Verify Euler’s theorem for the equation $f\left( x,y \right)=\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$
Answer
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Hint: Euler’s theorem for any homogenous $f\left( x,y \right)$ is given as
\[x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}=nf\left( x,y \right)\] , where n is the degree of the homogeneous function. Determine the value of ‘n’ by expression
$f\left( \lambda x,\lambda y \right)={{\lambda }^{n}}f\left( x,y \right)$
$\dfrac{\partial f}{\partial x}$ means differentiate the given function w.r.t $'x'$ by taking as $'y'$ a constant and $\dfrac{\partial f}{\partial y}$ means differentiation of the function w.r.t $'y'$ by taking $'x'$ as a constant. Use the relation $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ wherever necessary.
Complete step-by-step solution -
As, we know the Euler’s theorem for any homogeneous function \[f\left( x,y \right)\] of degree n is given as
\[x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}=nf\left( x,y \right)..........\left( i \right)\]
And we can verify any function to be homogeneous or not by the relation
$f\left( \lambda x,\lambda y \right)={{\lambda }^{n}}f\left( x,y \right)........\left( ii \right)$
Where ‘n’ is the degree of a homogeneous equation.
It means if any function follows equation(ii), then it will be a homogeneous function and ‘n’ is the degree of the equation.
Now, coming to the question, we are given the function $f\left( x,y \right)$ in the problem as
$f\left( x,y \right)=\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}.........\left( iii \right)$
Now, we can calculate value of $f\left( \lambda x,\lambda y \right)$ from the equation (iii) as
$\begin{align}
& f\left( \lambda x,\lambda y \right)=\dfrac{1}{\sqrt{{{\left( \lambda x \right)}^{2}}+{{\left( \lambda y \right)}^{2}}}} \\
& f\left( \lambda x,\lambda y \right)=\dfrac{1}{\sqrt{{{\lambda }^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)}} \\
& f\left( \lambda x,\lambda y \right)=\dfrac{1}{\lambda }\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \\
& f\left( \lambda x,\lambda y \right)={{\lambda }^{-1}}f\left( x,y \right) \\
\end{align}$
Now, we can compare the above relation by equation (ii) and hence, we get degree of the given function as
n = -1 ………..(iv)
Now, we have to verify the relation of equation (i), so let us calculate $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ by following approach
We have $f\left( x,y \right)=\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}={{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-1}{2}}}$
Now, we can calculate $\dfrac{\partial f}{\partial x}$ by differentiating the above function with respect to $'x'$ by taking $'y'$ as a constant.
We know
$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}........\left( v \right)$
So, we get $\dfrac{\partial f}{\partial x}$ as
\[\begin{align}
& \dfrac{\partial f}{\partial x}=\dfrac{\partial }{\partial x}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-1}{2}}} \\
& \dfrac{\partial f}{\partial x}=-\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-1}{2}-1}}\times \dfrac{\partial }{\partial x}\left( {{x}^{2}}+{{y}^{2}} \right) \\
\end{align}\]
Now, $'y'$ is acting as a constant. So, we know
$\dfrac{d}{dx}\text{constant = 0}$ . Hence, we get
\[\begin{align}
& \dfrac{\partial f}{\partial x}=\dfrac{-1}{2}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-3}{2}}}\left( 2x \right) \\
& \dfrac{\partial f}{\partial x}=\dfrac{-1}{2{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}\times 2x \\
& \dfrac{\partial f}{\partial x}=\dfrac{-x}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}..............\left( vi \right) \\
\end{align}\]
Now, we can calculate $\dfrac{\partial f}{\partial y}$ by differentiating $f\left( x,y \right)$ with respect to $y$ by taking $'x'$ as a constant. So, we get
$\begin{align}
& \dfrac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}\left( \dfrac{1}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{1}{2}}}} \right) \\
& \dfrac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-1}{2}}} \\
\end{align}$
Now, using the relation (v) , we get
$\dfrac{\partial f}{\partial y}=-\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{-\dfrac{1}{2}-1}}\times \dfrac{\partial }{\partial y}\left( {{x}^{2}}+{{y}^{2}} \right)$
Now as $'x'$ is acting as a constant, so we know $\dfrac{d}{dy}\text{constant}=0$ .Hence we get
\[\begin{align}
& \dfrac{\partial f}{\partial y}=\dfrac{-1}{2}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-3}{2}}}\times \left( 2y \right) \\
& =\dfrac{-1}{2{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}\times 2y \\
& \dfrac{\partial f}{\partial y}=\dfrac{-y}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}...........\left( vii \right)
\end{align}\]
Now, left hand side of the equation(i) is given with the help of equation(vi) and (vii) as
$\begin{align}
& L.H.S.=x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y} \\
& L.H.S.=x\left( \dfrac{-x}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}} \right)+y\left( \dfrac{-y}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}} \right) \\
& L.H.S.=\dfrac{-{{x}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}+\dfrac{-{{y}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}} \\
& L.H.S.=\dfrac{-1}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}\left( {{x}^{2}}+{{y}^{2}} \right) \\
\end{align}$
Now, we know the property of surds
$\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
So, we get the value of L.H.S. as
$\begin{align}
& L.H.S.=-1\times {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{1-\dfrac{3}{2}}} \\
& L.H.S.=-1{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{-\dfrac{1}{2}}} \\
& L.H.S.=\dfrac{-1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}..........\left( ix \right) \\
\end{align}$
Now, we can get value of Right-hand side of the equation(i) as
$\begin{align}
& R.H.S.=nf\left( x,y \right) \\
& R.H.S.=-1\times \dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \\
\end{align}$
Where, we know the value of ‘n’ is -1 from the equation(iv). So, we get
$R.H.S.=\dfrac{-1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}..........\left( x \right)$
Hence, we get the equation (ix) and (x) such that the RHS and LHS of the equation(i) are equal to each other. So, we get
$L.H.S.=R.H.S.=\dfrac{-1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$
Therefore, the Euler theorem is verified with the given equation.
Note: One may go wrong with the calculation of $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ . As both are confusing i.e. one needs to know where we need to differentiate the function by taking $'x'$ as a constant and where we need to take $'y'$ as a constant. So, one needs to be clear with all these concepts.
One may go wrong if he/she put the degree of the given equation as ‘2’ by confusing with the powers of $'x'$ and $'y'$ in the given function i.e. $f\left( x,y \right)=\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$ . So, always try to verify any homogeneous relation by equation.
$f\left( \lambda x,\lambda y \right)={{\lambda }^{n}}\left( f\left( x,y \right) \right)$ and get the degree of given equation by calculating the ‘n’ from above mentioned equation.
\[x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}=nf\left( x,y \right)\] , where n is the degree of the homogeneous function. Determine the value of ‘n’ by expression
$f\left( \lambda x,\lambda y \right)={{\lambda }^{n}}f\left( x,y \right)$
$\dfrac{\partial f}{\partial x}$ means differentiate the given function w.r.t $'x'$ by taking as $'y'$ a constant and $\dfrac{\partial f}{\partial y}$ means differentiation of the function w.r.t $'y'$ by taking $'x'$ as a constant. Use the relation $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ wherever necessary.
Complete step-by-step solution -
As, we know the Euler’s theorem for any homogeneous function \[f\left( x,y \right)\] of degree n is given as
\[x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}=nf\left( x,y \right)..........\left( i \right)\]
And we can verify any function to be homogeneous or not by the relation
$f\left( \lambda x,\lambda y \right)={{\lambda }^{n}}f\left( x,y \right)........\left( ii \right)$
Where ‘n’ is the degree of a homogeneous equation.
It means if any function follows equation(ii), then it will be a homogeneous function and ‘n’ is the degree of the equation.
Now, coming to the question, we are given the function $f\left( x,y \right)$ in the problem as
$f\left( x,y \right)=\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}.........\left( iii \right)$
Now, we can calculate value of $f\left( \lambda x,\lambda y \right)$ from the equation (iii) as
$\begin{align}
& f\left( \lambda x,\lambda y \right)=\dfrac{1}{\sqrt{{{\left( \lambda x \right)}^{2}}+{{\left( \lambda y \right)}^{2}}}} \\
& f\left( \lambda x,\lambda y \right)=\dfrac{1}{\sqrt{{{\lambda }^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)}} \\
& f\left( \lambda x,\lambda y \right)=\dfrac{1}{\lambda }\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \\
& f\left( \lambda x,\lambda y \right)={{\lambda }^{-1}}f\left( x,y \right) \\
\end{align}$
Now, we can compare the above relation by equation (ii) and hence, we get degree of the given function as
n = -1 ………..(iv)
Now, we have to verify the relation of equation (i), so let us calculate $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ by following approach
We have $f\left( x,y \right)=\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}={{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-1}{2}}}$
Now, we can calculate $\dfrac{\partial f}{\partial x}$ by differentiating the above function with respect to $'x'$ by taking $'y'$ as a constant.
We know
$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}........\left( v \right)$
So, we get $\dfrac{\partial f}{\partial x}$ as
\[\begin{align}
& \dfrac{\partial f}{\partial x}=\dfrac{\partial }{\partial x}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-1}{2}}} \\
& \dfrac{\partial f}{\partial x}=-\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-1}{2}-1}}\times \dfrac{\partial }{\partial x}\left( {{x}^{2}}+{{y}^{2}} \right) \\
\end{align}\]
Now, $'y'$ is acting as a constant. So, we know
$\dfrac{d}{dx}\text{constant = 0}$ . Hence, we get
\[\begin{align}
& \dfrac{\partial f}{\partial x}=\dfrac{-1}{2}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-3}{2}}}\left( 2x \right) \\
& \dfrac{\partial f}{\partial x}=\dfrac{-1}{2{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}\times 2x \\
& \dfrac{\partial f}{\partial x}=\dfrac{-x}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}..............\left( vi \right) \\
\end{align}\]
Now, we can calculate $\dfrac{\partial f}{\partial y}$ by differentiating $f\left( x,y \right)$ with respect to $y$ by taking $'x'$ as a constant. So, we get
$\begin{align}
& \dfrac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}\left( \dfrac{1}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{1}{2}}}} \right) \\
& \dfrac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-1}{2}}} \\
\end{align}$
Now, using the relation (v) , we get
$\dfrac{\partial f}{\partial y}=-\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{-\dfrac{1}{2}-1}}\times \dfrac{\partial }{\partial y}\left( {{x}^{2}}+{{y}^{2}} \right)$
Now as $'x'$ is acting as a constant, so we know $\dfrac{d}{dy}\text{constant}=0$ .Hence we get
\[\begin{align}
& \dfrac{\partial f}{\partial y}=\dfrac{-1}{2}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{-3}{2}}}\times \left( 2y \right) \\
& =\dfrac{-1}{2{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}\times 2y \\
& \dfrac{\partial f}{\partial y}=\dfrac{-y}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}...........\left( vii \right)
\end{align}\]
Now, left hand side of the equation(i) is given with the help of equation(vi) and (vii) as
$\begin{align}
& L.H.S.=x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y} \\
& L.H.S.=x\left( \dfrac{-x}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}} \right)+y\left( \dfrac{-y}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}} \right) \\
& L.H.S.=\dfrac{-{{x}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}+\dfrac{-{{y}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}} \\
& L.H.S.=\dfrac{-1}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{\dfrac{3}{2}}}}\left( {{x}^{2}}+{{y}^{2}} \right) \\
\end{align}$
Now, we know the property of surds
$\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
So, we get the value of L.H.S. as
$\begin{align}
& L.H.S.=-1\times {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{1-\dfrac{3}{2}}} \\
& L.H.S.=-1{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{-\dfrac{1}{2}}} \\
& L.H.S.=\dfrac{-1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}..........\left( ix \right) \\
\end{align}$
Now, we can get value of Right-hand side of the equation(i) as
$\begin{align}
& R.H.S.=nf\left( x,y \right) \\
& R.H.S.=-1\times \dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \\
\end{align}$
Where, we know the value of ‘n’ is -1 from the equation(iv). So, we get
$R.H.S.=\dfrac{-1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}..........\left( x \right)$
Hence, we get the equation (ix) and (x) such that the RHS and LHS of the equation(i) are equal to each other. So, we get
$L.H.S.=R.H.S.=\dfrac{-1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$
Therefore, the Euler theorem is verified with the given equation.
Note: One may go wrong with the calculation of $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ . As both are confusing i.e. one needs to know where we need to differentiate the function by taking $'x'$ as a constant and where we need to take $'y'$ as a constant. So, one needs to be clear with all these concepts.
One may go wrong if he/she put the degree of the given equation as ‘2’ by confusing with the powers of $'x'$ and $'y'$ in the given function i.e. $f\left( x,y \right)=\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$ . So, always try to verify any homogeneous relation by equation.
$f\left( \lambda x,\lambda y \right)={{\lambda }^{n}}\left( f\left( x,y \right) \right)$ and get the degree of given equation by calculating the ‘n’ from above mentioned equation.
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