Euler’s Theorem Statement: If u=f(x,y) is a homogeneous function of x,y of degree n, then
x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial u}=nu.
Proof: we have
u=f(x,y)=x^n\;g\left(\frac yx\right)\;\;,Therefore
\frac{\partial u}{\partial x}=n\;x^{n-1\;}g\left(\frac yx\right)\;+\;x^{n\;}g'\left(\frac yx\right)\left(\frac{-y}{x^2}\right).
\frac{\partial u}{\partial x}=n\;x^{n-1\;}g\left(\frac yx\right)\;-y\;x^{n-2\;}g'\left(\frac yx\right).....................(1).
and
\frac{\partial u}{\partial y}=\;x^{n\;}g'\left(\frac yx\right)\;\frac1x.
\frac{\partial u}{\partial y}=\;x^{n-1\;}g'\left(\frac yx\right).....................(2).
Thus from (1) and (2), we have
\frac{\displaystyle\partial u}{\displaystyle\partial x}+\frac{\displaystyle\partial u}{\displaystyle\partial y}=n\;x^{n-1\;}g\left(\frac{\displaystyle y}{\displaystyle x}\right)\;-y\;x^{n-2\;}g'\left(\frac{\displaystyle y}{\displaystyle x}\right)+\;x^{n-1\;}g'\left(\frac{\displaystyle y}{\displaystyle x}\right).
\frac{\displaystyle\partial u}{\displaystyle\partial x}+\frac{\displaystyle\partial u}{\displaystyle\partial y}=n\;x^n\;g\left(\frac yx\right).
\frac{\displaystyle\partial u}{\displaystyle\partial x}+\frac{\displaystyle\partial u}{\displaystyle\partial y}=nu.
\frac{\displaystyle\partial u}{\displaystyle\partial x}+\frac{\displaystyle\partial u}{\displaystyle\partial y}=nf(x,y)
Hence proved.
Example:
If u=arc\tan\left(\frac{x^3+y^3}{x-y}\right), show that
x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\sin 2u.
Solution:
we have
z =tan u.
\Rightarrow\;\;z(x,y)=\frac{x^3+y^3}{x-y}.
\Rightarrow\;\;z(tx,ty)=\frac{\left(tx\right)^3+\left(ty\right)^3}{tx-ty}.
\Rightarrow\;\;z(tx,ty)=\frac{t^3\left(x^3+y^3\right)}{t\left(x-y\right)}.
\Rightarrow\;\;z(tx,ty)=\frac{t^2\left(x^3+y^3\right)}{\left(x-y\right)}.
Thus z is homogeneous function of x,y of degree 2.
Therefore, by Euler’s theorem, we have
x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2z \;\;\;.................(1).
But
\frac{\partial z}{\partial x}=sec^2u\;\frac{\partial u}{\partial x}\;\;\;\;\;and\;\;\;\;\frac{\partial z}{\partial y}=sec^2u\;\frac{\displaystyle\partial u}{\displaystyle\partial y}.
Substituting into (1), we get
sec^2u\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2z.
\Rightarrow \;\;\sec^2u\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2\;\tan\;u.
or\;\;\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2\;\tan\;u\;.\frac{\displaystyle1}{sec^2u}.
\Rightarrow\;\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2\;\frac{\displaystyle\sin\;u}{\cos{\displaystyle\;}{\displaystyle u}}\;.\cos^2u.
\Rightarrow\;\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2\;\sin\;u\;.\;\cos u.
\Rightarrow\;\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=\sin\;2u.
Hence Euler’s theorem verified.
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