2%2dhomogeneous equation and euler theorem

8/14/2019 2%2DHomogeneous Equation and Euler Theorem

1/22

(25 Solved problems and 00 Home assignments)

Homogeneous Expression:

An expression of the form nn22n

211n

1n

0 ya...yxayxaxa ++++ ,

where each term of degree n , is called Homogeneous expression in x and y and of

degree or order n.

Homogeneous Function:

If this expression equal to some quantity u , then u is called Homogeneous

Function in x and y of degree n .

Now nn22n

211n

1n

0 ya...yxayxaxau ++++=

++

+

+=n

n

2

210n

xy

a.....xy

axy

aax

=xy

f xu n .

Also, we can write a Homogeneous Function in x and y of degree 'n' as

=

yx

f yu n .

Similarly, a Homogeneous Function in x ,y and z of degree n can be written as

Differential Calculus

Partial Differentiation

(Homogeneous Functions and Eulers Theorem)

Prepared by:

Dr. Sunil

NIT Hamirpur (HP)(Last updated on 26-08-2008)

Latest update available at: http://www.freewebs.com/sunilnit/


2/22

Partial Differentiation: Homogeneous Equation and Eulers Theorem

Prepared by: Dr. Sunil, NIT Hamirpur

2

=xz

,xy

Fxu n or

=

yz

,yx

Fyu n or

=zy

,zx

Fzu n .

Here 'u' is dependent variable and x, y, z are independent variables.

Eulers Theorem:

Statement: If u is a homogeneous function of x and y of degree n, then

nuyu

yxu

x =

+

.

Proof: Given u is a homogeneous function in x and y of degree n.

Then we may write

=xy

f xu n . (i)

Differentiating (i) partially w.r.t. x [keeping y as constant], we get

n n 12

n 2 n 1

u y y yx f nx f

x x x x

y yx yf nx f .

x x

= +

= +

Similarly, differentiating (i) partially w.r.t. y [keeping x as constant], we get

=

=

xy

f xx1

xy

f xyu 1nn .

Nown 2 n 1 n 1u u y y y

x y x x yf nx f y x f x y x x x

+ = + +

n yn x f nux

= =

.

This completes the proof .

Extension of Eulers Theorem:

Statement: If u is a homogeneous function of x and y of degree n, then show that

( )u1nnyu

yyxu

xy2x

ux 2

22

2

2

22

=

+

+

.

Proof: Since u is a homogeneous function in x and y of degree n then

nuyu

yxu

x =

+

. [by Eulers Theorem] ..... (i)

Differentiating (i) partially w. r. t. x [keeping y as constant], we get


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3

2 2

2

u u u ux y n .

x x x y x

+ + =

Multiplying by x, we get

xu

nxyxu

xyxu

xx

ux

2

2

2

2

=

+

+

( ) xu

x1nyxu

xyx

ux

2

2

2

2

=

+

. ......(ii)

Again, Differentiating (i) partially w. r. t. y [keeping x as constant], we get

yu

nyu

y

uy

xyu

x 2

22

=

+

+

.

Multiplying by y, we get

yu

nyyu

yy

uy

xyu

xy 2

22

2

=

+

+

( )

yu

y1ny

uy

xyu

xy 2

22

2

=

+

. .....(iii)

Adding (ii) and (iii), we get

( ) ( ) ( )u1nnnu1nyu

yxu

x1ny

uy

yxu

xy2x

ux 2

22

2

2

22 ==

+

=

+

+

.

This completes the proof.

Now let us solve some problems related to the above-mentioned topics:

Q.No.1.: Verify Eulers theorem, when 323 yyx3xz = .

Sol.: Since323

yyx3xz =

xy6x3xz 2 =

and 22 y3x3yz

=

.

( ) ( ) ( )z3yyx3x3y3x3yxy6x3xyz

yxz

x 323222 ==+=

+

.

Also

=

=xy

f xxy

xy

31xz 33

3 .

z is a homogeneous function of x and y of degree 3.

By Eulers theorem, we get z3nzyz

yxz

x ==

+

.

Hence, Eulers theorem is verified .


4/22



4

Q.No.2.: If 2 / 1

2 / 12 / 1

3 / 13 / 1

yx

yxu

++

= , then prove that u121

yu

yxu

x =

+

.

Sol.: Here

2 / 1

2 / 1

3 / 1

6 / 1

2 / 1

2 / 1

2 / 12 / 1

3 / 1

3 / 13 / 1

2 / 1

2 / 12 / 1

3 / 13 / 1

xy

1

xy

1x

x

y1x

x

y

1x

yx

yxu

+

+=

+

+=

++

=

=

+

+=

xy

f x

xy

1

xy

1x 12 / 1

2 / 1

2 / 1

3 / 1

12 / 1 .

u is a homogeneous function of x and y of degree121

.

By Eulers theorem, we have u121

nuyu

yxu

x ==

+

.

Hence the result .

Q.No.3.: If

=xy

f u , then show that 0yu

yxu

x =

+

.

Sol.: Here

=

= xyf xxyf u0 .

u is a homogeneous function of x and y of degree 0.

Hence by Eulers theorem, we have

0u.0nuyu

yxu

x ===

+

.

Hence the result .

Q.No.4.: If

= xy

xyf u , then show that

=

+

x

y

xyf 2y

u

yx

u

x .

Sol.: Here

=

=

=xy

Fxxy

f xy

xxy

xyf u 22 .




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5

===

+

xy

xyf 2u.2nuyu

yxu

x .

Hence the result .

Q.No.5.: If

+

= xytanyxsinu

11 , find the value of yuyxux + .

Sol.: Here

=

+=

+

=

xy

f xxy

tan

xy1

sinxxy

tanyx

sinu 011011 .



0u0nuyuy

xux ===+

.

Hence the result .

Q.No.6.: Verify Eulers Theorem on homogeneous functions in the following cases:-

(i) ( ) ( )5 / 15 / 14 / 14 / 1

yx

yxy,xf

++

= , (ii) ( ) zyxz,y,xf u ++==

(iii)

+=

y

yxtanz

221 , (iv)

+

=

x

ytan

y

xsinu 11

(v)

=xy

logxz 4 .

Sol.: (i) Here ( ) ( )( )

=

+

+=

++

=xy

f x

xy

1

xy

1x

yx

yxy,xf 20 / 1

5 / 1

4 / 1

20 / 15 / 15 / 1

4 / 14 / 1

( )y,xf is a homogeneous function of x and y of degree 201

.


( )5 / 15 / 14 / 14 / 1

yx

yx201

nf yf

yxf

x++

==

+

. (i)


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6

Again since ( ) ( )5 / 15 / 14 / 14 / 1

yx

yxy,xf

++

=

Differentiating partially w. r. t x and y respectively, we get

( ) ( )( )25 / 15 / 1

5 / 44 / 14 / 14 / 35 / 15 / 1

yx

x51yxx41yx

xf

+

+

+=


( ) ( )( )25 / 15 / 1

5 / 14 / 14 / 14 / 15 / 15 / 1

yx

x51

yxx41

yx

xf

x+

+

+=

(ii)

( ) ( )( )25 / 15 / 1

5 / 44 / 14 / 14 / 35 / 15 / 1

yx

y5

1

yxy4

1

yxyf

+

+

+=


( ) ( )( )25 / 15 / 1

5 / 14 / 14 / 14 / 15 / 15 / 1

yx

y51

yxy41

yx

yf

y+

+

+=

(iii)


( ) ( ) ( ) ( )( )25 / 15 / 1

5 / 15 / 14 / 14 / 14 / 14 / 15 / 15 / 1

yx

yx51yxyx

41yx

yf

yxf

x+

++++=

+

( )5 / 15 / 14 / 14 / 1

yx

yx201

++

= (iv)

Hence, the Eulers Theorem is verified .

(ii) Here ( )

=

+

+=++==

x

z,

x

yf x

x

z

x

y1xzyxz,y,xf u 2 / 1

2 / 12 / 12 / 1

( )z,y,xf u = is a homogeneous function of x ,y and z of degree21

.


( )zyx21

u21

nuzu

zyu

yxu

x ++===

+

+

.


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7

Again since ( ) zyxz,y,xf u ++==

Differentiating partially w. r. t x , y and z respectively, we get

x2

1xu

=

,y2

1yu

=

andz2

1zu

=

( )zyx21

z2

1z

y2

1y

x2

1x

zu

zyu

yxu

x ++=++=

+

+

.


(iii) Here

=+

=

+=

yx

f y1yx

tanyy

yxtanz 0

210

221 .

z is a homogeneous function of x ,y and z of degree 0.


0z.0nzyz

yxz

x ===

+

.

Again since

+=

y

yxtanz

221 .


( ) 2222222

22yx

1.y2x

xyx2.yx2

1y1.

y

yx1

1xz

++=

+++=

( )( )

22

222

222

22

22

2

22yx

yxy.

y2x

1

y

yxy2.yx

12y

.

y

yx1

1yz

+

++

=

+

+

++

=

( ) 222

22yx

x.y2x

1

+

+= .

( ) ( )( )

0yx

x.

y2x

1y

yx

1.

y2x

xyx

yz

yxz

x22

2

222222=

+

+

+++

=

+

.



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8

(iv) Here

=

+=

+

=

xy

f xxy

tan

xy1

sinxxy

tanyx

sinu 011011 .

u is a homogeneous function of x and y of degree 0.Hence by Euler's theorem, we have

0u0nuyu

yxu

x ===

+

.

Again since

+

=

xy

tanyx

sinu 11 .


22222

2

22 yxy

xy1

xy.

x

y1

1y1.

yx

1

1xu +=

++

= .

2222

2

222 yx

x

xyy

xx1

.

x

y1

1

y

x.

yx

1

1yu

++

=

++

=

.

0yx

x

xyy

xy

yx

y

xy

1x

y

uy

x

ux

22222222=

++

+

+

=

+

.


(v) Here

=

=xy

f xxy

logxz 44 .


Hence by Euler's theorem, we have

===

+

x

ylogx4z4nz

y

zy

x

zx 4 .

Again since

=xy

logxz 4 .



9/22



9

332

43 xxy

log.x4x

y

xy1

.xxy

log.x4xz

=

+

=

andy

xx1

xy1

.xyz 44 =

=


Q.No.7.: If yxyx

logu44

++

= , show that 3yu

yxu

x =

+

.

Sol.: Hereyxyx

logu44

++

=yxyx

e44

u

++

= .

ue is a homogeneous function of x and y of degree 3.

Hence by Eulers theorem, we have uuuu

e3neye

yxe

x ==

+

.

uuu e3yu

yexu

xe =

+

.

Hence 3yu

yxu

x =

+

.

Hence the result .

Q.No.8.: If

+

= yx

yxsinu 1 , show that

yu

xy

xu

=

.

Sol.: Here

+

= yx

yxsinu 1

=

+

=

+

=xy

f x

xy

1

xy

1x

yx

yxusin 02 / 1

2 / 1

0 .

usin is a homogeneous function of x and y of degree 0.


( ) ( )0usin.0usinn

yusin

yx

usinx ===

+

.

0yu

ucosyxu

ucosx =

+

0yu

yxu

x =

+

.

Henceyu

xy

xu

=

.

Hence the result .


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10

Q.No.9.: If u xyx

yx

= +

, show that

(i) xux

yuy

xyx

+ = .

(ii) 0y

uy

yxu

xy2x

ux

2

22

2

2

22 =

+

+

.

Sol.:(i) Here 21 uuxy

xy

xu +=

+

= (say)

Theny

uy

xu

xyu

yxu

xyu

yxu

x 2211

+

+

+

=

+

.

Now since 1u and 2u are homogeneous function of x and y of degree 1 and 0

respectively. Then by Eulers theorem, we have

==

+

xy

xnuyu

yxu

x 111 and 0nu

yu

yx

ux 2

22 ==

+

.

Hence 0xy

xy

uy

xu

xyu

yxu

xyu

yxu

x 2211 +

=

+

+

+

=

+

.

=

+

xy

xyu

yxu

x .

Hence the result . (ii) Again, since 1u and 2u are homogeneous function of x and y of degree 1 and 0

respectively. Then, by extension of Eulers theorem, we have

( ) ( ) 0xy

x.111u1nny

uy

yxu

xy2x

ux 2

22

2

2

22 =

==

+

+

.

Hence xu

xxy

ux y

yu

y2

2

2

22

2

22 0

+ + = .

Hence the result .

Q.No.10.: If ( ) 3 / 122 yxu += , prove that u92

y

uy

yxu

xy2x

ux

2

22

2

2

22 =

+

+

.

Sol.: Here ( )

=

+=+=xy

f xxy

1xyxu 3 / 23 / 12

3 / 23 / 122 .


11/22



11

u is a homogeneous function of x and y of degree32

.

Hence by extension of Eulers theorem, we have

( ) u92

132

32

u1nny

uyyx

uxy2x

ux 2

22

2

2

22

=

==

+

+

.

Hence the result .

Q.No.11.: If

++

= yxyx

sinu22

1 , then show that utanyu

yxu

x =

+

.

Sol.: Here

++

= yxyx

sinu22

1

=+

+=

++

==xy

f x

x

y1

x

y1

xyxyx

)say,z(usin 12

2

22.


Then, by Eulers theorem, we have znzyz

yxz

x ==

+

.

+ =x uux

y uuy

ucos cos sin

.

utanyu

yxu

x =

+

.

Hence the result .

Q.No.12.: If 2 / 1

2 / 12 / 1

3 / 13 / 11

yx

yxsinu

++

= , then prove that

(i) utan121

yu

yxu

x =

+

.

(ii) ( )utan13144

utanuyxyu2ux 2yy

2xyxx

2 +=++ .

Sol.: (i) Here2 / 1

2 / 12 / 1

3 / 13 / 11

yxyx

sinu++

=


12/22



12

2 / 1

2 / 1

3 / 1

6 / 1

2 / 1

2 / 1

2 / 12 / 1

3 / 1

3 / 13 / 1

xy

1

xy

1x

xy

1x

xy

1x

usin

+

+=

+

+

=

=

+

+=

xy

f x

xy

1

xy

1x 12 / 1

2 / 1

2 / 1

3 / 1

12 / 1 .

usin is a homogeneous function of x and y of degree121

.

Then, by Eulers theorem, we have

usin121

usinny

)u(siny

x)u(sin

x ==

+

usin121

yu

ucosyxu

ucosx =

+

utan121

yuy

xux =+ . ...(i)

Hence the result .

(ii) Differentiating (i) w. r. t. x partially, we get ( )x2xyxxx uusec121

yuuxu =++

Multiplying by x , we get

( )x2xyxxx2 xuusec121

xyuxuux =++ . .....(ii)

Differentiating (i) w. r. t. y partially, we get

( )y2yyyxy uusec121

yuuxu =++ .

Multiplying by y , we get

( )y2yy2yxy yuusec121

uyyuyxu =++ . ......(iii)


( ) ( ) ( )yx2

yxyx2

yy2

xyxx2

yuxu1usec121

yuxuyuxuusec121

uyxyu2ux +

=++=++

utan121

utanusec144

1utan

121

1usec121 22 +=

+=

( ) ( )12utan1utan144

1utan

121

utanutan1144

1 22 ++=++=


13/22



13

( )utan13utan144

1 2+= .

Hence the result .

Q.No.13.: If ( )

+

++

= xy

xy

x1m2yx

u

m22

, show that

( )m222

22

2

2

22 yxm2

y

uy

yxu

xy2x

ux +=

+

+

.

Sol.: Here( )

321

m22uuu

xy

xy

x1m2

yxu ++=

+

+

+= (say)

Theny

uy

xu

xy

uy

xu

xyu

yxu

xyu

yxu

x 332211

+

+

+

+

+

=

+

.

Now since 1u , 2u and 3u are homogeneous function of x and y of degree 2m , 1 and 0

respectively. Then by Eulers theorem, we have

1111 mu2nu

yu

yxu

x ==

+

, 2222 unu

yu

yx

ux ==

+

and 0nu

yu

yx

ux 3

33 ==

+

.

Hence 21332211 u.1u.m2

yu

yx

ux

yu

yx

ux

yu

yxu

xyu

yxu

x +=

+

+

+

+

+

=

+

.

Again, since 1u , 2u and 3u are homogeneous function of x and y of degree 2m, 1 and 0

respectively.

Then, by extension of Eulers theorem, we have

( ) ( ) ( ) ( )( )1m2

yx1m2m2u111u1m2m2u1nn

y

uy

yxu

xy2x

ux

m22

212

22

2

2

22

+

=+==

+

+

Hence ( )m222

22

2

2

22 yxm2

y

uy

yxu

xy2x

ux +=

+

+

.

Hence the result .

Q.No.14.: If ( )yxsinu += , show that

( ) ( )yxcosyx21

yu

yxu

x ++=

+

.

Sol.: Here ( )yxsinu += .


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14

( ) ( )

=

+=+== xy

f xxy

1xyxsay,zusin 2 / 12 / 1

2 / 11 .

z is a homogeneous function of x and y of degree

2

1.

Then, by Eulers theorem, we have xzx

yzy

nz z

+ = =12

.

( )yx21

usin21

yu

u1

1y

xu

u1

1x 1

22+==

+

.

( ) ( ) ( ) ( )yxcosyx21

yxsin1yx21

yu

yxu

x 22 ++=++=

+

.

Hence ( ) ( )yxcosyx21

y

u

yx

u

x ++=

+

.

Hence the result .

Q.No.15.: If

+= 221 yxsinz , then show that ztanzyxyz2zx 3yy

2xyxx

2 =++ .

Sol.: Here

+= 221 yxsinz ( )

=+=+==xy

xf x

y1xyxsay,uzsin 2

222 .


Then by Eulers theorem, we have unuyuy

xux ==

+

.

zsinyz

zcosyxz

zcosx =

+

ztanyz

yxz

x =

+

. (i)

Differentiating (i) w. r. t. x partially, we get

( )x2xyxxx zzsecyzzxz =++


( )x2xyxxx2 xzzsecxyzxzzx =++ . ......(ii)


( )y2yyyxy zzsecyzzxz =++ Multiplying by y , we get


15/22



15

( )y2yy2yxy yzzseczyyzyxz =++ . ......(iii)Adding (ii) and (iii), we get

( ) ( ) ztan)z(tanzsecyzxzyzxzzseczyxyz2zx 2yxyx2yy2xyxx2 =++=++ .

( ( ztanztanztan1zsecztan 322 === .Hence the result .

Q.No.16.: If ( )2iyxivu =+ , andvu

w = , prove that 0yw

yxw

x =

+

.

Sol.: Here ( )2iyxivu =+ ixy2yxivu 22 =+ .

Thus

=

=

== xy

f x

xy

2

xy

1

xxy2yx

vu

w0

2

022

.

w is a homogeneous function of x and y of degree 0.


0w.0nwyw

yxw

x ===

+

.

Hence 0y

wy

x

wx =

+

.


Q.No.17.: If ( )3ibyaxivu =+ ,show that

(i) xux

yuy

u

+ = 3 ,

(ii) xvx

yvy

v

+ = 3 ,

(iii) ( 0ywy

xwx =

+

where

vuw = .

Sol.: (i) Here ( )3ibyaxivu =+ )ybxa3yb(i)xyab3xa(ivu 22332233 +=+ .

Thus

=

==xy

f xxy

ab3ax)xyab3xa(u 32

2332233 .


16/22



16



w.3nwyw

yxw

x ==

+

.

Hence w3yw

yxw

x =

+

.


(ii) Here ( )3ibyaxivu =+ )ybxa3yb(i)xyab3xa(ivu 22332233 +=+ .

Thus

=

==xy

f xxy

ba3xy

bx)ybxa3yb(v 323

332233 .

v is a homogeneous function of x and y of degree 3.Then, by Eulers theorem, we have

v.3nvyv

yxv

x ==

+

.

Hence v3yv

yxv

x =

+

.


(iii) Here ( )3ibyaxivu =+ )ybxa3yb(i)xyab3xa(ivu 22332233 +=+ .

Thus

=

=

==xy

f x

xy

ba3xy

b

xy

ab3ax

ybxa3yb

xyab3xavu

w 0

23

3

223

02233

2233.

w is a homogeneous function of x and y of degree 0.


0w.0nwy

w

yx

w

x ===

+

.

Hence 0yw

yxw

x =

+

.



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17

Q.No.18.: If ( ) 2 / 1222 zyxv ++= ,then show that vzv

zyv

yxv

x =

+

+

.

Sol.: Here ( ) 2 / 1222 zyxv ++=

=

+

+=

xz

,xy

f xxz

xy

1xv 12 / 122

1

v is a homogeneous function of x ,y and z of degree 1 .


vv).1(nuzv

zyv

yxv

x ===

+

+

.

Hence the result .

Q.No.19.: If ( )yxsinu 1 += , prove that

ucos4u2cos.usin

yuy

yxuxy2

xux

32

222

2

22 =+

+

.

Sol.: Here ( )yxsinu 1 += ( )

=

+=+=xy

f xxy

1xyxusin 2 / 12 / 1

2 / 1 .

usin is a homogeneous function of x and y of degree21

By Eulers theorem, usin2

1usinn

y

)u(siny

x

)u(sinx ==

+

+ =x uux

y uuy

ucos cos sin

12

. + =xux

yuy

u

12

tan . ...(i)

Differentiating (i) w. r. t. x partially, we get

( )x2xyxxx uusec21

yuuxu =++


( )x2

xyxxx2

xuusec21

xyuxuux =++ ......(ii)


( )y2yyyxy uusec21

yuuxu =++

Multiplying by y , we get


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18

( )y2yy2yxy yuusec21

uyyuyxu =++ ......(iii)


( ) ( ) ( )yx2

yxyx

2

yy

2

xyxx

2

yuxu1usec2

1

yuxuyuxuusec2

1

uyxyu2ux +

=++=++

utan21

utanusec41

utan21

1usec21 22 =

=

( )ucos

ucos214

utan2

ucos

14

utan2usecutan

41

2

2

22 =

==

ucos4

u2cos.usin

y

uy

yxu

xy2x

ux

32

22

2

2

22 =

+

+

.

Hence the result .

Q.No.20.: If

++

= y3x2

yxtanV

331 , prove that

V2sinV4siny

Vy

yxV

xy2x

Vx

2

22

2

2

22 =

+

+

.

Sol.: Here

+

+=

y3x2

yxtanV

331 z

x

yf x

xy32

xy

1x

y3x2

yxVtan 2

3

233

=

=

+

+=

+

+= (say).


Then by Eulers theorem, we have z2nzyz

yxz

x ==

+

.

Vtan2yV

VsecyxV

Vsecx 22 =

+

V2sinVcosVsin2Vcos.VcosVsin2

VsecVtan2

yVy

xVx 22 ====+ . ....(i)

Differentiating (i) partially w. r. t. x [keeping y as constant], we get

xV

2.V2cosyx

Vy

xV

x

Vx

2

2

2

=

+

+

.



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19

xV

V2cosx2yx

Vxy

xV

xx

Vx

2

2

22

=

+

+

( )xV

x1V2cos2yx

Vxy

x

Vx

2

2

22

=

+

. ..(ii)

Again, Differentiating (i) partially w. r. t. y [keeping x as constant], we get

yV

2.V2cosyV

y

Vy

xyV

x 2

22

=

+

+

.


yV

V2cosy2yV

yy

Vy

xyV

xy2

22

2

=

+

+

( ) yVy1V2cos2

yVy

xyVxy 2

22

2

=

+

. .....(iii)


( ) ( ) V2sin1V2cos2yV

yxV

x1V2cos2y

Vy

yxV

xy2x

Vx 2

22

2

2

22 =

+

=

+

+

V2sinV4sinV2sinV2cosV2sin2 == .

Hence V2sinV4sin

y

Vy

yxV

xy2

x

Vx

2

22

2

2

22 =

+

+

.

Hence the result .

Q.No.21.: If

=xz

,xy

f xu n , show that nuzu

zyu

yxu

x =

+

+

.

Sol.: Here

=xz

,xy

f xu n .Let 1txy

= , 2txz

=

( ) f xt,tf xu n21n ==

Differentiating partially w. r. t x , y and z respectively, we get

+

+=

+

+=

22

21

n1n2

2

1

1

n1n

x

ztf

x

ytf

xf nxxt

.tf

xt

.tf

xf nxxu

2

1n

1

1n1n

tf

zxtf

yxf nxxu

=

,


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20

1

1n

1

n2

2

1

1

n

tf

xx1

tf

xyt

.tf

yt

.tf

xyu

=

=

+

= ,

2

1n

2

n2

2

1

1

n

tf

xx1

tf

xzt

.tf

zt

.tf

xzu

=

=

+

=

+

+

=

+

+

2

1n

1

1n

2

1n

1

1n1n

tf

xztf

xytf

zxtf

yxf nxxzu

zyu

yxu

x .

Hence ( ) nut,tf nxzu

zyu

yxu

x 21n ==

+

+

.

Hence the result .

Q.No.22.: If 222 yx

ylogxlogxy1

x

1u

+

++= show that 0u2yu

yxu

x =+

+

.

Sol.: Here

=+

+=+

++= xy

f x

x

y1

xy

log

xy1

1xyx

ylogxlogxy1

x

1u 2

2

22

222 .

u is a homogeneous function of x and y of degree 2 .

Hence by Eulers theorem, we have u2yu

yxu

x =

+

.

Hence 0u2yu

yxu

x =+

+

.

Q.No.23.: If ( )

++++

=zyxzyx

logz,y,xf 555

, show that 4zf

zyf

yxf

x =

+

+

.

Sol.: Here ( )

++++

=zyxzyx

logz,y,xf 555

zyxzyx

e555

f

++++

= ...(i)

=

++

+

+

=xz,

xyf x

xz

xy

1x

z

x

y1

xe 4

55

4f .

f e is a homogeneous function of x, y and z of degree 4.



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21

f f f f

e4ze

zye

yxe

x =

+

+

.

f f f f e4zf

zeyf

yexf

xe =

+

+

4zf

zyf

yxf

x =

+

+

.

Hence the result .

Q.No.24.: If

+

=zxyzxyzyx

sinu222

, show that 0zu

zyu

yxu

x =

+

+

.

Sol.: Here

=

+

=

+

=xz

,xy

f x

xz

xz

.xy

xy

xz

xy

1sinx

zxyzxyzyx

sinu 0

22

0222

.

u is a homogeneous function of x ,y and z of degree 0.

Then by Eulers theorem, we have

0u.0nuzu

zyu

yxu

x ===

+

+

.

Hence the result .

Q.No.25: Given that F(u) = V(x , y , z), where V is a homogeneous function of x , y , z of

degree n , then prove that)u(F

)u(Fn

z

uz

y

uy

x

ux / =

+

+

.

Sol.: Here V(x , y , z) is a homogeneous function of x , y , z of degree n , then by Euler's

theorem nVzV

zyV

yxV

x =

+

+

)u(nFz

)u(Fz

y)u(F

yx

)u(Fx =

+

+

)u(nFzu

)u(Fzyu

)u(Fyxu

)u(Fx =+

+

)u(F

)u(Fn

zu

zyu

yxu

x / =

+

+

.

Hence the result .


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22

NEXT TOPIC

Total Differentials,

Explicit Function, Implicit Functions

and

Total Differential Coefficient

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2%2dhomogeneous equation and euler theorem

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