Question

How would you determine if $f\left( x,y \right)=\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$ is homogeneous and what would it’s degree be?

Answer 1

Hint: Now to check if the given function is Homogeneous or not we will substitute $x=\alpha x$ and $y=\alpha y$ in the equation. Hence we will further simplify the equation and try to write it in the form \[f\left( x,y \right)={{\alpha }^{n}}f\left( x,y \right)\] . If we write it in this form then the function is Homogeneous function with degree n.

Complete step by step solution:
Now first let us understand the meaning of a Homogeneous function and Homogeneous differential equation.
A differential equation is an equation with derivative terms such as $\dfrac{dy}{dx}$ . Now we know that the solution of a differential equation is a function which satisfies the given equation.
Now any Differential equation $f\left( x,y \right)dy=g\left( x,y \right)dx$ is said to be Homogenous if the degree of $f\left( x,y \right)$ and $g\left( x,y \right)$ is same. Now let us understand when a function is said to be Homogenous. A function is called Homogeneous function of degree n when $f\left( x,y \right)$ can be written in the form of ${{k}^{n}}f\left( x,y \right)$
If the two functions are homogeneous and are of same degree then we can easily solve the differential equation by substitution of variables.
Now to check if the function is Homogeneous we will substitute $\alpha x$ in place of x and check if we can take $\alpha $ out. Now consider the given function $f\left( x,y \right)=\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$
Now substituting $x=\alpha x$ and $y=\alpha y$ we get,
$\begin{align}
  & \Rightarrow f\left( x,y \right)=\dfrac{\left( \alpha x \right)\left( \alpha y \right)}{\sqrt{{{\alpha }^{2}}{{x}^{2}}+{{\alpha }^{2}}{{y}^{2}}}} \\
 & \Rightarrow f\left( x,y \right)=\dfrac{{{\alpha }^{2}}xy}{\sqrt{{{\alpha }^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)}} \\
 & \Rightarrow f\left( x,y \right)=\dfrac{{{\alpha }^{2}}xy}{\alpha \sqrt{{{x}^{2}}+{{y}^{2}}}} \\
 & \Rightarrow f\left( x,y \right)=\alpha \dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \\
 & \Rightarrow f\left( x,y \right)=\alpha f\left( x,y \right) \\
\end{align}$

Hence the given function is Homogeneous function of degree 1.

Note: Now note that any linear differential equation is said to be a homogenous equation if there is no constant in the equation. Since there is no constant we can easily write the equation in the form $f\left( x,y \right)={{k}^{n}}f\left( x,y \right)$