + (Here variables have been seperated)


2 2dy dx

1 y 1 x=

+ +∫ ∫

or 1 1tan y tan x C− −= +

where C is an arbitrary constant.This is the required solution.

Example 28.12 Solve

( )2 2 2 2dyx yx y xy 0dx

− + + =

Solution : The given differential equation



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In equation (1), x's and y's have been separated from one another. Therefore, this equation is

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In equation (1), x's and y's have been separated from one another. Therefore, this equation is

To solve such differential equations, we integrate both sides and add an arbitrary constant on

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To solve such differential equations, we integrate both sides and add an arbitrary constant on

(

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( )

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)2 2

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2 21 y

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1 y2 21 y2 2

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2 21 y2 2 dx

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dx= +

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= +1 y= +1 y

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1 y= +1 y2 21 y2 2= +2 21 y2 2

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2 21 y2 2= +2 21 y2 2

)

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) (

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(2 2

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2 2(2 2(

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(2 2(d

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d2 2d2 2

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2 2d2 2y

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y2 2y2 2

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2 2y2 21 y

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1 y2 21 y2 2

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2 21 y2 2= +

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= +(= +(

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(= +(2 2= +2 2

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2 2= +2 2(2 2(= +(2 2(

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(2 2(= +(2 2(1 y= +1 y

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1 y= +1 y2 21 y2 2= +2 21 y2 2

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2 21 y2 2= +2 21 y2 2

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2 2

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2 2d

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dy

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ydyd

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dyd1

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1 y

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y

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+ +

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+ +2 2+ +2 2

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2 2+ +2 2y+ +y

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y+ +y


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MATHEMATICS

Notes

MODULE - VCalculus

460


( )2 2 2 2dyx yx y xy 0dx

− + + =

can be written as ( ) ( )2 2dyx 1 y y 1 x 0dx

− + + =

or ( ) ( )2 2

1 y 1 xdy dx

y x− − +

=

(Variables separated) ....(1)If we integrate both sides of (1), we get

2 21 1 1 1

dy dxy xy x

− = − −

∫ ∫where C is an arbitrary constant.

or1 1log y log x Cy x

− − = − +

orx 1 1log Cy x y

= + +

Which is the required general solution.

Example 28.13 Find the particular solution of

2dy 2xdx 3y 1

=+

when y (0) = 3 (i.e. when x = 0, y = 3).Solution : The given differential equation is

2dy 2xdx 3y 1

=+ or ( )23y 1 dy 2xdx+ = (Variables separated) ....(1)

If we integrate both sides of (1), we get

( )23y 1 dy 2xdx+ =∫ ∫ ,

where C is an arbitrary constant.3 2y y x C+ = + ....(2)

It is given that, ( )y 0 3.=

∴on substituting y = 3 and x = 0 in (2), we get

27 3 C+ =

∴ C = 30Thus, the required particular solution is

3 2y y x 30+ = +



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g

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g x C

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x Cx C

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x C− +

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− +− +

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− +g− +g

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g− +g x C− +x C

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x C− +x Cx C− +x C

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x C− +x C

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1 1

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1 1g C

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g C1 1g C1 1

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1 1g C1 1x y

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x yg C= + +g C

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g C= + +g Cg C= + +g C

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g C= + +g Cg C= + +g C

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g C= + +g C

Find the particular solution of

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Find the particular solution of

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2

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22x

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2x3

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3y 1

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y 12y 12

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2y 12=

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=y 1+y 1

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y 1+y 1

when y (0) = 3 (i.e. when x = 0, y = 3).

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when y (0) = 3 (i.e. when x = 0, y = 3). The given differential equation is

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2

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22x

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2x3

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3y 1

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y 12y 12

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2y 12y 1+y 1

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y 1+y 1 or

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or


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(

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(∫ ∫

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∫ ∫(∫ ∫(

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(∫ ∫( 2∫ ∫2

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2∫ ∫23∫ ∫3

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3∫ ∫3y∫ ∫y

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y∫ ∫y

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where C is an arbitrary constant.www.pick

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where C is an arbitrary constant.

MATHEMATICS 461

Notes

MODULE - VCalculus


28.6.2 Homogeneous Differential EquationsConsider the following differential equations :

(i) 2 2 dy dyy x xydx dx

+ = (ii) ( )3 3 2x y dx 3xy dy 0+ − =

(iii)3 2

2dy x xydx y x

+=

In equation (i) above, we see that each term except dydx

is of degree 2

[as degree of 2y is 2, degree of 2x is 2 and degree of xy is 1 + 1 =2]

In equation (ii) each term except dydx

is of degree 3.

In equation (iii) each term except dydx

is of degree 3, as it can be rewritten as

2 3 2dyy x x x ydx

= +

Such equations are called homogeneous equations .

Remarks Homogeneous equations do not have constant terms.

For example, differential equation

( ) ( )2 3x 3yx dx x x dy 0+ − + =

is not a homogeneous equation as the degree of the function except dydx

in each term is not the

same. [degree of 2x is 2, that of 3yx is 2, of 3x is 3, and of x is 1]

28.6.3 Solution of Homogeneous Differential Equation :To solve such equations, we proceed in the following manner :(i) write one variable = v. (the other variable).

( i.e. either y = vx or x = vy )(ii) reduce the equation to separable form(iii) solve the equation as we had done earlier.

Example 28.14 Solve

( )2 2 2x 3xy y dx x dy 0+ + − =


( )2 2 2x 3xy y dx x dy 0+ + − =



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Homogeneous equations do not have constant terms.

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Homogeneous equations do not have constant terms.

(

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( )

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)2 3

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2 3x x

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x x(x x(

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(x x(2 3x x2 3

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2 3x x2 3(2 3(x x(2 3(

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(2 3(x x(2 3( d

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d

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+ =

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+ =)+ =)

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)+ =)x+ =x

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x+ =x d+ =d

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d+ =d

is not a homogeneous equation as the degree of the function except

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is not a homogeneous equation as the degree of the function except

is 2, that of 3yx is 2, of

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is 2, that of 3yx is 2, of

28.6.3 Solution of Homogeneous Differential Equation :

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28.6.3 Solution of Homogeneous Differential Equation :To solve such equations, we proceed in the following manner :

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To solve such equations, we proceed in the following manner :write one variable = v. (the other variable).

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write one variable = v. (the other variable).( i.e. either y = vx or x = vy )www.pi

ckMyC

oach

ing.co

m

( i.e. either y = vx or x = vy )reduce the equation to separable form

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reduce the equation to separable form

MATHEMATICS

Notes

MODULE - VCalculus

462


or2 2

2dy x 3xy ydx x

+ += .....(1)

It is a homogeneous equation of degree two. (Why?)Let y vx= . Then

dy dvv xdx dx

= +

∴ From (1), we have

( )22

2x 3x.vx vxdvv x

dx x+ +

+ = or2

22

dv 1 3v vv x x

dx x + +

+ =

or 2dvv x 1 3v vdx

+ = + + or 2dvx 1 3v v vdx

= + + −

or 2dvx v 2v 1dx

= + + or 2dv dx

xv 2v 1=

+ +

or ( )2dv dx

xv 1=

+ .....(2)

Further on integrating both sides of (2), we get

1 C log xv 1−

+ =+

, where C is an arbitrary constant.

On substituting the value of v, we getx log x C

y x+ =

+ which is the required solution.

28.6.4 Differential EquationConsider the equation

dy Py Qdx

+ = .....(1)

where P and Q are functions of x. This is linear equation of order one.

To solve equation (1), we first multiply both sides of equation (1) by Pdxe∫ (called integrating factor) and get

Pdx Pdx Pdxdye Pye Qedx

∫ ∫ ∫+ =

or ( )Pdx Pdxd ye Qedx

∫ ∫= .....(2)

( )Pdx Pdx Pdxd dyye e Py.e

dx dx ∫ ∫ ∫= + ∵



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dv

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dvx

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dx

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=

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2

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2d

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v

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v 2

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2v 1

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v 1+ +

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+ +2+ +2

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2+ +2v 1+ +v 1

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v 1+ +v 1


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g x

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g xg x

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g x ,

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,

On substituting the value of v, we get

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On substituting the value of v, we get

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lo

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lo x C

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x Cx C

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x C

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lo

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log

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g x C

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x Cx C

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x C

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y x

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y x+ =

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+ =+ =

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+ =lo+ =lo

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lo+ =log+ =g

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g+ =g x C+ =x C

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x C+ =x Cx C+ =x C

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x C+ =x C

28.6.4 Differential Equation

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28.6.4 Differential EquationConsider the equation

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Consider the equation

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To solve equation (1), we first multiply both sides of equation (1) bywww.pick

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To solve equation (1), we first multiply both sides of equation (1) bywww.pick

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(called integrating factor) and getwww.pick

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(called integrating factor) and get

MATHEMATICS 463

Notes

MODULE - VCalculus


On integrating, we getPdx Pdxye Qe dx C∫ ∫= +∫ .....(3)

where C is an arbitrary constant,

orPdx Pdxy e Qe dx C− ∫ ∫= + ∫

Note : ∫ Pdxe is called the integrating factor of the equation and is written as I.F in short.

Remarks

(i) We observe that the left hand side of the linear differential equation (1) has become

( )Pdxd yedx

∫ after the equation has been multiplied by the factor Pdxe∫ .

(ii) The solution of the linear differential equationdy Py Qdx

+ =

P and Q being functions of x only is given by

( )Pdx Pdxye Q e dx C∫ ∫= +∫

(iii) The coefficient of dydx

, if not unity, must be made unity by dividing the equation by it

throughout.(iv) Some differential equations become linear differential equations if y is treated as the

independent variable and x is treated as the dependent variable.

For example, dx Px Qdy

+ = , where P and Q are functions of y only, is also a linear

differential equation of the first order.

In this case PdyI.F. e∫=and the solution is given by

( ) ( )x I.F. Q. I.F. dy C= +∫

Example 28.15 Solve

xdy y edx x

−+ =

Solution : Here x1P , Q ex

−= = (Note that both P an Q are functions of x)

I.F. (Integrating Factor)1dxPdx logxxe e e x∫∫ = = = ( )x 0>

On multiplying both sides of the equation by I.F., we get



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Pdx

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Pdx .

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.


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Some differential equations become linear differential equations if y is treated as the

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Some differential equations become linear differential equations if y is treated as theindependent variable and x is treated as the dependent variable.

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independent variable and x is treated as the dependent variable.

, where P and Q are functions of y only, is also a linear

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, where P and Q are functions of y only, is also a linear

differential equation of the first order.

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differential equation of the first order.Pdy

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PdyI.F

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I.F. e

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. e∫

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∫. e=. e

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. e=. eand the solution is given by

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and the solution is given by

)

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) (

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(Q

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Q.

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. I.F

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I.F= +

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= +(= +(

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(= +(Q= +Q

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Q= +Q.= +.

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.= +. I.F= +I.F

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I.F= +I.F= +

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= +∫

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∫= +∫= +

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= +∫= +

Solvewww.pick

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Solve

MATHEMATICS

Notes

MODULE - VCalculus

464


xdyx y x edx

−⋅ + = ⋅ or ( ) xd y x xedx

−⋅ =

Integrating both sides, we have

xyx xe dx C−= +∫where C is an arbitrary constant

or x xxy xe e dx C− −= − + +∫or x xxy xe e C− −= − − +

or ( )xxy e x 1 C−= − + +

or xx 1 Cy e

x x−+ = − +

Note: In the solution x > 0.

Example 28.16 Solve :

2dysinx ycosx 2sin x cosxdx

+ =


2dysinx ycosx 2sin x cosxdx

+ =

ordy yco tx 2 s i n x c o s xdx

+ = .....(1)

Here P cotx,Q 2 s i n x c o s x= =

Pdx cotxdx logsinxI.F. e e e s inx∫ ∫= = = =On multiplying both sides of equation (1) by I.F., we get (sin x >0)

( ) 2d ys inx 2sin x c o s xdx

=

Further on integrating both sides, we have2ysinx 2sin x cos x d x C= +∫

where C is an arbitrary constant (sin x >0)

or 32ysinx sin x C3

= + , which is the required solution.

Example 28.17 Solve ( )2 1dx1 y tan y xdy

−+ = −




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2

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2n

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n x

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x cosx

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cosx


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2

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2ycos

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ycos 2si

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2sin

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n+ =

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+ =ycos+ =ycos

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ycos+ =ycosx+ =x

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x+ =x

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dy

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dy yco t

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yco tdx

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dx+ =

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+ =yco t+ =yco t

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yco t+ =yco tx+ =x

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x+ =x

P

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P cotx,

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cotx,

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= =

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= =cotx,= =cotx,

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cotx,= =cotx,

I.F

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I.F.

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.On multiplying both sides of equation (1) by I.F., we get (sin x >0)

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On multiplying both sides of equation (1) by I.F., we get (sin x >0)

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( )

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( )d

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d ( )ys in( )

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( )ys in( )dx

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dx

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Further on integrating both sides, we have

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Further on integrating both sides, we have

MATHEMATICS 465

Notes

MODULE - VCalculus


( )2 1dx1 y tan y xdy

−+ = −

or1

2 2dx tan y xdy 1 y 1 y

−= −

+ +

or1

2 2dx x tan ydy 1 y 1 y

−+ =

+ +....(1)

which is of the form dx Px Qdy

+ = , where P and Q are the functions of y only..

1 dy 12Pdy tan y1 yI.F. e e e−+

∫∫= = =

Multiplying both sides of equation (1) by I.F., we get

( )11 1tan y tan y2

d tan yxe e

dy 1 y

−− − = +

On integrating both sides, we get

or ( )1tan y te x e .tdt C−

= +∫ ,

where C is an arbitrary constant and 12

1t tan yanddt dy

1 y−= =

+

or ( )1tan y t te x te e C−

= − +∫or ( )1tan y t te x te e C

−= − +

or ( )1 1 1tan y 1 tan y tan ye x tan y e e C− − −−= − + (on putting 1t tan y−= )

or 11 tan yx tan y 1 Ce−− −= − +

CHECK YOUR PROGRESS 28.2

1. (i) Is2

2d y

y sin x a solution of y 0 ?dx

= + =

(ii) Is 3 dyy x a solution of x 4y 0 ?dx

= − =

2. Given below are some solutions of the differential equationdy 3xdx

= .

State which are particular solutions and which are general solutions.



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1

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1yandd

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yanddt

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t1 y

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1 y= =

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= =yandd= =yandd

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yandd= =yanddt= =t

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t= =t

1 1

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1 1ta

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tan

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ntanta

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tanta y

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y1 1y1 1

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1 1y1 1ta

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ta1 1ta1 1

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1 1ta1 1e e C

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e e Ctae e Cta

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tae e Ctane e Cn

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ne e Cntantae e Ctanta

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tantae e Ctanta ye e Cy

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ye e Cy tae e Cta

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tae e Cta− −

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− −1 1− −1 1

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1 1− −1 1e e C− +e e C

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e e C− +e e C

n y 1

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n y 1 Ce

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Ce− −

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− −n y 1− −n y 1

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n y 1− −n y 1 Ce− −Ce

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Ce− −Ce− +

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− +n y 1− +n y 1

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n y 1− +n y 1


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n x a

MATHEMATICS

Notes

MODULE - VCalculus

466


(i) 22y 3x= (ii) 23y x 22

= +

(iii) 22y 3x C= + (iv) 23y x 32

= +

3. State whether the following differential equations are homogeneous or not ?

(i)2

2dy xdx 1 y

=+

(ii) ( ) ( )2 23xy y dx x xy dy 0+ + + =

(iii) ( ) 2dyx 2 x 4x 9dx

+ = + − (iv) ( ) ( )3 2 3 3x yx dy y x dx 0− + + =

4. (a) Show that y = a sin 2x is a solution of2

2d y

4y 0dx

+ =

(b) Verify that 3

3 23

d yy x ax c is a solution of 6

dx= + + =

5. The general solution of the differential equation

2dy sec x is y tan x Cdx

= = + .

Find the particular solution when

(a) x , y 14π

= = (b) 2x , y 03π

= =

6. Solve the following differential equations :

(a) ( )5 1 3dy x tan xdx

−= (b) 3 2 xdy sin x cos x xedx

= +

(c) ( )2 dy1 x xdx

+ = (d) 2dy x sin3xdx

= +

7. Find the particular solution of the equation x dye 4dx

= , given that y = 3 , when x = 0


(a) ( )2 2 2 2dyx yx y xy 0dx

− + + = (b)dy xy x y 1dx

= + + +

(c) 2 2sec x tanydx sec ytanxdy 0+ = (d) x y y 2dy e e xdx

− −= +

9. Find the particular solution of the differential equationdy

log 3x 4y,dx

= +



(a) ( )2 2x y dx 2xydy 0+ − = (b)2dy yx y

dx x+ =



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4

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4y 0

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y 0+ =

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+ =4+ =4

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4+ =4y 0+ =y 0

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y 0+ =y 0

3

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d y3d y3

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solutio

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solutio o

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of 6

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f 6d y

f 6d y

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d yf 6

d y+ =

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+ =solutio+ =solutio

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solutio+ =solution+ =n

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n+ =n o+ =o

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o+ =of 6+ =f 6

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f 6+ =f 6

The general solution of the differential equation

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The general solution of the differential equation

.

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.


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(b)

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(b)

Solve the following differential equations :

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( )

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( )1 3

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1 3( )1 3( )

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( )1 3( )ta

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tan x

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n x( )n x( )

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( )n x( )1 3n x1 3

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1 3n x1 3( )1 3( )n x( )1 3( )

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( )1 3( )n x( )1 3( )tan xta

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tan xta −

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−

( )

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( )( )2( )

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( )2( ) dy

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dyx x

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x x

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( )

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( )

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dx

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dx+ =

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+ =( )+ =( )

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( )+ =( )( )2( )+ =( )2( )

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( )2( )+ =( )2( ) dy+ =

dy

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dy+ =

dyx x+ =x x

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x x+ =x x( )x x( )+ =( )x x( )

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( )x x( )+ =( )x x( )x x+ =x x

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x x+ =x x( )2( )x x( )2( )+ =( )2( )x x( )2( )

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( )2( )x x( )2( )+ =( )2( )x x( )2( ) dyx xdy+ =

dyx xdy

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dyx xdy+ =

dyx xdy

Find the particular solution of the equation

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Find the particular solution of the equation


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(a)www.pick

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(a) ( )www.pick

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( )

MATHEMATICS 467

Notes

MODULE - VCalculus


(c)2 2dy x y y

dx x− +

= (d)dy y ysindx x x

= +

11. Solve :dy ysecx tan xdx

+ =


(a) ( )2 1dy1 x y tan xdx

−+ + = (b) 2 dycos x y t a n xdx

+ =

(c)dyxlogx y 2logxdx

+ = , x > 1

13. Solve the following differential equations:

(a) ( ) dyx y 1 1dx

+ + =

[Hint:dx x y 1dy

= + + ordx x y 1dy

− = + which is of the form dx Px Qdy

+ = ]

(b) ( )2 dyx 2y ydx

+ = , y > 0 [Hint: 2dxy x 2ydy

= + or dx x 2ydy y

− = ]

28.7 DIFFERENTIAL EQUATIONS OF HIGHER ORDERTill now we were dealing with the differential equations of first order. In this section, simpledifferential equations of second order and third order will be discussed.

28.7.1 Differential Equations of the Type 2

2d y

f(x)dx

=

Consider the differential equation2

2d y

f(x)dx

=

It may be noted that it is a differential equation of second order. So its general solution willcontain two arbitrary constants.

Now we have,2

2d y

f(x)dx

= ....(1)

ord dy

f(x)dx dx

=

Integrating both sides of (1) , we get

( ) 1dy f x dx Cdx

= +∫ , where 1C is an arbitrary constant

Let ( ) ( )f x dx x= φ∫



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which is of the form

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which is of the form

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dx

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dx P

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Px Q

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x Qdy

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dy+ =

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+ =P+ =P

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P+ =Px Q+ =x Q

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x Q+ =x Q

2

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22y

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2y= +

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= +x= +x

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x= +x or

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or

OF HIGHER ORDER

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OF HIGHER ORDERTill now we were dealing with the differential equations of first order. In this section, simple

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Till now we were dealing with the differential equations of first order. In this section, simple

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differential equations of second order and third order will be discussed.

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differential equations of second order and third order will be discussed.

28.7.1 Differential Equations of the Type

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28.7.1 Differential Equations of the Type

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2

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2

2

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2d y

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d y2d y2

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2d y2f(x)

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f(x)dx

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dx=

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=

f(x)

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f(x)

It may be noted that it is a differential equation of second order. So its general solution will

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It may be noted that it is a differential equation of second order. So its general solution willcontain two arbitrary constants.

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contain two arbitrary constants.

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2 www.pick

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2d y

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d yf(x)www.pi

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m

f(x)=www.pick

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=

MATHEMATICS

Notes

MODULE - VCalculus

468


Then ( ) 1dy x Cdx

= φ + .....(2)

Again on integrating both sides of (2), we get

( ) 1 2y x dx C x C= φ + +∫ ,

where 2C is another arbitrary constant. Therefore in order to find the particular solution weneed two conditions. [See Example 28.19]

Example 28.18 Solve 2

x2

d yxe

dx=


x2

d yxe

dx= .....(1)

Now integrating both sides of (1), we have

x1

dy xe dx Cdx

= +∫ ,

where 1C is an arbitrary constant

or x x1

dy xe e dx Cdx

= − +∫

or x x1

dy xe e Cdx

= − + .....(2)


( )x x1 2y xe e C dx C= − + +∫ ,

where 2C is another arbitrary constant.

or x x x1 2y xe e dx e C x C= − − + +∫

or x x1 2y xe 2e C x C= − + +

which is the required general solution.

Example 28.19 Find the particular solution of the differential equation2

22

d yx sin 3 x

dx= +

for whichdyy(0) 0and 0 when x 0dx

= = =


22

d yx sin 3 x

dx= + .....(1)



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x x

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x xe

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e d

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dx C

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x C− +

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− +x x− +x x

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x x− +x xe− +e

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e− +ex xex x− +x xex x

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x xex x− +x xex xd− +d

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d− +dx C− +x C

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x C− +x C∫

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∫− +∫− +

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− +∫− +

x x

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x xx

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xe e C

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e e Cx xe e Cx x

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x xe e Cx x

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e e C− +e e C

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e e C− +e e Cx xe e Cx x− +x xe e Cx x

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x xe e Cx x− +x xe e Cx x


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(

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(y

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y x

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x=

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= ∫

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∫

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is another arbitrary constant.

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is another arbitrary constant.


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Example 28.19www.pick

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Example 28.19

differential equations - pick my...

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