Question

If $u=\frac{\sqrt[3]{3y}-\sqrt[3]{3x}}{x+y}$ then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=?$

• A. $\frac{2}{3}u$
• B. $-\frac{2}{3}u$
• C. $\frac{3}{2}u$
• D. $-\frac{3}{2}u$

Answer 1

The correct option is B $-\frac{2}{3}u$

$u$ is a homogeneous eqn of degree $n=\frac{-2}{3}$

$\sum x\frac{\partial u}{\partial x}=nu$

$\Rightarrow \sum x\frac{\partial u}{\partial x}=\frac{-2}{3}u$

Alternate method

$\frac{\partial u}{\partial x}=\frac{\frac{-1}{3}{x}^{-2/3}(x+y)-\left(\right(y{)}^{1/3}-x^{1/3})}{(x+y{)}^2}$

$x\frac{\partial u}{\partial x}=\frac{\frac{-1}{3}{x}^{1/3}(x+y)-x(y^{1/3}-x^{1/3})}{(x+y{)}^2}$

$y\frac{\partial u}{\partial y}=\frac{\frac{1}{3}{y}^{1/3}(x+y)-y(y^{1/3}-x^{1/3})}{(x+y{)}^2}$

$\therefore \sum x\frac{\partial u}{\partial x}=\frac{\frac{1}{3}(y^{1/3}-x^{1/3})(x+y)-(x+y)(y^{1/3}-x^{1/3})}{(x+y{)}^2}$

$=(\frac{1}{3}-1)u=\frac{-2}{3}u$