Question

If $u=ln\left(\frac{x^3+y^3}{x^2+y^2}\right)$, then the value of $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$ is

• A. $e^u$
• B. 1
• C. u
• D. eu

Answer 1

The correct option is B 1

$u=ln\left(\frac{x^3+y^3}{x^2+y^2}\right)$

let, $V=e^u=\left(\frac{x^3+y^3}{x^2+y^2}\right)$

$\because V(\lambda x,\lambda y)=\lambda \times V(x,y),$ so V is homogenous function of degree 1.

$\therefore$ By Euler's theorem

$x\frac{\partial V}{\partial x}+y\frac{\partial V}{\partial y}=n\times V,$ n = degree of homogeneous function = 1

So, by Euler's theorem

$x\times \frac{\partial \left(e^u\right)}{\partial x}+y\times \frac{\partial \left(e^u\right)}{\partial y}=1\times e^u$

$\Rightarrow x\times e^u\times \frac{\partial u}{\partial x}+y\times e^u\times \frac{\partial u}{\partial y}=1\times e^u$

$\Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=1$