Question

If $u=\frac{x-y}{\sqrt{x^2+y^2}}$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=$

• A. $u$
• B. $-u$
• C. $3u$
• D. $0$

Answer 1

The correct option is D $0$

$u$ is homogeneous equation of degree $n=0$.
$\sum x\frac{du}{dx}=nu$
$\sum x\frac{du}{dx}=0$
Alternate solution,
$\frac{du}{dx}=\frac{\sqrt{x^2+y^2}-(n-y)}{(x^2+y^2)}\frac{x}{\sqrt{x^2+y^2}}$
$x\frac{du}{dx}=\frac{x\sqrt{x^2+y^2}-\frac{(x-y)x^2}{\sqrt{x^2+y^2}}}{x^2+y^2}$
$y\frac{du}{dy}=\frac{-y\sqrt{x^2+y^2}-\frac{(x-y)y^2}{\sqrt{x^2+y^2}}}{x^2+y^2}$
$\sum x\frac{du}{dx}=\frac{(x-y)\sqrt{x^2+y^2}-(x-y)\sqrt{x^2+y^2}}{x^2+y^2}$
$=0$