Solution of the differential equation $\frac{\sqrt{x} d x+\sqrt{y} d y}{\sqrt{x} d x-\sqrt{y} d y}=\sqrt{\frac{y^3}{x^3}}$ is given by

none of these

We have,

$\frac{\sqrt{x} d x+\sqrt{y} d y}{\sqrt{x} d x-\sqrt{y} d y}=\sqrt{\frac{y^3}{x^3}}$

$\Rightarrow \frac{d\left(x^{3 / 2}\right)+d\left(y^{3 / 2}\right)}{d\left(x^{3 / 2}\right)-d\left(y^{3 / 2}\right)}=\frac{y^{3 / 2}}{x^{3 / 2}}$

$\Rightarrow \frac{d u+d v}{d u-d v}=\frac{v}{u}$, where $u=x^{3 / 2}$ and $v=y^{3 / 2}$

$\Rightarrow u d u+u d v=v d u-v d v$

$\Rightarrow u d u+v d v=v d u-u d v$

$\Rightarrow \frac{u d u+v d v}{u^2+v^2}=\frac{v d u-u d v}{u^2+v^2}$

$\Rightarrow \frac{d\left(u^2+v^2\right)}{u^2+v^2}=-2 d \tan ^{-1}\left(\frac{v}{u}\right)$

On integrating, we get

$\log \left(u^2+v^2\right)=-2 \tan ^{-1}\left(\frac{v}{u}\right)+C$

$\Rightarrow \frac{1}{2} \log \left(x^3+y^3\right)+\tan ^{-1}\left(\frac{y}{x}\right)^{3 / 2}=\frac{C}{2}$