Part I :

Euler's theorem : If 'u' is a homogenous function of two variables x & y of degree 'n' then Euler's theorem states that $x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=n u$

Proof :

Let u=f(x,y) be the homogenous function of degree 'n'.

Let X=xt, Y=yt

$\therefore \dfrac{\partial X}{\partial t}=x\ and\ \dfrac{\partial y}{\partial t}=y\rightarrow (1)$

At $t=1,\rightarrow (2)$

X=x and Y=y

$\dfrac{\partial f}{\partial X}=\dfrac{\partial f}{\partial x}\ and\ \dfrac{\partial f}{\partial Y}=\dfrac{\partial f}{\partial y}\rightarrow (3)$

Now, $f(X,Y)=t^{n}f(x,y)\rightarrow (4)$

$\therefore f\rightarrow X,Y\rightarrow x,y,t$

Differentiate (4) partially w.r.t. 't'

$\dfrac{\partial f}{\partial X}\cdot\dfrac{\partial X}{\partial t}+\dfrac{\partial f}{\partial Y}\cdot\dfrac{\partial Y}{\partial t}=nt^{n-1}f(x,y)$

$\therefore \dfrac{\partial f}{\partial x}\cdot x +\dfrac{\partial f}{\partial y}\cdot y=n(1)^{n-1}f(x,y)$

(From 1,2 & 3)

$\therefore x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=n u \rightarrow (5)$

Part II : Verification

Method 1:

$u(x,y)=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\rightarrow (6)$

$\therefore u(X,Y)=\dfrac{\sqrt{XY}}{\sqrt{X}+\sqrt{Y}}$

Now, Put X=xt, Y=yt,

$\therefore u(X,Y)=\dfrac{\sqrt{(xt)(yt)}}{\sqrt{(xt)}+\sqrt{(yt)}}$

$=\dfrac{t\sqrt{xy}}{\sqrt{t}\left ( \sqrt{x}+\sqrt{y} \right )}$

$=t^{1/2}u(x,y)$

$\therefore$ u is homogenous function of degree 1/2.

$\therefore From (5),x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=\dfrac{1}{2}u$

$\therefore x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=\dfrac{\sqrt{xy}}{2\left ( \sqrt{x}+\sqrt{y} \right )}\rightarrow (7)$

Method 2:

From (6), $u=\dfrac{\sqrt{x}\sqrt{y}}{\sqrt{x}+\sqrt{y}}$

Partially differentiate w.r.t.'x'

$\dfrac{\partial u}{\partial x}=\sqrt{y}\times \dfrac{\left ( \sqrt{x}+\sqrt{y} \right )\cdot\dfrac{1}{2\sqrt{x}}-\sqrt{x}\cdot\dfrac{1}{2\sqrt{x}}}{\left ( \sqrt{x}+\sqrt{y} \right )^{2}}$

$=\dfrac{\sqrt{y}}{\left ( \sqrt{x}+\sqrt{y} \right )^{2}}\times\dfrac{1}{2\sqrt{x}}\left [ \sqrt{x}+\sqrt{y}-\sqrt{x} \right ]$

$=\dfrac{\sqrt{y}}{\left ( \sqrt{x}+\sqrt{y} \right )^{2}}\times\dfrac{1}{2\sqrt{x}}\times\sqrt{y}$

$\therefore x\dfrac{\partial u}{\partial x}=\dfrac{xy}{2\sqrt{x}\left ( \sqrt{x}+\sqrt{y} \right )^{2}}\rightarrow(8)$

Similarly, interchanging 'x' and 'y',

$y\dfrac{\partial u}{\partial y}=\dfrac{yx}{2\sqrt{y}\left ( \sqrt{y}+\sqrt{x} \right )^{2}}\rightarrow (9)$

On adding (8) and (9),

$x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=\dfrac{xy}{2\sqrt{x}\left ( \sqrt{x}+\sqrt{y} \right )^{2}}+\dfrac{xy}{2\sqrt{y}\left ( \sqrt{y}+\sqrt{x} \right )^{2}}$

$=\dfrac{xy}{2\left ( \sqrt{x}+\sqrt{y} \right )^{2}}\left [ \dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}} \right ]$

$=\dfrac{(\sqrt{x})^{2}(\sqrt{y})^{2}}{2(\sqrt{x}+\sqrt{y})^{2}}\times\dfrac{(\sqrt{y}+\sqrt{x})}{\sqrt{x}\sqrt{y}}$

$=\dfrac{\sqrt{xy}}{2(\sqrt{x}+\sqrt{y})}\rightarrow (10)$

From (7) and (10), Euler's theorem is verified.