Question

Euler's theorem on homogeneous function

x³+y³/x+y, prove that x(du/dx) +y(du/dy)=2u​

Answer 1

$\textbf{Given:}$

$\mathsf{u(x,y)=\dfrac{x^3+y^3}{x+y}}$

$\textbf{To prove:}$

$\mathsf{x\,\dfrac{{\partial}u}{\partial}x}+y\,\dfrac{{\partial}u}{\partial}y}=2u}$

$\textbf{Solution:}$

$\textbf{Euler's theorem:}$

$\boxed{\begin{minipage}{7cm} \\\textsf{If u(x,y) is a homogeneous function of degree n,}\\\\\mathsf{then,\;x\,\dfrac{{\partial}u}{{\partial}x}+y\,\dfrac{{\partial}u}{{\partial}y}=n\,u}\\ \end{minipage}}$

$\mathsf{Consider,}$

$\mathsf{u(tx,ty)=\dfrac{(tx)^3+(ty)^3}{tx+ty}}$

$\mathsf{u(tx,ty)=\dfrac{t^3x^3+t^3y^3}{tx+ty}}$

$\mathsf{u(tx,ty)=\dfrac{t^3(x^3+y^3)}{t(x+y)}}$

$\mathsf{u(tx,ty)=t^2\,\dfrac{(x^3+y^3)}{x+y}}$

$\mathsf{u(tx,ty)=t^2\,u(x,y)}$

$\therefore\textsf{u is a homogeneous function of degree 2}$

$\mathsf{By\;Euler's\;theorem}$

$\mathsf{x\,\dfrac{{\partial}u}{\partial}x}+y\,\dfrac{{\partial}u}{\partial}y}=2u}$

$\mathsf{Hence\;proved}$

$\textbf{Find more:}$

If u=sin^-1(x2+y2/x+y)prove that xdu/dx)+ydu/dy)=tan u

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