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Question

Question: Which of the following functions are homogeneous? A) \[x\sin y + y\sin x\] B) \[x{e^{y/x}} + y{...

Which of the following functions are homogeneous?
A) xsiny+ysinxx\sin y + y\sin x
B) xey/x+yex/yx{e^{y/x}} + y{e^{x/y}}
C) x2xy{x^2} - xy
D) arcsin(xy)\arcsin (xy)

Explanation

Solution

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

A) f(x,y)=xsiny+ysinxf\left( {x,y} \right) = x\sin y + y\sin x
Now, by using the homogeneity condition, we will get
f(tx,ty)=txsin(ty)+tysin(tx)\Rightarrow f\left( {tx,ty} \right) = tx\sin \left( {ty} \right) + ty\sin \left( {tx} \right)
Factoring out common terms, we get
f(tx,ty)==t(xsinty+ysintx)\Rightarrow f\left( {tx,ty} \right) = = t\left( {x\sin ty + y\sin tx} \right)
But xsinty+ysintxf(x,y)x\sin ty + y\sin tx \ne f\left( {x,y} \right)
Hence it is not a homogeneous function.

B) f(x,y)=xey/x+yex/yf\left( {x,y} \right) = x{e^{y/x}} + y{e^{x/y}}
Now, by using the homogeneity condition, we will get
f(tx,ty)=txety/tx+tyetx/ty\Rightarrow f\left( {tx,ty} \right) = tx{e^{ty/tx}} + ty{e^{tx/ty}}
Factoring out common terms, we get
f(tx,ty)=t(xey/x+yex/y)\Rightarrow f\left( {tx,ty} \right) = t\left( {x{e^{y/x}} + y{e^{x/y}}} \right)
f(tx,ty)=tf(x,y)\Rightarrow f\left( {tx,ty} \right) = tf\left( {x,y} \right)
Hence it is a homogeneous function.

C) f(x,y)=x2xyf\left( {x,y} \right) = {x^2} - xy
Now, by using the homogeneity condition, we will get
f(tx,ty)=(tx)2txty\Rightarrow f\left( {tx,ty} \right) = {(tx)^2} - txty
f(tx,ty)=t2x2t2xy\Rightarrow f\left( {tx,ty} \right) = {t^2}{x^2} - {t^2}xy
Factoring out common terms, we get
f(tx,ty)=t2(x2xy)\Rightarrow f\left( {tx,ty} \right) = {t^2}({x^2} - xy)

f(tx,ty)=t2f(x,y) \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)=sin1(xy)\arcsin (xy) = {\sin ^{ - 1}}(xy)
f(x,y)=sin1(xy)\Rightarrow f\left( {x,y} \right) = {\sin ^{ - 1}}\left( {xy} \right)
Now, by using the homogeneity condition, we will get
f(tx,ty)=sin1(txty)\Rightarrow f\left( {tx,ty} \right) = {\sin ^{ - 1}}\left( {txty} \right)
Factoring out common terms, we get
f(tx,ty)=sin1(t2(xy))\Rightarrow f\left( {tx,ty} \right) = {\sin ^{ - 1}}\left( {{t^2}(xy)} \right)
But sin1(t2(xy))f(x,y){\sin ^{ - 1}}\left( {{t^2}(xy)} \right) \ne f\left( {x,y} \right)
Hence it is not a homogeneous equation.

Therefore, the functions xey/x+yex/yx{e^{y/x}} + y{e^{x/y}} and x2xy{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 xx by multiplying the variable with any constant txtx. 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 d1{d_1} and d2{d_2} is homogeneous of degree d1+d2{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 dd is homogeneous of degree kdkd . These are the rules of homogeneous function.