VERIFY EULERS THEOREM FOR THE HOMOGENOUS FUNCTION

Let A = {(x, y) a < x < b, c < y < d} ∈ 2, F : A-> ℝ, we say that F is a homogeneous function on A, if there exists a constant P such that F(λx, λy) = λp f(x, y) for all λℝ such that (λx, λy)∈A. This constant is called degree of F.

Step 1 :

In the given function, apply x = λx and y = λy.

Step 2 :

Do the possible simplification.

Step 3 :

Get the function in the form of λp f(x).

P is the degree of the polynomial.

Then,

x(∂F/∂x) (x, y) + y(∂F/∂y) (x, y) = pF(x, y)

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(λxλy) = λx log (λy/λx)

g(λxλy) = 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(λxλy) = ((λx)2+(λy)2)/√((λx)+(λy))

u(λxλy) = ((λ2x2+λ2y)2)/λ(x+y)

u(λxλy) = λ2(x2+y)2/√λ(x+y)

u(λxλy) = λ2-1/2(x2+y)2/(x+y)

u(λxλy) = λ3/2(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 = ev[(x2+y2)/(x+y)]

Applying x = λx and y = λy

v(λxλy) = [(λx)2+(λy)2]/((λx)+(λy))

v(λxλy) = [λ2(x2+y2)/λ(x+y)]

v(λxλy) = λ[(x2+y2)/(x+y)]

It is a homogenous function of degree 1.

F = ev

 ∂F/∂x = ∂(ev)/∂x∂F/∂x = ev(∂v/∂x) --(1) ∂F/∂x = ∂(ev)/∂y∂F/∂x = ev(∂v/∂y)--(2)

(1)+(2)

x(∂F/∂x) + y(∂F/∂y) = nF

xev(∂v/∂x) + yev(∂v/∂y) = (1)ev

ev[x(∂v/∂x) + y(∂v/∂y)] = ev

[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 = ew

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(∂(ew)/∂x)  ==> xew(∂F/∂x) ----(1)

y(∂F/∂y) = y(∂(ew)/∂y)  ==> yew(∂F/∂y) ----(2)

z(∂F/∂z) = z(∂(ew)/∂z)  ==> zew(∂F/∂z) ----(3)

(1)+(2)+(3)

xew(∂F/∂x) + yew(∂F/∂y) + zew(∂F/∂z) = 5ew

x(∂F/∂x) + y(∂F/∂y) + z(∂F/∂z) = 5 Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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