Question

Solution of the equation $xdx+ydy+\frac{xdy-ydx}{x^2+y^2}=0$

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

Answer 1

Answer: (C)

$y=x\tan \left(\frac{c-x^2-y^2}{2}\right)$

We have, $xdx+ydy+\frac{xdy-ydx}{x^2+y^2}=0$
$\frac{1}{2}d(x^2+y^2)+d\left({\tan }^{-1}\left(\frac{y}{x}\right)\right)=0$
Integrating both sides, we get

$\frac{1}{2}(x^2+y^2)+{\tan }^{-1}\left(\frac{y}{x}\right)=\frac{c}{2}$
$\Rightarrow x^2+y^2+2{\tan }^{-1}\frac{y}{x}=c$
$\therefore y=x\tan \left(\frac{c+x^2+y^2}{2}\right)$ is the required solution.