Euler's Theorem, Proof and Examples

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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Euler’s Theorem, Proof and Examples

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