Question

The solution of the differential equation
$xdx+ydy+\frac{xdy-ydx}{x^{2}+y^{2}}=0$, is

• A. $y=x\,tan\left(\frac{x^{2}+y^{2}+C}{2}\right)$
• B. $x=y\,tan\left(\frac{x^{2}+y^{2}+C}{2}\right)$
• C. $y=x\,tan\left(\frac{C-x^{2}-y^{2}}{2}\right)$
• D. none of these

Answer 1

Answer: (C)

Given, $xdx+ydy+\frac{xdy-ydx}{x^{2}+y^{2}}=0$
$\Rightarrow \frac{1}{2}d\left(x^{2}+y^{2}\right)+d\left(tan^{-1} \frac{y}{x}\right)=0$
$\Rightarrow \frac{1}{2}\left(x^{2}+y^{2}\right)+tan^{-1} \frac{y}{x}=\frac{C}{2}\quad$ (Integrating)
$\Rightarrow \frac{C-x^{2}-y^{2}}{2}=tan^{-1} \frac{y}{x}$
$\Rightarrow y=x\,tan\left(\frac{C-x^{2}-y^{2}}{2}\right)$