Question

the degree of homogeneous function z=√x+√y/x+y​

Answer 1

xe  

y/x

+ye  

x/y

 

C

x  

2

−xy

Homogeneous equation by definition is

f(λx)=λ  

n

f(x)

A)f(x,y)=xsiny+ysinx

Now

f(λx,λy)=λxsin(λy)+λysin(λx)

=λ(xsinλy+ysin(λx))

But

xsinλy+ysin(λx)



=f(x,y)

Hence it is not a homogeneous function.

B)f(x,y)=xe  

y/x

+ye  

x/y

 

f(λx,λy)=λxe  

λy/λx

+λye  

λx/λy

 

=λ(e  

x/y

+e  

y/x

)

=λf(x,y)

Hence it is a homogeneous function.

C)f(x,y)=x  

2

−xy

f(λx,λy)=λ  

2

x  

2

−λ  

2

(xy)

=λ  

2

(x  

2

−xy)

=λ  

2

f(x,y)

Hence it is a homogeneous function.

D)f(x,y)=sin  

−1

(xy)

f(λx,λy)=sin  

−1

(λ  

2

(xy))



=f(x,y)

Hence it is not a homogeneous equation

Answer 2

Answer:

The degree of homogeneous function is -$\frac{1}{2}$.

Step-by-step explanation:

Let a function is f(x,y). If all the x and y are replaced by kx and ky respectively and the function will come $k^{n} f(x,y)$ then the function is called a homogenious function.

Here the n is called the degree of the homogenious function.

$z=\frac{\sqrt{x} +\sqrt{y} }{x+y}$

$z' = \frac{\sqrt{kx} +\sqrt{ky} }{kx+ky}\\ or , z' = \frac{k^{\frac{1}{2} } (\sqrt{x} +\sqrt{y} )}{k(x+y)}\\ or , z' = k^{-\frac{1}{2} } \frac{\sqrt{x} +\sqrt{y} }{x+y}\\\\or , z' = k^{-\frac{1}{2} } z$

So, The degree of homogeneous function is -$\frac{1}{2}$.

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