VERIFY EULERS THEOREM FOR THE HOMOGENOUS FUNCTION

Problem 1 :

Prove that

f (x, y) = x³ − 2x²y +3xy² + y³

is homogeneous; what is the degree? Verify Euler’s Theorem for f .

Solution :

f (x, y) = x³ − 2x²y +3xy² + y³

Apply x = λx and y = λy

f (λx, λy) = (λx)³ − 2(λx)²(λy) +3(λx)(λy)² + (λy)³

f (λx, λy) = λ³x³ − 2λ³(x²y) +3λ³xy² + λ³y³

f (λx, λy) = λ³(x³ − 2x²y +3xy² + y³)

So it is not a homogeneous function of degree 3.

Given : x³ − 2x²y +3xy² + y³

∂F/∂x = 3x²-2(2x)y+3(1)y²+0

∂F/∂x = 3x²-4xy+3y²

x(∂F/∂x) = x(3x²-4xy+3y²) ---(1)

Given : x³ − 2x²y +3xy² + y³

∂F/∂y = 0-2x²(1)+3x(2y)+3y²

∂F/∂y = -2x²+6xy+3y²

y(∂F/∂y) = y(-2x²+6xy+3y²) ---(2)

(1) + (2)

x(∂F/∂x) + y(∂F/∂y) = x(3x²-4xy+3y²)+y(-2x²+6xy+3y²)

= 3x³-4x²y+3xy²-2x²y+6xy²+3y³

= 3x³-6x²y+9xy²+3y³

= 3(x³-2x²y+3xy²+y³)

= P F(x, y)

Hence it is proved.

Problem 2 :

Prove that

g(x, y) = x log (y/x)

is homogenous, what is the degree ? Verify Euler's  theorem for g.

Solution :

g(x, y) = x log (y/x)

Applying x = λx and y = λy

g(λx, λy) = λx log (λy/λx)

g(λx, λy) = λx log (y/x)

It is a homogenous function of degree 1.

∂F/∂x = x (x/y)(-yx^-2) + log(y/x)(1)

∂F/∂x = x (x/y)(-y/x²) + log(y/x)

∂F/∂x = -1 + log (y/x)  -----(1)

∂F/∂y = x (x/y)(1/x)

∂F/∂y = (x/y) -----(2)

(1) + (2)

x(∂F/∂x) + y(∂F/∂y) = x[-1 + log (y/x)] + y(x/y)

x(∂F/∂x) + y(∂F/∂y) = -x + xlog (y/x) + x

x(∂F/∂x) + y(∂F/∂y) = x log (y/x)

Hence it is proved.

Problem 3 :

If u(x, y) = (x²+y²)/√(x+y)

prove that x(∂u/∂x) + y(∂u/∂y) =  (3/2)u

Solution :

u(x, y) = (x²+y²)/√(x+y)

Applying x = λx and y = λy

u(λx, λy) = ((λx)²+(λy)²)/√((λx)+(λy))

u(λx, λy) = ((λ²x²+λ²y)²)/√λ(x+y)

u(λx, λy) = λ²(x²+y)²/√λ(x+y)

u(λx, λy) = λ^2-1/2(x²+y)²/√(x+y)

u(λx, λy) = λ^3/2(x²+y)²/√(x+y)

It is a homogenous function of degree 3/2.

x(∂u/∂x) + y(∂u/∂y) =  (3/2)u

Problem 4 :

If

v(x, y) = log [(x²+y²)/(x+y)]

prove that x(∂v/∂x) + y(∂v/∂y) =  1

Solution :

F = v = log [(x²+y²)/(x+y)]

F = e^v = [(x²+y²)/(x+y)]

Applying x = λx and y = λy

v(λx, λy) = [(λx)²+(λy)²]/((λx)+(λy))

v(λx, λy) = [λ²(x²+y²)/λ(x+y)]

v(λx, λy) = λ[(x²+y²)/(x+y)]

It is a homogenous function of degree 1.

F = e^v

(1)+(2)

x(∂F/∂x) + y(∂F/∂y) = nF

xe^v(∂v/∂x) + ye^v(∂v/∂y) = (1)e^v

e^v[x(∂v/∂x) + y(∂v/∂y)] = e^v

[x(∂v/∂x) + y(∂v/∂y)] = 1

Hence it is proved.

Problem 4 :

If

w(x, y, z) = log[(5x³y⁴+7y²xz⁴-75y³z⁴)/(x²+y²)]

find x(∂w/∂x) + y(∂w/∂y) + z(∂w/∂z).

Solution :

F = w(x, y, z) = log[(5x³y⁴+7y²xz⁴-75y³z⁴)/(x²+y²)]

F = w = log[(5x³y⁴+7y²xz⁴-75y³z⁴)/(x²+y²)]

Let F = e^w

 w(λx, λy, λz) = [λ⁷(5x³y⁴+7y²xz⁴-75y³z⁴)/λ²(x²+y²)]

= λ⁵(5x³y⁴+7y²xz⁴-75y³z⁴)/(x²+y²)

It is a homogenous function of degree 5.

x(∂F/∂x) = x(∂(e^w)/∂x)  ==> xe^w(∂F/∂x) ----(1)

y(∂F/∂y) = y(∂(e^w)/∂y)  ==> ye^w(∂F/∂y) ----(2)

z(∂F/∂z) = z(∂(e^w)/∂z)  ==> ze^w(∂F/∂z) ----(3)

(1)+(2)+(3)

xe^w(∂F/∂x) + ye^w(∂F/∂y) + ze^w(∂F/∂z) = 5e^w

x(∂F/∂x) + y(∂F/∂y) + z(∂F/∂z) = 5

Hence the answer is 5.

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