maths unit i

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Differential Equations

UNIT 4 DIFFERENTIAL EQUATIONS

Structure

4.1 IntroductionObjectives

4.2 Preliminaries4.3 Formation of Differential Equations

4.4 Methods of Solving Differential Equations of the First Order and FirstDegree

4.4.1 Separation of Variables4.4.2 Homogeneous Differential Equations4.4.3 Exact Differential Equations4.4.4 Linear Differential Equations

4.5 Summary

4.6 Answers to SAQs

4.1 INTRODUCTION

Analysis has been dominant branch of mathematics for 300 years, and differentialequations is the heart of analysis. Differential equation is an important part of mathematics for understanding the physical sciences. It is the source of most of the ideasand theories which constitute higher analysis. Many interesting geometrical and physicalproblems are proposed as problems in differential equations, and solutions of theseequations give complete picture of the state of these problems.

Differential equations work as a powerful tool for solving many practical problems of

science as well as wide range of purely mathematical problems. Earlier we defined firstand higher order ordinary and partial derivatives of a given function. In this unit, wemake use of these derivatives and we first introduce some basic definitions. Then wediscuss the formation of differential equations. We also discuss various techniques of solving some important types of differential equations of the first order and first degree.The method of separation of variables together with methods of solving homogeneous,exact and linear differential equations have been taken up in this unit. Differentialequations which are reducible to homogeneous form or equations reducible to linear formare also considered in this unit.

Objectives

After studying this unit, you should be able to

distinguish between ordinary and partial differential equations and betweenthe order and the degree of an equation,

form a differential equation, use the method of separation of variables, identify homogeneous or linear equations and solve them, examine the differential equations for reduction to homogeneous form or

earlier form,

verify whether a given differential equation is exact or not, and solve exact

equations, and identify an integrating factor in some simple cases, which makes the given

equation exact.


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Calculus : Basic Concepts 4.2 PRELIMINARIES

In this section, we shall define and explain the basic concept in differential equations andillustrate them through examples. Recall that given an equation or relation of the type f ( x, y) = 0, involving two variables x and y, where y = y ( x), we call x the independentvariable and y the dependent variable . Any equation which gives the relation between

the independent variable and the derivative of the dependent variable with respect to theindependent variable is a differential equation. In general we have the followingdefinition.

Definition

A differential equation is an equation that involves derivatives of dependent variable with respect to one or more independent variables.

For example,

,0=+ ydxdy

x . . . (4.1)

,nz y z

y x z

x =

+

. . . (4.2)

,02

22

2

2

=+

y

za

x

z. . . (4.3)

are differential equations.

Note that x x x x xdxd

cossin)sin( += is not a differential equation. It is an

identity.

Differential equations are classified into various types. The most obvious

classification of differential equations is based on the nature of the dependentvariable and its derivative (or derivatives) in the equations. The followingdefinitions give the various types of equations.

Definition

A differential equation involving derivatives with respect to a single independent variable is called an ordinary differential equation (abbreviated as ODE) .

For example, Eq. (4.1) namely,

0=+ ydxdy

x

is an ordinary differential equation.

Definition

A differential equation containing partial derivatives, that is, the derivative of adependent variable with respect to two or more independent variables is called a

partial differential equation (abbreviated as PDE) .

For example, Eqs. (4.2) and (4.3), namely

,nz y z

y x z

x =+

,02

22

2

2

=+

y

za

x

z

are partial differential equations.

Differential equations can be further classified by their order and degree.


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Differential EquationsDefinition

The n th derivative of a dependent variable with respect to one or more independent variables is called a derivative of order n, or simply an n th order derivative.

For example, ,,,2

2

2

2

2

y x z

x

z

x

y

are second order derivatives and, y x

z

x

y

2

3

3

3

,

are third order derivatives.

Definition

The order of a differential equation is the order of the highest order derivativeappearing in the equation.

For example,

02

=+

xdxdy

is of order one . . . . (4.4)

(Because the highest order derivative isdxdy

which is of first order.)

22 22

22 +=++ x ydxdy x

dx yd x , is of order two . . . (4.5)

(Highest order derivative is 2

2

dx

yd .)

023

22

22

3

3=

++

dxdy

xdxdy

dx

yd

dx

yd is of order three . . . . (4.6)

Definition

The degree of a differential equation is the highest exponent of the derivative of the

highest order appearing in it after the equation has been expressed in the form free from radicals and fractions as far as derivatives are concerned.

For example, Eqs. (4.1), (4.2), (4.3) and (4.5) are of first degree and Eqs. (4.4) and(4.6) are of second degree.

Equation 2

223

2

1dx

yd k

dxdy =

+ . . . (4.7)

is of second degree. To determine its order, we make the equation free fromradicals. We need to square both the sides so that

2

2

22

32

1

=

+

dx

yd k

dxdy

Since the highest exponent of the highest derivative, that is, 2

2

dx

yd is two and, by

definition, the degree of Eq. (4.7) is two.

And now an exercise for you.

SAQ 1

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

(i) y x

dxdy

432 =


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Calculus : Basic Concepts

(ii)23

1

+=

dxdy

dxdy

(iii) xdxdy

dx

yd += 322

(iv)dxdy

dx

yd =3

3

(v)25

231

2

2

1

+=

dxdy

k dx

yd

In this unit, we will only be concerned with the study of certain types of ordinarydifferential equations which we shall simply refer to as differential equations, droppingthe word ordinary.

The principal task of the theory of differential equations is to find all the solutions of agiven differential equation. But then it is natural to ask as to what exactly is the meaningof a solution of a differential equation. The answer to this question is given in thefollowing definition.

Definition

A solution or an integral of a differential equation is a relation between thevariables, not involving the derivatives, such that relation and the derivativesobtained from it satisfy the given differential equation.

To illustrate this let us do the following examples.

Example 4.1

Prove that y = cx2 . . . (4.8)

is a solution of xy = 2 y. . . . (4.9)

Solution

Step 1

Differentiating both sides of Eq. (4.8) with respect to x, we get

y = 2 cx . . . (4.10)

where y stands fordxdy

.

Step 2

Substituting in Eq. (4.9) the values of y and y obtained from Eqs. (4.8) and(4.10), respectively, we get an identity

x . 2cx = 2 cx2


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Differential EquationsIt should also be noted that a differential equation may have many solutions.

For instance, each of the functions54

sin,3sin,sin =+== x y x y x y is

a solution of the differential equation x y cos= . But we also know fromour knowledge of calculus that any solution of the equation is of the form

y = sin x + c, . . . (4.11)

where c is a constant. If we regard c as arbitrary, then Eq. (4.11) representsthe totality of all solutions of the equation.

If you have understood the above example, then you may now try thefollowing exercise.

SAQ 2

Verify that each function is a solution of the differential equation written next to it.

(i) 12;2 +=++= x yc x x y

(ii) y x y x xc x y +=+= 22 ;

(iii) 025;5cos5sin =++= y y x B x A y

(iv) x x e y yec x y =++= ;)(

(v) . y y y y yecec x x In;Iny 2221 =+=

As illustrated above, a differential equation may have more than one solution. It mayeven have infinitely many solutions, which can be represented by a single formulainvolving arbitrary constants. Accordingly, we classify various types of solutions of adifferential equation as follows.

Definition

The solution of the order differential equation which contains n arbitraryconstants is called its general solution .

n th

Definition Any solution which is obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution .

For example, involving two arbitrary constants c xc xc y 2cos2sin 21 += 1 and c2

is the general solution of second order equation 042

2=+ y

dx

yd whereas

is a particular solution (taking x x y 2cos2sin2 += 1and2 21 == cc ).

In some cases, there may be solutions of a given equation which cannot be obtained byassigning a definite value to the arbitrary constant in the general solution. Such a solution

is called a singular solution of the equation. For example, the equation. . . (4.12)02 =+ y y x y


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Calculus : Basic Conceptshas the general solution, A further solution of Eq. (4.12) is2ccx y =

4

2 x y = . Since

the solution cannot be obtained by assigning a definite value to c in the general solution,it is a singular solution of Eq. (4.12).

After going through the definitions and illustrations given in Section 4.2, you would havedefinitely got the clue that a differential equation can also be derived from its solution bythe process of differential, algebraic processes of elimination, etc. Now we shall discussthe method of finding the differential equation when its general solution is given.

4.3 FORMATION OF DIFFERENTIAL EQUATIONS

Consider the relation

. . . (4.13)3bxax y +=

Differentiating Eq. (4.13) with respect to x, we get

. . . (4.14)23bxa y +=

Again differentiating Eq. (4.14) with respect to x, we obtain

bx y 6= . . . (4.15)

Solving Eqs. (4.14) and (4.15) for a and b, we get

y x y x y

x ya x y

b =

=

=

21

61

3,61 2 .

Substituting these values of a and b in Eq. (4.13), we get

y x y x y x y += 2261

21

or y x y x y = 231

or 0332 = y y x y x

It is a differential equation, which has been obtained from Eq. (4.13). Generalising theabove procedure we obtain the following rule to find the differential equation when itsgeneral solution is given.

Rule

(a) Differentiate the general solution. Let us refer to the resulting equation as

the derived equation.(b) Differentiate the derived equation and obtain the second derived equation.

(c) Differentiate the second derived equation and continue the process until thenumber of derived equations is equal to the number of independent arbitraryconstants involved in the general solutions.

(d) Eliminate the arbitrary constants and obtain relation between derivatives anddependent and independent variables.

Example 4.2

By eliminating the constants h and k , find the differential equation of which

222)()( ak yh x =+ . . . (4.16)

is a solution.

Solution


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Differential EquationsStep 1

Differentiating Eq. (4.16), we get the first derived equation as

0)()( =+dxdy

k yh x . . . (4.17)

Step 2

Differentiating Eq. (4.17), we get the second derived equation as

0)(12

2

2=

++

dxdy

dx

yd k y . . . (4.18)

Step 3

Finally, we eliminate h and k from Eq. (4.16), Eqs. (4.17) and (4.18) to yield

2

2

2

1

)(

dx

yd

dxdy

k y

+

=

and

2

2

2

1

)(

dx

yd

dxdy

dxdy

h x

+

=

Substituting these values in Eq. (4.16), we obtain

,1

2

2

22

32

=

+ dx yd a

dxdy

as the required differential equation.

How about doing some exercises now?

SAQ 3

(a) Find the differential equations having the following as solutions.

(i) c x y += 3

(ii) x x ecec y +=21

(iii) 322

1 c xc xc y ++=

(b) Find the differential equations whose solution is given by

)sincos( x B x Ae y x +=

where A and B are arbitrary constants.


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Any differential equation of the first order and first degree may be written in theform.

Calculus : Basic Concepts

),( y x f dxdy =

In the next section, we shall discuss various methods of solving first order and first

degree differential equations.

4.4 METHODS OF SOLVING DIFFERENTIAL EQUATIONSOF THE FIRST ORDER AND FIRST DEGREE

In general, it may not be very easy to solve even the apparently simple equation

),( y x f dxdy = . This is because no formulas exist for obtaining its solution in all cases.

But, there are certain standard types of first order equations for which routine methods of solution are available. We now discuss four of them that have many applications. Let usdiscuss them one by one.

4.4.1 Separation of Variables

This is a technique for solving an important class of differential equations, namely, thoseof the form

,)()(

yq x p

dxdy = . . . (4.19)

where p ( x) is a function of x only and q ( y) is a function of y only. Such an equation iscalled an equation with separable variables, or a separable equation . For example, the

differential equation ye

xdxdy )12( += is of the same type as Eq. (4.19) with

ye yq x x p =+= )(and12)( .

In order to solve the differential equation given by Eq. (4.19), we may proceed asfollows :

Step 1

Rewrite Eq. (4.19) as

dx x pdy yq )()( = . . . (4.20)

Step 2

Integrate both sides of Eq. (4.20) and getcdx x pdy yq += )()(

or c xP yQ += )()( . . . (4.21)

where dx x p xPdy yq yQ )()(and)()( ==Note that in Step 2, there is no need to write two constants since these can becombined into one as in Eq. (4.21).

We now take up an example to illustrate these steps.

Example 4.3

Solve 2

23

y

xdxdy =


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Differential EquationsSolution

Step 1

The given differential equation is

2

23

y

xdxdy = . . . (4.22)

Multiplying both sides of Eq. (4.22) by y2, we get

22 3 xdxdy

y =

or, . . . (4.23)dx xdy y 22 3=

Step 2

Integrating both sides of Eq. (4.23), we get

cdx xdy y += 22 3

c x y += 333

, where c is a constant

or, . . . (4.24),3 133 c x y +=

where is a constant.cc 31 =

Step 3

Solving Eq. (4.24) for y and we get

31

13 ]3[ c x y += .

It is the required solution of the differential Eq. (4.22).

Let us look at another example.

Example 4.4

Solve 223 y yt dt dy += . . . (4.25)

Solution

Step 1

Note that right hand side of Eq. (4.25) is not of the form . )()(

yqt p

However, if we write Eq. (4.25) as

,)1(

)1( 2

323

+=+=

y

t yt

dt dy

. . . (4.26)

we see that the right hand side of Eq. (4.26) is of the form

23 )(and1)(with)()( =+= y yqt t p

t q

t p

Separating the variables in Eq. (4.26), we get

dt t dy y

)1(1 3

2+= . . . (4.27)

Step 2


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Integrating both sides of Eq. (4.27), we getCalculus : Basic Concepts

cdt t dy y

++= )1(1 32

or, ,411 4 ct t

y++= . . . (4.28)

where c is the constant of integration.

Step 3

Solving Eq. (4.28) for y, we get

,4

4

14 ct t

y++

= with c1 = 4c, as the required solution of Eq. (4.25).

You may now try this exercise yourself.

SAQ 4

Solve the following differential equations :

(i) 25

y

t dt dy +=

(ii) 222 . ye ydt dy t =

(iii) )3(4 = ydt dy

(iv)

t

y

dt

dy =

Not all differential equations can be solved by the method of separation of variables. Forexample, we cannot solve,

122 += yt dt dy

by the method of separation of variables because the expression cannot be

written in the form

122 + yt

.)()(

yqt p

In such cases we have to look for other techniques of solving

differential equations. In the next subsections, we shall take up another technique of solving first order equations.

4.4.2 Homogeneous Differential Equations

You are already familiar with the definition of a homogeneous function. Recall that afunction is said to be homogeneous of degree n, if ),( y x f

. . . (4.29),),(),( y x f y x f n

=for all values of x and y


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We also say that any function of the form x y

is homogeneous of degree zero since

.0

=

x y

x y

However, the function is not homogeneous for

for any n.

234),( y x x y xg +=),()()()(),( 234 y x f x x x y xg n+=

From Eq. (4.29), a useful relation is obtained by letting .1 x

=

This gives for a homogeneous expression of the degreenth

=

= x y

x y

f y x f xn

,1),(1

(say)

or

=

=

x y

x x y

f x y x f nn ,1),( . . . (4.30)

A homogeneous equations of the first order is of the form

0),(),(or),(),( == dy y xgdx y x f

y xg y x f

dxdy

,

where f ( x, y) and g ( x, y) are homogeneous functions of the same degree.

For example, xy

y xdxdy

2

22 = is a homogeneous differential equation. Here

are both homogeneous functions of degree two.

You must have noticed above that

xy y xg y x y x f 2),(and),( 22 ==

x

y

plays an important role in a homogeneous function.Thus we would expect that the substitution

xv yv x y == or . . . (4.31)

might be effective in solving a homogeneous equation.

In fact, it will be seen that the substitution (4.31) in a homogeneous equation of the firstorder and first degree reduces the equation to an equation with separable variables. Letus, now, take up an example.

Example 4.5

Solve the differential equation. . . (4.32)0)2( =+ dy ydx y x

Solution

Step 1

Clearly the given equation can be expressed as, y

x ydxdy = 2 , which is a

homogeneous differential equation. Putting xv y = in Eq. (4.32), we get

)since(0)()2( xdvvdxdy xdvvdxvxdxvx x +==++

0)21( 22 =++ vdv xdxvv x

0)1( 2

=

+ dvv

v xdx


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Step 2 Calculus : Basic Concepts

In Eq. (4.33) the variables are now separated and, therefore, integrating, we

get 112 ,)1(

1ccdv

v

vdx

x=

+ being a constant.

,)1(

||ln12

cdvv

v x =

+

. . . (4.33)

For finding the integral ,)1( 2

dvv

v

we substitute v 1 = t so thatdv = dt and Eq. (4.33) becomes

121

||ln cdt t

t x =++

1211

||ln cdt t

dt t

x =++

1|)1(|ln1

||ln ct t x =+

1|)1(|ln11

||ln cvv

x =+

. . . (4.34)

(since ( v 1) = t ).

It is convenient to replace the constant c1 by Inc, where c > 0. ThenEq. (4.34) becomes,

11)1(

ln =

vcv x

. . . (4.35)

Step 3

Replacing v by x y

in Eq. (4.35), we get

x y x

c x y

=ln

which is the required solution of Eq. (4.32).

Let us consider another example.

Example 4.6

Solve 222

y x

xydxdy

= . . . (4.36)

Solution

Step 1

The given Eq. (4.36) is homogeneous, since both numerator xy y x f 2),( = and the denominator are homogeneous functions of degree 2.

22),( y x y xg =

We may rewrite Eq. (4.36) as

2

1

2

=

x y

x y

dxdy . . . (4.37)


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or, 21

2

v

vdxdv

xv

=+ , where , x y

v =

or, 2

2

2 1

)1(

1

2

v

vvv

v

vdxdv

x+=

=

Therefore,

c xdx

dvvv

v +=+

)1()1(2

2

. . . (4.38)

where c is a constant of integration. To integrate Eq. (4.38), we write it as

c xdx

dvv

vv

+=

+ 21

21

|,|ln||ln|)1(|ln||ln 2 A xvv +=+

where A is another constant, such that c = ln | A |

Thus, we get | Axv

v|ln

1ln

2=

+

Axv

v =+ 21

Step 3

Replacing v by x y

in the last equation, we obtain

y A y x Ax y x

xy 1or

2222 =+=+

as the required solution of Eq. (4.36).

You may now try the following exercises.

SAQ 5

(a) Solve the following differential equations :

(i) xy

y xdxdy 22 +=

(ii) y x y x

dxdy x += tan

(iii) dx y xdx ydy x 22 +=

(iv) 0sinsin =

+ dy x y

xdx x y

y x

(b) Solve the following equations :

(i) dx y xdy y x x )()( 222 +=+

(ii) 0)2(32232

=++ dx y xy y x xdy y x(iii) 0)2( =+ dx ydy x xy

(iv) 0)4()26( 222 =++ dy xy xdx y x


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Equations Reducible to Homogeneous Form Calculus : Basic Concepts

Sometimes it may happen that a given equation is not homogeneous but can bereduced to homogeneous form by considering certain change of variables. Forexample, consider the equation of the form

D By Ax

cbyax

dx

dy

++

++= . . . (4.39)

This can be reduced to homogeneous form by changing the variables x, y to X , Y respectively, where

x = X + h,

y = Y + k,

with h and k as constants to be so chosen as to make the given equationhomogeneous. With these new variables we have

)1because(. ===dxdX

dX dY

dxdX

dX dY

dxdy

Therefore, Eq. (4.39) becomes

)()(

D Bk Ah BY AX cbk ahbY aX

dX dY

++++++++= . . . (4.40)

which will be homogeneous provided h and k are so chosen so that

ah bk c+ + =0 Ah Bk D+ + =0

Solving the last two equations for h and k , we get

AbaBaD Ac

k AbaB BcaD

h

=

= and

which always have meanings except when

aB Ab = 0 that is when Bb

Aa = .

So, with these values of h and k , the given equation can be reduced to ahomogeneous equation and the resulting equation is

BY AX bY aX

dX dY

++=

It can now easily be solved by means of the substitution Y = vX .

But what happens to the equation if Bb

Aa = ?

In such cases we let r Bb

Aa == say.

Then a = Ar and b = Br and the given equation becomes

D By AxC By Axr

dxdy

++++= )( . . . (4.41)

We, then, put Z By Ax =+ and its differential with respect to x, that is,

dxdZ

dxdy

B A =+ in Eq. (4.41) to get


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A D Z C rZ

BdxdZ

D Z crZ

AdxdZ

B+

++=

++=

or1

so that the variables are separated and hence the equation can be solved. Let usillustrate the above discussion with the help of the following examples.

Example 4.7

Solve the differential equation

dx y xdy y x )726()632( =+ . . . (4.42)Solution

Step 0

Here33

26

Step 1

Putting x = X + h, y = Y + k , in the given equation,632726

+=

y x y x

dxdy

, we

get

Y X Y X

k Y h X k Y h X

dX dY

3226

6)(3)(27)(2)(6

+=

+++++= . . . (4.43)

provided that

. . . (4.44)0726 = k hand . . . (4.45)0632 =+ k h

Step 2

Solving Eqs. (4.44) and (4.45) for h and k , we get

.1and23 == k h

Step 3

Eq. (4.43) is in homogeneous form; thus putting Y = vX in Eq. (4.43), we get

dvvv

v X dX

vX X vX X

dX dv

X v)643(2

46or

3226

2 ++=

+=+

Since the variables have been separated on integration, we get

|)643(|ln|ln2||ln 2 += vv| X c

where c is a constant of integration.Note that the integral on the right hand side is of the form

dx x f x f

)()(

which has led us to ||ln|)643(|ln||ln2 2 cvv X =++

. . . (4.46)cvv X =+ )643( 22

Substituting back the value of

= X Y

v in Eq. (4.46), we get

c X

Y

X

Y X =

+= 643 2

22


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Calculus : Basic Concepts or . . . (4.47)c X XY Y =+ 22 643

Step 4

But2

32and1

=== x X yk yY .

Thus, replacing these values of X and Y in Eq. (4.47), we get

c x x

y y =

+

22

232

62

32)1(4)1(3

or29

1812643 22 +=++ c x y x xy y

is the required solution.

Example 4.8

Solve 0)124()12( =++++ dy y xdx y x

Solution

Step 0

Here21

42 = . Hence exceptional case.

Step 1

Rewrite the given equation as

1)2(212+

++= y x

y xdxdy

. . . (4.48)

Step 2

Puttingdxdt

dxdy

t y x =+=+ 2thatso2 , we get

121

2+==

t t

dxdt

dxdy

or,12

)1(3=

t t

dxdt

or, dxdt t

t =

)1(3

12. . . (4.49)

Step 3

Since Eq. (4.49) is in variable separable form, we integrate both sides andget

dt t

dt t

t dt

t t

dxc1

131

132

)1(312

=

=+

dt t

dt t 1

131

11

132

+=

313

132 1c

t dt dt x +

+= , where 31cc = .

1|)1(|ln23 ct t x ++= . . . (4.50)


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Differential EquationsStep 4

Replacing back the value of t in Eq. (4.50), we get

c y x y x x ++++= |)12(|ln)2(23

or 0|)12(|ln2 =++++ c y x y x

as the required solution.

You may now try this exercise.

SAQ 6

Solve the following differential equations:

(i) 0)22()42( =+++ dy y xdx y x

(ii) 0)1(4)132( =++ dy xdx y x

(iii) 0)2()32( =+++ dy ydx y x

(iv) 0)164(3)232( =++ dy y xdx y x

You have studied the total differential of a given function. Now, in the next sub-section,we shall make use of this to define an exact differential equation. We shall also discussthe method of solving it.

4.4.3 Exact Differential Equations

Suppose that we are given a function g ( x, y) = c. Then its total differential is given by

0=+

= dy

yg

dx xg

dg .

For instance, the equation xy = c has the total differential

ydx + xdy = 0,

Now, we consider the reverse situation, that is, given the differential equation

M ( x, y) dx + N ( x, y) dy = 0 . . . (4.51)

can we find a function g ( x, y) = c, such that

dg = Mdx + Ndy ,

where N yg

M xg =

=

and ?

If so, we say that Eq. (4.51) is an exact differential equation.

It can be shown that the necessary and sufficient condition for the differential equation

Mdx + Ndy = 0 to be exact is x N

y M

=

.

We will not be proving this result here. However, we shall be making use of this to verify

whether a given differential equation of type (4.51) is exact or not.Various steps involved in solving an exact differential equation Mdx + Ndy = 0 are asfollows.

Step 1


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Integrate M with respect to x, regarding y as a constant.Calculus : Basic Concepts

Step 2

Integrate with respect to y those terms in N which do not involve x.

Step 3

The sum of the two expressions obtained in Steps 1 and 2 equated to a constant isthe required solution.

Let us illustrate this method with the help of an example.

Example 4.9

Solve the equation

0)sec(cos)tansin1( 2 =+ dy x xdx y x . . . (4.52)

Solution

The given equation is of the form

Mdx + Ndy = 0 with M = 1 sin x tan y, N = cos x sec2

y.

Now, y x y

M 2secsin=

,

and y x x N 2secsin=

,

so that x N

y M

=

,

which shows that Eq. (4.52) is an exact equation.

To solve Eq. (4.52), we proceed as follows :Step 1

Integrating M w.r.t. x regarding y as a constant, we have

x y xdx y x costan)tansin1( += Step 2

We integrate those terms in N w.r.t. y which do not involve x. But there is

no such term because . y x N 2seccos=

Step 3

The required solution is the sum of expressions obtained from steps 1 and 2.That is,

c x y x =+ costan .

In order to check your calculations you can find the total differential of theequation obtained as a result of Step 3, and get back the original differentialequation.

In practice differential equations are rarely exact but can often easily betransformed into exact equations on multiplications by some suitable functionknown as integrating factor . In other words, the integrating factor is a functionwhich when multiplied with a non-exact differential equation makes it exact. Ingeneral such a function exists but is difficult to obtain, except for certain cases.

Consider for example, the equation


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0= ydx xdy Differential Equations

which is not exact, since x N

y M

. Here M = y, N = x.

1)(and)( ==

=

x x x

N y

y y M

.

x N

y M

, thus given equation is not exact.

On multiplication by 21

ythroughout the given equation, then, we get

01

2= x

ydy

y

x,

which is exact, as can be readily verified by using the condition

x

N

y

M

=

Let us consider another illustration.

Example 4.10

Show that the differential equation 011 =

+ dx y x

xdy

xbecomes exact on

multiplication by xy, and solve it.

Solution

The differential equation

011 =

+ dx y x

ydy

y. . . (4.53)

is not exact for x N

y M

Multiplying Eq. (4.53) by xy, we get

. . . (4.54)0)( 2 =+ dx x y xdy

This is now exact because

)(1)(2

x x x N

x y y y M

=

==

=

.

It can, then, be verified easily that the solution of Eq. (4.54) is given by

constant3

2=

x xy .

Remark

In some cases, integrating factor can be found by inspection as illustrated above bytwo simple examples. Obviously guessing comes from practice only. However,rules for finding the integrating factor do exist for some other cases. But we do not

intent to consider these rules for determining the integrating factor in this course.Here we will restrict our discussion only to some simple type of equations forwhich we can find the integrating factor by inspection.

You may now try the following exercises.


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SAQ 7Calculus : Basic Concepts

(a) Show that becomes exact on

multiplication by

0)1()2( 3 = dy xy xdx x y

21

xand solve it.

(b) Solve the following differential equation :

(i) 0)1()1( =++ dx y yxdy x yx

(ii) 0)1(2 2 =++ dy xdx xy

(iii) 0ln

ln32)(lnln 223 =

++

+ dy y x y y

xdx xy

x y

In the next sub-section we introduce you to the method of solving the most importanttype of differential equations namely, the linear differential equations. These equationsare important because of their wide range of applications.

4.4.4 Linear Differential Equations

We begin with the definition of linear differential equation.

Definition Linear differential equation is a differential equation which contains thedependent variables and derivatives in first degree only.

For example, the equation 22

x x y

dxdy =+ is a linear differential equation; however

52 =+ xdxdy

y is not linear because of the presence of the termdxdy

y , the product

of depend variables and its derivative.

The general form of the first order linear differential equation is

)()( xq y x pdxdy =+ . . . (4.55)

where p ( x) and q ( x) are functions of the independent variable x alone, or areconstants. From our discussion in Sub-section 4.4.3 you will at once see that theEq. (4.55) can be solved by the use of an integrating factor. If, on multiplicationwith a suitable integrating factor, the given equation can be expressed as an exactderivative, then the resulting equation can be integrated directly, In general, tosolve such differential equation we proceed as follows.

Multiply both sides of this equation by , (an integrating factor), we get pdx

e

pdx pdx

eq pydxdy

e =

+ . . . (4.56)


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The left hand side, now, is the differential coefficient of , that is,Eq. (4.56) is nothing but

pdxe y

pdx pdx

eqe ydxd =][ . . . (4.57)

Integrating both sides of Eq. (4.57) with respect to x, we get

cdxeqe y pdx pdx += where c is a constant of integration.

They may, again, be written after multiplying both sides by as pdx

e

[ ],cdxeqe y pdx pdx += . . . . (4.58)which is the required solution of Eq. (4.55).

The factor , on multiplying, by which Eq. (4.55) becomes an exactdifferential, serves the purpose of integrating factor of the differential Eq. (4.55).

pdxe

Let us now solve a few examples.

Example 4.11

Solve the differential equation

1sincos =+ x ydxdy

x . . . (4.59)

Solution

Step 1

Dividing both sides of Eq. (4.59) by cos x, we get

x x x

ydxdy

cos1

cossin

=+

or x x ydxdy

sectan =+ . . . (4.60)

which is of the form

)()( xq y x pdxdy =+ ,

with x xq x x p sec)(andtan)( == .Step 2

Now, I. F. = xeee xdx x pdx secseclntan === Step 3

Multiply both sides of Eq. (4.60) by sec x, we obtain

x x x ydxdy

x 2sectansecsec =+

or x x ydxd 2sec)sec( = . . . (4.61)

Step 4

Integrating both sides of Eq. (4.61), we getcdx x x y += 2secsec

where c is an arbitrary constant.


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Thus, c x x y += tansec Calculus : Basic Concepts

or, xc xc x x

x y cossin

sec1

sectan +=+=

is the required solution of Eq. (4.59).

Let us consider another example.Example 4.12

Solve the equation ydxdy

y x =+ )2( 3 . . . (4.62)

Solution

Step 1

The equation involves y3 and these hence it is not linear, but if we view y asthe independent variable, and rewrite Eq. (4.62) as

32 y xdydx y +=

or 22 y y x

dydx = . . . (4.63)

then Eq. (4.63) is linear with x as the dependent variable and is of the form

22)(and1

)(with)()( y yq y

y p yq x y pdydx ===+

Step 2

The procedure for solving Eq. (4.63) is exactly same as before. Integrating

factor of Eq. (4.63) isdy y p

e)(

y yee

ydy y 11ln1

====

Step 3

Multiplying both sides of Eq. (4.63) by y1

, we get

y y x

dydx

y2

1 =

or y y

xdyd

21

. =

. . . (4.64)

Step 4

Now integrating both sides of Eq. (4.64), we get

c y

y

x += 21 , where c is a constant.

Thus, is the required solution of Eq. (4.62).)( 2 yc y x +=

You may now try this exercise.


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185

Differential EquationsSAQ 8


(i) 32 =+ y xdx

dy

(ii) xe y

x

x

dx

dy =

+1

22

(iii) xe x x ydxdy x costan 2=+

(iv) x ydxdy

x x ln2ln =+

(v) 0)102( 3 =+ ydxdy

y x

Equations Reducible to Linear Form

Sometimes, equations which are not linear can be reduced to the linear form bysuitable transformations of the variables. One such equation is

n y xQ y xPdxdy

)()( =+ . . . (4.65)

It is called Bernoullis equations after James Bernoulli, who studied it in 1695.Clearly, Eq. (4.65) is not linear as it contains a power of y (dependent variables)which is not unity. However, this equation can be reduced to the linear form asfollows :

Dividing Eq. (4.65) by yn, we get

Q yPdxdy

y nn =+ + 1 . . . (4.66)

Put , z y n =+ 1

so thatdxdz

dxdy

yn n =+ )1(

or dxdz

ndxdy

yn

11+=

Substitute in Eq. (4.66) the values of 1and + nn ydxdy

y in terms of z, we get

)1()1( nQ znPdxdz =+ ,

which is linear with z as the new dependent variable.

We now illustrate this method with the help of an example.

Example 4.13

Solve the equation

dx y xedx xydy x 32

2 =+ . . . (4.67)

Solution


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Step 1Calculus : Basic Concepts

Eq. (4.67) can be written as

322 ye x xydxdy x=+ . . . (4.68)

which is of the form

n y xQ y xPdxdy

)()( =+

with 3and)(,2)(2

=== ne x xQ x xP x

Step 2

Divide Eq. (4.68) throughout by y3, we get

223 2 xe x xydxdy

y =+ . . . (4.69)

Let 2= y z

Then,dxdy

ydxdz 12)2( = . . . (4.70)

dxdy

y 32 =

Therefore,dxdz

dxdy

y213 = . . . (4.71)

Substituting in Eq. (4.69) values of and2 ydxdy

y 3 from Eqs. (4.70) and

(4.71) respectively, we have

2

221 xe x xz

dxdz =+

or2

24 xe x xzdxdz = . . . (4.72)

which is linear with z as the dependent variable.

Step 3

To solve Eq. (4.72), we multiply both sides of Eq. (4.72) by

(integrating Factor) to get224 xdx x ee

=

2222 222 24 x x x x e xe xze

dxdz

e =

232 x xe =

or22 32 2)( x x xe ze

dxd =

Integrating both sides, we get

. . . (4.73)cdx xe ze x x += 22 32 2For finding 23 3let,2 2 xt dx xe x = so that

32

dt dx x =


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Therefore, in terms of t , becomesdx xe x232

23

31

31

3 xt t ee

dt e ==

Thus, from Eq. (4.73), we get

ce ze x x

+= 22 32

31

or222 232

31 x x x ceee z +=

22 2

31 x x cee +=

Step 4

But 2= y z

Therefore,22 22

3

1 x x cee y +=

or, where c22 2

123 x x ece y += 1 = 3c

is the required solution of Eq. (4.67).

SAQ 9


(i) dxe xydx ydy x22=+

(ii) dy y xdy x

y

dx 2222 =+

(iii) 3352 y x x y

dxdy =

(iv) 0cos423 =+ x y y xdxdy

x

(v) 0)( 2 =+ dyedxe y x y x x

We now conclude this unit by giving summary of what we have covered in it.

4.5 SUMMARY

In this unit, we have learnt that

(i) (a) An equation involving derivatives of dependent variable w.r.t. to one ormore independent variables is called a differential equation .

(b) The order a differential equation is the order of highest derivative appearingin it.

(c) The degree of a differential equation is the highest exponent of the highestorder derivative in it, after removing radicals/fractions.


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Calculus : Basic Concepts(ii) A differential equation of the form

)()(

yq x p

dxdy = is called a separable equation and

solution is obtained by direct integration.

(iii) The equation),(),(

y xg y x f

dxdy = , where f ( x, y) and g ( x, y) are homogeneous functions

of the same degree is called homogeneous equation and to solve it, we put y = vx and the equation becomes separable equation.

(iv) The differential equation M dx + N dy = 0 is said to be exact if x N

y M

=

and its

solution is = constant.)of tindependenterms()constantastreating( x y

dy N dx M +

(v) (a) Equation of the form )()( xq y x pdxdy =+ is called a linear differential

equation and is its integrating factor.

pdx

e

(b) Equations of the type n y xq y x pdxdy

)()( =+ are reduced to linear form by

the substitution . z y n =1

4.6 ANSWERS TO SAQs

SAQ 1

(i) 1, 2

(ii) 1, 6

Hint : First square both the sides, the highest order derivative isdxdy

, and

its power is 6.

(iii) 2, 1

(iv) 3, 2

(v) 2, 2

SAQ 3

(a) (i) dxdy y x y == ,32

(ii) 2

2

,0dx

yd y y y ==

(iii) 3

3

,0dx

yd y y ==

(b) Here ]sin)(cos)[( x B A x B Ae y x +=

y x A x Be x += ]sincos[

]sincos[2 x A x Be y x =

][2 y y =


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Differential EquationsHence, 022 =+ y y y

SAQ 4

(i)ct t

y++=

2315

32

(ii) 0or2

22 =+= yct e y t

(iii) t ce y 43 +=

(iv) 0or)( 2 =+= yct y

Hint : ct ydt t dy y +== 2 / 12 / 12 / 12 / 1 thus,

SAQ 5

(a) (i) c x x y += ||ln2

(ii) xc x y = =

sin

(iii) 222 cx y x y =+=

(iv) c x y

x =

+ cos||ln

(b) (i) y/xe xc x y = 2)(

(ii) x y

x

x y x

x yc+

=+ 2

)(

)(ln 4

(iii) c y x

y =+ln Hint : Put y = v x

therefore, 0)()1( =++ dxvx xdvvdxv x x

or, 0)12(2 2 / 3 =+ dvv xdxv

or, 0)12(2

2 / 3=+ dv

v

vdx

x

022 2 / 3

=

+

dvvvdx x

12 / 12ln2ln2 cvv x =++

or, cv xv =+ 2 / 1ln

Putting x y

v =

c y x

y ++ln

(iv) cx x y y x =+ )32()2(

SAQ 6

(i) )2()2( 3 +=+ y xc y x


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Calculus : Basic Concepts (ii) 34 )1()32( += xc x y

(iii) )1()42( 2 +=+ y xc x y

(iv) c y x y x =++ )32(ln6

SAQ 7

(a) 32 22 x ycx xy ++=

(b) (i) c xy

y = 1ln

Hint : Use the substitution xy = t then variables are separated.

(ii) 2)1( x

c y

+=

(iii) c y x y x =+ 32

3

1)(lnln)(ln

Since 0ln

ln32)(lnln 223 =

++

+ dy y x y y

xdx xy

x y

032

lnln)(lnln 223 =

++

+ dy y xdx xydy y y

xdx

x y

0)(31

)(lnln[ln 32 =+ y x y xd

or c y x y x =+ 3231)(lnln)ln(

SAQ 8

(i) 2 x

c x y +=

(ii)11

12

=+=

x

ce

x x

y x

(iii) xc x x xe y x coscos)22( 2 ++=

(iv) 2)(lnln xc x y +=

(v) c y xy += 52 2

Hint : Given equation can be written as 2102

y x ydy

dx =+ , which is linear

with x as dependent variable.

SAQ 9

(i) ( Hint : Use the substitution)(1 2 xc ye x = z y

=1 )

(ii) ( Hint : Put31 2 yc y x += z y

=1 )


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Differential Equations(iii) c x xy =+ 52

(iv) 33 )sin3( y xc x +=

(v) cye y x x =+ 33


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