Answer

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**Hint:**

Here, we have to find the functions which are homogeneous. We will find the functions which are homogeneous by using the condition of homogeneity. A homogeneous function is one with multiplicative scaling behavior i.e., if all its arguments are multiplied by a factor, then its value is multiplied by some power of this factor.

**Formula Used:**

A homogeneous function is a function which satisfies the condition,

\[f(tx,ty) = {t^n}f(x,y)\] for some \[n > 0\]

**Complete step by step solution:**

We are given functions, to check whether it is a homogeneous function.

A homogeneous function is a function which satisfies the condition,

\[f(tx,ty) = {t^n}f(x,y)\] for some \[n > 0\].

A) \[f\left( {x,y} \right) = x\sin y + y\sin x\]

Now, by using the homogeneity condition, we will get

\[ \Rightarrow f\left( {tx,ty} \right) = tx\sin \left( {ty} \right) + ty\sin \left( {tx} \right)\]

Factoring out common terms, we get

\[ \Rightarrow f\left( {tx,ty} \right) = = t\left( {x\sin ty + y\sin tx} \right)\]

But \[x\sin ty + y\sin tx \ne f\left( {x,y} \right)\]

Hence it is not a homogeneous function.

B) \[f\left( {x,y} \right) = x{e^{y/x}} + y{e^{x/y}}\]

Now, by using the homogeneity condition, we will get

\[ \Rightarrow f\left( {tx,ty} \right) = tx{e^{ty/tx}} + ty{e^{tx/ty}}\]

Factoring out common terms, we get

\[ \Rightarrow f\left( {tx,ty} \right) = t\left( {x{e^{y/x}} + y{e^{x/y}}} \right)\]

\[ \Rightarrow f\left( {tx,ty} \right) = tf\left( {x,y} \right)\]

Hence it is a homogeneous function.

C) \[f\left( {x,y} \right) = {x^2} - xy\]

Now, by using the homogeneity condition, we will get

\[ \Rightarrow f\left( {tx,ty} \right) = {(tx)^2} - txty\]

\[ \Rightarrow f\left( {tx,ty} \right) = {t^2}{x^2} - {t^2}xy\]

Factoring out common terms, we get

\[ \Rightarrow f\left( {tx,ty} \right) = {t^2}({x^2} - xy)\]

\[ \Rightarrow f\left( {tx,ty} \right) = {t^2}f\left( {x,y} \right)\]

Hence it is a homogeneous function.

D) We know that \[\arcsin (xy) = {\sin ^{ - 1}}(xy)\]

\[ \Rightarrow f\left( {x,y} \right) = {\sin ^{ - 1}}\left( {xy} \right)\]

Now, by using the homogeneity condition, we will get

\[ \Rightarrow f\left( {tx,ty} \right) = {\sin ^{ - 1}}\left( {txty} \right)\]

Factoring out common terms, we get

\[ \Rightarrow f\left( {tx,ty} \right) = {\sin ^{ - 1}}\left( {{t^2}(xy)} \right)\]

But \[{\sin ^{ - 1}}\left( {{t^2}(xy)} \right) \ne f\left( {x,y} \right)\]

Hence it is not a homogeneous equation.

**Therefore, the functions \[x{e^{y/x}} + y{e^{x/y}}\] and \[{x^2} - xy\] are homogeneous functions.**

**Note:**

We should be very conscious in finding whether the given functions are homogeneous or not. We should replace the variable \[x\] by multiplying the variable with any constant \[tx\]. The zero function is homogeneous of any degree. The sum of homogeneous functions of the same homogeneous degree is also homogeneous of the same degree, unless it is identically the zero function. The product of homogeneous functions of degrees \[{d_1}\] and \[{d_2}\] is homogeneous of degree \[{d_1} + {d_2}\] . The reciprocal of a homogeneous function of degree d is homogeneous of degree −d. The k th power of a homogeneous function of degree \[d\] is homogeneous of degree \[kd\] . These are the rules of homogeneous function.

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