Question

Solution of the differential equation $\left(y+x\sqrt{xy}\left(x+y\right)\right)dx+\left(y\sqrt{xy}\left(x+y\right)-x\right)dy=0$ is

• A. $\frac{x^{2}+y^{2}}{2}+tan^{-1} \sqrt{\frac{y}{x}}=c$
• B. $\frac{x^{2}+y^{2}}{2}+2\,tan^{-1} \sqrt{\frac{x}{y}}=c$
• C. $\frac{x^{2}+y^{2}}{2}+2\,cot^{-1} \sqrt{\frac{x}{y}}=c$
• D. None of these

Answer 1

Answer: (B)

The given equation is written as
$y\,dx -x\,dy+x\sqrt{xy} \left(x+y\right)dx+y\sqrt{xy}\left(x+y\right)dy=0$
or $ ydx-xdy+\left(x+y\right)\sqrt{xy}\left(xdx+ydx\right)=0$
or $\frac{ydx-xdy}{y^{2}}+\left(\frac{x}{y}+1\right)\sqrt{\frac{x}{y}}\left(d\left(\frac{x^{2}+y^{2}}{2}\right)\right)=0$
or $d\left(\frac{x^{2}+y^{2}}{2}\right)+\frac{d \left(\frac{x}{y}\right)}{\left(\frac{x}{y}+1\right)\sqrt{\frac{x}{y}}}=0$
or $\frac{x^{2}+y^{2}}{2}+2\,tan^{-1} \sqrt{\frac{x}{y}}=c$