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1. INFOMATICA ENGG.ACADEMY CONTACT: 9821131002/9076931776
49
THEOREMS
Homogeneous Function in two variables: A function u is said to be a homogeneous function in x and y
of degree n if it is expressible as
u = xn
𝛟 (y/x) where 𝛟 (y/x) is a function of y/x
1) Euler’s Theorem for u :
If u is a homogeneous function in x and y of degree ‘n’, then show that x
𝝏𝒖
𝝏𝒙
+ y
𝝏𝒖
𝝏𝒚
= nu.
Proof: Since u is a homogeneous function in x and y of degree n,
Let u = xn
𝛟 (y/x) ……….(I)
Differentiating (I) partially with respect to x,
𝝏𝒖
𝝏𝒙
= nxn- 1
𝛟 (y/x) + xn
𝛟’ (y/x) (- y/x2
)
= nxn- 1
𝛟 (y/x) – y xn-2
𝛟’ (y/x)……….. (i)
Also, differentiating (I) partially with respect to y,
𝝏𝒖
𝝏𝒚
= xn
𝛟’ (y/x) 1/x
= xn- 1
𝛟’ (y/x) …………(ii)
Multiplying eq. (i) by eq. (ii) by y and adding we get,
x
𝝏𝒖
𝝏𝒙
+ y
𝝏𝒖
𝝏𝒚
= nxn
𝛟 (y/x) – yxn-1
𝛟’ (y/x) + yxn- 1
𝛟’ (y/x)
= n xn
𝛟 (y/x)
= nu [ from (I) ]
Hence x
𝜕𝑢
𝜕𝑥
+ y
𝜕𝑢
𝜕𝑦
= nu
2) Deduction:
If u is an homogeneous function in x and y of degree n,
then show that x2 𝝏 𝟐 𝒖
𝝏𝒙 𝟐 + 2xy
𝝏 𝟐 𝒖
𝝏𝒙𝝏𝒚
+ y2 𝝏 𝟐 𝒖
𝝏𝒚 𝟐 = n (n-1) u
Proof : Since, u is an homogeneous function in x and y of degree n,
by Euler’s Theorem we have,
x
𝜕𝑢
𝜕𝑥
+ y
𝜕𝑢
𝜕𝑦
= nu …………… (I)
Differentiating (I) partially with respect to x we get,
x
𝜕2 𝑢
𝜕𝑥2 +
𝜕𝑢
𝜕𝑥
+ y
𝜕2 𝑢
𝜕𝑥𝜕𝑦
= n
𝜕𝑢
𝜕𝑥
 x
𝜕2 𝑢
𝜕𝑥2 + y
𝜕2 𝑢
𝜕𝑥𝜕𝑦
= (n- 1)
𝜕𝑢
𝜕𝑥
…………….. (II)
Similarly differentiating (I) partially w. r. t. y we get,
y
𝜕2 𝑢
𝜕𝑦2 + x
𝜕2 𝑢
𝜕𝑦𝜕𝑥
= (n- 1)
𝜕𝑢
𝜕𝑦
………… (III)
Multiplying eq. (II) by x and eq. (III) by y and adding we get,
x2 𝜕2 𝑢
𝜕𝑥2 + 2xy
𝜕2 𝑢
𝜕𝑥𝜕𝑦
+ y2 𝜕2 𝑢
𝜕𝑦2 = (n- 1) x
𝜕𝑢
𝜕𝑥
+ (n- 1) y
= (n- 1) (x
𝜕𝑢
𝜕𝑥
+ y
𝜕𝑢
𝜕𝑦
)
= (n- 1) nu [from (I) ]
 x2 𝜕2 𝑢
𝜕𝑥2 + 2xy
𝜕2 𝑢
𝜕𝑥𝜕𝑦
+ y2 𝜕2 𝑢
𝜕𝑦2 = n (n- 1) u

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3) Euler’s Theorem for f(u):
If f (u) is an homogeneous function in x and y of degree n then show that x
𝝏𝒖
𝝏𝒙
+ y
𝝏𝒖
𝝏𝒚
= n
𝒇(𝒖)
𝒇′(𝒖)
Proof : Let z = f(u)
Now f(u) is an homogeneous function in x and y of degree n.
 z is an homogeneous function in x and y of degree n.
 By Euler’s Theorem we have,
x
𝝏𝒛
𝝏𝒙
+ y
𝝏𝒛
𝝏𝒚
= nz……………………(I)
But
𝝏𝒛
𝝏𝒙
= f ’(u)
𝝏𝒖
𝝏𝒙
and
𝝏𝒛
𝝏𝒚
= f ’(u)
𝝏𝒖
𝝏𝒚
Substituting in (I) we get,
xf ’(u)
𝝏𝒖
𝝏𝒙
+ yf ’(u)
𝝏𝒖
𝝏𝒚
= nf(u) [Since z = f(u) ]
 x
𝝏𝒖
𝝏𝒙
+ y
𝝏𝒖
𝝏𝒚
= n
𝒇𝒖
𝒇′(𝒖)
4) Deduction:
If f (u) is an homogeneous function in x and y of degree n then,
show that, x2 𝝏 𝟐 𝒖
𝝏𝒙 𝟐 + 2xy
𝝏 𝟐 𝒖
𝝏𝒙𝝏𝒚
+ y2 𝝏 𝟐 𝒖
𝝏𝒚 𝟐 = G(u) [G ’(u) – 1] where G ’(u) = n
𝒇𝒖
𝒇′(𝒖)
Proof : Since f(u) is an homogeneous function in x and y of degree n,
by Euler’s Theorem we have
x
𝜕𝑢
𝜕𝑥
+ y
𝜕𝑢
𝜕𝑦
= n
𝑓(𝑢)
𝑓′(𝑢)
= G(u) (say) ………..(I)
Differentiating (I) partially with respect to x,
x
𝜕2 𝑢
𝜕𝑥2 +
𝜕𝑢
𝜕𝑦
+ y
𝜕2 𝑢
𝜕𝑥𝜕𝑦
= G ’(u).
𝜕𝑢
𝜕𝑥
 x
𝜕2 𝑢
𝜕𝑥2 + y
𝜕2 𝑢
𝜕𝑥 𝜕𝑦 2 = [ G ’(u) – 1 ]
𝜕𝑢
𝜕𝑥
…………(II)
Similarly differentiating (I) partially with respect to y,
x
𝜕2 𝑢
𝜕𝑦𝜕𝑥
+ y
𝜕2 𝑢
𝜕𝑦2 = [ G ’ (u) – 1 ]
𝜕𝑢
𝜕𝑦
…………(III)
Multiplying eq. (II) by x and eq. (III) by y and adding we get,
x2 𝜕2 𝑢
𝜕𝑥2 + 2xy
𝜕2 𝑢
𝜕𝑥𝜕𝑦
+ y2 𝜕2 𝑢
𝜕𝑦2 = [ G ’(u) – 1 ] x
𝜕𝑢
𝜕𝑥
+ [ G ’(u) – 1 ] y
𝜕𝑢
𝜕𝑦
= [ G ’(u) – 1 ] (x
𝜕𝑢
𝜕𝑥
+ y
𝜕𝑢
𝜕𝑦
)
= [ G ’(u) – 1 ] G(u) …….[ From (I)]
Hence x2 𝜕2 𝑢
𝜕𝑥2 + 2xy
𝜕2 𝑢
𝜕𝑥𝜕𝑦
+ y2 𝜕2 𝑢
𝜕𝑦2 = G(u) [ G ’(u) – 1 ] where G ’(u) = n
𝑓𝑢
𝑓′(𝑢)
Homogeneous function in three variables: A function u is said to be an homogeneous function in x, y, z
degree n if it expressible as u = xn
𝛟 (y/x, z/x)
5) Euler’s Theorem in three variables
If u is an homogeneous function in x, y, z of degree n, then show that x
𝝏𝒖
𝝏𝒙
+ 𝒚
𝝏𝒖
𝝏𝒚
+ z
𝝏𝒖
𝝏𝒛
= nu
Proof: Since u is an homogeneous function in x, y, z of degree n,
Let u = xn
𝛟 (y/x, z/x ) ……..(I)
If p = y/x and q = z/x

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51
Then u = xn
𝛟 (p, q) where p and q are functions of x, y, z
Now
𝝏𝒑
𝝏𝒙
= -
𝒚
𝒙 𝟐 ,
𝝏𝒑
𝝏𝒚
= 1/x,
𝝏𝒑
𝝏𝒛
= 0
𝜕𝑞
𝜕𝑥
= -
𝑧
𝑥2 ,
𝜕𝑞
𝜕𝑦
= 0,
𝜕𝑞
𝜕𝑧
= 1/x
Then
𝝏𝒖
𝝏𝒙
=
𝝏
𝝏𝒙
[ xn
𝛟 (p, q) ]
=
𝝏
𝝏𝒙
[ xn
]. 𝛟 (p, q) + xn
.
𝝏
𝝏𝒙
[ (p, q) ]
= nxn-1
𝛟 (p, q) + xn
(
𝜕𝜙
𝜕𝑝
.
𝜕𝑝
𝜕𝑥
+
𝜕𝜙
𝜕𝑞
.
𝜕𝑞
𝜕𝑥
)
= n xn-1
𝛟 (p, q) + xn
(−
𝑦
𝑥2 .
𝜕𝜙
𝜕𝑝
-
𝑧
𝑥2 .
𝜕𝜙
𝜕𝑞
)
= n xn-1
𝛟 (p, q) – y xn- 2 𝜕𝜙
𝜕𝑝
– zxn- 2 𝜕𝜙
𝜕𝑞
…………….(i)
Also
𝜕𝑢
𝜕𝑦
= xn 𝜕
𝜕𝑦
[𝛟 (p, q) ]
= xn
(
𝜕𝜙
𝜕𝑝
.
𝜕𝑝
𝜕𝑦
+
𝜕𝜙
𝜕𝑞
.
𝜕𝑞
𝜕𝑦
)
= xn
(1/x
𝜕𝜙
𝜕𝑝
+0
𝜕𝜙
𝜕𝑞
)
= xn-1 𝜕𝜙
𝜕𝑝
………………(ii)
Similarly
𝜕𝑢
𝜕𝑧
= xn-1 𝜕𝜙
𝜕𝑞
………………(iii)
Multiplying eq. (i) by x, eq. (ii) by y, eq. (iii) by z and adding we get,
x
𝜕𝑢
𝜕𝑥
+ 𝑦
𝜕𝑢
𝜕𝑦
+ z
𝜕𝑢
𝜕𝑧
= nxn
𝛟 (p, q) – yxn-1 𝜕𝜙
𝜕𝑝
- zxn-1 𝜕𝜙
𝜕𝑞
+ yxn-1 𝜕𝜙
𝜕𝑝
+zxn-1 𝜕𝜙
𝜕𝑞
= n xn
(p, q)
= n u [From (I)]
Hence x
𝜕𝑢
𝜕𝑥
+ 𝑦
𝜕𝑢
𝜕𝑦
+ z
𝜕𝑢
𝜕𝑧
= n u