2%2DHomogeneous Equation and Euler Theorem
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Transcript of 2%2DHomogeneous Equation and Euler Theorem
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8/14/2019 2%2DHomogeneous Equation and Euler Theorem
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(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/
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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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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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 .
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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 .
u is a homogeneous function of x and y of degree 2.
Hence by Eulers theorem, we have
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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 .
u is a homogeneous function of x and y of degree 0.
Hence by Eulers theorem, we have
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
.
Hence by Eulers theorem, we have
( )5 / 15 / 14 / 14 / 1
yx
yx201
nf yf
yxf
x++
==
+
. (i)
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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
+
+
+=
Multiplying by x, we get
( ) ( )( )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
+
+
+=
Multiplying by y, we get
( ) ( )( )25 / 15 / 1
5 / 14 / 14 / 14 / 15 / 15 / 1
yx
y51
yxy41
yx
yf
y+
+
+=
(iii)
Adding (ii) and (iii), we get
( ) ( ) ( ) ( )( )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
.
Hence by Eulers theorem, we have
( )zyx21
u21
nuzu
zyu
yxu
x ++===
+
+
.
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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 ++=++=
+
+
.
Hence, the Eulers Theorem is verified .
(iii) Here
=+
=
+=
yx
f y1yx
tanyy
yxtanz 0
210
221 .
z is a homogeneous function of x ,y and z of degree 0.
Hence by Eulers theorem, we have
0z.0nzyz
yxz
x ===
+
.
Again since
+=
y
yxtanz
221 .
Differentiating partially w. r. t x and y respectively, we get
( ) 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=
+
+
+++
=
+
.
Hence, the Eulers Theorem is verified .
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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 .
Differentiating partially w. r. t x and y respectively, we get
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=
++
+
+
=
+
.
Hence, the Eulers Theorem is verified .
(v) Here
=
=xy
f xxy
logxz 44 .
z is a homogeneous function of x and y of degree 4.
Hence by Euler's theorem, we have
===
+
x
ylogx4z4nz
y
zy
x
zx 4 .
Again since
=xy
logxz 4 .
Differentiating partially w. r. t x and y respectively, we get
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
9
332
43 xxy
log.x4x
y
xy1
.xxy
log.x4xz
=
+
=
andy
xx1
xy1
.xyz 44 =
=
Hence, the Eulers Theorem is verified .
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.
Hence by Eulers theorem, we have
( ) ( )0usin.0usinn
yusin
yx
usinx ===
+
.
0yu
ucosyxu
ucosx =
+
0yu
yxu
x =
+
.
Henceyu
xy
xu
=
.
Hence the result .
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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 .
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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.
z is a homogeneous function of x and y of degree 1.
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++
=
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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)
Adding (ii) and (iii), we get
( ) ( ) ( )yx2
yxyx2
yy2
xyxx2
yuxu1usec121
yuxuyuxuusec121
uyxyu2ux +
=++=++
utan121
utanusec144
1utan
121
1usec121 22 +=
+=
( ) ( )12utan1utan144
1utan
121
utanutan1144
1 22 ++=++=
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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 .
u is a homogeneous function of x and y of degree 1.
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 =++
Multiplying by x , we get
( )x2xyxxx2 xzzsecxyzxzzx =++ . ......(ii)
Differentiating (i) w. r. t. y partially, we get
( )y2yyyxy zzsecyzzxz =++ Multiplying by y , we get
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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.
Then, by Eulers theorem, we have
0w.0nwyw
yxw
x ===
+
.
Hence 0y
wy
x
wx =
+
.
This completes the proof .
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 .
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
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16
u is a homogeneous function of x and y of degree 3.
Then, by Eulers theorem, we have
w.3nwyw
yxw
x ==
+
.
Hence w3yw
yxw
x =
+
.
This completes the proof .
(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 =
+
.
This completes the proof .
(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.
Then, by Eulers theorem, we have
0w.0nwy
w
yx
w
x ===
+
.
Hence 0yw
yxw
x =
+
.
This completes the proof .
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
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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 .
Hence by Eulers theorem, we have
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 =++
Multiplying by x , we get
( )x2
xyxxx2
xuusec21
xyuxuux =++ ......(ii)
Differentiating (i) w. r. t. y partially, we get
( )y2yyyxy uusec21
yuuxu =++
Multiplying by y , we get
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
18
( )y2yy2yxy yuusec21
uyyuyxu =++ ......(iii)
Adding (ii) and (iii), we get
( ) ( ) ( )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).
z is a homogeneous function of x and y of degree 2.
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
=
+
+
.
Multiplying by x, we get
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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
=
+
+
.
Multiplying by y, we get
yV
V2cosy2yV
yy
Vy
xyV
xy2
22
2
=
+
+
( ) yVy1V2cos2
yVy
xyVxy 2
22
2
=
+
. .....(iii)
Adding (ii) and (iii), we get
( ) ( ) 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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Partial Differentiation: Homogeneous Equation and Eulers Theorem
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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.
Hence by Eulers theorem, we have
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Partial Differentiation: Homogeneous Equation and Eulers Theorem
Prepared by: Dr. Sunil, NIT Hamirpur
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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Partial Differentiation: Homogeneous Equation and Eulers Theorem
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22
NEXT TOPIC
Total Differentials,
Explicit Function, Implicit Functions
and
Total Differential Coefficient
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