Engineering Mathematics 1

ENGINEERING MATHEMATICS -ISECOND EDITION

P.B. Bhaskar RaoM.Sc., Ph.D. Retd. Professor, Former Chairman, Board of Studies, Department of Mathematics Osmania University Hyderabad

S.K.V.S.SriramacharyM.A., M.Phil., B.Ed. Professor & Head (Retd.) Department of Mathematics University College of Engineering (Autonomous) Osmania University Hyderabad

M. Bhujanga RaoM.Sc., Ph.D. Professor, Dept. of Mathematics University College of Engineering (Autonomous) Director of Centre for Distance Education Osmania University Hyderabad

BSP BS Publications4-4-309, Giriraj Lane, Sultan Bazar, Hyderabad - 500 095 A.P. Phone: 040 - 23445688 e-mail: [email protected]

Copyright 2008, by Publisher

All rights reserved. No part of this book or parts thereof may be reproduced, stored in a retrieval syste'm or transmitted in any language or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publishers,i

Published by :

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Printed atAdithya Art PrintersHyderabad.

ISBN: 978-81-7800-151-7

Contents

CHAPTER -1 Ordinary Differential Equations of First Order and First Degree ..................................................... 1 CHAPTER -2 Linear Differential Equations with Constant Coefficients and Laplace Transforms ...................... 69 CHAPTER-3 Mean Value Theorems and Functions of Several Variables .............................................. 111 CHAPTER-4 Curvature and Curve Tracing ................................................ 213 CHAPTER-5 Application of Integration to Areas, Lengths, Volumes and Surface areas ........................ 313 CHAPTER-6 Sequences of Series .............................................................. 385 _ CHAPTER-7 Vector Differentiation ............................................................. 475 CHAPTER-8 Laplace Transforms ............................................................... 623

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1Ordinary Differential Equations of First Order and First Degree1.1 IntroductionDifferential euqtions play an important role in many applications in the field of science and engineering, such as (i) problems relating to motion of particles (ii) problems involving bending of beams (iii) stability of electric system, etc. For example, Newton's law of cooling states that the rate of change of temperature of a body varies as the excess temperature of the body to that of its surroundings. If 8(t) is the temperature of the body at time 't' and 8 0 is the temperature of the room in which the body is kept, then dt gives the rate of change of temperature with time.

de

dedt

= K(8 - 8 0) ; K is constant

Similarly Newton's second law of motion for a particle of mass m moving in a straight line can be written as

d 2xm dt 2

=F

Where m is the mass, x is the distance of the particle at time 't' measured from a fixed origin and F the external impressed force.

2

Engineering Mathematics - I

A differential equation is an equation involving an unknown function and its derivatives. Ifthere is only one independent variable and one dependent variable the equation is called (Ill ordinary differential equation. If there are more than one independent variable the equation is called a partial differential equation as this involves partial derivatives. For example:

4

d3y dx3

+3x

dy dx

_ y=e(

.... (a)

12 + Y 6 -x 3 d y J4 + (d2 YJ8 + (dY) _ 8 (dx;3 dx 2 dx

.... (b)

.... (c)

.... (d)

.... (e)

.... (t)

.... (g)

The first four equations (a), (b), (c) and (d) are ordinary differential equations and the remaining three are partial differential equations.

Order 0/ a differential equation: The order of a differential equation is the order of the highest ordered derivative appearing in the equation. DegreeThe degree of a differential equation is the power to which the highest ordered derivative appears in the equation after clearing the radicals if any.

0/ a differential equation:

Ordinary Differential Equations of First Order and First Degree

3

In the above examples:

Example: Example: Example: Example:

1.1(a) is a differential equation of order 3 and degree I. 1.1(b) is of third order and fourth degree differential equation. 1.1(c) is a second order, first degree differential equation. 1.1(d) is a second order, second degree ditferential equation.

1.2

ExampleFormation of an ordinary D.E : The differential equations ar~ formed by eliminating all the arbitrary constants th~t are involved in the functional relationship between the dependent and independent variables. For example:

y

=

cx2 + c 2 where c is an arbitrary constant.

.... (I)

To eliminate 'c': (only one constant) From(l)dv_0

dx

=

c.2x+ 0

I (~V c= 2x dx Substitution of c in (I) gives

y

=

_I dy x2+ 2x d\: 4x 2

_1_(dy )2d);=

d ~ ( dx

)2 + 2x 3 2 d

dx

- 4x2y

0

is the required D.E and y = cx2 + c 2 is called the solution of the D.E.

Note:Depending on the number of constants in the given equation differentiate it as many number oftimes successively. Then the elimination of the arbitrary constants from the resulting equations and the given equation gives the required differential equation whose order is equal to the number of constants.

1.3

ExampleEliminate the arbitrary constants a, b from xy + x 2 = aeX + be-X and form the differential equation.

4


Solution:The given equation is xy + .x2 = ae-'" + be-x..... ( I)

The number of arbitrary constants is two. Differentiating (I) w.r., to 'x' two times successively.

dyx dx

+ Y + 2x = ae-' - be-x=

_

.....(2)

d2y

X ---2

ely dy + - + - + 2.1 d-c dx dx

aex + be-x

.... (3)

From (I), (2) and (3) el imination of a, b gives the D.E. from (I) and (3) we get

x

--?

d l )' 2dy . . + - - + 2 = xy + x 2 IS the requIred D_E. dxdx

1.4

Example

a.x2 + by =

Form the differential equation by eliminating the constants a and b from 1

Solution:Differentiating ax2 +

by =

I w.r.t 'x'=

.... (I)

dy 2ax+ 2byd-c

0

.... (2)

Again differentiating wr.t., 'x' 2a+2by-?

d 2y dy dy +2b-.- =0 dxdx dx

.... (3)

Elimination of a, b from (I), (2) and (3) gives? x-

iyYl (Y.h + yl2

-I

x

0 =0 0

Expanding the determinant we get

2 y x d y +x(d )2 2 dx dx

_ y(dY)=odx


5

1.5

ExampleForm the differential equation by eliminating the constants from

y

=

a secx + b tan x

SolutionGiven equation is

y= asecx + btanxDifferentiating w.r. to 'x'

.... (I)

dy dx dy dx= =

asecx tanx + bsec 2x

.... (2)

secx[a tan x + b sec xl

.... (3)

Further differentiation gives

d 2y-, 2

(X

= a sec x tan 2x

+ a sec3 x + h2sec 2x tan x

I.e.,

--1

d Ydx-

2

=

asecx tan x

2

1 + bsec-xtanx + bsec 2xtanx + asec 3x

I.e.,Substituting

d 2y

dx 2

= secx t'lnx(atanx + bsecx) + sec 2x(btanx + asecx) ..... (4)

asecx + btanx

=y

from (I)

and in (3) we get

atanx + bsecx

= -secx

(~) from (2)-2

[-f)d2dxi.e.,

=

dY) ( secx tanx ~ + sec x(y) secxdx=

d 2y dy -- tanx - - ysec2x 2dx

0

6


1.6

ExampleForm the differential equation of all circles passing through the origin and having their centres on the x - axis. 3

-2

Substituting c = -2 in (I)

ysec 2x

=

secx - 2

is the required solution

1.12.6

Exampledy Solve (x + 2.v) -,(X=

y

SolutionX

+

2.vx y

=

y yd dx

--- =

dx dy

2.v

is in the form of

P

I

= -

-1

Y

'

q

1

=

2,1Y

y

28


1 x-

y

=

f2y

2

1 x-dy+c y

- =y+CY

x

1.13.7 ExampleSolve(x + y+ I) -

dy dx

=

1

Solutiondy dx- X =y+l PI=-lq\,=y+1

IF ==

e

f-1dy

e- Y

Solution is given byx(IF)

= fql(/F)dy+cf(y + I)e-Y dy + c

xe-Y

=

i.e.,

or

x +y + 2

=

ce Y

Exercise 1(e)

1.13 Solve the Following Differential EquationsI. (I + y)dx= (t~n-Iy

- x)dx [ADS: xetan- I y = tan-Iyetan-Iy --etan- I y + c]

dy 2. cos2x - + Y

dx

= tanx[ADS: Y = ce-tanx - tanx -1]


29

dy 3. x - +2y-x2 1og=0 dx(ADS: y

c=-?

x-

I .r2 +- x210gx --I4 16

4.

dy dx + ycot x = 4x cosec.x, if y

=

n/ 0, when x = ~2n2 (ADS: ysinx = 2x2 - 2

5. yeYdx

=

(y

+ 2x&)dy( ADS : xy~2

=c-

e~Y

I

6. (x + 31) dy dx

=

y

( ADS : x

=

21 + cy ]

3

7. (xy -I )

dy dx+ y3

+ y3

= 0

I

I ADS: xdy dx

= ce Y

.

+ - + II Y

I

8.

= x 3 - 2xy if y = 2 when x =

I

[ADS: 2y - x 2 + 1 = 4 el~~ ]

dy x+ ycosx 10. dx = I +sinx[ADS: y(l+sinx)=

c - -] 2

x2

30


1.13.1 Non-linear Differential Equation of First OrderBer noulli j. equation:The differential equation of the formely dx + PY=

llyn

.... (I)

where p, q are functions ofx alone is said to be a Bernoulli's differential equation. Dividing (I) throughout by yn

dy =q dx Substitutingynt-l(coefficient ofp) "" vy-n_ + py-n+l

.... (2)

1 dy dv (1-n) - - = y" dx dx(2) reduces toI dv - - - +pv=q

(I-n)dl'

dv - + (1-n)pv = (I-n)q dxwhich is linear in v. The avove D.E can be solved by using the method given in 1.12.1 example.

1.13.2 ExampleSolve dy - ytanx = ysecx dx Solutiondx - ytanx = ysecx

dy

.... (I)

dy 1 y-2 -d - - tan x = sec x x YSubstituting

-y =v,


31

+--=/ dx dx

1 dy

dv

(1) reduces todv dx + vtan x=

sec x

is linear in v. Here p=

tanx, qe e

=

secx

IF

=

fpd cosx dx = II in the RHS

siny

e;lIlt

= [te t =

el] + c

sinyeSlI1\"

eSItlX [sinx-l] + c

Exercise - 4(f)

1.14 Solve the Following Differential EquationsI. (ylogx -I )ydx =x(~y

1 (Ans: y2.

=

1 + logx + ex )

dy . - cosx + YSlnx = dx

...; ysecx

r.::::::::(Ans :2y y12

-Jsecx

=

tanx + 2c )

( Ans :dy tany ---Ix l+x

~y-

= -

sin 2x -sinx -~ + ce 2SIIl\"

2

)

4.

=

(I+x)

~secy

[Ans: siny = (1 + x) (eX + c) )

36


dy 5. x - + ylogy = xyeX dx

Idy 2 x3 6. 3-+--y=-) dx x+ I y

ADs : xlogy = (x - I)e"

+c

J

I ADs:7.dy + ytanx dx dy dx=

(x +

Iii =

x6-

6

+-

2x 5

+- +C 5 4

X4

isecx

I ADs:8.X2

cos2x

=

y(c + 2sin x) J

Y +xy

1

1.14.1 Exact Differential EquationsLet us consider the differential equation Mdx + Ndy = 0 where M, N are functions ofx,y. If this equation is to be exact, then it must have been derived by directly differentiating some function ofx,y. Hence

Mdx + Ndy = du, say

.... (I)

But from differential calculus

du = -dr:+-dy ax ayFrom (I) and (2) we get M= ax'

au

au

.... (2)

au

N=

au

ay

Now

-=--

aMay

a2u

ayax

and -ax- - -ax-ay-


37

-

aM

oy

= -

oN.IS tIle con d ItlOn .. lor exactness.

AX

:. The differential equation Mdx +

N~v =

0

. 'f aM aN IS exact I - = -

oy

ax

Then the solution is expressed in the form

+(treatingyas constant integrate w.r.t x) (integrate w.r.t y those terms that are independent of x)

Note:IfN has no term independent ofx then the solution is fMdx=

e

1.14.2 ExampleSolve (x + 2y - 3) dy - (2x - Y + \ )dx = 0

Solution(x+2y - 3)dy - (2x - Y + I) dx = 0 M = -(2x - Y+ \) N = (x + 2y - 3)-=\

aM

oy

-=\

aN

ax

The given differential equation is exact The solution is - f(2x - y + \ )dx +

fe x + 2y - 3)=

=

(ry =e

e

_2X2 2i - -2- + yx - x + 2

- 3y =

=>

I - x 2 + xy -

x - 3y

e

38


1.14.3 ExampleSolve (x 2+ y.)dr: + 2xy dy = 0

Solution

(x 2 + .V) dx + 2xy dy M=x2+y. -=2y

=

0N=

2xy =2y

aM av

-

aM ax

The given differential equation is exact Solution is f(x + y22

}h + f2xydy = c

1.14.4 ExampleSolve (I + e

-,:~, )dx + /Y

[I.'... ;

1(

dy = 0

SolutionY (I + e'x Y ) dx + e x'

J -;;, ) dy = 0

t Y x'y M = J + e: , N = e,i

[

J- Y

x)

-=-

aM~

-x

,y'e'Y

i

oM oNoy oxThe given differential equation is exact


39

Solution is

X

+ e

x

Y

(y)

=

c

Exercise - 4(g) 1.15 Solve the Following Differential Equations1. (el' + 1)cosxdx +eJ'sinx~v =

0

lADs: (oY + 1)sinx = c2. (vcosx + siny + y)tb' + (sinx + xcosy + x)dy=

I

0

I ADs: ysinx + (siny + y)x = c I3. (x 2-

ay)dx = (ax - ;l)(~v

ADS: x 3 + ),3 - 3ll.\y

= C

I

4. (ax + hy + g)d'C + (hx + hy + j)dy"= 0ax I ADs. - 2 5. (x 2 +2

+ (liy + g)x + .!. ':'

hi

+ fy

=

c

I

1- a2 )xdx + (x 2 -.v - P)ydy =

0

I ADS1.15.1 Integrating factors

: x4 + 2x21

- 2a2x2 - l - 2b 2;l =

4c

I

If the differential equation Mdx + Ndy = 0 is not exact, it can be made exact by multiplying it with some function of x, y. Such a function is called an integrating

jactor.Rule!t' for fillt/illg the illtegrtltillg factors :1. Integrating factors found by inspection:

ExampleSolve x dy- ydx=

0

Solution

xdy- ydx

=

0

40


Dividing by x2---'-,=-=--

xdy- ydxx2

=

0

On integration

YxMx + Ny7:-

=c

First method /0 find an integrating jac/or : If the differential equation Mdx + Ndy is not exact, but is homogenous and0, then the integrating factor is 1 . Multiply the differential Mx+Ny

equation by IF. The DE becomes exact.

1.15.2 ExampleSolve

(x 2y - 2xy2)dx - (x 3

-

3x2y)dy

=0 =0(x 3-

Solution(x 2y - 2xy)dx - (x 3-

3x2y)dy

.... (I)

The differential equation (I) is homogeneous

M -

= x 2y- 2xy=x2-4xy

N-

= -

3x2y)

aMoy

aN

ax

=-3x2 + 6xy

The DE is not exact and

Mx + Ny = x 2yIF=

7:-

0

) 1 Mx+ Ny - x 2 y2

Multiplying the DE by the integrating factor ~ x y

(

y2 X2Y -2X2

X

Y

2

)d _(X2 -3X3Y )dYX2

=

X

Y

2

0

.... (2)


41

write

MI

= ---

1

2

Y

X

-x 3 N =-+I y2 Y

then

--=-=--

aMI

aNI

ry

ax

y2

DE (2) is exact Solution is

x - - 210gx + 3logy + c yi.e.,x

or

1.15.3 Second Method to Find the Integrating FactorIf the differential equation Mdx + Ndy = 0 is not exact and is of the form

j(xy)ydx + g(xy)xdyand then

=

0

Mx- Ny *- 01 Mx _ Ny is an integrating factor

1.15.4 ExampleSolve (x 2 y2 + xy + 1)ydx + (x2y2 - xy + 1)xdy

=0.... (1)

Solution

(x2y2 + xy + 1)ydx + (x2y2 - xy + 1)xdy = 0M

= x2j3 + xy2 + y, N = x2y2 - x2y + x Mx - Ny = 2x2y2 *- 0IF=-- Mx- Ny 2x 2y2

1

1

Hence the

42


Multiplying (I) by IF

2 (X y 2 +xy+ 2x 2y2

1)+ (X

2

y 2 -xy+ 2x 2y2

I) = 0dy=O .... (2)

- y+-+dx+- x - - + 1 2 x x 2y 2 y xy-

I{ I 2} I{ II}

DE (2) is exact Solution of the DE (I) is

~ 'y+~+-!-tx+~ 'x_J.-+_I_ )dY = c2

Jl

x

x- y

r

2

Jl

y

xy

1

- xy + log x - - - -logy = c 2 xy 2

1 [

I] I

1.15.5 ExampleSolve (xysin xy + cos xy)ydx + (xysin xy - cos xy)xdy = 0

Solutiou(xysin xy + cos xy)dx + (xysinxy - cosxy)xdy = 0 M .... (1)

= (xysin xy + cos x,v)y, N = (xysin xy - cos xy)x=

Mx - Ny

2xycosxy

=F-

0

IF=---

Mx- NY

2xycosxy

Multiplyingthe DE by 2xycosxy The DE reduces to

~(ytanXY+~)dX+~( xtanxy- ~) dy

=

c

.... (2)

Ordinary Differential Equations of First Order and First Degreewhich is exact (verify) .. Solution is

43

-Iogsecxy + -Iogx - - log y 2 2 2I.e.,~

1

1

1

=

c

logxsecxy = 2c + logy. Taking 2c as log A xsecxy

= AyExercise - 4(h)

1.16 Solve the Following Differential EquationsI. (x 3;

+ x 2y2 + xy + 1)ydx + (x 3;

-

xli - xy +

1)xdy

=

01 - - 210gy = c xy

r ADS: xy 2.

I

(xli + xy +

I )ydx + (xli

-

xy + 1)xdy = 0

r ADS:

1 xy + logx - logy - xy

=

c]

[ ADS: logx2 - logy - _I = c ] xy

4. (x4y4 + x 2

i

+ xy)y dx + (x 4y4 - x 2i + xy)xdy

=

0

I ADS : ~ x 2i2

- _I xy

logx - logy

=

c ]

1.16.1 Third Method to Find the Integrating Factor

aMIf Mdx + Ndy = 0 is not exact and

aN Nis a function of x alone say fix),

_a-=-~_ _ ax_

then the IF

=e

J

j (x)dr

44


1.16.2 ExampleSolve

(x 2 + y + 6x )dx + yxdy = 0 (x 2 + Y + 6x)dx + yxdy = 0

Solution..... (1 )

M = x2 +

Y + 6x

N = yx

aM aN ay= 3y, fu = y2 aM aN

ay-iliN IF = e f~2tJx

3y2 - y2y x2

2X

is a function of x alone

= x2

Multiplying the given DE by x2

x2(x2 + y + 6x)dx + yi3dy = 0

..... (2)

-

aMI

ay

=3x2y

DE (2) is exact Solutionis

f(x + x

4

2

/

+ 6x 3 }tx + fy x dY = c

2 2

1.16.3 ExampleSolve

(x2 + y) dx - 2xydy = 0

Solution(x2 +

y)dx - 2xydy = 0

..... (1 )


45

M

= .xl + .0, N = - 2xyaN=

aMay

2y, ox

=

-2y

aM aNaxN

ax

2y+2y -2xy

-2 x

is a function of x alone

IF

= e

f -~dx 1 x =x21

MuItiplyingthe DE by

IF= -

x2

The given DE reduces to

[ l)x

2y 1+dx,.--dy=O 2x

..... (2)

-a --2' y X(2) is exact

aM, _ 2y

-=-

oN,

2y

Solution of (1) is

J

l+-dx+ J--dy=c x2 x

y2

2y

x- -

y2x

=c

1.16.4 Fourth Method to Find the Integrating Factor

aN aMIf the differential equation Mdx + Ndy

= 0 is not exact and

ax M By

1

is a

[function of y alone, say fly)

46


Then

IF

=

e

f

t(y)

log5 = 0 - c c=-log5

SubstitutilllJ; c value in (1) log m = kt-log5

I.e.,

kt= 10g(

~)

.... (2)

When t = 30 minutes mass deposit increases by , I ' gram m = 5 + 1 = 6 grams Substituting t=

1 "2 (hours), m =6

Substituting in (2) log(1Y t= 109; Now to find the mass after 10 hours (i.e t = 10) from (3) we get.... (3)

IOg(H IO~ log 7~ log (7) ~ IOg(%)"x

m = 5"5

(

6)20

grams

541.17.5 Example


The rate at whi~h a certain substance decomposes in a certain solution at any instant is proportional to the amount of it present in the solution at that instant. Initially, there are 27 grams and three hours later, it is found that 8 grams are lett. How much substance will be left after one more hour.

SolutionIf m grams is the amount of the substance left in the solution at time 't', then the rate at which it decomposes is dm , which is proportional to m. dl By law of decay

dt

dm

= -

km (k> 0)

f~

=-k fdt+c.... (I)

logm = -kt+ c Initially when t = 0, m = 27 From (1) we have log27 = - k.O +c

=>

c= log27

Substitution of'c' in (I) gives log m = - kt + log27

10g(;; )= -

kt

.... (2)

It is given that m = 8 when t = 3 .. From (2) 8 log (2 ) = - k, 3 7

8 -k= log ( 272 -k=log3

)X


55

Then (2) becomes

10g~ =IOg(%}when t = 4

... (3)

III

log 27

= log ( "3 )

2

4

m=27 x (%r grams

m=

3

16

grams.

1.17.6 Example The number x of bacteria in a culture grow at a rate proportional to x. The value ofx was initially 50 and increased to 150 in one hour what will be the value ofx after 12" hour. Solution

1

dx -=/0:dt

-

dx =kdt x.... (I)

logx = kt + c c is the constant of integration whent= O,x= 50

..or(1) =>

log 50 = k.O + c c= log50 logx = kt + log50

56


x log- =kl 50 x = 150, when 1= 1.. or 150 log 50

= k.I

k= log3

(2) then gives log (

5O) = flog3=

x

we want to find x when I

"2

3

X

3

50X

=3

(3)2

= 50 (3)2 grams

1.17.7 Example The rate of cooling of a body is proportional to the difference between the temperature of the body and the surrounding air. If the air temperature is 20C and the body cools for 20 minutes from 140C to 80C, find when the temperature will be 35C. Solution If 8 is the temperature of the body at time '1' then from Newton's law of cooling

-d8 -a(8-20) dl= f~ 8-20

~

- =-k(8-20) dl

de

k rldl + c

J'

log(8 -20) = - kt + c Initially when t = 0, 8 = 140 log(8 - 20) = 0 + c,

.... (I)


57

c = log( 120), (I) reduces to log(140 - 20) = - kt + log 120 or +kt = log( 120) - 10g(S - 20).... (2)

It is given that q = 80 when t = 20 minutes k. 20= logI20-log(80-20) I (120) k= 20 log 60I k= -log2 20

Substituting in (2) 1 (2 0 IOg2)t= logI20-log(S-20)It is required to find t when

0= 35c0 20 10g( \25 ) log2

logI20-log(35 - 20) t= I -log2 20

10g(8) 2010g23 60 10g2 . t= 20 - - = - - = - - =60mmutes log2 log2 log2

Exercise - 4(k)I. In a certain reaction, the rate of conversion of a substance at time "t' is proportional to the quantity of the substance still untransformed at that instant. At the end of one hour 60 grams while at the end offour hours 21 grams remain. How many grams of the first substance was there initially? ( ADs: 85 grams approximately I The rate of growth of a bacteria is proportional to the number present. If initially there were 100 bacteria and the amount doubles in '1' hour, how many bacteria will

2.

1 be there after 2"2 hours.(ADs: 564

I

58


3. Under certain conditions cane sugar in water is converted into dextrose at a rate which is proportional to the amount unconverted at any time. 1f75 grams was there at time t = 0.0 and 8 grams are converted during the first 30 minutes find the amountI converted in 12 hour.[ADS: 21.5 gms]

4. The rate of cooling of a body is proportional to the difference between the temperature of the body and the surrounding air. If the surrounding air is kept at 30c and the body cools from 80c to 60c in 20 minutes. Find the temperature ofthe body after 40 minutes.[ADS: 48c J

5. If the air is maintained at 30c and the temperature of the body cools from 80c to 60c in 20 minutes. Find the temperature of the body after 40 minutes.[ ADS: 48c]

6. The rate at which a heated body cools in air is proportional to the difference between the temperature of the body and that of the surrounding air. A body originally at 80 cools down to 60c in 20 minutes the temperature of the air being 40c what will be the temperature of the body after 40 minutes from the original temperature.[ADS: 50c J

7. The rate at which bacteria multiply is proportional to the instantaneous number present. If the original number doubles in 2 hours in how many hours will it triple.[ ADS: 210g3 J

log2 8. Water at temperature 100C cools in 10 minutes to 80C in a room of temperature 25C. Find the temperature of water after 20 minutes.[ ADS: 65.5c]

9. A cup of coffee at temperature 100C is placed in a room whose temperature is 15C and it cools to 60C in 5 minutes, find its temperature after a further interval of 5 minutes. [ADS: 38.8c ]


59

Orthogonal Trajectories1.18.0 Definition (1)A trajectory of a family of curves is a curve cutting all th.e members of the system according to some law. For example a curve cutting a family of curves at a constant angle is a trajectory.

Definition (2)If a curve cuts every member of a given family of curves at right angles, it is called an orthogonal Trajectory. The orthogonal trajectories of a given family of curves _ themselves form a family of Curves. If the two families of curves are such that each member of either family cuts each member of the other family at right angles then the members of one family are known as the orthogonal trajectories of the other. In two dimensional problems in the flow of heat, the curves along which the heat flow takes place and the isothermal curves or loci of points at the same temperature are orthogonal trajectories. In hydrodynamics, the flow of water from a lake into narrow channel produces a family of streamlines which are orthogonal trajectories to the curves of equal Velocity Potential. In the flow of electricity in thin conducting sheets, the paths along which the current flows are the orthogonal trajectories of the equipotential curves and vice - versa.

1.18.1 Orthogonal Trajectories: Cartesian CoordinatesLet

f(x,y,c)=O

..... (1)

be a family of curves, where c is a parameter. We can form a first order differential equation by eliminating 'c' from (1) \ i.e.,

F(X,y,:)=o

..... (2)

is the differential equation whose general solution is (1 ) If the two curves are orthogonal (curves intersecting at right angles) the product of the slopes of the tangents at their point of intersection must be equal to -I. Suppose

(x, y)

is the point of intersection of the curve (I) and its orthogonal

trajectory .

60


At this point slope of the tangent to the curve (I) is dy and -dx is the slope of

dx

dy. dx

the tangent to the orthogonal trajectory. Therefore on replacing dy by - dx in (2)

dy

, the equation thus obtained is the differential equation of the family of orthogonal trajectories of the family(l) .

. . (Suppose

xJI,

-

:)=0

...

(3)

is the differential equation of the system of orthogonal trajectories, and its solution is the fami Iy of orthogonal trajectories of (I) .

1.18.2 Orthogonal Trajectories - Polar Coordinates

f(r,O,c) = 0

..... (1)

is the family of curves where c is the arbitrary constant. We can form a differential equation

F(r,o, :~ )= 0

..... (2)

of the family (I), after elimination of the constant 'c'. Let be the angle between the radius vector and the tangent at any point on a member of the family of curves. Then

(r, 0) ..... (3)

dO tan=rdr

Let I be the angle between the radius vector and the tangent at any point (lj, ~ ) on the trajectory. Then tan;/, Y'I

dB = r, _I 1 dr,I

..... (4)

At a point of intersection of the given curve and the orthogonal trajectory

lj=r,From (3) and (4),lj

01 =0

Hence tan1

= -cot

dOl dr l

= --; dO

1 dr

1 dri.e., -; dO

= -lj dlj

dOl


61

Hence the differential equation of the orthogonal trajectory is obtained by

. dB fi 1 dr. J dB fi dr Sll bstltutmg - r - or - - , I.e., -r- ordr r dB dr dBTherefore the differential equation F

(r, B, _r2 ~~) = 0

..... (5)

gives the differential equation of orthogonal trajectory of the family of cllrves(l) Solution of (5) is the orthogonal trajectory of the family of curves.

1.18.3 ExampleFind the orthogonal trajectory of the family of curves a/ parameter.

= x 3 , where a is variable

Solution:Given family of curves is

a/ = x

3

..... ( I)

Differentiating (I) w.r.t. 'x'; 2ay dy

dx

= 3x 2

..... (2)

Eliminating 'a' from (1) and (2)

.. 2 /

. (X3) .y dy _ dx -3x2xdy =3y dx

2

..... (3)

is the differential equation of the family (I). Now replacing :

by - : in(3),

gives the differential equation of the orthogonal trajectory to (I) as -2x dx = 3y

dy

Integrating both sides2X2

23 2 -2~ = L + c . Therefore 2 2

+ 3/ = c is the equation of orthogonal trajectory of (I)

62


1.18.4 ExampleFind the orthogonal trajectory of the family of parabolas y2 parameter.

= 4ax

where 'a' is the

Solution:

y2 =4axdifferentiating (I) w.r.t. 'x';

..... (1)

y dy dx

= 2ay2 = 2X(Y:)

..... (2)

Eliminating 'a' from (I) and (2).

.. y=2x

dy dx

..... (3)

is the differential equation of the family (1) Replacing

dy by _ dx in(3) dx dy

Y=2X( -: Jis the differential equation of the orthogonal trajectory to (I). Integrating (4)

..... (4)

fydy = -2 fxdx + c; 2X2 + y2 = c is the orthogonal trajectory of (I)1.18.5 ExampleFind the orthogonal trajectories of ~ + ( 22 2

a

a +A

= 1 whereA is the parameter.

Solution:

-+ =1 a2 a2 +ADifferentiating (I) w.r.t. 'x';

X2

y2

..... (1)

2x + 2y dy a 2 a2 +A dx

=0..... (2)


63

Eliminating A from (I) and (2)

xxy -2- - - = 1 a a 2 -dy dx..... (3) ---

)

:. ( x

2

-

a 2 ) : = xyY ) = xy (2 X -a 2)( -d dx

is the differential equation of the family (I) dy.111 (3) -dx - fOor S ub stltutmg;

dy

dx2

. I.e,

y

dY~-( x' x a

)dx22

is the differential equation of the orthogonal trajectory to (1) Integrating we get-=a logx---

Y 2

2

X

2+c

x +l2

= 2athe

2

log x + C is the orthogonal trajectory of the family of curves( 1)

1.18.6 ExampleFind2

orthogonal

trajectories

of

the

family

of

coaxial

circles

x + l + 2gx + C = 0 where g is the parameter.Solution:Given Differentiating (I) w.r.t. 'x' ; Eliminating 'g' from (I) and (2)

x 2 + y2 + 2gx + C = 0dy 2x+2y-+2g+0=O dx x + y2 - 2x ( x + y : ) + c = 0 y2 _ x 2 _ 2xy dy + c = 0 dx2

..... (1 ) ..... (2)

..... (3)

is the differential equation of the family of (1 ) Substituting - dx for dy in (3) we get,

dy

dx dx y2 _ x 2 + 2xy - + C = 0 dy..... (4)

64


Which is the differential equation of orthogonal traject{)ry. SimplifYing

dx 1 2 c+ / 2x---x = - - dy y y

..... (5)

is a linear equation of the Bernoulli's form (non - linear differential equation of first order and first degree) Substituting x 2 = t,

2x dx = dt in (5) dy dy(c+ / ) y..... (6)

dt t ---= dyis linear in t.

y

:.I.F=e

I 1 I--dy Y =-

YGeneral solution of(6) is

t.! = yt

f +yy2 .!dy+k yC

-=-y+-+k.y y

C

x 2 + Y - ky -

C

=0

( since t = x 2 )

is the equation of the orthogonal trajectory of (I)

1.18.7 ExampleFind the equation of the system of orthogonal trajectories of the parabolas

r

=

2a , where 'a' is the parameter. l+cosO 2a = r (1 + cos 0) 2a = 2r cos 2 0/2~ a = rcos 0/22

Solution:..... (1)

log a = logr + 21ogcosO/2.Differentiating w.r .to '0 '

o=! dr + 2 (-sinO/2).(l/2)=O i.e, r dO cosO/2


65

1 dr ---tanB/2 = 0 r dBSubstituting -r 2

..... (2)

10 elr. for m (2) we get dr dB

?dB) - -r-tanB/ 2 =0 r dr-

1(r

dr

=

-cotB/2dB

..... (3)

is the differential equation of the 0I1hogonai trajectory . Integrating (3)

log r

= -2 log sin B/2 + log c2

log(~) = logsinc .2 / (

B/2

-=sm B 2=--'--------'r 2

1- cos B)

:. 2c = r (1 - cos D) is the equation of orthogonal trajectory of the fami Iy of ( I )1.18.8 ExampleFind the orthogonal trajectory of

rill

= alii

cosmB where 'a' is the parameter...... (1)

Solution:Given

rill = a cosmBln

mlogr = mloga+ log cos mBDifferentiating w.r.to 'B' ;

mdr =

1

r dB

cosmB

( -msmmB . )

~ drr

=

dB

-tan(mB)

..... (2)

is the differential equation of the family of curves (I) Replacing -

dr

dB

by -r -

2

dB . m(2) d,.

66


~.(_r2 da) = -tan (rna)r dr dr = cot (rna)dar

..... (3)

is the differential equation of the orthogonal trajectory. Integrating (3)

rn log rln = log e sin (rna) ; rilllll

'log r = J.-Iogsin (rna) + loge= e lll sin (rna) is the orthogonal trajectory of the

family (I)Exercise 4 (s)1.

Find the orthogonal trajectories of the following family of curves, where a variable parameter.(i) (ii) (iii) (iv) (v) (vi)

IS

a

y=ax 2 xy=a 2 x 2 -xy+ l =a 2 2X2 + y2

[ADS: x +21 =e] [ADS: x - y2?

2

=e ]

[ADs: (x-y)=e(x+y)2] [ ADs: x

= kx

2

= - y2 log ( ;

)]

x 2 _ y2 = a2 X2 /3 + y2/3 = a2/3

[ADs: xy = e] [ADs: y4/3 _ X4 /3 = e 4 /3}

2.

Show that the system of confocal and coaxial Parabolas orthogonal.

l

=

4a (x + a)

is self

3.

3 Find the orthogonal trajectories of the curves 3xy = x 3 - a , a being the parameter.

4.

Prove that orthogonal trajectories of a system of circles x 2 + y2 - ay = 0 is

x 2 + y2 -bx =0

Ordinary Differential Equations of First Order and First Degree 5. Find the orthogonal trajectories of the following family of curves.

67

(i) (ii)(iii)

r=aBr =e aO[ ADS:

[ADS: r

= ce 2 I (log)2 + B2 = c II

-

0' --

rn sinnB = a"r = a(l +eosB)

[ADS: rll eosnB = e"

2 (iv) reosB = asin B(v) (vi)

[ ADS: r = e (3 + eos 2B) ] [ADS: r = c(l-eosB) I [ADS: r [ADS: r ?

r = aeosB 2 (vii) r2 = a eos2B

= esinO I?

= C- Sin 20 )

"This page is Intentionally Left Blank"

2Linear Differential Equations with Constant Coefficients and Laplace Transforms2.1.1 Let(i) the differential operator" ~" be donoted by 'D'

dx

(ii) PI' P2' P3' ............. Pn be either functions of x or constants (iii) R be a function of x then the general form of a linear differential equation (L.D.E) of order n is given by

- R Dn Y + PI Dn- I Y + P2 Dn- 2Y + ................. PrJ! -

.... (I)

or simply (Dn + ppn-I + ............ +Pn)Y = 0, If PI' P2' ..................., Pn are constants then (I) is called a L D E with constant coefficients. Denoting the differential operator (Dn + PI Dn - I + .............. + Pn) by f (D), (I ) can be written asf(D)y=

R

.... (2)

2.1.2 Consld er the D.E, dy dx + Py = 0

Separating the variables dy = - P(x) dx y

70


Integration yields logy= fp(x)dx => y =Ce -fp(x)dx + logc Solution isy= Ce-

..... (3)

fpdx= Cernx.... (4)

Suppose p is a constant = -m, say, then solution is y (i.e.) the solution of (D - m)y = 0 is y = Ce rnx

2.1.3

In this chapter the attention is mostly confined to L.D.E.S with constant coefficients. If R = 0 then equation (2) becomesf(O)y=

0

We shall take, here afterwards, f(m)

= 0 as the auxiliary equation, (A.E)...... (6)

2.1.4 Consider a second order L.O.E(D2 + a\ D + ~)y

=0

A.E. is m + aim + a2 Now four cases arise.

2

= o.

Let m\, m 2 be the roots of this equation.(i) m\, m2 are real and distinct

(ii) m\, m 2 are real and equal (iii) m\, m2 are complex and distinct (iv) m\, m2 are complex and equal

2.1.5 Case i : (6) can be written as[0 2 - (m\ + m 2 ) D + m\m 2 ] y

= 0 or (D - m2 ) (0 - m\) y = 0

.... (7)

-

Call (0 - m l ) y

=

y

(7) now becomes (0 - m 2) Y = 0

From (4) it follows that Y = Ce D12X From (7) (D - m\)y = Ce D12X I.F.=

e-rn\X .... (8)

Solution is y . e-m\x = C fe(m,-m')x dx + C\

y.e-rn\x=

Cm 2 -m!

e(rnrrn\)x dx + CI' Call

Cm 2 -m!

as C 2 .

Linear Differential Equations with Constant Coefficients ...

71

(i.e.,) y

=

C2 e-1Il 2x + C, elll,x where C, and C 2 are arbitary constants

Similarly ifm" m2, .......... I11n are real and dinstinct roots off(l11) = 0 then y = C, e'n,~ -I- C 2 el11 2x -I- .............. + C n elllnx

.... (I)

(I) is the solution of f(O)y = 0 where C" C 2, ........... , C n are arbitrary constants.

2.1.6 (Case ii) : m,

=

m2,

From (8) it follows that the solution of(6) is given by

:. y

= (Cx

+ C,) elll,x=

Similarly if m, is repeated say 'r' times then solution of(D - m,Y y

0 is ..... (II)

y = (C, + C 2x + C 3x2 + .............................. + c;-r-') elll,x(II) is the solution corresponding to a root m, repeated r times

2.1.7 Case (iii) : Ill" m2 are complex, say (a if3) Then solution of(6) is given by

y

=

A eta I~)X + B e(a=

IfJJX

where A and B are arbitrary constants

(i.e.,) y=

Ae ax . e Vlx + Beax.e-1jJx = eax [A(cosjJx + isinjJx) + B (cosjJx' - isinjJx)]..... (III)

eax [C, cos/lr + C 2 sin,lk]

where A + B = C, and i(A - B) = C 2(III) gives the solution corresponding to two complex conjugate roots (a ifJ)

2.1.8 Case (iv) : If (a ifJ) are repeated say's' times then the corresponding solution (follows from case (ii) and (iii) is given byy = ~[(C, + C 2x + C 3x2 + ........................ + CsXS- ' ) ] cosf3x

+ (d, + d 2x + d 3x2 + ....................... + dsxS- ' ) + sinf3x]Example 2.1.9Solve- 2 +5-+6y=O

.... (IV)

d2 y

dy

dx

dx

Sol. A.E is m2 + 5m + 6 => (m + 2) (m + 3) = 0:.m=-2,m=-3. Solution is y=

Ae-2x -I- B e-3x

72


Example 2.1.10Solve dx 2 -11 dx +30y = 0

d2y

dy

Sol. A.E is m2 - 11 m

+ 30 => (m -

5) (m - 6) = 0

Solution is y = Ae 5x + B e6x

Example 2.1.11Solve2 dx 2 +5 dx -12y=0

d2y

dy

Sol. A.E is 2m 2 + 5m - 12 => (2m - 3)(m + 4) = 0m =3/2, m=-4 :. Solution is y = C 1 e2x + c 2 e--4x3

Example 2.1.12d 3y d2y Solve -+4--6y=0 dx3 dx 2

Sol. m3 + 4m2 + m - 6 = 0(m - I) (m + 2) (m + 3) = 0 :. Solution is y = cl~ + c 2 e-2x + C 3 e-3x

Exercise - 2(a)Solve the following differences:

1.

d 2y dy -+2--3y=0 2dx2

dx

Ans : y

= A~ + Be-3xA~

2.

d y _ 3y = 0 dx 2

Ans: y =

+ Be-X

3.

d 3y d2y dy -+2--5--6y=0 2 3dx dx dx

Ans : y = Ae-X + Be2x + Ce-3x

4.5.

d2 y dy 9 -+18--16y=0 2dx dx

d2y dy -+3-+2y=O 2dx dx

Ans : y

= Ae-X + Be-2x


73

Solved ExamplesExample 2.1.13

dZy dy Solve -z +6-+9y=0 dx dX'Sol. A.E is mZ + 6m + 9 =

i.e. (m + 3)z = 0, m = -3,-3 :. Solution is y = e-3x (A + Bx)Example 2.1.14

Solve~+~=O 2 3dx dx

d3

dZ

Sol. A.E is m + mZ =3

=> m (m

Z

+ I) =

:. m=O,O,-1 Solution is y = (A + Bx)e oX + C e-x = (A + Bx) + Ce-XExample 2.1.15

Solve

- 4 + 181

d4ydx

dZydx~

+ 81y=0

Sol. A.E. is m4 -18m 2 + 81 =:. (mz - 9) (m z - 9) =

:. m = 3, 3, -3,-3Solution is y=

(A + Bx) e-3x + (C + xD)e 3x = (C(x + Cz)e-3x + (C 3x -t- C 4 )e 3x Exercise 2 (b) d2y dy +10- +25y=0dx

I.

Solve

-Z

dx

Ans : y = e-5x (Ax + B) Ans : y(Ax + B)e2x + cex

2.3.

=

Ans : y

=

(Ax + B) + (Cx +

D)~

4.

Ans : y =

(M + Bx + C) e-t

74Solved Examples Example 2.1.16 Solve


d2 y-2

dx

+4- +9y=0dx

dy

Sol. A.E. is m2 + 4m + 9 =0

m=-2i/5a=

-2 and

f3

=

/5=

:. Solution is y Example 2.1.17

e-2x (Acos x

/5 + Bsin /5 x)

d 3y Solve dx 3 + y= 0

Sol. A.E. is m3 + 1 = 0 :. (m + 1)( m2 - m + I) = 0

1+J3i :.m=-I,m=-2

Example 2.1.18 Solve -+2-+3-+2-+y=0 dx 4 dx' dx 2 dx Sol: A.E. is m4 + 2m 3 + 3m 2 + 2m + 1 = 0(m 2 + m + 1)2 = 0

d4 y

d3 y

d2 y

dy

:. m=

-1J3i 2

-1J3i 2


75

Exercise 2 (c)I.

Ans: y

= Acos

J7x

+ Bsin

J7x

2.

d 2y dy Solve22 +4- +3y=O dx dx

x x Ans : y = e-x (Acos.fi + Bsin .fi )

3.

4.

d4 y Solve dx4 -64y = 0Ans : y=

C\e2x + C 2e-2x + e-x (C 3cosfix+c 4 sin fix) + ~(C3cos.J3x+C6sinfix)

2.2.1 Consider the O.E.f{O) y

=

R

.... (I)

Operating both sides with f(IO) (called inverse operator)

1 we have f(O) [f(O)]yI

=

I f(O) R

or y = f(O) R This solution is called the particular Integral of(l) while the solution off(O) y = 0 is called the complementary function (C.F.) Suppose y\ is C.F and Y2 is P.I for (I) Thenf{O) y\ andf{O) Y2..... (2)

= 0 from (2)=

= R from (1)f(0) y\ +f(O) Y2

Hencef(O) [y\ + Y2]

=O+R=R

76


Which shows thatYI + Y2 is also a solution of(l)'Yl + Y2 (i.e.) C.F. + P.I is called the most general solution of(l).

2.2.2 Calculation of

ID-I11,

R

D _ nl, R = y, say, Operating both sides with (0-m 1), we get (0-m 1) Y = R(i.e.)

dy

dx - m1y= R f-mjtU

I.F.lse =e-ml x :. Solution is given by

y.e-mf = fRe-lIl/ dx+c(i.e) y = Ce-ml X + ell/IX fRe-1Il IX dx = C.r. + P.I P.I.=e mI X fHlIllx dx If

m

I

=O~ '0

R= fRdx

I Thus the operator 0 stands for integration

Example 2.2.3

Sol. D 5

~

Fe" fe".xdx ~ m = aiC.F. is y = C I cos ax + C 2sin ax

P.I =

I - I- - -I tan ax = (0+ ai)(O- ai) 2ai [ O-ai D+ ai

I

J tanax

2ai [e lax J e

-a/\

tan axdx - e -IllX J eat" tan axdxJ2

r Je -

alX

tanaxdx

=

J. . sinax (l-cos ax) (cos ax -Ismax) - - dx = JSin ax- i dx cos ax cos ax cosax ai isin ax -\og(sec at' + tan ax) + - a a

= - -- -

lilly

Je

an

tanax

dx

=

cosax i isinax - - - + - \og(sec ax + tan ax) - - a a a

II

=

(Cosax+aisinax) [-cos ax + isin ax -ilog (sec ax + tan ax]

a

[-\- icos ax log(sec ax + tan ax) + sin ax \og(sec ax + tan ax)]\

:. P.l. = - 2 ? [-2icos ax log(sec ax + tan ax)]

a-,

cos ax - - - - log(sec ax + tan ax) a :. Most general solution isy

= C 1 cos ax + C 2 SIl1

.

ax -

- - 2-

cos ax log(sec ax + tan ax) a

78


Example 2.2.6Solve (02-50+6)y = ~

Sol. A.E. is m2-5m+6 = 0C.F. is y

= C)e 2x + C2 e3xI=

P.1. isy= 0 2 -50+6 (eX)

2

eX(from 5.2.4)

:. Complete solution (C.S.) is y

= C.F. + P.I.

Exercise 2 (d)I I. 0 cosx I 4. 0-2 sinx

3. 0 (x 2 )

I

Ans: (I)+sillx (2) e-x (3)Solve the following: 6.d2y dx 2-

3

x3

(4)

(2sinx+ cos x)

5

dy 4 dx + 3y = e2x

7.

d2 y-2

dx

+y=5x+3dy 3 dx + 2y = e-2x

8.

d2 ydx 2-

9.

10.


79

II.

xsin ax Ans : y = C I cos ax + C 2sin ax + C 2sin ax + - - -

a

+ 12. (D 2 + 9) Y = tan 3x Ans : y=

cos ax-2-

a

log cos ax

C1cos 3x + C 2sin 3x

cos3x - - - log(sec 3x + tan 3x)

9

Methods of finding P.1.We shall now consider the methods for finding P.1. where R is of some special form

2.3.1 Particular integralWhere R is of the form eax Case (i) iff(a) Since=F

0

De ax = aeax ; D2e ar = a2 ear ...... Dn(eax) = an.earj{D) ear=

(Dn + K1D n- 1 + K 2 Dn- 2 + ...... K,)e ax

= (an + K1a n- 1 + .......... + Kn)e an = j{a) earOperating with _I- on both sides

feD)- I

feD)

[f(D)e a,]

= - (-)

I f D

[f(a)elU]

(i.e)

eax

=

1 f(a) eax = f(a). [ feD) I e ax] feD)1 at --e f(a)

(or)

1 a, --e feD)

=

1 1 1 Note: If K is a constant then feD) (K) = feD) K.eax = f(m)K3x

for example -D-:0 2-_-2-D-+-l e

I 3 I 3 9 _ 6 + I = e x = "4 e x

80


2.3.2 Case ii) If./{a) = 0 then it is possible to write./{O) as (0) (O-ay where (a) i:- 0 Suppose r = 1 thenP.1. = - - eax = - - - f(O) O-a (0)

1

I

1

e(lX

1 1 1 - - eax = - - - - eax O-a $(a) $(a) O-a = (la) eax

Je~a\ .eaxdx = ~a) ea\dx = ~a) ell B = 0:. y(x) = e-2x (-cosx) + 2e-2x

= 2e-2x - e-2x cosxExample 2.3.5

SolveSol.

d3 y d2 y dy - 3 +2+ - =e 2x 2 dx dx dxA.E. is m3 + 2m2 + m = 0 ; m(m+ 1)2 = 0

m = 0, -I, -I.C.F. = C1eOx + (C 2 +C 3x)e-XI e 2x 2x P.l. = 03 + 202 + 0 e = 18

82


Example 2.3.6_Solve

- , -3 dx-

d2 y

dy

dx

+2y=e-~

Sol.

f(O) = 0 2-30+2 A.E. is m2 _ 3m + 2 = 0 => (m - 2) (m - I) = 0; m = 1,2here f(l)I I I I=

0I

.. (0-1) (0-2) eX= (0-1) (1-2) e = (0-1) eX=-eX:. Most general solution is:. y

X

~

(using 5.3.2.(3)

= CteX + C2 e2x - xe n

2.4.1 R is of the form sin ax or cos a\"0 2 (sin ax) =:' (~a2i~ sin ax

(02f (sin ax) = (_a2)2 sin ax

(0 2)3 (sin ax)

=

(_a2)3 sin ax

(02)n (sin ax) = (_a2)n sin ax

Hence j(02) sin ax = j(-a2) sin ax, provided f( -a2)

:;i:

0

1

1 (0 )2

j(O )

2'SIl1

ax -

_

1

1 2 . (,)j(-a )SIl1 ax0-

Qr sin ax =

1(02 )I.

j(-a

2

)

sin ax sinax

1(02 )

SIl1

ax

=

f(-a 2 )I 2 cos ax ax = I, cos ax f(-O ) 1(-a-)

Similarly it can be shown that


83

Ifj{O) vanishes when 0 2 is replaced by _a2 then we write

j(O)

I

[sin ax] =

j(O)

1

. 1ar [Imaginary part of e ]

1 . = I.P of - - e 1ar

j(D)1

Similarly

j(O)

1

[cosar] = R.P. of

j(O)

e 'ar

.

Solved Examples Example 2.4.2

SolveSol.

-

d2 ydx2

-3 -

dydx

+ 2y = sm3x

.

A.E. is m2 -3m+2 = 0; m = 1,2

I. 1 . -1. (3 D - 7) . P.I., (sm3x)= 9 30 2 (sm3x)= (3 7 (sm3x)= 9D 2 49 (sm3x) 0- - 3D + 2 - + D + .) +

(3D - 7 ) . 9cos3x + 7 sin 3x = -81-49 (sm3x)=--0-2=-_-3D-+-2Hence complete solution is

84


Example 2.4.3Solve (D3 -D) Y = sillX

Sol.

A.E. is m 3 -m m=O, m =1

=0

C.F. is C1e.x +

c2eX + C3e-x =

C 1 + C2e x + C3e-x

I.Sit1X I I P.1. = D3 _ D = D2 _ 1 . D (sitlX)I -cosx x = - - - . cosx= = --SillX D2_1 -1-1 2

The complete solution is

y = C + C eX + C e-x -~SillX I 2 3 2 Example 2.4.4Solve (D2 - D + 1) y = cos2x

Sol. A.E. is m2 - m + 1 = 0 ~ m =~~

1J3i 2

:. C.F. = e 2 (Acos-x + B sin- x)

fi2

fi2

I P.I. = D2 _ D + 1 cos2x =

I

- 4- D+1I

cos2x =

1

-3 - D

cos2x = -

(3-D)

9 - D2

cos2x

(3 - D )cos2x13x

=-

13 (2sin2x + 3cos2x)

C.F. + P.1. = e 2

fi BSIl1-x . fi ( Acos-x+ 2 2

1

1 (2sin2x + 3cos2x) _13

Example 2.4.5Solve D2(D2+9) = sin2x + 5 A.E. is m2(m 2+9) = 0 ~ m = 0, 0, + 3i C.F.

Sol.

= (C I + C2x) + C3 cos3x + C4sin3x


85

P.1.

=

D2 (D2 + 9) sin2x + (D2 + 9) D2 + ~._I_(I) -4(-4+9z) 9 D2 sin 2x

= _ sin 2x20

+ 5x 18

2

Example 2.4.6Solve

(D 4 -2D 3 + 2D2 - 2D + 1 ) y 2D 3

= eX + sin2(x/2)

Sol.

A.E. is f(D) = D4 + 2D2 - 2D + 1 m4 -2m 3 + 2m2 - 2m + 1 = 0

= (m-l )2 (m2+1) = 0:. m= I, I,m=i,-iC.F. is (C t +C 2x)eX + (C 3cosx + C 4sinx) 1 1 P.1. = - - eX + - - sin2(x/2) f(D) f(D)

_1_ f(D)

.Sin

2

(x/2) -

_

(? ) Sin (x/2) (D-It D-+l0

1

.

2

_?

1

(D-It D +1

(2

)

~~= -

[(0_1),1(0' + I) (I-COSX)]--.--

~ ~ [-I -20~(0'

1)(0' + I) COS X

1

1 21

1 1 (cosx) - 4D D2 + 1lID (D2) (cosx)

=

"2 + "4 . D2 + 1 .2

= ~+~._l_.~(-sinx)4 D2 + 1 (-I)1 1

=

"2 + "4 . D2 + 1 . sinx

1

86


[)2

1 1 1 + 1 sinx = I.P of 02 + 1 ex = I.P of (D +-i)---:(-D----:-i)

ex

= I.P of 2i

I

XCiX

= "2 I.P. of -i[cosx + isinx] = -~COSX

x

:. P.1.

=

1 1 "2 - "8 x cos.\"

:. Ilence the complete solution is

Y =(C +CI

2

x)~+C

1 x r 1 x cosx+C sinx+ -.-e +---cosx 3' 4 2 2! 2 8 Exercise 2(e)

2

Solve the following(1)

(2)

(D2+D+ l)y = sin2\" Ans:r ( C,cos-x+ C Jj J X )] e" o' SI / 1 - X -1 - ('J .... cos 2x+ 3' SIll ' .... 2 2 13

J3

(3)

(D6 + 1) y

=

3 x sin"2 x.sin "2jj-,

Ans.

(C1COSX + C 2sinx) + e- 2 (C3cos"2 +C 4 SIIl"2)+ e2- (Cscos"2 +C 6SIIl"2)x 1 + -sinx+ -cos2x 12 126 (D3 + 4Di y = Sin2x

x

. x

.fh

x

. x

(4)

Ans.

y

=

xsinx C 1 + C2 cos2x + C]sin2x --8-

-16

cos2x

I 2.5.1 P.1. is of form feD) xm

I - - xm feD)

=

[f(D)rl.x m


87

Now expand [f(0)r 1 in ascending powers of 0 and retain as far as 0 111 and then operate on xl11

Solved Examples Example 2.5.2Solve (0 2 + 50 + 6) y = x

Sol. A.E. is m2 + 5m + 6 = 0 => m = -2,-3C.F. is C 1e- 2x + C 2e- 3x

P.1.

=

1 0 2 + 5D + 6 x

="6

1 (I

0 + 50)_1 + 8 x= -

2

= -

I 50 I 5 [1--] x= - [x--] 6 6 6 6I 2

x

---

6x

5 36

Solution is therefore)' = C e- 2x + C e-3x + - - -

5

6

36

Example 2.5.3Solve

d3 v dy _ . - 3 - -2y=x2 dx' dx=

Sol. A.E is m3 - 3m -2

0 => (m-2) (m+ 1)2

=

0

C.F. is C 1e 2x + (C 2 + C3x)e-X

P.1.

=

x 0- -30-2,

2

2

(1_(D -3D1,.2 2 .,

3

=--

0 3 -3D 02 -30 , ., [1+ +( t+ ....... ]x222 1

_ -I 3D 9 2 2 _ 1 2 6x 9 _ I 2 - - [ I - - + - D ] x ---[x --+-] - - [2x -6x+9] 224 2 224

Complete Solution is y = C 1e 2x + (C 2 + C 3x)e-X

-

~ (2x 2 4

6x

+ 9)

Exercise 2(f)I. Solve (03 - 302 +2)y = x

88


2.

Solve (D2 + 2D + 3)y = x +.~ Ans. y=

e-x (C I cos fix + C2 sin J;:) +

14-6x

27

3.

Solve (D2 + 2D + I) Y

=

cos2x + x 2=

Ans. y 4. (D2 - 4D + 4) y=

(y + C e-> -~(3COS2X 2

2sin 2x) + (x 2 - 2x)

x 2 + sin2x + e3t"

Ans. y2.6.1

=

(C I +

, C2x)c~X

2X2 + X + 3 cos 2x 3 + - - 8-8- + e x

To find - - (e 2m2 + 4m + 3 = 0 =>m= C.F. is y~2J2i1

2

=-l+J2i

=

e-Z [Acos

~ z + Bsin ~ z]

Linear Differential Equations with Constant Coefficients

000

103

Pol. =

I 1 z= 40- +80+6 61

40 2D2'Z = -6 [1+- +-3-rl z 3 1+-- + ---3 34 3

1

I

40

:m 2

= -

1 6

40 r1-]z= 3

- [z- - ] = - --

1 6

z 6

2 9

Complete solution is y = C.F. + P.1.1 B' 1 ] + z -2 =e-Z[ Acos r;::z+ SIl1-Z ..;2 ,fi 6 9

= -2 3 [Acos( r;:: log(2x+3+ Bsin(

e x+..;2..;2

I

1

1

log(2x+3))] + -6 log (2x+3)- ::..9

1

')

Exercise 2(h)Solve the following differential equations I. (x 202 - 3xO + 4) Y = 2x2 Ans. (C I + C2logx~ + x 2 (Iogx)2 (x 2D2 - 3xD + 5)y = sin(log(x Ans. x 2(C Icoslogx + C2sinlogx) + (coslogx + sinlogx) (x 2D2 - xD +2)y = xlogx Ans. x(Clcoslogx + C 2sinlogx) + xlogx (x 2D2-3xD+5)y = x 2 sinlogx Ans.5.

2.

3.

4.

x2(C)coslogx +

C2sinlog~) - ~x

x2 1ogxcoslogr

(x D -3xD+l)yAns.

2 2 ( _,-s_in_l--,og~x-.:.)_+_1 = -

x2[C Icosh (filogx) + C.,sinh( J3log~) + _I + _1- [5sin(logx) - 6cos(\ogx)]6x 61x

Ans.

104


8.

(x20 2+xO+I)y= (I0gx)2 + sin(logx)

Ans. 9. (2x-I)2

C I coslogx + C2sinlogx + (Iogx? -x[2coslogx - sinlogx]-2 +

d 2y

dx

dy (2x-l) - +2y = 0

dx

Ans.

y= C 1(2x-l)+ ~ 2x+ I

C

10.

d2 dy (5+2x)2 dx; - 6 (5+2x) dx + 8y = 6x

Ans.

y

=

C (5+2x)2+.fi + C (5+2xi t fi _ 3x _ 45I 2

2

8

2

11.

(2x+I)2 d y _ 6 (2x+l) dy + 16y = 8 (l+2x)2

dx 2

dx

Ans. 12.

y

= C 1 + C2 log(2x+ 1) + [log(2x+ 1)2](2x+ 1)2

d2y dy (2x+3) - 2 -(2x+3) --12y=6x dx dx

Ans.

C 1(2x+3)2 +

Ci2X+3f~

- 2x +

%

2.9.1 Linear Differential equations of second orderd 2y dy The general form is dx 2 + P dx + Qy = R

where P, Q, R are functions of x.

Method of variation of ParametersLet the C.F. of the equation2

d Y

dx 2

+ P dy + Qy = Rdx

be

y=Au+ Bv

where A and B are constants and u, v are two independent solutions-,ofd 2y

dx2 + P dx +Qy = R

dy

.... (I)


105

be

y

=

Au + Bv

.... (2)

where A and B are constants and u, v are two independent solutions ofd 2y dy - 2 +P-+QllJ= 0dx dx.J'

.... (3)

such that where

1I2

+ PU I + Qu = 0 and v2 + PV I + Qv = 0d 2u dx 2'

tl2 =

ul

=

du dx etc.

In order to obtain the solution of the equation (I) the arbitrary constants A and Bare treated as arbitrary functions of x and are chosen in such a way that

y = A(x)u + B(x)v satisfies (1)Differentiation of(2) gives= All,

.... (4)

+ lIAI + BVI + vB, (suffixes indicating the order of the derivative).... (5)

Now we chose A and B such that Alu + BI V = 0 So, Again differentiatingd2y dx 2 = (Au 2 + lIIAI) + (Bv2 + vIB,)

.... (6)

substituting (4), (5) and (6) in (I) we get[AlI 2 + BV2 + Alu, + Blv,] + P[Au, + Bvd + Q(Au + Bv] = R

(or)

A(u2 + Pu, + Qlt) + B(v2 + PV I + Qv) + Allt, + BI vI = R

Alu, + Biv i = R (u and v are solutions of(3) Solving (5) and (7) we get dA -vR dB uR

.... (7)

dx

=A =I

uv, --u, v

and -

dx

= B = --I1/V,

-II, v

integrating we obtainA(x)=

J

-vRdx +CI;B(x)= (uv, -1I,v)

J

uR +C 2 lIV, -u,v

.... (8)

where C I and C 2 are arbitrary constants Substituting (8) in (4) we get the complete general solution of the equation (I)

106

Engineering Mathematics - I1I

Working Rule: First find two independent solutions

and v of(3). Then C.F. is given by

y = Au+Bv where A and B are arbitrary constants. Treating A and B as functions of x, we have the solution of (1) as y = AI(x) 1I + BI(x) u where A(x) and B(x) are given by (8)

Solved Examples Example 2.9.2Solve by the method of variation of parameters ~- y =dx 2

d2

2--x

1+c

Sol. A.E. is m2 - I = 0 => m = IC.F. is y= A~

+ Bc-x

Assuming A and B as functions of x such that the given equation is satisfied by

y

=

Ac-'" + Bc-X we have=

dydx

AcX - Be-x + ~-+e-xdxdx

dA

dB

= A~

- Be--x

.... (I)

dA

dB

Choosing A and B so that ~~ + e--x dx = 0

.... (2)

.... (3)

Substituting (I) to (3) in the given equation, we get~-

dAdx

+e-x -

dB dx

=--

2

l+c x

.... (4)dA-x

Solving (2) and (4) we get -dx and dBdxCX

=

_c_ eX + I

.... (5).... (6)

=---

eX + I

= log (~+ I) -e-x-xIntegrating (5) and (6) A = _e- x + log( 1+e-') - x + C" B = -Iog( 1+~) + C 2


107

Hence the solution is

y=[-e-x+log(ex +(or)

1)-x+C,1~+[C2-log(1 +~)]e-X

y=C,ex +C 2e-x -1 +elllog(e'+I)-x]-e-'\'log(el+I)J+P+Q=Otheny=~(ii)

Remarks: If(i) I

I-P+Q=Otheny=e-'x

and (iii) P + Qx = 0 then y = x are solutions of D2y + POy + Qv = 0

Example 2.9.3Solve by the method of variat ion of parameters

- \ + (I 3

d 3y (ix"

cOlt") ~ dx

d

cotx = sin 2x

.... (I)

Sol:

d y dv dx 3 + (I - cotx)

dx - cotx = 0Q = -cou comparing with original eqn.

.... (2)

C.F.P

=

I - cott

:. I - P + Q = I - I + colx - cotx = 0showing that v = e-X is a solution or (2) To find another independent solution of(2) let11

= eO-x . w

By this substitution, the equation (2) reduces tod 2 w dw 2 -x -+-(l-cotx--e )=0 dx 2 dr c- x

or

d2w dw - 2 = (l+cotx)dx dx

~

i(~) dwdX

= I

+ cou

log - = x + logsinx dx

dw

dw or - =eTsinx dx

:. w= JeXsinxdx=- e; (cosx-sinx)u = e-x [--(cosx-smx)] = - - (cosx - Slnx)eX .

I

.

2

2

The second independent solution can be taken as cosx - sinx

108


Thus solution of(2) is

y

=

A (COS X

-

silu) + Be-x

For finding the solution of (I) we treat A and B as functions of x

i.e. y = A(x) (cosx - sinx) + B(x).e-x is the solution of (I) where A(x) and B(x) are obtained by solving (cosx - sinx) A I + e-x . BI anddA solving (4) and (5) ~= =

0

.... (4) .... (5)

-2 sinx--'-----'4

I

and

dB sin2x = ~--exdx4

(I-cos2x)

integrating we getI A = --

2

f' cosx slflxdx+C 1 =--+C 12

.... (6)

and Hence the complete solution is

.... (7)

y

=

C 1(cosx - sinx) + C2e-x -~ (sin2x - 2cos2x)10

Example 2.9.4Solve by the method of variation of parameters(I -

x)

d y + x dy _ Y dx 2 dx

2

= (I

_ x)2

The given equation can be written asd 2y dx 2

x

+ I-xxI-x'

dy I dx - l-x Y = I-xI

.... (I)

p= _ . Q= Clearly P + Qx = 0 Hence y

I-x

andX= I-x

= x IS a-solutIon of - 2 +dx

.

.

d2y

-I-

X

-x

dy I dx - --y = 0I-x

.... (2)


109

To obtain the second independent solution of(2) take y = vx Then (2) reduces to -'} + dx (-+-.1)dxI-x

d 2v

dv

x

2

x

=

0

d . dv dv x 2 - [-]+- [-+-] = 0 dx dx dx I-x x d dv ~~ dx

- - =- - - - =+( ) -- =+1---d,,1- x x I- x x I- x x

x

2

I-x-I

2

1

2

lo.g(:) =x+ log(x-I)-logx2dv

dx

= -.e-t= e-t [-+-, ] 2xX

x-I

1

(-I)

x-

:. v

= e"'" .x

1

The second independent solution is xv = e"'" solution of equation (2) is Y = Ae"'" + Bx To find the solution of (I) treat A and B as functions of x such that.... (3)

dA dB e"'" - + x dt"dx

=

0 I -x

.... (4)

and,

dA dB e"'" - + x dx dx

=

.... (5)

dA dB Solving (4) and (5) we get d; = -xe-X and dx = I

Hence the complete solution is y

=

A e"'" + B.x

= [C1+e-x(I+x)]e"'" + (x+C 2 ) x = C1e"'" + Cr + x 2 + (I+x)

110


Exercise (i)Solve the following by the method of variation of parametersI.d 2y dx 2

-

+ a2y

=

sec ax

x . I Ans. y = C,cos a:r + C '2sm ax + -SIl1 ax + -cos ax Iog(cos a,) a a

2.

d2 y

dx 2 + Y = tanx=

Ans. y3.

-[Iog(secx + tanx)] cosx + C,cosx + C2sinx

--T + 4y = cosec 2x dxAns. y = C,cos 2x + C2sin2x + xcos 2x + sin 2~ log(sin 2x)

d2 y

4.

x2~-2x(l+x) -+2(1 +x)y=x3 2

d2

dy

dx

dx

Ans. y = C, xe 2x + C

r - 4-"4

x2

x

3Mean Value Theorems and Functions of Several Variables3.1.0 This chapter deals with (i) Rolle's theorem (ii) Lagrange's mean value theorem also called as first mean value theorem. (iii) Cauchy's Mean Value theorem (iv) Higher Mean Value theorems. (v) Curvature (vi) Centre of curvature. (vii) Evolutes (viii) Envelopes 3.1.1

Rolle's TheoremIfj{x) is (i) continuous in [a, b], (ii) differentiahle in (a, b) and (iii)j{a) = j{b), then there exists a'c'E(a,b) Proof: Suppose (i)

3f'(C)=0

(ii)

f (x) is a constant function throughout the interval [ a, b ], then f' (x) = 0 VX E ( (I, h) Hence theorem is proved ... f (x) is not a constant function in [ (I ,h ]. As f(x) is continuous in [a,b],there exists a maximum value say, at 'c' sayat'{f (a~d~h) for

(a ~ C ~ b)

and a minimum value,

f(x)

in(a,b).

112


f (c) 7:- f (d)Suppose

and at least one of them is different from and

f ((l) == f (b)

f( c) 7:- f( a)then

'c + h' be a point in the neighborhood of 'c' ,.$;

/(c+II)- /(c)h

0 when II> O.

.... ( 1)

and Further

/(c+I1)- /(c)Iz

;::: 0 when h < O.

.... (2)

f(x)

is differentiable in (a, h). As h -t 0 we get from (I) and (2) that -

. / ,(c) == h~O Lt /(c + h) - /(c) .. h /(c+h)- /(c) h-~O hLI$;

0 and

/(c+h)- /(c) h~O hLt

;::: 0 respectively.

.

i.e.,

1'( c) $; 0

and

1'( c) ~ 0 simultaneously => 1'( c) == 0f (d) 7:- f (a)

Similarly the theorem can be proved when

3.1.2 Geometrical interpretation of Rolle's theoremLet P and Q be two points on the curve y=f(x). AP==BQ ordinates f(a)=f(b and the curve is continuous from P to Q. It can be shown that there is at least one point on the curve y = f(x) between x = a and x = b at which the tangent to the curve is parallel to x-axis. .y

x=a P f(a)

y=aQ

.i(b)

x'

0

A

B

x

y'

fix) is a constant function

Mean Value Theorems and Functions of Several Variables

113

x=cy

Qf(a) f(d)f(b)

o

A

B

x

y'

j(c) andj(d) are both different fromj(a)

=

j(b)

x=cy

x=a

x=b

f(a)

f(c)

f(d)

oy'

A

B

x

j(c)

-:F j(a)

andj(d)

=

j(a)

=

j(b)

x=ay

x=b

x=d f(a) f(d) f(b)

x'

0y'

A

B

x

j(d)

* j(a) and j(c) = j(a) = j(b)

114

Eng~neering

Mathematics - I

3.1.3

Verify Rolle's theorem for /Solution:Consider the function

(x) = log { :;a:a:) }in ( a,b )x a+b

J (x) = 10g{ x(2 + ah)} in the interval ( a, b ).

J(x+h)- J(x) h~O h

Lt

=2

I[ (x+h)2 +llh I x +abj - log - og--+ h~O h ( a + b) ( x + h) ( a + b) x

2

Lt

=

+h~O

Lt

~rlOg (x +_~h) + 2xll+h2 -log x+ hj

hl

(x 2 +ah)1

x

= Lt -1 [ log {'1+ 2xh 7 +h r"~O h x- + ab I =

}

-log {h}] 1+x

=

+h~O x 2 + ab

Lt [2X

1 ] --+O(h) x

where O(h) indicates terms of order h and higher powers of h.I(

X)

2x - ~ x +ab x2

which indicates that Further

I (x)2

is differentiable in ( a,b ) and hence continuous also.

I

(a +ab) (a ) = log ( ) a a+b2

= log 1 = 0) = 0. Thus

I (b) = log

(b +ab) b a+b(

I (a ) = 0 = I (b)

All the conditions of Rolle's theorem are satisfied. Hence 3c( a < c < b) such that

I

I(

C)

=0II () C =0. gIves

c +ab

2

2c

--=O~cE

1 c

2

=ab or c=ab

Clearly c

= +ab

is the G.M of a and b and so

(a, 1)

The theorem is thus verified.


115

3. .4 1

. Ven'fy Rolle's t I1eorem '" lor the function Solution:

j"() S1l1X.10 x = --eX

(0 ,Tl )

f(x) =~lIl(x) is continuous in [a, b]j(x) and A x are differential in (a, b),

..... (3)

Hence j>(x) is differential in (a, b) andj(a)=

..... (4)..... (5)

j>(b)

118


(3), (4), (5) show that ~(x) satisfies all the conditions of Rolle's theorem.

3 atleast one value c in (a, b) ;)~'

~' =

0

(x)

=

f' (x) + A

~' (c)

= f' (c) + A = 0f'(c)..... (6)

A

= -

From (2) and (6) it follows that

f'{c) = j'(b)- f{a) b-a3.2.2 Geometrical Interpretation of Langrange's Mean Value TheoremP and Q are two points on the continuous curve y = j(x) corresponding to x = a and x = b respectively... P[a, j(a)],

Q [b,j(b)] are two points on the curve.

... h . PdQ SI ope on t he Ime JOll1lng t e pomts an IS

f{b)f{a) ' RIS a pomt . h b_a on t e

curve between P and Q corresponding to x = c, so that f'{c) is the slope of the tangent line at R [c,j(c)].

f'{c) =

f{b)- f{a)b_ay

means that the tangent at R is parallel to the chord P Q.

Q

p

f(a)

f(c)

f(b)

x'

0y'

x=a

x=c

x=b

x

Fig. 3.2 Hence this theorem tells that there is at least one point R on the curve PQ where the tangent to the curve is parallel to the chord PQ.


119

3.2.3 ExampleVerity Lagrange's theorem for the functionf(x) = (x - I) (x- 2) (x- 3) in (0, 4).

Solution j(x)=

(x - I)(x - 2)(x - 3) and a

=

0, b

=

4

j(x) = x 3 - 6 Xl + II x - 6 is an algebric polynomial and (0, 4) is a finite interval.Hencej(x) is differentiable in (0, 4) and is continuous in [0,4] showing that the conditions of Lagrange's Mean Value theorem are satisfied.

:3 atleast one value 'c' in (0,4), such thatf{b)- f{a) b-a j(0) = - 6, f(4) = 6=

f'{c)

..... ( I)

f'{x) = 3 x2 - 12 x + IIf'{c) = 3c2 - 12 c + 11From (I)

3c2 _ 12 c + II 3c2 - 12 c + 8 =c=62fj3

=

6 - (- 6) 4-0

The point

c = 6 2fj

3

W h ICh CIear Iy

IIe . 111 ( ,4 .

)= ~

3.2.4 ExampleVerify Langrange's Mean Value theorem for the functionj(x) in (0, I) :

Solution j(x) = ..~

is differentiable in (0, I) and continuous in [0, I] :3 atleast one value 'c' in (0, 1) such that3

l'{c)

=

f{b)- f{a) b-a f(l) - f(O) 1-0

..... ( I)

3

'{ ) f c

=

..... ( 1)

j(0) = eO

=

I, f(l) = e

f'{x)

=

~ gives

f'{c)

=

e

C

120


From (I)

eC =

-

e-I ' 1-0

c=log(e-I), thisc EO, I)

3.2.5 ExampleVerify Langrange's Mean Value theorem for the functionj(x) in (3,4).

= 5x2 + 7x + 6

Solutionj(x) is an algebraic polynomial and the interval (3,4) is finite, j(x) is differentiable in (3,4) and continuous in [3,4]. . . 3 atleast one value c r.(3, 4) such that j'(e) = f(ll) - f(3) 4-33

..... (1 )

f'{c}

=

f{h}- f{a} b-a

j(3) = 72,j(4) = 114, f'{x} = lOx + 7IOc+7= 114-72

4-3(3, 4)

c = 3, 5

E,

Exercise - 3(8) I. Verify Langrange's Mean Value theorem for the following functions :I. j{x)=

x(x - I) (x - 2) in (0,

X)

2. j(x)=logxin(1,e)3. j(x) 4. j(x)=

x 2 _ 3x _

I

in ( -~

1, I;)

= a2 - 7x + lOin (2, 5)

3.3.1

Cauchy's Mean Value TheoremIf two functionsj(x) and g(x) are (i) continuous in [a, b] (ii) differential in (a, b) and (iii) g'(x) =F- 0 in (a, b) then 3 atleast one value 'c' in (a, b) 3

f'{c} g'{c}

f{b}- f{a}

= g{b}-g{a}


121

Proof:Define a new functionj(x)

= j(x) + A g (x)

.....( I)

Where A is a constant such that 4>(a) + 4>(b) j(a) + A g (a) = j(h) + A g (h)

..... (2)

[j{b}- j{a}]A=- [g{b}-g{a}]j{x), A g (x) are continuous in [a, b)

..... (3)

Hence 4>(x) is continuous in [a, b) j(x), A g (x) are differentiable in (a, h) Hence 4>(x) is differentiable in (a, b) and 4>(a) = 4>(b) ..... (4) ..... (5)

..... (6)

(4), (5), (6) show that 4>(x) satisfies all the conditions of Rolle's theorem3 atleast one value c in (a, b) ;) 4>' (c)=

0

But

4>'{x} = f'(x} + Ag' (x) 4>'{c} = f'{c} + Ag' (c) = 0

=>

A= -

/'{c} g'{c} /'{c} g'(c}/{b}- /(o)=

..... (7)

From (3), (7) we get

g(h}- /(o)

3.3.2 ExampleVerify Cauchy's Mean Value theorem for j(x)=-?

1 and g(x) x-

= -

1 in (0, h) x

Solutionj(x), g(x) are differentiable in (a, b) and continuous in [a, b)

122


::3 atleast one value 'c' in (a, b) :)

f'{c)

f{b)- f{a)=

g'{c)Here

g{b)- f{a)

f'{x) = ~ X3g'{x)

=7

-I

-2C 3

=--IT//c 2

=

1b

a

2

c c

a+b ab 2ab a+

= --b which is the Harmonic Mean of , a' and 'b'

c E(a, b)

3.3.3 ExampleVerify Cauchy'sMea~

Valve Theorem for fix)

=

e, g(x) =

e-X in (3, 7)

Solutionfix), g(x) are differentiable in (3, 7) and continuous in [3, 7]

::3 atleast one value 'c' in (a, b) :)

f'{c) f{b)- f{a) g'{c) = g{b)-g{a)Here

f'{x)g'{x)

= =

ee-x

c = 5 E (3, 7)


123

Exercise - 3(C)I. Considering the functionsj{x):::: x 2 , g (x):::: (x) in Cauchy's Mean value theorem for (a, b) prove that 'c' is the arithmetic mean between a and b. 2. Verify Cauchy's Mean Value theorem for j{x):::: sin x, g(x):::: cos x in (a, b) 3. Verify Cauchy's Mean Value theorem for j{x)::::

fx, &>(x):::: fx

I

in (a, b)

3.4.1

Higher Mean Value Theorem with Lagrange's form of remainder (Taylor's theorem with Lagrange's form of remainder):If a functionj{x) is such that

(i) j(x); f'{x}, r{x} ..... fn - I (x) are continuous in [a, a + h](ii)

fW{x} exists in (a, a + h), then :3 at least one number '8' between '0' and' I' h hn - ' h" 3j{a + h):::: j{a) + ,f'{a} + ..... + - ( \"f(n-I)(a) + -F (ll + 9h) 1. n -I,. n!

Proof: Define a new function.j{x):::: j{x) + (a+h-x) f'{x} + (a+h-xY r(x) + ... J! 2!

+

(a + h - X ),,-1 /"-1 (a + h - x)" (-I) (x) + .A n . n.,3

..... (1 )

where A is a constant

cp (a) :::: cp (a + h)2

h j" (a) + 2! h fW (a)+ cp(a) :::: j{a) + 1! hn - I n- J! hn n!A

..... + - ( I",,r-I(a) + and cp (a + h) :::: j{a + h) cp (a) :::: cp (a + h)

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

butThus

j(a + h) :::: cp (a + h) :::: cp (a)

124


Hence using (2) we getj(a + h) = lea) + - j(a) + -

h J!

h2 I" (a) + ..... . 2!..... (5)

j(x), I'(x) , I" (x ), ... .fn-I(x) and (a + h - x) (a + h - x)2 etc continllolls in [a, a + 17] and differentiable in (a, a + h)

Hence

~(x)

is continuous in

ra, a + 17] and differentiable in (a, a + 17)~' (c) =

..... (6)

(4) and (6) show that ~(x) satisfies all the conditions of Rolle's theorem.

:.

:3 atleast one value 'c' (a < c < 0 + 11) ;)

0

Write c = a + e 17 where 0 < e < I

:.

:3

e E (0,

I) ;)

~'

(0 +

e 17) = 0

..... (7)

Differentiating (I) with respect to 'x'

.,.

+ [(a+h-x)"J

(n-I)

I" (x)- (n-IXa+h-xt(n-I)

2

In-J/(x)j_ n(a+h-xtF;'n!

J

A ..... (8)

From (7) we get~' (0

+ e 17) =

(h - eh),,-J

(n-I)

[rea + eh]

=0..... (9)

From (5) and (9) it follows thatj(a + h)=

hi' (a) + -, h I j(a) + -1'. 2.

2

"

(a) + ...

(0 < e < 1)


125

The last term

~ .!' (a + 0 h) is called Lagrange'sform of remainder.n!

3.5.1

Higher Mean Value Theorem with Cauchy's form of remainder (Taylor's theorem with Cauchy's form of remainder):If a functionf{x) is such that (i)f(x), f'{x) , r{x) ..... fll(X) arc continuous in

[a, a + 11] and (ii),f'1(x) exists in (a, a + h) then 3 atleast one number '0' between '0' and' 1' such that2

f(a + h)

=

h '" (I) + ... .f{a) + h f' () a + j!f

I!

Proof: Define a new fUllction~(x) = .f{x) +

(a+h-x) () (a+h-x)2 () f' X + f" x + .... I! 2! '

(a + h - x)"+where A is constant

I

{n-l}!3 ~

fll-l(x)+(a+h-x)A

..... (1)..... (2)

(a + h) = ~ (a)

From (1)

~

(a + h)

= f(a + 11)h .'I

~(a) = f(a) + -1'.1From (2), (3), (4)f{a + h)=

{a} +

f 2!

h2

"

h"-' , {a} + ... + -(-1)'.1 11-1 (a) + II

n- .

A

..... (4)

.f{a) + -"

h " j

,J

.

(0) + -,

n- f"

2.

(a) + .......

.. ... (5)f(x), f'{x) , r(x) ..... jil- I (X) and (a + h - x), (a + h - x)2 ...... are continuousin [a, a + h] and differentiable in (a, a + 1/)

126

Engineering Mathematics - I Hence (x) is continuolls in [a, a + h]

..... (6) ..... (7)

(x) is differentiable in (a, a + h)(2), (6) (7) show that (x) satisfies all the conditions of Rolle's theorem. :. 3 atleast one number '0' in between '0' and' I' Differentiating (I)W.f.3 '

(a

+ 0 h) = 0

..... (8)

to 'x'

'{x) = I'{x) + [(a+h-x) IW{x) - f'{x) (a+h-x f I"'{x)- 2(a+h-x) f"{x)] 2! 2 + ... + ... +llu1 )

[

+ [(a+h-x

(11-1).

F(x)_{n_I){a+h-x)"2 f"'{X)] -A)

(n -I).

I.e., '{x) =

(a+h-x)"-' {n-I}. f"(x)-A

..... (9)

From (8) and (9) we get

{h-Oh)"-I'

(a + 0 h)

=

{n -I}.

fll (a + 0 h) -

A=0

..

A =

h,,-I (I - 0),,-1 (n-I)' fll(a+Oh)

..... (10)

From (9) and (10)

flo + h)

=

hi' h j.w j(a) + -" (a) + -, (a) + .... . 2.

2

The last term

h"{I-OY--1(

n-I!

)

fll (a + 0 h) is called Cauchy's form of remainder.


127

3.5.2

Alternate form of Lagrange's Mean Value theoremIf f(x) is i) continuous in the closed interval [a. a + h] and ii) derivable in the open interval ( a . a + h ) then there exists at least one number B, 0 such that f(a+h)=f(a)+f'(a+Bh).

< B f( a + h) = f(a)+ hf'(a+Bh) a+h-a 3.5.3 Example:If f(x+h)=f(x)+hf'(x+Bh), O -(-) = ()g' e

f'(c)

f(b)- f(a) ()g h - g a

where c

E

(a,b)

which

. .IS

C. M . V theorem.

3.5.7 Example: Apply Maclaurin's theorem with Lagrange's from of remainder for the function

f (x) = eX

and show that 1 + x + ~

~ e ~ 1 + x + ~ eX for every x ~ 0 . 2 2t

2

2

Solution: Maclaurin's theorem with L form of remainder is

f(X) = f(o)+~r'(o)+~ f"(O)+ ... +1~_1 in-I) (0)+ ~~ f" (Ox)~ ~ ~

2

n-I

n

X + -X eX = 1+ X + T=)~

2

3

lJ

X Ox (SInCe . Dn ( eX) = eX) + ...... + IX ~ _ 1 +- e ~ ~

n-I

n

........ (1) where 0 < (} < 1 Taking n = 2 in (1) we geteX

= 1+ X + ~ eOx

X2

........ (2)

For every X ~ 0 we have eOx ~ eX where 0 < (} < 1

Mean Value Theorems and Functions of Several Variables1 1

131

1+x+-e S +x+-e

x-

Ox

1

x-

x

II2

~

........ (3)

Further e

Ox

> 1 whenever x ~ 0 and 0 < 0 < 1

l+x+-sl+x+-e

X

X

2

fh

II

II

........ (4) From (2), (3) and (4) it follows thatX 1 x Ox x x, l+x+-s +x+-e =e sl+x+-e2 2 2

II

II

II

3.5.8 Example

. Taylor's theorem to prove that loge ( 1 + x ) < 1- 2 x +3 x whenever x> 0 . ApplySolution: According to Taylor's theorem

2

3

h2 h"- I I I( a+h) = l(a)+11 1'( a)+ ~ f"(a) + ... + In-l f"- (a)+ R" hwhere

Rn

=

h" (1-0),,-" ~ n- .p

I" (a+Oh)

, 0(a,b)

[f g][fg]=

=

Im

Lt(x,y)->(a,b)

/m

and

(iii)

Lt(x,y)->(a,h)

I

1III

g

Note: 2(x,y) LI ~ (a, b) exists iffLI [ LI

(x-~a) y~b

I(x,b) ==

1

Lt

[

Lt

(y~b) x~a

f(x,h)

1

134


3.6.4 Examplef(x, y)=

x- + V . 2 ' find Lt as (x, v) 2x+ y .

~

~

(2, I)

Solution

y~ x~2

LI [Lt

x +y LI + 5 2X+;2 . = y~ 4+ y2 =5=1..... (i)

2

1 [4Lt=

y]

x~2 y~l

Lt [Lt

x

2

+y ]

2x+ y2

x~2

[X2 +

2x+1 =5=1

I] 5

..... (ii)

The two limits (i) and (ii) are equal.Ltx' + Y ---=-;;-=1'J

Henee

(x,y) ~ (2,1) 2x + y2

3.6.5 Example

f(x, v).

=

x- - Y' ? J find whether the limit exits as (x, y) x- + y-

J

)

~

(0, 0)

Solution

Thus

L/ x - Y J ') [ X ~ 0 Y ~ 0 x- + y-

I,I

2 2]

7:-

y

~ 0 ly ~ 0 x- + y?

LI

I [""'

I,/

x] - Y

2])

lIenee the limit does not exist.

3.6.6

Example

LtFind~

xy

(x,y)

~

(0,0) y- - x-

')


135

SolutionPuty =11/X2

Lt(X,y)~(O,O)

xy

LtX2

mx

/

__

X~O m 2 x 2 __

X2

It is clear that limits will be different for different values of m i.e., the limit depends upon the slope of the path along which (x. y) approaches (0,0). Hence the limit does not exist.

Exercise - 3(0)I. Examine whether the following limits exist. Find them if they exist.Lt(i)(x.y)~

x 2 ~ y2 + 4

Lt(ii)(x, J~-~

x 2 + 4y2

(2,1)

3xy2yx~2y

(0,0) y2X

~ 2X2

Lt(iii)(x.y)

~(2,2) xy~2x

(iv)

Y4 J J (x,y) ~ (0,0 ) x~ yLt~

4

Lt(v)

x-y-

J

J

(x,y) ~ (0,0) x 2 + y2

3.7.1

Concept of ContinuitySuppose(i)

Lt(~,y) ~(a,b)

j(x. y) exists

and

(ii)

Lt(x,y)~(a,b)

f(.~, y)

=

j(a, b)

thenj(x. y) is said to be continuous at (a. b).

Note: 1Ifj(x. y) is said to be continuous at every point of a region R, then it is said to be continuous in R.

Note: 2Letj(x. y) and g(x, y) be continuous at (a, b) then f g,fg, and are all continuous at (~ b).

f

g

(g -:t 0)

136


3.7.2 ExampleConsider the function/ex, y)Lt

= x 2 + Y - 2x when (x,

y) 7= (0, 0) and.l( I, I)

=0

(x,y)

~(I,I)

f(x,y) =

Lt [Ltx~

1 y

~

I

(x 2

J + y-- 2x) ]

=/(I, I)Hence the function is continuolls at (1, 1)

3.7.3 ExampleConsider the function f(x, y) = ~ y J when (x, y) 7= (0,0) and.l(O, 0) = 0 . . x + yLet (x, y)~2 2

(0, 0) along the path yLt(.Y,y)~

=

mx Lt2 2

(0,0)

f(x,y) =

~y 2 (x, y) ~ (0,0) x- + yLt

x 2 m 2 y2

(x, y) ~ (0,0) x 2 + 1112 x 2

m2x 2 -(.r, y) ~ (0,0) I + m 2Lt

=

=.1(0 0),

lienee .l(x, y) is continuous at (0, 0)

3.7.4 ExampleConsider the function.l(x, y)Lt

=xy

xyx2 _ y2 (x, y 7= 0, 0), .1(0, 0)

=

Lt

mx

2

(x, y) ~ (0,0) x 2 _ y2 along the path y= niX

(x, y) ~ (0,0) x 2 (I - m 2 )

= - - 7=

III

1-JIl

2

.1(0, 0) except when

III

=

Hence the given function is discontinuous at (0, 0)


137

3.7.5 ExampleConsider the functionj(x, y) ==X

2

Y

2

~X2 _ y2

(x, y) cf. (0, 0)

2

at (x, y)

=

(0, 0)

In this case it is convenient to introduce (polar co-ordinates). Substitutions x = rcosB,y=

rsinO.

x 2 y2

r4cos20sin2B

l"(2 + y2 - Jr2 (cosr'

0 + sin 20)

-(sin 2 20) 4

4=

{x2 + y24

y2

Now

, )3 2 E . I (x-, +y~ - provided /x/ < E 34 2

thus when

/x - 0/

Engineering Mathematics 1

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