Question

Which of the following functions are homogeneous?

• A. $f\left(x,y\right)=\frac{x-y}{x^2+y^2}$
• B. $f\left(x,y\right)=x^{\frac{1}{3}}{y}^{-\frac{2}{3}}{\tan }^{-1}\left(\frac{x}{y}\right)$
• C. $f\left(x,y\right)=x\left(\ln \sqrt{x^2+y^2}-\mathrm{lny}\right)+{\mathrm{ye}}^{\frac{x}{y}}$
• D. $f\left(x,y\right)=x\left(\ln \frac{2x^2+y^2}{x}-\ln (x+y)\right]+y^2\tan \left(\frac{x+2y}{3x-y}\right)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(x,y\right)=\frac{x-y}{x^2+y^2}$
B $f\left(x,y\right)=x^{\frac{1}{3}}{y}^{-\frac{2}{3}}{\tan }^{-1}\left(\frac{x}{y}\right)$
C $f\left(x,y\right)=x\left(\ln \sqrt{x^2+y^2}-\mathrm{lny}\right)+{\mathrm{ye}}^{\frac{x}{y}}$

Substitute kx and ky in place of x and y.
$i.f(\mathrm{kx},\mathrm{ky})=\frac{\mathrm{kx}-\mathrm{ky}}{k^2{x}^2+k^2{y}^2}=k^{-1}\left(\frac{x-y}{x^2+y^2}\right)=k^{-1}f\left(x,y\right)$
$\mathrm{ii}.f\left(\mathrm{kx},\mathrm{ky}\right)={\left(\mathrm{kx}\right)}^{\frac{1}{3}}{\left(\mathrm{ky}\right)}^{-\frac{2}{3}}{\tan }^{-1}\frac{\mathrm{kx}}{\mathrm{ky}}=k^{\left(\frac{1}{3}-\frac{2}{3}\right)}{x}^{\frac{1}{3}}{y}^{-\frac{2}{3}}{\tan }^{-1}\left(\frac{x}{y}\right)=k^{\left(-\frac{1}{3}\right)}{x}^{\frac{1}{3}}{y}^{-\frac{2}{3}}{\tan }^{-1}\left(\frac{x}{y}\right)=k^{\left(-\frac{1}{3}\right)}f\left(x,y\right)$
$\mathrm{iii}.f\left(\mathrm{kx},\mathrm{ky}\right)=\mathrm{kx}\left(\ln \sqrt{{\left(\mathrm{kx}\right)}^2+{\left(\mathrm{ky}\right)}^2}-\mathrm{lnky}\right)+{\mathrm{kye}}^{\frac{\mathrm{kx}}{\mathrm{ky}}}=\mathrm{kx}\left(\ln \frac{\sqrt{{\left(\mathrm{kx}\right)}^2+{\left(\mathrm{ky}\right)}^2}}{\mathrm{ky}}\right)+{\mathrm{kye}}^{\frac{\mathrm{kx}}{\mathrm{ky}}}=\mathrm{kx}\left(\ln \frac{k\sqrt{x^2+y^2}}{\mathrm{ky}}\right)+{\mathrm{kye}}^{\frac{x}{y}}(\mathrm{because}k>0)=\mathrm{kx}\left(\ln \frac{\sqrt{x^2+y^2}}{y}\right)+{\mathrm{kye}}^{\frac{x}{y}}=k\left(x\left(\ln \sqrt{x^2+y^2}-\mathrm{lny}\right)+{\mathrm{ye}}^{\frac{x}{y}}\right)=kf(x,y)$$\mathrm{iv}.f\left(\mathrm{kx},\mathrm{ky}\right)=\mathrm{kx}\left(\ln \frac{2{\left(\mathrm{kx}\right)}^2+{\left(\mathrm{ky}\right)}^2}{\mathrm{kx}}-\ln (\mathrm{kx}+\mathrm{ky})\right]+{\left(\mathrm{ky}\right)}^2\tan \left(\frac{\mathrm{kx}+2\mathrm{ky}}{3\mathrm{kx}-\mathrm{ky}}\right)=\mathrm{kx}\left(\ln \frac{k\left(2x^2+y^2\right)}{x}-\ln (\mathrm{kx}+\mathrm{ky})\right]+{\left(\mathrm{ky}\right)}^2\tan \left(\frac{x+2y}{3x-y}\right)=\mathrm{kx}\left(\ln \frac{k\left(2x^2+y^2\right)}{x(\mathrm{kx}+\mathrm{ky})}\right]+{\left(\mathrm{ky}\right)}^2\tan \left(\frac{x+2y}{3x-y}\right)=\mathrm{kx}\left(\ln \frac{\left(}{2x^2+y^2}\right]+{\left(\mathrm{ky}\right)}^2\tan \left(\frac{x+2y}{3x-y}\right)=\mathrm{kx}\left(\ln \left(\frac{2x^2+y^2}{x}\right)-\ln \left(x+y\right)\right]+{\left(\mathrm{ky}\right)}^2\tan \left(\frac{x+2y}{3x-y}\right)\mathrm{This}\mathrm{cannot}\mathrm{be}\mathrm{expressed}\mathrm{in}\mathrm{the}\mathrm{desired}\mathrm{form}.$