Euler’s Theorem, Proof and Examples

Euler Theorem

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.

To read about Euler click here.

Spread the love
Azhar Ali

Azhar Ali

I graduated in Mathematics from the University of Sargodha, having master degree in Mathematics.

Leave a Reply

Your email address will not be published.

Mathematics is generally known as Math in US and Maths in the UK.

Contact Us

Copyright by Double Math. All Right Reserved 2019 to 2022