Question

If $u=(x^3+y^3+z^3)(x+y+z)$

then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=?$

• A. $u$
• B. $-3u$
• C. $12u$
• D. $4u$

Answer 1

The correct option is D $4u$

$u$ is homogenous equation of degree $n=4$

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

$\Rightarrow \sum x\frac{\partial u}{\partial x}=4u$

Alternate solution,

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

$x\frac{\partial u}{\partial x}$ $=$ $3x^3(x+y+z)+x(x^3+y^3+z^3)$

$y\frac{\partial u}{\partial y}$ $=$ $3y^3(x+y+z)+y(x^3+y^3+z^3)$

$y\frac{\partial u}{\partial z}$ $=$ $3z^3(x+y+z)+z(x^3+y^3+z^3)$

$\sum x\frac{\partial u}{\partial x}=3(x^3+y^3+z^3)(x+y+z)+(x+y+z)(x^3+y^3+z^3)=4u$