Question

5. Verify Euler's Theorem when u = x2 – 2xy + 5y2​

Answer 1

$\large\underline{\sf{Solution-}}$

Euler's Theorem states that

If u(x, y) is a homogeneous function of degree n, then

$\sf \: x \: \dfrac{\partial \: u}{\partial \: x} + y \: \dfrac{\partial \: u}{\partial \: y} = n \: u$

Now,

Given function is

$\rm :\longmapsto\:u = {x}^{2} - 2xy + {5y}^{2} - - - (1)$

$\rm :\longmapsto\:u = {x}^{2}{\bigg(1 - \dfrac{2y}{x} + 5 \: \dfrac{ {y}^{2} }{ {x}^{2} } \bigg) }$

$\rm :\implies\:u \: is \: a \: homogeneous \: function \: of \: degree \: 2$

So,

we have to verify that

$\: \: \: \: \: \: \: \: \bull \: \sf \: \: \: \boxed{ \sf \: x \: \dfrac{\partial \: u}{\partial \: x} + y \dfrac{\partial \: u}{\partial \: y} = 2u}$

Now,

Differentiate partially equation (1), w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{\partial \: u}{\partial \: x} = 2x - 2y - - - (2)$

Differentiate partially equation (1) w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{\partial \: u}{\partial \: y} = - 2x + 10y - - - (3)$

Now,

Consider,

$\rm :\longmapsto\: x \: \dfrac{\partial \: u}{\partial \: x} + y \dfrac{\partial \: u}{\partial \: y}$

$\sf \: \: \: = \: \: \: x(2x - 2y) + y( - 2x + 10y)$

$\sf \: \: \: = \: \: \: {2x}^{2} - 2xy - 2xy + {10y}^{2}$

$\sf \: \: \: = \: \: \: {2x}^{2} - 4xy + {10y}^{2}$

$\sf \: \: \: = \: \: \: 2({x}^{2} - 2xy + {5y}^{2})$

$\sf \: \: \: = \: \: \: 2u$

$\bf\implies \: x \: \dfrac{\partial \: u}{\partial \: x} + y \dfrac{\partial \: u}{\partial \: y} = 2u$

${\boxed{\boxed{\bf{Hence, verified}}}}$

Additional Information :-

If u(x, y) is a homogeneous function of degree n, then

$\boxed{ \sf \: x\dfrac{ {\partial}^{2}u }{ {\partial \: x}^{2}} + y\dfrac{ {\partial}^{2}u}{\partial \: x \: \partial \: y} = (n - 1)\dfrac{\partial \: u}{\partial \: x} }$

$\boxed{ \sf \: y\dfrac{ {\partial}^{2}u }{ {\partial \: y}^{2}} + x\dfrac{ {\partial}^{2}u}{\partial \: x \: \partial \: y} = (n - 1)\dfrac{\partial \: u}{\partial \: y} }$

$\boxed{ \sf \: {x}^{2}\dfrac{ {\partial}^{2}u }{ {\partial \: x}^{2}} + 2xy\dfrac{ {\partial}^{2}u}{\partial \: x \: \partial \: y} + {y}^{2}\dfrac{ {\partial}^{2}u }{ {\partial \: y}^{2}} = n(n - 1)u}$