Question

If \[u = \log \dfrac{{{x^2} + {y^2}}}{{xy}}\], then find the value of \[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}\].
A. 0
B. \[u\]
C. \[2u\]
D. \[u - 1\]

Answer 1

Hint: First we will change the given equation into exponential form. To find the degree we will substitute \[x = xt\] and \[y = yt\] in the equation. Then apply Euler’s theorem to calculate \[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}\].

Formula used:
Euler’s theorem: If \[f\] is a homogeneous function of degree \[n\] of the variables \[x,y\]; then \[x\dfrac{{\partial f}}{{\partial x}} + y\dfrac{{\partial f}}{{\partial y}} = nf\].
If \[x = {\log _e}y\] then \[y = {e^x}\]

Complete step by step solution:
Given equation is:
\[u = \log \dfrac{{{x^2} + {y^2}}}{{xy}}\]
Rewrite the above equation as an exponential formula:
\[{e^u} = \dfrac{{{x^2} + {y^2}}}{{xy}}\]
To find the degree of the equation, we will put \[x = xt\] and \[y = yt\] in the above equation:
\[{e^u} = \dfrac{{{{\left( {xt} \right)}^2} + {{\left( {yt} \right)}^2}}}{{xt \cdot yt}}\]
\[ \Rightarrow {e^u} = \dfrac{{{x^2}{t^2} + {y^2}{t^2}}}{{xy{t^2}}}\]
Take common \[{t^2}\] from the numerator of the equation
\[ \Rightarrow {e^u} = \dfrac{{{t^2}\left( {{x^2} + {y^2}} \right)}}{{xy{t^2}}}\]
Cancel out \[{t^2}\] from denominator and numerator
\[ \Rightarrow {e^u} = \dfrac{{\left( {{x^2} + {y^2}} \right)}}{{xy}}\]
Rewrite the equation
\[ \Rightarrow {e^u} = {t^0}\dfrac{{\left( {{x^2} + {y^2}} \right)}}{{xy}}\]
Since the power of \[t\] is zero, so the degree of the equation is zero.
Apply the Euler’s theorem:
\[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = 0 \cdot u\]
\[ \Rightarrow x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = 0\]
Hence the option A is correct.

Note: Students often do a common mistake to solve the given question. They calculate the partial derivatives of \[u = \log \dfrac{{{x^2} + {y^2}}}{{xy}}\] with respect to \[x\] and \[y\] to compute the value of \[x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}\] which is a lengthy process.