Question

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

• A. $\frac{3}{2}log\left(\frac{y}{x}\right)+log\left|\frac{x^{3/2}+y^{3/2}}{x^{3/2}}\right|+ta{n}^{-1}{\left(\frac{y}{x}\right)}^{3/2}+c=0$
• B. $\frac{2}{3}log\left(\frac{y}{x}\right)+log\left|\frac{x^{3/2}+y^{3/2}}{x^{3/2}}\right|+ta{n}^{-1}\frac{y}{x}+c=0$
• C. $\frac{2}{3}log\left(\frac{y}{x}\right)+log\left(\frac{x+y}{x}\right)+ta{n}^{-1}\left(\frac{y^{3/2}}{x^{3/2}}\right)+c=0$
• D. None of the above

Answer 1

The correct option is D

None of the above

We have, $\frac{\sqrt{x}dx+\sqrt{y}dy}{\sqrt{x}dx-\sqrt{y}dy}=\sqrt{\frac{y^3}{x^3}}$
$\Rightarrow \frac{d\left(x^{3/2}\right)+d\left(y^{3/2}\right)}{d\left(x^{3/2}\right)-d\left(y^{3/2}\right)}=\frac{y^{3/2}}{x^{3/2}}$
$\Rightarrow \frac{du+dv}{du-dv}=\frac{v}{u}$, where $u=x^{3/2}$ and $v=y^{3/2}$
$\Rightarrow udu+udv=vdu-vdv$
$\Rightarrow udv+vdv=vdu-udv$
$\Rightarrow \frac{udu+vdv}{u^2+v^2}=\frac{vdu-udv}{u^2+v^2}$
$\Rightarrow \frac{d(u^2+v^2)}{u^2+v^2}=-2d\left(ta{n}^{-1}\left(\frac{u}{v}\right)\right)$
On integrating, we get
$log(u^2+v^2)=-2ta{n}^{-1}\left(\frac{u}{v}\right)+c$
$\Rightarrow \frac{1}{2}log(x^3+y^3)+ta{n}^{-1}{\left(\frac{x}{y}\right)}^{3/2}=\frac{c}{2}=C$
Where, $c$ and $C$ are constants.