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

Answer 1

The correct option is D

None of the above

we have $\frac{\sqrt{x}dx+\sqrt{y}dy}{\sqrt{x}dx-\sqrt{y}dy}=\sqrt{\frac{y^3}{x^3}}$
$\Rightarrow \frac{d\left(x^{\frac{3}{2}}\right)+d\left(y^{\frac{3}{2}}\right)}{d\left(x^{\frac{3}{2}}\right)-d\left(y^{\frac{3}{2}}\right)}=\frac{y^{\frac{3}{2}}}{x^{\frac{3}{2}}}\Rightarrow \frac{du+dv}{du-dv}=\frac{v}{u},whereu=x^{\frac{3}{2}}andv=y^{\frac{3}{2}}\Rightarrow udu+udv=vdu-vdu\Rightarrow udu+vdv=vdu-udv\Rightarrow \frac{udu+vdv}{u^2+v^2}=\frac{vdu-udv}{u^2+v^2}\Rightarrow \frac{d(u^2+v^2)}{u^2+v^2}=-2dta{n}^{-1}\left(\frac{v}{u}\right)$
On integrating, we get
$log(u^2+v^2)=-2ta{n}^{-1}\left(\frac{v}{u}\right)+c\Rightarrow \frac{1}{2}log(x^3+y^3)+ta{n}^{-1}{\left(\frac{y}{x}\right)}^{\frac{3}{2}}=c$