Question

Find the solution of $xdx+ydy+\frac{xdy-ydx}{x^2+y^2}=0$.

• A. $x^2+y^2-2{\tan }^{-1}\left(y/x\right)=c$
• B. $x^2+y^2-{\tan }^{-1}\left(y/x\right)=c$
• C. $x^2-y^2+2{\tan }^{-1}\left(y/x\right)=c$
• D. $x^2+y^2+2{\tan }^{-1}\left(y/x\right)=c$

Answer 1

The correct option is D $x^2+y^2+2{\tan }^{-1}\left(y/x\right)=c$

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