Answer
(a) $f(t x, t y)=t^{2} f(x, y)$ The function is homogeneous of degree 2
(b) $f(t x, t y)=t f(x, y)$ The function is homogeneous of degree 1
(c) $f(t x, t y)=t^{3} f(x, y)$ The function is homogeneous of degree 3
(a) $f(t x, t y)=t^{-4} f(x, y)$ The function is homogeneous of degree (-4)
Work Step by Step
We know that a function $f(x, y)$ is said to be homogeneous of degree $n$ if
\[
f(t x, t y)=t^{n} f(x, y) \text { for } t>0
\]
$(\mathrm{a}) f(x, y)=y^{2}+3 x^{2}$
\[
\therefore f(t x, t y)=3(t x)^{2}+(t y)^{2}=3 t^{2} x^{2}+t^{2} y^{2}=t^{2}\left(3 x^{2}+y^{2}\right)=t^{2} f(x, y)
\]
So, the function is homogeneous of degree 2
(b) $f(x, y)=\sqrt{x^{2}+y^{2}}$
\[
\therefore f(t x, t y)=\sqrt{(t x)^{2}+(t y)^{2}}=\sqrt{t^{2}\left(x^{2}+y^{2}\right)}=t \sqrt{x^{2}+y^{2}}=t f(x, y)
\]
So, the function is homogeneous of degree 1
c) $2 y^{3}+x^{2} y=f(x, y)$
\[
\therefore f(t x, t y)=(t x)^{2}(t y)+2(t y)^{3}=3 t^{3} x^{2} y^{2}+2 t^{3} y^{3}=t^{3}\left(x^{2} y+2 y^{3}\right)=t^{3} f(x, y)
\]
So, the function is homogeneous of degree 3.
$(\mathrm{d}) \mathrm{a}) f(x, y)=\frac{5}{\left(x^{2}+2 y^{2}\right)^{2}}$
\[
\therefore f(t x, t y)=\frac{5}{\left((t x)^{2}+2(t y)^{2}\right)^{2}}=\frac{5}{t^{4}\left(x^{2}+2 y^{2}\right)^{2}}=t^{-4} f(x, y)
\]
So, the function is homogeneous of degree (-4).
