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Differential Equation - 1 1

10 MAT 21 Dr. V. Lokesha 2012

Engineering Mathematics – II

(10 MAT21)

LECTURE NOTES (FOR II SEMESTER B E OF VTU)

VTU-EDUSAT Programme-16

Dr. V. Lokesha

Professor of Mathematics

ACHARYA INSTITUTE OF TECNOLOGY

Soldevanahalli, Bangalore – 90



ENGNEERING MATHEMATICS – II

Content

CHAPTERS

UNIT I DIFFERENTIAL EQUATIONS – I

Unit-1



Differential Equation- I

Equations of first order and higher degree

Overview:

In this unit, we shall study differential equations of the first order and higher degree,

We study the differential equations solvable for and the problems involving in it, differential

equation solvable for and the problems involving in it, Differential equation solvable for and

some problems involving in it. We discuss the problems on special type called Clairaut’s

Equation and reducible to clairaut’s form involving both general solution and singular solution

and we discuss the application of the first order and first degree differential equation with

illustrative examples.

Objective:

At the end of this unit he will be able to understand

• To obtain the solution of non-linear differential equation.

• To obtain the solution of the differential equation of the form .

• The method of the solution is simple involving well known methods.

• Singular solution exists for higher degree equations of first order.

• Clairaut’s equation has numerous engineering applications like geodesics.

• Non linear equation of first order differential equation is reduced to linear differential

equations of first order.

• Mathematical models for some of the applications like Kirchoff’s law, Newton’s law of

cooling etc.

Equations solvable for .



Introduction:

We are already familiar with differential equations of the first order and first degree , Now

we shall study differential equations of first order and degree higher than the first. For

convenience, we denote by . Such equations are of the form .

A differential equation of the first order but of the nth degree is of the form

Where are functions of and .

In several cases (2) can be solved by reducing (2) to first order and first degree (n)

equations by solving for p(b), solving for y(c), solving for x.

In this unit we discuss the following cases.

• Equations solvable for .

• Equations solvable for y.

• Equations solvable for x.

• Clairaut’s equation

Now we discuss the first case Equations solvable for p.

Splitting up the left hand side of (2) into n linear factors, we have

.

Equating each of the factors to Zero,

.

Solving each of these equations of the first order of first degree, we get the solutions

. These n solutions constitute the general

solution of (1).

Otherwise, the general solution of (1) may be written as

.



Problems:

1. Solve :

Sol: Given equation is

From (1)

On integration, we get

From (2)


Thus, , constitute the required solution.

On combining these into one, the required solution can be written as



.

2. Solve:

Sol: We have,

Adding on both sides,

=

From (1):

From (2):



Thus combining (3) and (4), the required general solution is

3. Solve:

Sol: we have,

On Integration, we get





2

2 2

2

2

2

Given that 6 0 =

( 3 )( 2 ) 0

4. + 6 =0

3 0 or 2 0

3 0

Sol :

dy d

dyx p xyp y where p

d

yx xy y

dx d

xxp y xp yxp y xp y

dyx

x

x

yd

+ − =

⇒ +

−

− =

⇒ + = − =

⇒ + =

( ) ( )3 2

2

1 2

1 2

3

1 2

or 2 0

3 log or 2 log

log log or log log

The general solution yis y 0

dyx y

dxdy dx dy dx

C Cy x y x

yyx C C

x C Cxx

− =

⇒ + = −

− −

=

⇒ =

∴ =

=

2

2

Given that 5 6 0 =

( 6)( 1) 0 6 0 or 1 0

5. So

Sol

lv

6 0 or 1 0

e

:

5 6 0 dy

p p where pdx

p pp pdy dy

dx dx

p p

− − =

⇒ − + =

⇒ − = + =

=

⇒ = + =

− −

−

( )( )1 2

1 2

6 =0 or 0 6 = or

The general solution 0 6 is

dy dx dy dxy x C y x C

y x C y x C

⇒ − + =

⇒ − + =

∴ − − + − =

( )( )

2 ( 4)( 3) 0 4 =0 or 3 0

6. Solve

The ge

7 1

ner 4 3al solu

So

tion

l

is

:

0

2 0 p p

dy dx dy dx

y x C y x C

p p− −

− − −

=

⇒ − − =

∴

+

−

−

=

=



Exercise:

Solve the following differential equations:

1.

2.

( ) ( )

2

2

Given that ( ) 0 =

Sol

(

7.

0

)( 1

( ) =0

) 0

The general solution i

:

s

dyxp y x p y where p

dxx

dy dyx y x y

dx

xy C

d

y x C

p y

x

p

− − − =

⇒−

− − −

+

∴ −

−

−

=

=

2 2 2

2

2 2

Given that ( ) 0 =

( ) ( ) 0 ( )( ) 0

S

0 or

ol

8. ( )

:

=0

dy dy

xy x y xy

dyxyp x y p xy where p

dxxp yp x y yp

dx dx

xxp y yp x

dyx y

dx

− + + =

⇒

− +

− − − =

⇒ − − =

⇒ −

=

+

( )( )

2 2

2

1 2

2

y 0

0 or 0

log log or

The general solutio 0is yn x

dyx

dxdy dx

ydy xdxy x

yC y x C

xC y x C

− =

⇒

− −

− = − =

⇒ −

∴ −

=

=

=

( )

( )2

2

2 2

2

Given that sinh sinh 1 0

sinh cosh

sinh cosh sinh cosh

sinh cosh or sinh

9.

cosh

Solve 2 sinh

Sol :

1 0

cosh

p x x

p x x

p x x or p x xdy dy

x x x xdx d

p x

xy

p

− − − =

⇒ − =

⇒ − = − = −

⇒ = + = −

⇒ =

+ − =

( ) ( )1 2

1 2

1 2

1 2

sinh or y cosh sinh

or y2

The required solut

2 2 2 or

ion is 0

x x x x x x x x

x

x

x

x

x x C x x C

e e e e e e e

y e C

ey C C

y e C y e

e

C

y C

− − − −

−

+ + = − +

+ − + −

∴

⇒ = + + = − +

⇒ = + = − +

− − + − =



Equations solvable for

Introduction:

It is possible to express explicitly as a function of and

Differentiating equation (1) w.r.t. and Let ,

We get an equation of the form

Let the solution of this equation be

Eliminating between (1) and (2) we get a relation in and which is the required solution.

Note: It is not possible always to eliminate between (1) and (2). In that case (1) and (2)

together constitute the solution giving and in terms of the parameter .

Problems:

1. Solve:

Sol:

Differentiating (1) w.r.t.



Consider

Integrating,

Substituting (3) in (1), which is the required solution.

2. Solve: .


Differentiating both sides w.r.t. ,

Or

Discarding the factor , we have

Integrating

Or .

Putting this value of in (1), we have which is the required solution.

3. Solve:





Or

Which is a linear equation.

or

Putting these values of in (1), we have

Or

Equations (3) and (4) together constitute the general solution of (1).

1

1

4. Solve 2 On differentiating w.r.to x, we get

2 2

2 2

This is a linear differential equation in

p

Integrati

Sol :

n

n

n

n n p

y px p

dy dp dpp p x np

dx dx dxdx dx x

p x np npdp dp p

−

− −

= +

= = + +

⇒ + = − ⇒ + = −

( )

2

2

2

g factor(IF) = =

the solution is ( ) ( )

dppdp p

n

e e p

x IF np IF dp c−

∫∫ =

∴ = − +∫( )

( )

2

12

1

( )

1

(1)1

Substituting this value of x in given equation, we get

12 y (2)

1Equations (1) and (

n

n

n

n

x p np dp c

npxp c

nnp

x cn

n pc

p n

+

−

⇒ = − +

−⇒ = +

+−

⇒ = + − − − −+

−= + − − − −

+

∫

2) together constitute the general solution.



Exercise:


1.

2.

3.

3

2

2

2

2

On differentiating w.r.to , we get

1 3

1 3 = =

3 1

1 1

6. Solve

So

3 = 1

l : x

dy dpp p

dx dx

dp p pdp dx

dx p p

pd x

x

p

p

p d

y

= = +

−⇒ ⇒

−

− +⇒

−

=

+

2

1 3 1 =

1

On integrating we get,

the solution is =3 log( 1) (1)2

This solution expresses x in terms of p and the given differential equation expresses y in terms of

p dp dxp

px p p c

⇒ + +

−

∴ + + − + − −

p. hence (1) and the given differential equation constitute itsgeneral solution.



Equations solvable for .

Introduction:

If the given equation on solving for , take the form then

differentiating w.r.t. gives an equationof the form = .

Now it may be solve the new differential equation in . Let its solution be

. The elimination of from (1) and (2) gives the required solution.

Problems:

1. Solve: .

Sol: Solving for , we have


Discarding the factor we have

Integrating

Putting this value of p in the given equation, we have

which is the required solution.



2. Solve:

Sol: Solving for , we have


Integrating,

Equations (1) and (2) together constitute the general solution.

3. Solve:

Sol: Writing the above equation in the form


⇒

Integrating,



2 2

2 2

4

2

2

2 3

3

1 Solving for , we get

2

Diffrentiating w.r.to

1 1 1

4.

Sol :

2 2 2

2 2 2

2

yx x y p

p

y

dx y dp dpyp y p

dy p p p dy dy

dp dpp p y yp y p

y px

y dy

p

d

y

= −

⇒ = = − − −

⇒

=

−

+

= − −

( ) ( )3 3 1+2 1+2 0dp

p yp y ypdy

⇒ + =

( )

2

3

33

1+2 0

Consider 0 0

log log log .

Putting this value of p in the given equation, we get

2 y+ 2

dpyp p y

dy

dp dy dpp y

dy y pc

y p c yp c py

cx c

ycy

ycx

⇒ + =

+ = ⇒ + =

⇒ + = ⇒ = ⇒ =

= += ⇒

( )

2

2

2

2

2

1

S o lving fo r , w e get 2 2

D ifferen tiating w .r.to ' '1 1

1 1 1

1

5 . 2 = 0

S ol :

1

x x y yp

ydx y dp dp

y pdy p p p dy dy

y px y

dpp y

p p d

p

y

pp

= −

⇒ = = − + +

⇒ − = −

⇒ −

−

+

2

11 0

dpy

p dy

+ − =



2 2

2

2

1 1 0

Consider 0 0

log log log .

Substituting this value of p in the given equation, we get

2 + 0 y 2=

dpp y

p dy

dp dy dpp y

dy y pc

y p c yp c py

cx cy cx c

y y

⇒ − + =

+ = ⇒ + =

⇒ + = ⇒ = ⇒ =

− +− ⇒ = 0

( )

( )

1

2

2

222

2

Solving for , we get

tan (1)1

Differentia

6. tan

ting w.r.to ' '

1 21 1

Sol :

1

1 1

xp

x ppy

dp dpp p p

dy dydx dp

dy p p dy p

pp x

p

−= + − −

= −

− −+

+ −

+

⇒ = = ++ +

( )

( )

( )

2 2

22

22

2

2 1 21

1

2

1

(2)

(1) 1

(2) .

p p dp

p dyp

pdy dp

p

and constitute the general sol

c

u

y cp

tion

= −+

+ −⇒ =

+

⇒ =+

⇒ − − − −



:


1.

2.

3.

2 2

2 2

2 2

1

Solving for , we get log

Differentiating w.r.to ' '1 1 1 1

log (1 log )

1 1 1 lo

7. log

g log

Sol :

0

y y xypp

x x y yp y

ydx dp p dp

y y ydy p p dy p y y dy

dp p dpy y y

p dy

p

p y y dy

= +

−⇒ = = + + − +

−⇒ + − +

=

=

+

2

2

1 log 1 1 0

1 log 1 0

1 0 0

log log log

Substituting this value in

y dp p y dpy

p p dy y p dy

p y dpy

p y p dy

y dp dp dy

p dy p y

p y cp

c p cy

o yr

⇒ − − − =

⇒ − − =

⇒ − = ⇒ − =

⇒ − = ⇒ = =

(1) gives the general solution.



CLAIRAUT’S EQUATION

Introduction:

An equation of the form is known as Clairaut’s equation.

Differentiating w.r.t , we get


Integrating,

Putting , in (1), the required solution is

Thus, the solution of (1) is obtained by writing for p.

Note: 1.If we eliminate from and (1), we get an equation involving no

constant. This is the singular solution of (1) which gives the envelope of the family of straight

lines (3).

2. Equations which are not in the Clairaut’s form can be reduced to Clairaut’s form by

suitable substitutions (transformation).

To obtain the singular solution, we proceed as follows:

i) Find the general solution by replacing by i.e., (3).

ii) Differentiate this w.r.t. giving .

iii) Eliminate from (3) and (4) which will be the singular solution.



Problems:

1. Solve:

Sol: The given equation can be written as

This is of Clairaut’s form. Hence putting c for p, the solution is .

2. Solve:

Sol: {In problems involving and ,

put }

Put and

So that

Or which is of Clairaut’s form.

Its solution is and hence .

3. Solve: .

Sol: Put

So that

Or which is of Clairaut’s form.



Its solution is and hence .

4. Solve: . Also find its singular solutions.

Sol. Given equation can be written as which is the Clairaut’s

equation.

(1)

To find the singular solution, differentiate w.r.t. giving

0 = x + → (2)

To eliminate c from (1) and (2), we rewrite (2) as

C = N ( -1)/x

Now substituting this value of c in (1), we get

y = N ( -1) + {N( -1)/x}

which is the desired singular solution.

5. Solve:

Sol: Clairaut’s equation is . Its general solution is obtained by replacing by . Thus

is the required complete integral.

To obtain the singular solution, differentiate the general solution w.r.t. . Then

From

Eliminating from the general solution, we get



Thus the singular solution (which is a parabola) is the envelope of the family of one

parameter family of straight lines (representing the general solution).

6) Solve the D.E

Sol:

Given

=> …..(1)

this is in the form

replacing ‘p’ by ‘c’ we get

is the general solution.

To find singular sol: consider

=>

Using this in (1) we get

is the singular solution.

7 : Solve

Sol :

….(1) is clairauts equation.

The general soln is

Now consider

Using this in (1) we get

tan( )p px y= −

tan( )p px y= −

1tan p px y−= − 1tany px p−

= −

( )y px f p= +

'[ ( )] 0x f p+ =

2

10

1x

p− =

+

1 xp

x

−=

( )1 1(1 ) tan

xy x x

x− −

= − −

1tany cx c−= −

log( ) 0p y px+ − =

log( )

p

p

y px p

y px e

y px e

−

−

− = −

⇒ − =

⇒ = +

cy cx e

−= +

( )

' ( ) 0

0

1log

p

p

x f p

x e

x e

px

−

−

+ =

⇒ − =

⇒ =

∴ =

( ) ( )

( )

( )

1log1log

1log

1log

xy x ex

y x xx

y x xx

−

= +

⇒ = +

⇒ − =



is the singular solution.

8) Solve use substitution

Sol : Let …..(1)

We have , where

(1) becomes

Divide by X we get ……(2)

is the clairauts Equ with

The solution of (2) becomes

The general Soln of (1) is

Consider

Using this in (2)

is singular solution of (1)

Exercise:


1.

2.

3.

2 3 2 0yp x p x y+ − =2 2;X x Y y= =

12

2

dy dy dY dX dY xp x P

dx dY dX dx y dX y

= = = =

( )2

2 3 22

2 2 4 2 2

2 2

0

0

0

x xy p x p x yyy

x p x p x y

XP X P XY

+ − =

⇒ + − =

⇒ + − =

2 3 2 0yp x p x y+ − =

2Y XP P= +

2( ) .f P P=

2Y Xc c= +2 2 2y x c c= +

dYP

dX=

'( ) 0

2 0

2

X f P

X P

XP

+ =

⇒ + =

−⇒ =

( )2

2

2 4 4

XX XY X−−= + =

42

4

xy

−=



Miscellaneous Problems

Problems:

1. Solve:


On factorizing, we get

From (1):


From (2):



2. Solve:

Sol:


Integrating,

3. Solve:

Sol: is of Clairaut’s type whose solution is got by replzcing p by c.

.



4. Solve:




Or

Or

Putting this value in (1), we get , which is the required solution.

Applications of Differential equations of first order

Overview:

In this section, we study the engineering applications in ordinary differential equation

of first order by illustrative examples.

Objective:

At the end of this section, you will be able to understand

• The physical problems in real situations can be mathematically modeled as a differential

equation.

• The application of differential equation is in various fields like research, Engineering,

Sciences, Management, Life science etc.



Application to Electric Circuits:

If be the electrical charge on a condenser of capacity C and be the current, then

(a) .

(b) The potential drop across the inductance is .

(c) The potential drop across the resistance is .

(d) The potential drop across the capacitance is .

Also, by Kirchhoff’s Law, the total potential drop (voltage drop) in the circuit is equal to

the applied voltage (E.M.F.).

Example 1. A constant electromotive force E volts is applied to a circuit

containing a constant resistance R ohms in series and a constant inductance

L henries. If the initial current is zero, show that the current builds up to half

its theoretical maximum in seconds.

Sol. Let be the current in the circuit at any time t

By Kirchhoff’s Law, we have

Which is Leibnitz’s linear equation,

Thus (2) becomes,

This equation gives the current in the circuit at ant time .

Clearly, increases with and attains the maximum value .

Let the current in the circuit be half its theoretical maximum after a time T seconds.



Then

Example 2. The equations of electromotive force in terms of current for an

electrical circuit having resistance R and a condenser of capacity C, in series,

is Find the current at any time , when .

Sol. The given equation can be written as .

Differentiating both sides w.r.t. t, we have

Or which is Leibnitz’s linear equation.

Which gives the current at any time .



Newton’s Law of cooling

Example: According to Newton,s Law of cooling, the rate at which a substance cools in

moving air is proportional to the difference between the temperature of the substance

and that of the air. If the temperature of the air is and the substance cools from

in 15 minutes, find when the temperature will be 4 .

Sol. Let the unit of time be a minute and T the temperature of the substance of any instant time

t. Then by Newton,s Law of cooling, we have

Integrating,

Initially, when

Substituting the value of c in (1), we have

Or

Also, when

Dividing (2) by (3), we have

Now, when T=40, we have from (4),

Hence the temperature will be of 4 after 52.2 minutes.



Conduction of Heat

The fundamental principle involved in the problems of heat conduction is that the

quantity of heat Q flowing per second across a slab of area A and thickness whose faces are

at temperature , is given by where k is the coefficient of thermal

conductivity and depends upon the material of the body.

Example: A long hollow pipe has an inner diameter of 10 cm and outer diameter of 20cm. The

inner surface is kept at 200 and the outer surface at 50 . The thermal conductivity is 0.12.

How much heat is lost per minute from a portion of the pipe 20 meters long? Find the

temperature at a distance x = 7.5cm from the centre of the pipe.

Sol. Here the isothermal surfaces are cylinders, the axis of each one of them is the axis of the

pipe. Consider one such cylinder of radius x cm and length 1 cm. The surface area of this

cylinder is . Let be quantity of heat flowing across this surface, then

Integrating, we have

Since



Also

Subtracting (3) from (2), we have

Or

Hence the heat lost per minute through 20 metre length of the

pipe=60*2000Q=120000*163=1956000 cal.

Now, let T=t, when x = 7.5

From (1),

Subtracting (2) from (5),

Dividing (6) by (4), we have

.



Multiple choice questions:

1. The general solution of the equations yp2+(X-Y)P-X=0 is

A) (x-y-c)(y2-x

2-c)=0 B) (yx-c)(x

2-y

2-c)=0

C) (y-x-c) )(y2+x

2-c)=0 D) (y-x

2-c)(x

2-y

2-c)=0

2.The given differential equation is solvable for x if it is possible to express x in term of

A) x & y B) x & p C) y & p D) None of these

3. The singular solution of the equation y=px+a/p is

A) y2=4ax B) x

2=4ay C)x

2=y D)y

2=x

4. The general solution of the clairaut’s equation is

A) y=cx + f(c) B) x=cy +f(c) c)y=c x-f(c ) D)none of these

5. The differential equation of first order but second degree ( solvable for p) has the general

solution as

A) f1(x,y,c)+f2(x,y,c)=0 B) f1(x,y,c)*f2(x,y,c)=0

C) f1(x,y,c) - f2(x,y,c)=0 D) f1(x,y,c)/f2(x,y,c)=0

6. If the given differential equation is solving for x then it is of the form

A) x= f(p/y) B) y= f( x,p) C) x= f(y/p) D) x=f( y,p)

7. Clairaut’s equation of P= sin(y-xp) is

A) y= p/x+ sin-1

p B) y=px+sinp C) y=px+sin-1

p D) y=x+sin-1

p

8. The differential equation of R.L series circuit is

A) +Ri =E B) L +I =E C) +Ri = D) L + Ri =E

9. The general solution of the equation - = -

A) (x2+y

2+c)(y-c)=0 B) (x

2+y

2+c)(xy) =0

C) (x2-y

2-c)(xy-c)=0 D) none of these

10. The general solution of the equation y= 2px+ y2p

3 is

A) x2 =2cx + c B) y

2=2cx+c

3 C) y=2cx+c

3 D) none of these



11. The generally clairaut’s equation in the form

A) xy= px+ F(p) B) y2= px+ f(P) C) y = p x +f(p) D) y= pc+f(c)

12. The general solution of the clairaut’s equation is obtained by replacing

A) x by c B) y by c C) x by y D) p by c

13. The given differential equation is solvable for y if it is possible to express y in term of

A) x & y B) x & p C) y & p D) None of these

14. The general solution of the equation p= sin(y-xp) is

A) x= cy + sin-1

c B) y = cx + sin-1

x C) y = cx + sin-1

c D) y = px + sin-1

p

15. The general solution of the equation y= 3x+logp

A) y= 3x+logc B) y= 3x+logy C) y= 3x+log(3+cey) D) y= 3x+log(3+e

x)

16. In general, singular solution of the equation in the form

A) f(x, p, c)=0 B) f( y,p,c)=0 C) f ( x,y,c) =0 D) f( x,y,p)=0

17. The singular solution of the equation Y=Px +

A) y-k=4ax B) (y+k)2=4ax C) (y- k)

2=4ax D) y+k=4ax

18. Clairaut’s equation of (px-y)(py+x)=2p by taking substitutions X=x2,Y=y

2 is

A) y2=cx

2- B) y=px- C) y=px+ D) none of these

19. The given differential equation is solvable for x then general solution of the equation in the

form

A) f(x, p, c)=0 B) f( y,p,c)=0 C) f ( x,y,c) =0 D) f( x,y)=0

20. The singular solution of the equation is y=cx+ is

A) x+y=4x B) (x+y)2=4x C) (x+y)

2=4 D) none of these

Answers:

1-C 2-C 3-A 4-A 5-B 6-D 7-C 8-D 9-C 10-C

11-C 12-D 13-B 14-C 15-A 16-D 17-B 18-A 19-C 20-B

engineering mathematics – ii - e-learning · engineering mathematics – ii (10 mat21) ... sol:...

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