Question

The solution of the differential equation $xdx+ydy+\frac{ydx-xdy}{x^2+y^2}=0$ is
(where $C$ is integration constant)

• A. $x^2-y^2=2{\tan }^{-1}\frac{y}{x}+C$
• B. $x^2+y^2=2{\tan }^{-1}\frac{y}{x}+C$
• C. $x^2+y^2={\cot }^{-1}\frac{y}{x}+C$
• D. $x^2-y^2=2{\cot }^{-1}\frac{y}{x}+C$

Answer 1

The correct option is B $x^2+y^2=2{\tan }^{-1}\frac{y}{x}+C$

Let $x=r\cos \theta ,y=r\sin \theta$
$\Rightarrow xdx+ydy=rdr$
Also $xdy-ydx=r^{2}d\theta$
$\therefore$ Given equation can be rewritten as
$rdr-\frac{r^{2}d\theta }{r^2}=0$
$\Rightarrow rdr=d\theta$
$\Rightarrow \frac{r^2}{2}=\theta +c$
$\Rightarrow x^2+y^2=2{\tan }^{-1}\frac{y}{x}+C;$
where $2c=C^2$