engineering mathematics

AE51/AC51/AT51 ENGG. MATHEMATICS - I JUNE 2013

© IETE 1

Q2 (a) If czyx zyx = , show that at x=y=z, ( ) 12

exlogxyxz −−=∂∂

∂

Answer Log of czyx zyx = , we get

)1......(..........loglogloglog czzyyxx =++

Diff both sides w.r.t x and 1y

)3(..........log1log1

yzor 0log

21log1

)2(..........log1log1

xzor 0log

21log1

zy

yzz

yzzy

yy

zx

xzz

xzzx

xx

++

−=∂∂

=∂∂

+∂∂

++

++

−=∂∂

=∂∂

+∂∂

++

Diff (3) paste all w.r.t. x, we get

[ ]

[ ] 1

12

2

2

2

log

)log(log)log1(

1log1log11

)log1(log1

))2(sin(2log1

log11)log1(

log1

1.)log1(

log1

−

−

−=

+−=+

−=++

++

−=

==++

++

−=

∂∂

++

=∂∂

∂

exx

xexxxx

xxx

xzyAtx

guxzz

yxz

zzy

yxz

Q2 (b) Expand ( ) ( )xytany,xf 1−= in powers of ( )1x − and ( )1y − upto second

degree terms. Answer Here,


© IETE 2

( ) ( )

−−−−−−−−−−−−−−−−+−+=

−−+

−−

∠+−−

∠+−+−+=

−−−−−−−−+−−+

−∠

+−∠

+−+−+==∴

=−=+−

−+

−=

−=+−

−=

−=+−

−=

=+

=

=∴+

=

==∴=

−

−−

eyxyx

yxynyx

fyxf

yfxfyfxfxyyxf

fyxyx

yxyxf

fyxyxyxf

fyx

xyyxf

fyx

xyxf

fyx

yyxf

fxyyxf

xyyy

xxyx

xyxy

yyyy

xxxx

yy

x

22

22

221

222

22

22

222

3

222

3

22

222

11

)1(41)1(

41)1(

21)1(

21

4

)....1)(1(21)1(

21)

21()1(

21

21)1(

21)1(

4

)1,1()1)(1()1,1(

)1(2

1)1,1()1(2

1)1,1()1()1,1()1()1,1()(tan),(

021

21)1,1('''''''''''''''''''''

)1(2

11),(

21)1,1('''''''''''''''''''''

)1(2),(

21)1,1('''''''''''''''''''''

)1(2),(

21)1,1(''''''''''''''''''''''''

1),(

21)1,1('''''''''''''''''''

1),(

41tan)1,1('''''''''''''''''''''tan,

π

π

π

Q3 (a) Change the order of integration and then evaluate it dydxxe yxx

00

2−∞

∫∫

Answer To change order of integration, we take strip parallel to x-axis to cover the each

bounded by .,0,,0 xyyxx ==∞== Strip moves parallel to itself from y=0 to y=∞ ,keeping its ends on x=y two x=∞

hence the integral in changed order in changed order is


© IETE 3

( )

[ ]21

21

21

210

2

2

0

0 00

0

0

0

2

2

=−=

+−==−−=

−=

∞−

∞ ∞−∞−

∞−−

∞∞

−

−∞=

=

∞

=

∫ ∫∫

∫

∫∫

y

yyyy

y

yx

yxx

yxy

e

dyeyedyyedyey

dyey

dxdyxe

Q3 (b) Find the volume of the tetrahedron bounded by the coordinate planes and

the plane 1cz

by

ax

=++

Answer Volume of the tetrahedran bunneded by the coordinate to plnes and the plain


© IETE 4

6622322

31

22

231

22

12

1112

1

dx )1(

dudy dx dy

valentire onedx taken dy

1

3

2

2322

0 0

22)/1(

0

2

0

)/1(

0

0

)1(

0 0

a

0

)1(

0

abcabababababababc

aa

aa

abaa

aab

aaabc

dxan

bb

an

anb

anbcdxy

by

axyc

dyby

axc

zdz

dz

ncz

by

ax

a Ganb

a anb

a nb z yb

=

−+−+−−=

+−−

−−

−=

−−−−

−=

−−=

−−=

=

=++

∫ ∫

∫ ∫

∫ ∫ ∫ ∫ ∫

∫∫∫

−

−

− +

Q4 (a) Solve the equation 04x84x61x43x63x4x21x31x21x=

++++++++

Answer

0=

21-or x 0 x

)12(30012032

1

232142322

1

4846143634213121

==

+−=+

−+

=+++

+=

++++++++

Hence

xxx

xxx

xxxxxx

xxx

xxxxxxxxx

Q4 (b) Find the values of λ for which the equations ( ) ,03y2x2 =++λ− ( ) ,07y4x2 =+λ−+ ( ) 06y5x2 =λ−++ are consistent and find the values of x

and y corresponding to each of these values of λ . Answer The given equation will be consistent if


© IETE 5

055y2x073y2x

032yx becomeequation 1,g

121,1.

01212

i.e 0652

742322

23

=++=++=++==+=−=

=+−−

=−

−−

λλ

λλλ

λλ

λ

soluting

ei

3rd equation is 4 times (1)-(2). :. 3rd n independent solution given by (1) and (2) and it is x=-5, y=1

Soluting y λ=-1, equation become

21x1,y

137

1

13 08- 2

1 0,32

673

5 25 2

2 3

2313121

11

==

−=

−=++=

+−−

=

Hence

yx

RRRRRRR

orzyx

Q5 (a) Use Regula-Falsi method to compute the real root of 2xex = correct to three decimal places.

Answer Equation can be written as

bygiven is.root approch second1.su and 73575888.0 464423228.0)73575888.0()(

73575888.0718281828.2

)2(1)()()()(x

bygiven is apporch.first , method falsi RegulaBy 1. and 0between line (1) ofRoot 80.71828182f(1) and 2)0(

)1....(..........02)(

1

1

betweenlineRootfxf

afbfabfbaf

fxexf x

∴−==

=−−

=−−

=

∴=−=

=−=


© IETE 6

bygiven isroot .approch thirdso 1. and 0.82952077between line 0562935.0)83952077.()(

83952077.0464423228.718281828.

)464423228.(1)718281828(.73585888.0x

2

2

rootfxf

∴−=+=

+=+

−−=

Q5 (b) Use Runge-Kutta method of order four to find y(0.2) for the equation

( ) 10y,xyxy

dxdy

=+−

= . Take h = 0.2.

Answer

Here

1.167836y(0.2)

167836.06

)22

14144.0)2.0()16619.1()2.0()16619.1(2.0),(

16619.0)1.0()08333.1()1.0()08333.1(2.0)

2,

2(

16666.02.12.0

)1.0()1.1()1.0()1.1(2.0)

2,

2(

2.001012.0),(

2.0,1,0,),(

4321

3004

003

002

001

00

=

=+++

=∴

=++++−+

=++=

=++++−+

=++=

==++++−+

=++=

=+−

==∴

===+−

==

Hence

kkkkk

kyhxhfk

kygyhxhfk

kyyhxhfk

xyxhfk

hyxxyxyyxf

dxdy

Q6 (a) Solve the equation

++

−=xsin1

xcosyxdxdy

Answer Given equation can be written as (x+y cos x)dx+(1+sin x)dy=0 Here M=x+y cos x, N=1+sin x

costant) (sin2

x issolution required henceeach in equ

cos

2

akyxy

GivenxNx

yM

=++

∴∂∂

==∂∂


© IETE 7

Q6 (b) Find the orthogonal trajectories of the family of coaxial circles 0Cy2yx 22 =+λ++ , λ being the parameter.

Answer Given family of coaxial circles is

xxcHence

x

dxxe

eexedx

xe

e

dxW

XW

Xxee

ey

Wronskiandxdy

dxdyyx

wegetdiffCyyx

x

x

xx

x

x

xx

x

logeee)(cyissolution compute

eloge -

xe-e

yydx y-yP.I

e3e 3

xey

y yW

by given is

)2....(..........0222

),1()1.......(..........02

3x3x3x21

3x3x

2

3

6

33

2

3

6

3x3x

12

21

6x33u3

3x3

12

11

21

22

−−+=

−=

+=

+=∴

=+

==

∴

=++

=+++

∫ ∫

∫∫

λ

λ

Q7 (a) Solve the differential equation x2cosxy4dx

yd 22

2+=+

Answer

Q7 (b) Use method of variation of parameters to solve 2

x3

2

2

xey9

dxdy6

dxyd

=+−

Answer Q8 (a) Show that

(i) ∫∫ππ

π=θθ

θθ2

0

2

0

dsin1dsin

(ii) ( ) ( ) ( )n,mn,1m1n,m β=+β++β


© IETE 8

Answer

ππ

π

ββθθ

θ

βθθ

βθθ

θθθ

ππ

π

π

π

==

==

++=

=

=

=

=

=∴

=

∫∫

∫

∫

∫ −−

21sin

41

41

41

41

43

21

41

.

45

21

43

41

21

41

21

41

.

21

43

21

43

41

21,

41

21,

43

41

sin1sin

are we

)1....(21,

41-m putting

21,

41

21

sin1

)1....(21,

43-m putting

21,

43

21sin

)1..(..........d cossin 2),(

2/

0

2/

0

2/

0

2/

0

122

0

12

gU

d

gmultiplyin

isxbyd

isxbyd

mmknowBWe xm

( ) ( )

RHSnmnmnm

nmnmnmnm

nmnmnm

nmnm

nmnmnmnmLHS

mn

==+

=++

+=

+++

=++

++

+++

=+++=

),()()(

111

11,11,

β

ββ

Q8 (b) Solve in series the equation ( ) 041219 2

2

=+−− ydxdy

dxydxx

Answer

Here x=0 is an singular point because coefficient of 2

2

dxyd become zero at x=0


© IETE 9

[ ][ ][ ]

[ ]

0212

01

0

2 0 1

01 1 0

0

0 0

22

110

1121

1 0

22

11

2 0

221

1 0

22

11 0

)1)(2(3)13)(43(

)2(343

)1(313

)13)(43(499a)43)(1(a3

04)1(12)1(9a)1(9a-get we x,of powersdifferent of coeffequation integer.an by .differ not do anddistiner are

0a 37,0

.012)1(9a

xof powersat low of coeff zex the to(equating isequation 0........................14

.......)(.......)2()1(a12

)1)((......................)1)(2()1()1(ax)-ax(1get we,equation given theis values

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

..........................ay

ammmma

mma

similarly

amma

mmammmm

oramammmm

Roots

becausem

eimamm

Indicialxaxaxaxa

nnmaxmaxmamnxnmnmaxmmamxmaxmm

ngSabstituti

nnmnmanmmanmanmdxdy

enanananLet

nmn

mmm

nmx

mmm

nmn

mmm

nmn

mmm

nmn

mmm

++++

=++

=

++

=∴

+−=−−=−+

=++−++−

∴

≠=

=−−

=++++++

++++++++−

−+++++++++−

−+++++++++=∴

+++=∴

+++

−++−

−+−−

−+−

+++

++++

∠+

∠++=

+=

+++=∴

∠+

∠++=∴

××===

=

.....16.13.1014.11.8

13.1011.8

1081

337.4.1

234.1

311

cy issolution compute

.....16.13.1014.11.8

13.1011.8

1081

.....33

4.123

4.1311

1.2.33147,

1.2.31.4,

3

02

3237

1

33

220

2211

3237

02

33

2201

0330220

1

exxxxA

exxxA

ycyHence

exxxxay

exxxay

aaaaa

a

fm


© IETE 10

Q9 (a) Show that ( ) ( ) ( )xJx241xJ

x8

x48xJ 02134

−+

−=

Answer We know that

( ) [ ]

( )

)(241)(848

)(J6)()(2124

)(J6)(J124

)(J)(J)(J46

)()(,6)(J

)()(,4)(J

)()(,2)(J

get we, 1,2,3n

||||2

||||2

0213

1012

122

212

234

123

012

11

11

xJx

xJxx

xx

xJxJxx

xx

xx

xxxxx

xJxJx

x

xJxJx

x

xJxJx

x

putting

nJxJxxxJ

nJxJx

xxJ

nnn

nnn

−+

−=

−

−

−=

−

−=

=−

−=

−=

−=

−=

=

−=∴

+=

−+

+−

Q9 (b) Show that ( )21

1

x1−∫−

)x(Pm′ )x(Pn′ dx = 0

Answer Page Number 622 of Textbook-I

Text Books

1. Higher Engineering Mathematics, Dr. B.S.Grewal, 40th edition 2007, Khanna publishers, Delhi. 2. Text book of Engineering Mathematics, N.P. Bali and Manish Goyal, 7th Edition 2007, Laxmi Publication (P) Ltd.

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