Question

If $u={\sec }^{-1}\left(\frac{x^3-y^3}{x+y}\right)$ then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=$?

Answer 1

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

To find $x\frac{du}{dx}+y\frac{du}{dy}=?$

Let $z=\sec u=\frac{x^3-y^3}{x+y}$

Now $\frac{x^3-y^3}{x+y}=\frac{{\left(\lambda x\right)}^3-{\left(\lambda y\right)}^3}{\lambda x+\lambda y}=\frac{{\lambda }^3(x^3-y^3)}{\lambda (x+y)}$

$\frac{{\lambda }^2(x^3-y^3)}{(x+y)}$ which is a homogeneous equation of degree $2$.

$\therefore$ By euler's theorem we have

$x\frac{dz}{dx}+y\frac{dz}{dy}=2z\frac{dz}{dx}=\frac{d\left(\sec u\right)}{dx}=\sec u\tan u\frac{du}{dx}\frac{dz}{dy}=\sec u\tan u\frac{du}{dy}\therefore x\frac{dz}{dx}+y\frac{du}{dy}=2z\Rightarrow x\sec u\tan u\frac{du}{dx}+y\sec u\tan u\frac{du}{dy}=2\sec u\Rightarrow x\frac{du}{dx}+y\frac{du}{dz}=2\cot u$