Differential Equations

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Mathematics / Differential equations

DifferentialEquations

01. Motivation

02. Solving Differential Equations

CONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOTESTESTESTESTES



Differential Equations

A differential equation can simply be said to be an equation involving derivatives of an unknown function. Forexample, consider the equation

2dy xy xdx

+ =

This is a differential equation since it involves the derivative of the function ( )y x which we may wish to determine.We must first understand why and how differential equations arise and why we need them at all. In general, we cansay that a differential equation describes the behaviour of some continuously varying quantity.

Scenario - 1 : A freely falling body

A body is released at rest from a height h. How do we described the motion of thisbody ?

The height x of the body is a function of time. Since the acceleration of the body is g, wehave

2

2

d x gdt

=

This is the differential equation describing the motion of the body. Along with the initialcondition (0) ,x h= it completely describes the motion of the body at all instants afterthe body starts falling.

Scenario - 2 : Radioactive disintegration

Experimental evidence shows that the rate of decay of any radioactive substance isproportional to the amount of the substance present, i.e.,

dm mdt

=

where m is the mass of the radioactive substance and is a function of t. If we knowm (0), the initial mass, we can use this differential equation to determine the mass of thesubstance remaining at any later time instant.

Section - 1 MOTIVATION



Scenario - 3 Population growth

The growth of population ( of say, a biological culture) in a closed environment isdependent on the birth and death rates. The birth rate will contribute to increasing thepopulation while the death rate will contribute to its decrease. It has been found that forlow populations, the birth rate is the dominant influence in population growth and thegrowth rate is linearly dependent on the current population. For high populations, thereis a competition among the population for the limited resources available, and thus thedeath rate becomes dominant. Also, the death rate shows a quadratic dependence onthe current population.

Thus, if ( )N t represents the population at time t, the differential equation describing thepopulation variation is of the form

21 2

dN N Ndt

=

where 1 and 2 are constants.

Along with the initial population N (0), this equation can tell us the population at anylater time instant.

____________________________________________________________________________________

These three examples should be sufficient for you to realise why and how differential equations arise and why theyare important.

In all the three equations mentioned above, there is only independent variable (the time t in all the three cases).Such equations are termed ordinary differential equations. We might have equations involving more than oneindependent variable:

2f fx xx y + =

where the notation x

stands for the partial derivative, i.e., the term fx

would imply that we differentiate the

function f with respect to the independent variable x as the variable (while treating the other independent variable

y as a constant). A similar interpretation can be attached to y

.

Such equations are termed partial differential equations but well not be concerned with them in this chapter.

Consider the ordinary differential equation2

22

d y dyx x cdx dx

+ + =

The order of the highest derivative present in this equation is two; thus, well call it a second order differentialequation (DE, for convenience).

The order of a DE is the order of the highest derivative that occurs in the equation



Again, consider the DE

232 2

3

d y dy x ydx dx

+ =

The degree of the highest order derivative in this DE is two, so this is a DE of degree two (and order three).

The degree of a DE is the degree of the highest order derivative that occurs in the equation, when all the derivatives in the equation are made of free of fractional powers.

For example, the DE

22 2

21dy d yx kdx dx

+ =

is not of degree two. When we make this equation free of fractional powers, by the following rearrangement,222 2

21dy d yk xdx dx

=

we see that the degree of the highest order derivative will become four. Thus, this is a DE of degree four(and order two).

Finally, an nth linear DE (degree one) is an equation of the form

1

0 1 11 ...n n

n nn nd y d y dya a a a y bdx dx dx

+ + + + =

where the ia s and b are functions of x.

Solving an nth order DE to evaluate the unknown function will essentially consist of doing n integrations on the DE.Each integration step will introduce an arbitrary constant. Thus, you can expect in general that the solution ofan nth order DE will contain n independent arbitrary constants.

By n independent constants, we mean to say that the most general solution of the DE cannot be expressed in fewerthan n constants. As an example, the second order DE

2

2 0d y ydx

+ =

has its most general solution of the form

cos sin .y A x B x= + ...(1)

(verify that this is a solution by explicit substitution).



Thus, two arbitrary and independent constants must be included in the general solution. We cannot reduce(1) to arelation containing only one arbitrary constant. On the other hand, it can be verified that the function

x by ae +=

is a solution to the second-order DE

2

2

d y ydx

=

but even through it (seems to)contain two arbitrary constants, it is not the general solution to this DE. This isbecause it can be reduced to a relation involving only one arbitrary constant :

x by ae +=

x bae e=

xce= (where )bc a e=

Let us summarise what weve seen till now : the most general solution of an nth order DE will consist of n arbitraryconstants; conversely, from a functional relation involving n arbitrary constants, an nth order DE can be generated(well soon see how to do this). We are generally interested in solutions of the DE satisfying some particularconstraints (say, some initial values).Since the most general solution of the DE involves n arbitrary constant, wesee that the maximum member of independent conditions which can be imposed on a solution of the DE is n.As a first example, consider the functional relation

2 2 31 2

x xy x c e c e= + + ...(1)

This curves equation contains two arbitrary constants; as we vary 1c and 2 ,c we obtain different curves; thosecurves constitute a family of curves. All members of this family will satisfy the DE that we can generate from thisgeneral relation; this DE will be second order since the relation contains two arbitrary constants.We now see how to generate the DE. Differentiate the given relation twice to obtain

2 31 22 2 3

x xy x c e c e = + + ...(2)

2 31 2" 2 4 9

x xy c e c e= + + ...(3)

From (1), (2) and (3), 1c and 2c can be eliminated to obtain2 3 2

2 3

2 3

2 3 2 04 9 2 "

x x

x x

x x

e e x ye e x ye e y

=

21 12 3 2 04 9 2 "

x yx y

y

=

26 3 " 18 9 8 4 4 2 " 6 6 0y x y x y y x y + + + + =2" 5 6 6 10 2y y y x x + = + ...(4)



This is the required DE; it corresponds to the family of curves given by (1). Differently put, the most generalsolution of this DE is given by (1).

As an exercise for the reader, show that the DE corresponding to the general equation

2x xy Ae Be C= + +

where A, B, C are arbitrary constants, is

3 2 0y y y + =

As expected, the three arbitrary constants cause the DE to be third order.

Find the DE corresponding to the family of rectangular hyperbolas 2.xy c=

Solution: Since the equation for a rectangular hyperbola contains only one arbitrary constants, the correspondingDE for the family of rectangular hyperbolas will be first order and can be obtained by differentiationonce.

2xy c=

0xy y + =

This is the required DE.

Find the DE associated with the family of circles of a fixed radius r.

Solution: The circles are of a fixed radius but their centres are not. Let the centre be denoted by the variablepoint ( , ).h k

Then the equation of an arbitrary circle of the family is

2 2 2( ) ( )x h y k r + = ...(1)

This contains two arbitrary constants and therefore will give rise to a second-order DE.

Differentiating (1), we have

( ) ( ) 0dyx h y kdx

+ = ...(2)

Example 1

Example 2



Differentiating (2) again, we have

22

21 ( ) 0d y dyy kdx dx

+ + = ...(3)

2

2

2

1( )

dydxy k

d ydx

+ = ...(4)

Using (4) in (2), we have

2

2

2

1( )

dy dydx dxx hd ydx

+ = ...(5)

Using (4) and (5) in (1), and simplifying, we have the required DE as3 22 2

221

dy d yrdx dx

+ =

which as expected, is second order.

____________________________________________________________________________________

If all this talk about arbitrary constants and solutions to DEs, confuses you, lets view the whole discussion from aslightly different perspective.

Referring to example -1, suppose we are given the DE

0xy y + = ...(1)This is a first-order DE and its most general solution will contain one arbitrary constant; in fact, the most generalsolution of this DE is

xy = where is an arbitrary constant. Now suppose we are told that the curve satisfying the DE in (1) passes through(2, 2). This additional constraint enables us to determine the value of for this particular curve :

(2)(2) 4= =

and thus the curve 4xy = is a particular solution of the DE in (1).

To emphasize once more,

xy = is a general solution to (1)

while 4xy = is a particular solution to (1)

which was obtained from the general solution by using the fact that the curve passes through (2, 2).



As another example, consider the DE obtained in example - 2.

( )( ) ( )32 221 y r y + =This is a second-order DE and its most general solution will contain two arbitrary constants; the most generalsolution can be found to be

( ) ( )22 2x y r + =

where and are arbitrary constants .

To determine a particular solution to this DE, we need two additional constraints which can enable us to evaluate and .

Find the DE associated with the family of straight lines, each of which is at a constant distance p from the origin.

Solution: Any such line has the equation

cos sinx y p + = ...(1)

where is a variable. Different values of give different lines belonging to this family. Since theequation representing this family contains only one arbitrary constant, its corresponding DE will befirst order .

Differentiating(1), we have

cos sin 0y + =

1tany

=

2 2

1sin , cos1 ( ) 1 ( )

yy y

= = + + ...(2)


2 21 ( ) 1 ( )xy y p

y y

= + +

2 2 2( ) (1 ( ) )xy y p y = +

As expected, this is a first order DE.

Example 3



Q. 1 Find the DE of all circles in the first quadrant which touch the coordinate axes.

Q. 2 Find the DE of all circles touching the x-axis at the origin.

Q. 3 Find the DE of all hyperbolas having their axes along the coordinate axes.

TRY YOURSELF - I



In this section, we consider how to evaluate the general solution of a DE. You must appreciate the fact thatevaluating the general solution of an arbitrary DE is not a simple task, in general. Over time, many methods havebeen developed to solve particular classes of DEs. Fortunately for us, at this level we are required to deal withonly the simplest of cases.

Well be considering only first order and first degree DEs. Note that any such DE can be written in the generalform

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

where M and N are functions of x and y.

TYPE - 1 : VARIABLE SEPARABLE FORM

This is by and large the simplest type of DE that well encounter. As the name suggests,in such an equation, M is a function of x only and N is a function of y only. Thus, such aDE is of the form

( ) ( ) 0f x dx g y dy+ =

which can be solved by straightforward integration to obtain

( ) ( )f x dx g y dy C+ = where C is an arbitrary constant.

Observe how the variables are separated in this type of DE and its general solution.

As a simple example, consider the DE

2 0xdx y dy+ =

This is obviously in variable -separable form. Integrating, we obtain

2xdx y dy C+ = 2 3

2 3x y C + =

This is the required general solution of the DE.

Solve the DE 2 .dy dyy x ydx dx

= +

Section - 2 SOLVING DIFFERENTIAL EQUATIONS

Example 4



Solution: A little observation will show you that the variables are separable in this DE:

(1 ) ( )dyy y xdx

= +

(1 )dy dx

y y x =

+

11

dxdyy y x

+ = + Integrating both sides, we have

ln ln(1 ) ln( )y y x C = + +

ln ln( ) ln1

y x Cy

= + + In the last step, we have written the arbitrary constant of integration C as C so that the wholeexpression can be combined now by taking antilog on both sides. (Theres no loss of generality indoing so and it is often done to make the final expression look simpler). Thus, we now have,

( )1

y C xy

= +

( )(1 )y C x y = + This is the required general solution of the DE; as expected it contains only one arbitrary constant.

Solve the DE 2 2 2 2 21 .dyxy x y x ydx

= +

Solution: Again, this DE is of the variable separable form as can be made evident by a slight rearrangement.

2 2 2(1 )(1 )dyxy x ydx

= +

2 2

2

11

y xdy dxy x

= +

2

1 111

dy x dxy x

= +

Integrating both sides, we have

21tan ln

2xy y x C = +

This is the required general solution. ___________________________________________________________________________________

Example 5



Sometimes, the DE might not be in the variable-separable (VS) form; however, some manipulations might be ableto transform it to a VS form. Lets see how this can be done. Consider the DE

( )cosdy x ydx

= +

This is obviously not in VS form. Observe what happens if we use the following substitution in this DE:

x y v+ =

1 dy dvdx dx

+ =

Thus, the DE transforms to

1 cosdv vdx

=

1 cosdv vdx

= +

1 cosdv dx

v =

+

which is clearly a VS form. Integrating both sides, we obtain

1 cosdv dx

v=

+

21 sec2 2

v dv dx =

tan2v x C = +

tan2

x y x C+ = +

This is the required general solution to the DE.

From this example, you might be able to infer that any DE of the form

( )dy f ax by cdx

= + +

is reducible to a VS form using the technique described. Let us confirm this explicitly:



Substitute

ax by c v+ + =

dy dva bdx dx

+ =

1dy dv adx b dx

=

Thus, our DE reduces to

1 ( )dv a f vb dx =

( )dv a bf vdx

= +

( )dv dx

a bf v =

+

which is obviously in VS form, and hence can be solved.

Solve the DE ( )

2

2 .dy rdx x y

=+

Solution: Substituting ,x y v+ = we have

1dy dvdx dx

=

and thus the DE reduces to

2

21dv rdx v

=

2

2 2

v dv dxr v

=+

2

2 21r dv dx

r v

= +

Example 6



Integrating, we have

1tan vv r x Cr

= +

1( ) tan x yx y r x Cr

+ + = +

Solve the DE ( ) ( ) ( )

( )1 ln

.ln

x y x y x ydydx x y

+ + + +=

+

Solution: Again, the substitution x y v+ = will reduce this DE to the following VS form:

( )1 ln1ln

dv v v vdx v

+ =

( )1 lnvvv

= +

lndv vvdx v

= +

( )ln

1 lnv dv dx

v v =

+


( )ln

1 lnv dv dx

v v=

+

To evaluate the integral on the LHS, we use the substitution ( )1 ln v t+ = which gives 1 .dv dtv

=

Thus,

1t dt dxt =

lnt t x C = +

( ) ( )1 ln ln 1 lnv v x C + + = +

( )( ) ( )( )1 ln ln 1 lnx y x y x C + + + + = +

Example 7



TYPE - 2 : HOMOGENEOUS DEsBy definition, a homogeneous function ( ),f x y of degree n satisfies the property

( ) ( ), ,nf x y f x y = For example, the functions

( ) 3 31 ,f x y x y= +( ) 2 22 ,f x y x xy y= + +( ) 3 / 23 , x yf x y x e xy= +

are all homogeneous functions, of degrees three, two and three respectively (verify thisassertion).Observe that any homogeneous function ( ),f x y of degree n can be equivalently writtenas follows:

( ), n ny xf x y x f y fx y

= = For example,

( ) 3 31 ,f x y x y= +

3 3

3 31 1y xx yx y

= + = + Having seen homogeneous functions we define homogeneous DEs as follows :

Any DE of the form M(x, y) dx +N(x, y) dy = 0 or ( )( )

,,

M x ydydx N x y

=

is called homogeneous if M(x, y) and N(x, y) are homogeneous functions of the same degree.

What is so special about homogeneous DEs ? Well, it turns out that they are extremely

simple to solve. To see how, we express both ( ),M x y and ( ),N x y as, say n yx Mx

and .nyx Nx

This can be done since ( ),M x y and ( ),N x y are both homogeneous

functions of degree n. Doing this reduces our DE to

( )( )

,,

n

n

y yx M MM x ydy yx x Py ydx N x y xx N Nx x

= = = =

(The function ( )P t stands for ( )( )

M tN t

)



Now, the simple substitution y vx= reduces this DE to a VS form :

y vx=

dy dvv xdx dx

= +

Thus, dy yPdx x

= transforms to

( )dvv x P vdx

+ =

( )dv dx

P v v x =

This can now be integrated directly since it is in VS form.

Let us see some examples of solving homogeneous DEs.

Solve the DE 2 .dy x ydx x y

=+

Solution: This is obviously a homogeneous DE of degree one since the RHS can be written as

2 22

11

y yxx y x xyyx y xxx

= =+ + +

Using the substitution y vx= reduces this DE to

21

dv vv xdx v

+ =+

21

dv vx vdx v

= +

22 2

1v v

v =+

( )23 11

vv

+=+

( )( )21

3 1v dxdv

xv+ =

+

Example 8



Using 1t v= + above, we have

23t dxdtt x

=


23t dxdtt x

=

21

1 ln 3 ln ln2

t x C = +

( )( )2 2 2ln 3x t C =

( )2 23x t C =

( )( )22 3 1x v C + =( )2 22 2x v v C =

Substituting yx for

,v we finally obtain the required general solution to the DE:

2 22 2 .x xy y C =

Solve the DE 2 2 .xdy ydx x y dx = +

Solution: Upon rearrangement, we have

2 2y x ydydx x

+ +=

2

1y yx x

= + +

This is obviously a first degree homogeneous DE. We substitute y vx= to obtain:

21dvv x v vdx

+ = + +

21dv dx

xv =

+

Example 9




2ln 1 ln lnv v x C+ + = +

ln Cx=

2

21y y Cxx x

+ + =

Solve the DE ( )/ /1 1 0.x y x y xe dx e dyy

+ + =

Solution: This DE can be rearranged as

/

/

1

1

x y

x y

xeydx

dy e

=+

Using the substituting x vy= (note : not )y vx= can reduce this DE to a VS form. (We did not usey vx= since that wouldve led to an expression involving complicated exponentials).

We now have

( )11

v

v

e vdvv ydy e

+ =

+

1vv

dy e dvy e v

+ = +


ln ln lnvy e v C= + +

( )vy e v C + =

/x y x Cey y

+ =

_____________________________________________________________________________________

This example should serve the show that y vx= will not always be the most appropriate substitution to solve ahomogeneous DE; x vy= could be more appropriate in such a scenario, as in the example above.

Example 10



Many a times, the DE specified may not be homogeneous but some suitable manipulation might reduce it to ahomogeneous form. Generally, such equations involve a function of a rational expression whose numerator anddenominator are linear functions of the variable, i.e., of the form

dy ax by cfdx dx cy f

+ += + + ...(1)

Note that the presence of the constant c and f causes this DE to be non-homogeneous.

To make it homogeneous, we use the substitutions

x X h +

y Y k +

and select h and k so that

00

ah bk cdh ek f+ + =

+ + = ...(2)

This can always be done if .a db e

The RHS of the DE in (1) now reduces to

( ) ( )( ) ( )

a X h b Y k cfd X h e Y k f

+ + + + + + + +

( )( )

aX bY ah bk cfdX eY dh ek f

+ + + += + + + +

aX bYfdX eY

+ = + (Using (2))

This expression is clearly homogeneous ! The LHS of (1) is dydx which equals

.dy dY dXdY dX dx

Since 1,dy dxdY dX

=

the LHS dydx equals

.dYdX Thus, our equation becomes

dY aX bYfdX dX eY

+ = + ... (3)

We have thus succeeded in transforming the non-homogeneous DE in (1) to the homogeneous DE in (3). This cannow be solved as described earlier.

Let us apply this technique in some examples.



Solve the DE 2 4 .

3 3dy y xdx y x

= +

Solution: We substitute x X h + and y Y k + where h, k need to be determined :

(2 ) (2 4)( 3 ) ( 3 3)

dy dY Y X k hdx dX Y X k h

+ = = + +

h and k must be chosen so that

2 4 0k h =

3 3 0k h + =This gives 2h = and 3.k = Thus,

2x X= +

3y Y= +

Our DE now reduces to

23

dY Y XdX Y X

=

Using the substitution ,Y vX= and simplifying, we have (verify),

2

35 1

v dXdvv v X

= +

We now integrate this DE which is VS; the left-hand side can be integrated by the techniques describedin the unit on Indefinite Integration.

Finally, we substitute YvX

= and

2X x=

3Y y=

to obtain the general solution. _____________________________________________________________________________________

Suppose our DE is of the form

dy ax by cfdx dx ey f

+ += + +

We try to find h, k so that

0ah bk c+ + =

0dh ek f+ + =

Example 11



What if this system does not yield a solution ? Recall that this will happen if .a db e= How do we reduce the DE to

a homogeneous one in such a case ?

Let a bd e= = (say). Thus,

( )ax by c dx ey cdx ey f dx ey f

+ + + +=+ + + +

This suggests the substitution ,dx ey v+ = whichll give

dy dvd edx dx

+ =

1dy dv ddx e dx

=

Thus, our DE reduces to

1 dv v cde dx v f

+ = +

dv ev ec ddx v f

+ = ++

( ) ( )e d v ec df

v f + + +=

+

( )( )

v f dv dxe d v ec df

+ = + + +

which is in VS form and hence can be solved.

Solve the DE 2 1.2 1

dy x ydx x y

+ =+ +

Solution: Note that h, k do not exist in this case which can reduce this DE to homogeneous form. Thus, we usethe substitution

2x y v+ =

1 2 dy dvdx dx

+ =

Example 12



Thus, our DE becomes

1 112 1

dv vdx v

= +

2 2 11

dv vdx v

= ++

3 1

1vv=+

13 1v dv dxv+ =

1 413 3 1

dv dxv

+ =


11 4 ln(3 1)3 3

v v x C + = +

Substituting 2 ,v x y= + we have

242 ln(3 6 1) 33

x y x y x C+ + + = +

2 ln(3 6 1)3

y x x y C + + =

____________________________________________________________________________________

TYPE - 3 FIRST ORDER DEs

We now come to a very important class of DEs : first-order linear DEs, their importancearising from the fact that many natural phenomena can be described using such DEs.

First order linear DEs take the form

( ) ( )dy P x y Q xdx

+ =

where P and Q are functions of x alone.

To solve such DEs, the method followed is as described below :

We multiply both sides of the DE by a quantity called the integrating factor (I.F.) where

. .Pdx

I F e=



Why this is chosen as the I.F. will soon become clear when we see what the I. F.actually does :

Pdx Pdxdye Py e Qdx

+ =

The left hand side now becomes exact, in the sense that it can be expressed as the exactdifferential of some expression :

Pdx Pdxdy de Py yedx dx

+ =

Now our DE becomes

Pdx Pdxd ye Q edx =

This can now easily be integrated to yield the required general solution:

Pdx Pdxye Qe dx C = +

You are urged to re-read this discussion until you fully understand its significance. Inparticular, you must understand why multiplying the DE by the I. F. Pdxe on both sidesreduces its left hand side to an exact differential.

Solve the DE tan cos .dy y x xdx

+ =

Solution: Comparing this DE with the standard form of the linear DE ,dy Py Qdx

+ = we see that

( ) tan , ( ) cosP x x Q x x= =

Thus, the I.F. is

tan ln(sec ). . secxdx xI F e e x= = =

Multiplying by sec x on both sides of the given DE, we obtain

sec tan sec 1dyx y x xdx

+ =

Example 13



The left hand side is an exact differential :

( sec ) 1d y xdx

=

( )secd y x dx =Integrating both sides, we obtain the solution to our DE as

secy x x C= +

Solve the DE 2.dx x y

dy y+ =

Solution: The I.F. is

( ). .

P y dyI F e= (y is the independent variable in this DE)

1 dyye=

ln ye=

y=

Multiplying by the I.F. on both sides, we have

3dxy x ydy

+ =

3( )d xy ydy

=

3( )d xy y dy =

Integrating both sides gives

4

4yxy C= +

Solve the DE 3 3 .dy x y xydx

=

Example 14

Example 15



Solution: We have,

3 3dy xy x ydx

+ =

Note that since the RHS contains the term 3,y this DE is not in the standard linear DE form. However,a little artifice can enable us to reduce this to the standard form.

Divide both sides of the equation by 3.y

33 2

1 dy x xy dx y

+ = ...(1)

Substitute 21 vy

=

3

2 dy dvy dx dx =

3

1 12

dy dvy dx dx

= ...(2)


312

dv xv xdx

+ =

3( 2 ) 2dv x v xdx

+ = ...(3)

This is now in the standard first-order linear DE form. The I.F. is

22. .

xdx xI F e e = =

Thus, the solution to (3) is

. . ( ) ( . )v I F Q x I F dx = 2 232x xve x e dx =

Performing the integration on the RHS by the substitution 2t x= and then using integration by parts,we obtain

2 2 2( 1)x xve e x C = + +

2 2 22

1 ( 1)x xe e x Cy

= + +

This is the required general solution to the DE.



____________________________________________________________________________________

This example also tells us how to solve a DE of the general form

( ) ( ) ndy P x y Q x ydx

+ = ...(4)

We divide by ny on both sides :

11 ( ) ( )nndy P x y Q x

y dx ++ =

and then substitute 1ny v + = and proceed as described in the solution above.

DEs that take the form in (4) are known as Bernoullis DEs.



EXACT DEs*EXACT DEs*EXACT DEs*EXACT DEs*EXACT DEs*

In the last section, we discussed how the multiplication of the . .Pdx

I F e= on both sides of the linear DE

( ) ( )dy P x y Q xdx

+ =

renders this into an exact DE. We now consider the general case of exact DEs. In particular, we want to see whatcondition must be satisfied in order that the DE

( , ) ( , ) 0M x y dx N x y dy+ = ...(1)

is exact.

In order for this DE to be exact, its LHS must be expressible as the complete differential of some function( , ),f x y i.e.

( )( , ) ( , ) ( , )M x y dx N x y dy d f x y+ = ...(2)Now since the function ( , )f x y is a function of both x and y, its total differential is a sum of partical differentialswith respect to x and y, i.e.,

f fdf dx dyx y = + ...(3)

Comparing (1) and (2), we have

,f fM Nx y = = ...(4)

This gives

2 2

,f M f Ny x y x y x = =

For continuous ( , ),f x y

2 2f fx y y x =

and thus for the DE in (1) to be exact, we see that the necessary (and in fact sufficient) condition isM Ny x

=

* This section contains advanced material and is optional.



If this condition is satisfied, the DE in (1) reduces to

( , ) 0df x y =

which upon integration leads to the required solution:

( , )f x y C=

As an example, consider the DE33 ( 2) ( 2 ) 0x xy dx x y dy + + = ...(5)

We have,

2( , ) 3 ( 2) 3MM x y x xy xy

= =

3 2( , ) 2 3NN x y x y xx

= + =

Since ,M Ny x

=

the DE is exact and hence we can find a function ( , )f x y such that the DE is expressible as

( , ) 0.df x y = Let us try to explicitly find this function.

From (4), we have

3 ( 2)f fM x xyx x = =

Integrating with respect to x, while treating y as a constant, we have3 2( , ) 3 ( )f x y x y x y= + ...(6)

The function ( )y acts as the arbitrary constant of integration, since y is constant for this integration process.

From (4) again we have

3 2f fN x yy y = = +

Evaluating fy from (6), we have

3 3( ) 2fx y x yy+ = = +

( ) 2y y =

2( )y y C = + ...(7)



Finally substituting (7) into (6), we have

3 2 2( , ) 3f x y x y x y C= + +

Thus, the solution to the DE in (5) is

( , ) constantf x y =

3 2 23 constantx y x y + = ...(8)

You are urged to verify that (8) is indeed the required solution by differentiating (8) and observing that (5)is obtained.

Solve the DE 2 22 ( 3 ) 0.xydx x y dy+ + =

Solution: First of all, notice that this DE is homogeneous :

2 2

23

dy xydx x y

= +

The substitution y vx= leads to

2

2 2 2 2

2 23 1 3

dv vx vv xdx x v x v

+ = =+ +

2

21 3

dv vx vdx v

= +

3

2

3 31 3

v vv

=+

2

2

3 (1 )1 3v v

v +=

+

2

2

1 3 3(1 )

v dxdvv v x+ = +

The substitution 2v t= leads to

1 3 1 31 2

t dxdtt t x

+ = +

1 1 31 2

dxdtt t x

+ = +

Example 16



Integrating both sides gives

1ln(1 ) ln 3ln ln2

t t x C+ + = +

3(1 )t t x C + =

2 3(1 )v v x C + =

2 2( )y x y C + =

We now solve this DE again using the exact differential approach since by observation this DE satisfiesthe required criterion for it to be exact.

We have,

2 22 , 3f fM xy N x yx y = = = = +

Integrating the first relation, we have

2( , ) ( )f x y x y y= + ...(1)

Differentiating this w.r.t. y and comparing with the expression for fy

above, we have

2 2 2( ) 3f x y x yy = + = +

2( ) 3y y =

3( ) 'y y C = + ...(2)

Thus, from (1) and (2),

2 3( , ) 'f x y x y y C= + +

The solution to the exact DE is

( , ) constantf x y =

2 3x y y C + =

2 2( )y x y C + =

which is the same as the one obtained earlier. Thus, the exact differential approach might lead to thesolution faster than the other approaches weve discussed earlier.

____________________________________________________________________________________



Sometimes, the fact that the DE is exact is evident merely be inspection. We list down such exact differentials(verify the truth of these relations):

2

2

12 2

2 2

( )

tan

2

xdy ydx d xyxdy ydx yd

x xydx xdy xd

y yxdy ydx yd

x y x

x yxdx ydy d

+

+ ++

Table -1

Solve the DE 2 2 2 2 1 .xdy y dx

x y x y

= + +

Solution: Upon rearrangement, this DE gives

2 2

xdy ydx dxx y = + ...(1)

From Table-1, the L.H.S of (1) is the exact differential 1tan .ydx

Thus, our DE reduces

to 1tan 0yd dxx

+ =

Integrating, we obtain the solution as

1tan y x Cx

+ =

However, it is very likely that we wont be able to make out just be inspection whether the DE is exactor not.

Example 17



If the DE is not exact, it can be rendered exact by multiplying it with an integrating factor I.F. In thecase of the first-order linear DE

dy Py Qdx

+ =

the I.F. Pdxe renders the DE exact:

Pdx Pdxd ye Qedx =

and the solution is now obtainable by integration.

If fact, a systematic approach exists to determine the I.F. in a general case (if such an I.F. is possibleat all.). However, well not be discussing that approach here since it is beyond our current scope.

Let us see another example, where the solution is easily obtained by the recognition of exact differentialspresent in the equation.

Solve the DE ( ) ( )cos sin .y yx ydx xdy y xdy ydxx x

+ =

Solution: Upon rearrangement, we have

( )tany yydx xdy xdy ydxx x

+ =

2tany xdy ydxxyx x

=

2tanydx xdy y xdy ydx

xy x x+ =

From Table-1 this can be written as

( ) tand xy y ydxy x x

=

The solution is now obtained simply by integrating both sides :

ln( ) ln sec lnyxy Cx

= +

sec yxy Cx

=

Example 18



Solve the following DEs:

Q. 1 2 2( ) .xdx ydy a x y dy+ = +

Q. 2 2 2(3 ) ( ) 0.xy y dx x xy dy+ + + =

Q. 32 3 .

2 4dy x ydx x y

+=+ +

Q. 4 cot cos .dy y x xdx

+ =

Q. 5 3 ln .dyx y y xdx

=

TRY YOURSELF - II



As described in the introduction, differential equations are so important for the very reason that they find a wideapplication in studying all sorts of scientific phenomena. The motion of a body in a force field, radioactive decayand population growth were some of the phenomena mentioned that must use DEs for analysis.

In some of the subsequent solved examples, we apply the DE- solving techniques that weve learnt in the previoussection, to solve practical and interesting problems.

Let :f + ! ! be a differentiable function such that

( ) ( ) ( )1

1 lnxxf x e x f x dx

e= +

Find a simple expression for f(x).

Solution: Differentiating the given relation, we have

( )1 lnxdf x fdx x e

= +

1 lndf f xdx x

=

This is evidently a first-order linear DE; the IF is .dx xe e = Multiplying it across both sides of the

DE renders the DE exact and its solution is given by

1 lnx xe f e x dxx

= lnxe x C= +

( ) ln xf x x C e = + ... (1)From the relation specified in the equation, note that

( ) ( ) ( )1

1

11 1 1 lnf e f x dxe

= + = e

From (1), f(1) = Ce. This gives C = 1.

Thus, the function f(x) has the simple form

( ) ln xf x x e= +

SOLVED EXAMPLES

Example 1



Solve the following DEs

(a) ( )2cos 2cos sin

sinx y xdy

dx y

=

(b) ( )2 2

2 22 1 2dy dyy x y xdx dx

+ = + +

Solution: (a) We have,

2sin 2cos cos sin cosdyy y x xdx

= ... (1)

Observe that the substitution cos y z = will reduce this DE to a standard linear DEcos y z =

sin dy dzydx dx

= ...(2)


( ) 22cos sin cosdz x z x xdx

+ =

The I.F for this DE is 2cos 2sinxdx xe e =Thus, the solution will be given by

2sin 2sin 2cos sinx xze e x x dx= ...(3)To integrate the R.H.S, we use the substitution sin cos .x t xdx dt= = Thus, the integralreduces to

2 2tt e dt2 2 2 2

'2 4 2

t t tt e e te C= + + (Integration by parts)

2 2sin 2sin 2sinsin sin '2 4 2

x x xx e e x e C = + +

Finally, the solution to the DE is, from (3)

22sinsin 1 sincos '

2 4 2xx xz y C e= = + +

2 2sin4cos 2sin 2sin 1 xy x x Ce = + +

Example 2



(b) Let .dy pdx

= Thus, this DE is

( ) ( )( )2 2 2 22 1 2py x p y x+ = + +2 2 2 22 4 2 0x p xyp y x + =

( )2 2 2 2 22

4 16 8 2

4

xy x y x y xp

x

=

2 2 4

2

4 8 164

xy x y xx

+=

2 2

2

4 2 2 24

xy x y xx

+=

2 22

2y xy

x x+

=

21 1

2y yx x

= +

The substitution y = vx reduces this DE to

2

12

dv vv x vdx

+ = +

2

122

dv dxxv

= +


2 1ln 2 ln '2

v v x C+ + = +

2ln 2 ln 2x v v C + + =

2 22ln 2 ln y v xx Cx x

+ + = ... (4)

Thus, we obtain two different solutions to the DE, one corresponding to the + and one to the sign in (4).



A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to itssurface area in contact with air (proportionality constant = k > 0). Find the time after which the cone is empty.

Solution: We need to form a differential equation which describes the variation of the amount of water left in thecone with time.

r t( )

h t( )

H

Let us denote the height of the water remaining in the cupat time by ( ). Denote the volume at time by ( )

t h tt v t

Fig - 1

R

From the geometry described in the figure above,

( )( )

r t Rh t H

=

( ) ( )Hh t r tR

=

The volume v(t) of the cone is

( ) ( ) ( )213

v t r t h t=

( )33

H r tR

= .... (1)

Now, it is specified that the rate of evaporation (the rate of decrease of the waters volume)is proportional to the surface area in contact with air:

2dv krdt

= ... (2)

Example 3



From (1) and (2), we have

2 2H drr krR dt

=

dr Rkdt H

=

This is the DE representing the variation in the radius of the water surface with time. The inital radiusis R and the final radius is 0. If the time taken for the entire water to evaporate is T, we have

0

0

T

R

Rkdr dtH

=

HTk

=

Note that the time taken is independent of the radius of the cone and depends only on its height. Thus,for example, two cones full of water, with the same height, but one of them having a radius saya 1000 times larger than the other, will become empty in the same amount of time!

A curve C has the property that if the tangent drawn at any point P on C meets the coordinate axes at A and B, thenP is the mid-point of AB. C passes through (1, 1). Determine its equation.

Solution: Let the curve be ( ).y f x=

The tangent at any point ( ),P x y has the equation

( )dyY y X xdx

=

This meets the axes in ,0dxA x ydy

and 0, .dyB y x

dx

Since P itself is the mid-point of AB,

we have

2 , 2dx dyx y x y x ydy dx

= =

dy ydx x

=

Example 4



This is in V.S form and can be solved by straight forward integration:

dy dxy x=

ln ln lny x k = + (k is an arbitrary constant)

xy k =

Since the curve C passes through (1, 1), we have k = 1. Thus, the equation of C is

1xy =

A and B are two separate reservoirs of water. The capacity of reservoir A is double the capacity of reservoir B.Both the reservoirs are filled completely with water, their inlets are closed and then water is released simultaneuouslyfrom both the reservoirs. The rate of flow of water out of each reservoir at any instant of time is proportinal to thequantity of water in the reservoir at that time. One hour after the water is released, the quantity of water in reservoir

A is 112 times the quantity of water in reservoir B. After how many hours do both the reservoirs have the same

quantity of water?

Solution: Assume the intial volumes of water in A and B to be 2 V and V.

Denote the volume of water in A and B by v1 and v2 respectively. We have,

1 21 2,A B

dv dvk v k vdt dt

= =

where kA and kB are constants of proportionality (not given). These two DEs are in VS form and thesolution can be obtained by simple integration.

( ) ( )1 21 2

1 22 0 0

,v t v tt t

A BV V

dv dvk dt k dtv v

= =

( ) ( )1 2ln , ln2 A B

v t v tk t k t

V V = =

( ) ( )1 22 ,A Bk t k tv t Ve v t Ve = =

Example 5



It is given that

( ) ( )1 231 12

v t v t= = =

322

A Bk kVe Ve =

( ) 34

B Ak ke =

3ln4B A

k k =

Let T be the time at which the volumes in the two reservoirs become equal. We thus have,

( ) ( )1 2v t T v t T= = =

2 A Bk T k TVe Ve =

( ) 2A Bk k Te =

( ) ln 2A Bk k T =

4ln ln 23

T =

( )ln 2

ln 4 / 3T = hours



ASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENT

[ LEVEL - I ]

Q. 1 Show that

(a)2

2 0d ydx

= represents all non-vertical lines.

(b)2

2 0d xdy

= represents all non-horizontal lines.

Q. 2 Find the DE corresponding to parabolas whose axis of symmetry is parallel to the x-axis.

Q. 3 Find the DE corresponding to the curve y = a sin (mx + b).

Q. 4 Solve the DE ( ) ( )1 3 3 0x y dx x y dy+ + + + =

Q. 5 Solve the DE sin siny yx dy y x dxx x

= Q. 6 Solve the DE ( ) ( )2 2 2 24 2 4 2 0x xy y dx y xy x dy + =

Q. 7 Solve the DE ( ) ( )cos sindy x y x ydx = + + +

Q. 8 Solve the DE ( )32x y dy ydx+ =Q. 9 Solve the DE ( )22ln ln

dy y yx xdx x x

+ =

Q. 10 Show that the curve for which the normal at every point passes through a fixed point is a circle.

[ LEVEL - II ]

Q 1. Find the general solution of the differential equation 2

2 32 3

dy x ydx x y

+ = + +

Q. 2 Let u(x) and v(x) satisfy the differential equations

( ) ( ) ( ) ( )anddu dvp x u f x p x v g xdx dx

+ = + =

respectively where p(x), f(x) and g(x) are continuous functions. If ( ) ( )1 1u x v x> for some( ) ( )1 1and for all ,x f x g x x x> > prove that any point ( ),x y , where x > x1, does not satisfy the

equations ( ) ( )andy u x y v x= = .



Q. 3 Determine the equation of the curve passing through the origin in the form y = f (x), which satisfies the

differential equation ( )sin 10 6 .dy x ydx = +

Q. 4 A normal is drawn at a point P(x, y) of a curve, It meets the x-axis at Q. If PQ is of constant length k,then show that the differential equation describing such curves is

2 2dyy k ydx

=

and find the equation of such a curve passing through (0, k)

Q. 5 A curve passing through the point (1, 1) has the property that the perpendicular distance of the normalat any point P on the curve from the origin is equal to the distance of P from the x-axis. Determine theequation to the curve.

Q. 6 A hemispherical tank of radius 2 meters is initially full of water and has an outlet of 12 cm2cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the output is

according to the law ( ) ( )0.6 2 ,v t gh t= where v(t) and h(t) are respectively the velocity of the flowthrough the outlet and the height of water level above the outlet at time t and g the acceleration due togravity. Find the time it takes to empty the tank.

Q. 7 A country has a food deficit of 10%. Its population grows continuously at a rate of 3% per year. Itsannual food production every year is 4% more than that of the last year. Assuming that the average foodrequirement per person remains constant, prove that the country will become self-sufficient in food aftern years, where n is the smallest integer bigger than or equal to

( )ln10 ln 9

ln 1.04 0.03

Q. 8 A curve f(x) passes through the point (0, 1). A curve ( ) ( )xg x f x dx

= passes through the point(0, 1/n). The tangents drawn to both the curves at points having the same abscissa, intersect on thex-axis. Find f(x).

Q. 9 Given the curves y = f(x) passing through the point (0, 1) and ( )x

y f t

= dt passing through the point

10, .2

The tangents drawn to both the curves at the points with equal abscissa intersect on the

x-axis. Find the curve y = f(x).

Q. 10 Solve the DE ( ) ( )21 x dy y x y dx+ =



ASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENT

( ANSWERS )

[ LEVEL - I ]

2.3

3 0d xdy

=

3.3 2

3 2

d y dy d yydx dx dx

=

4.33 ln 2 2 12

x y x y C + + + = 5. ln cos

yx Cx

=

6.3 3

2 22 23 3x y x y xy C

+ =

7. ln 1 tan 2x y x C

+ + = +

8.2x y C

y

= +

9. 21 1ln 2

Cx y x

= +

[ LEVEL - II ]

1. ( ) ( )3 43( 3) 3 6x y C x y + = + 3.11 5 tan 4tan 5

3 4 3tan 4xy x

x =

4. 2 2 2x y k + = 5.2 2 2x y x + =

6.47 10 sec

135T

g =

8. ( ) nxf x e =

9. ( ) 2 xf x e = 10. ( )2

21 ln 1x x x Cy

+ = + + +



TRY YOURSELF - I

1. ( )2 2

2 1 dy dyx y x ydx dx

+ = +

2. ( )22 2dyx y xydx =

3. ( )2' ' "yy x y xyy = +

TRY YOURSELF - II

1. ( )2 2ln 2x y ay C + = +

2. ( )2 2 2x y xy C + =

3. ( ) ( )( ) ( )2 21 2 1 2y x y x C + + + + + =

4. 21sin sin2

y x x C = +

5.2 2

22 ln 2

x xx x Cy

= + +

ANSWERS

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Differential Equations