Double Math

# 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.