Question

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

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

Answer 1

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

$xdy+ydx=\frac{xdy-ydx}{x^2+y^2}$
If we apply exact differential method, we get
$d\left(xy\right)=d\left({\tan }^{-1}\frac{y}{x}\right)$
Integrating both sides, we get
$xy={\tan }^{-1}\left(\frac{y}{x}\right)+c$