Question

The solution of, $\frac{xdy}{x^2+y^2}=(\frac{y}{x^2+y^2}-1)dx$, is given by

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

Answer 1

Answer: (B)

${\tan }^{-1}\left(\frac{y}{x}\right)+x=C$

$\frac{xdy}{x^2+y^2}=(\frac{y}{x^2+y^2}-1)dx$

$\frac{xdy}{x^2+y^2}=\frac{ydx}{x^2+y^2}-dx$

$\frac{xdy-ydx}{x^2+y^2}=-dx$

$\frac{d}{dx}{\tan }^{-1}\frac{y}{x}=-dx$

Integrating both sides, we get

${\tan }^{-1}\frac{y}{x}=-x+c$

${\tan }^{-1}\frac{y}{x}+x=c$.