Question

Verify Euler’s theorem when \[f\left( x,y \right)=\dfrac{{{x}^{4}}+{{y}^{4}}}{x+y}\].

Answer 1

Hint: First, before proceeding for this, we must know the following Euler’s theorem is stated by the equation for the function f(x,y) of degree n as $x\dfrac{df}{dx}+y\dfrac{dy}{dx}=nf\left( x,y \right)$. Then, we can see clearly that we get the function f(x,y) as homogenous function order degree 3 which gives the value of n as 3. Then, for the given question as \[f\left( x,y \right)=\dfrac{{{x}^{4}}+{{y}^{4}}}{x+y}\], the formula to be proved for Euler’s theorem is as $x\dfrac{df}{dx}+y\dfrac{dy}{dx}=3f\left( x,y \right)$, we can verify the theorem.

Complete step-by-step solution:
In this question, we are supposed to verify the Euler’s theorem for the function \[f\left( x,y \right)=\dfrac{{{x}^{4}}+{{y}^{4}}}{x+y}\].
So, before proceeding for this, we must know the following Euler’s theorem is stated by the equation for the function f(x,y) of degree n as:
$x\dfrac{df}{dx}+y\dfrac{dy}{dx}=nf\left( x,y \right)$
So, we have the equation of the function f(x,y) and firstly we need to prove it as homogenous to apply Euler’s theorem by taking common from numerator and denominator as:
\[\begin{align}
  & f\left( x,y \right)=\dfrac{{{x}^{4}}\left( 1+\dfrac{{{y}^{4}}}{{{x}^{4}}} \right)}{x\left( 1+\dfrac{y}{x} \right)} \\
 & \Rightarrow f\left( x,y \right)=\dfrac{{{x}^{3}}\left( 1+\dfrac{{{y}^{4}}}{{{x}^{4}}} \right)}{\left( 1+\dfrac{y}{x} \right)} \\
\end{align}\]
So, we can see clearly that we get the function f(x,y) as homogenous function order degree 3 which gives the value of n as 3.
Now, for the given question as \[f\left( x,y \right)=\dfrac{{{x}^{4}}+{{y}^{4}}}{x+y}\], the formula to be proved for Euler’s theorem is as:
$x\dfrac{df}{dx}+y\dfrac{dy}{dx}=3f\left( x,y \right)$
So, firstly we need to calculate the partial derivative of the function f(x,y) with respect to x as:
$\dfrac{\partial f\left( x,y \right)}{\partial x}=\dfrac{\partial }{\partial x}\left( \dfrac{{{x}^{4}}+{{y}^{4}}}{x+y} \right)$
Now, by using the property of the differentiation, we get:
$\begin{align}
  & \dfrac{\partial f\left( x,y \right)}{\partial x}=\dfrac{\left( x+y \right)\dfrac{\partial \left( {{x}^{4}}+{{y}^{4}} \right)}{\partial x}-\left( {{x}^{4}}+{{y}^{4}} \right)\dfrac{\partial \left( x+y \right)}{\partial x}}{{{\left( x+y \right)}^{2}}} \\
 & \Rightarrow \dfrac{\partial f\left( x,y \right)}{\partial x}=\dfrac{\left( x+y \right)\left( 4{{x}^{3}} \right)-\left( {{x}^{4}}+{{y}^{4}} \right)\times 1}{{{\left( x+y \right)}^{2}}} \\
 & \Rightarrow \dfrac{\partial f\left( x,y \right)}{\partial x}=\dfrac{\left( 4{{x}^{4}}+4{{x}^{3}}y \right)-{{x}^{4}}-{{y}^{4}}}{{{\left( x+y \right)}^{2}}} \\
 & \Rightarrow \dfrac{\partial f\left( x,y \right)}{\partial x}=\dfrac{3{{x}^{4}}+4{{x}^{3}}y-{{y}^{4}}}{{{\left( x+y \right)}^{2}}}.....\left( i \right) \\
\end{align}$
Similarly, we need to calculate the partial derivative of the function f(x,y) with respect to y as:
$\dfrac{\partial f\left( x,y \right)}{\partial y}=\dfrac{\partial }{\partial y}\left( \dfrac{{{x}^{4}}+{{y}^{4}}}{x+y} \right)$
Now, by using the property of the differentiation, we get:
$\begin{align}
  & \dfrac{\partial f\left( x,y \right)}{\partial y}=\dfrac{\left( x+y \right)\dfrac{\partial \left( {{x}^{4}}+{{y}^{4}} \right)}{\partial y}-\left( {{x}^{4}}+{{y}^{4}} \right)\dfrac{\partial \left( x+y \right)}{\partial y}}{{{\left( x+y \right)}^{2}}} \\
 & \Rightarrow \dfrac{\partial f\left( x,y \right)}{\partial y}=\dfrac{\left( x+y \right)\left( 4{{y}^{3}} \right)-\left( {{x}^{4}}+{{y}^{4}} \right)\times 1}{{{\left( x+y \right)}^{2}}} \\
 & \Rightarrow \dfrac{\partial f\left( x,y \right)}{\partial y}=\dfrac{\left( 4{{y}^{4}}+4{{y}^{3}}x \right)-{{x}^{4}}-{{y}^{4}}}{{{\left( x+y \right)}^{2}}} \\
 & \Rightarrow \dfrac{\partial f\left( x,y \right)}{\partial y}=\dfrac{3{{y}^{4}}+4{{y}^{3}}x-{{x}^{4}}}{{{\left( x+y \right)}^{2}}}.....\left( ii \right) \\
\end{align}$
Now, by multiplying equation (i) and (ii) with x and y respectively, and then adding these two equations, we get:
$\left( \dfrac{3{{x}^{4}}+4{{x}^{3}}y-{{y}^{4}}}{{{\left( x+y \right)}^{2}}} \right)\times x+\left( \dfrac{3{{y}^{4}}+4{{y}^{3}}x-{{x}^{4}}}{{{\left( x+y \right)}^{2}}} \right)\times y$
Now, by solving the above expression, we get:
$\begin{align}
  & \left( \dfrac{3{{x}^{5}}+4{{x}^{4}}y-x{{y}^{4}}}{{{\left( x+y \right)}^{2}}} \right)+\left( \dfrac{3{{y}^{5}}+4{{y}^{4}}x-y{{x}^{4}}}{{{\left( x+y \right)}^{2}}} \right) \\
 & \Rightarrow \left( \dfrac{3{{x}^{5}}+4{{x}^{4}}y-x{{y}^{4}}+3{{y}^{5}}+4{{y}^{4}}x-y{{x}^{4}}}{{{\left( x+y \right)}^{2}}} \right) \\
 & \Rightarrow \left( \dfrac{3{{x}^{5}}+3{{x}^{4}}y+3{{y}^{5}}+3{{y}^{4}}x}{{{\left( x+y \right)}^{2}}} \right) \\
 & \Rightarrow \left( \dfrac{3{{x}^{4}}\left( x+y \right)+3{{y}^{4}}\left( x+y \right)}{{{\left( x+y \right)}^{2}}} \right) \\
 & \Rightarrow \left( \dfrac{3{{x}^{4}}+3{{y}^{4}}}{\left( x+y \right)} \right) \\
 & \Rightarrow 3\left( \dfrac{{{x}^{4}}+{{y}^{4}}}{\left( x+y \right)} \right) \\
\end{align}$
Then, we can see that, we get the value of the expression as:
$\begin{align}
  & x\dfrac{df}{dx}+y\dfrac{dy}{dx}=3\left( \dfrac{{{x}^{4}}+{{y}^{4}}}{\left( x+y \right)} \right) \\
 & \Rightarrow x\dfrac{df}{dx}+y\dfrac{dy}{dx}=3f\left( x,y \right) \\
\end{align}$
Hence, the Euler’s theorem is verified for the function \[f\left( x,y \right)=\dfrac{{{x}^{4}}+{{y}^{4}}}{x+y}\]which is homogenous in nature.

Note: Now, to solve these type of the questions we need to know some of the basic differentiation properties to get the answer easily. So, the property required here for the differentiation is as:
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}$.
Here we are doing partial differentiation so the value of $\dfrac{\partial x }{\partial y}$ and $\dfrac{\partial y }{\partial x}$ will be zero this should be keep in mind while doing the partial differentiation.