Question

The differential equation $xdx+ydy=xdy-ydx$ has solution

• A. $\log (x^2+y^2)=2{\tan }^{-1}\frac{y}{x}+C$
• B. $\log (x^2+y^2)={\tan }^{-1}\frac{y}{x}+C$
• C. $\log (x^2+y^2)+{\tan }^{-1}\frac{y}{x}+C$
• D. none of these

Answer 1

Answer: (A)

$xdx+ydy=\frac{1}{2}d\left(x^2+y^2\right)$
$xdy-ydx=x^{2}d\left(\frac{y}{x}\right)$
$\therefore$ given differential equation can be written as
$\frac{d\left(x^2+y^2\right)}{x^2+y^2}=\frac{2x^{2}d\left(\frac{y}{x}\right)}{x^2+y^2}$
$=\frac{2d\left(\frac{y}{x}\right)}{1+{\left(\frac{y}{x}\right)}^2}$
$\Rightarrow log\left(x^2+y^2\right)$
$=2ta{n}^{-1}\frac{y}{x}+C$