Question

state and prove Euler's theorem for homogeneous function of two variables.

Answer 1

Answer:

To prove :

 x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial x}=nz    

Step-by-step explanation:

Let z be a function dependent on two variable x and y.

Let z = f(x,y)

Using property of homogeneous function,

z = x^{n}Ф($\frac{y}{x}$    .................(1)

Differentiating (1) with respect to x

$\frac{\partial z}{\partial x}= \phi (\frac{y}{x})nx^{n-1}+x^{n}\phi '(\frac{y}{x})\frac{\partial \frac{y}{x}}{\partial x}$

$\frac{\partial z}{\partial x} = \phi (\frac{y}{x})nx^{n-1}+x^{n}\phi '(\frac{y}{x})\frac{-y}{x^{2}}$ ................(2)

Differentiating (1) with respect to y

$\frac{\partial z}{\partial y} = \phi '(\frac{y}{x})\frac{\partial }{\partial y}\frac{y}{x}x^{n}+\frac{\partial }{\partial y}x^{n}\phi (\frac{y}{x})$

$\frac{\partial z}{\partial y} = \phi '(\frac{y}{x})\frac{1}{x}x^{n}+\frac{\partial }{\partial y}x^{n}\phi (\frac{y}{x})$

$\frac{\partial z}{\partial y} = \phi '(\frac{y}{x})x^{n-1}$ ...........(3)

Multiply (2) by x add(3) by y and then adding we get

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} =\phi (\frac{y}{x})nx^{n}-x^{n-1}y\phi '(\frac{y}{x})+y  \phi '(\frac{y}{x})x^{n-1}$

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nx^{n}\phi \frac{y}{x}$

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz$

Hence, proved.

Answer 2

Answer:

f(x,y)=(x^+y^3)(x-y)