The document discusses homogeneous functions in two and three variables, focusing on Euler's theorem for these functions. Key deductions and proofs are provided regarding the relationship between derivatives and the degree of homogeneity. Specific theorems are derived for both two-variable and three-variable homogeneous functions, detailing the mathematical relationships that govern their behavior.

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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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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