Explanation

(b)
Given, $\frac{dy}{dx} = \frac{x + y}{x - y}$
Put $y = vx\ \Rightarrow \frac{dy}{dx} = v + x\ \frac{dv}{dx}$
$\therefore$ $v + x\ \frac{dv}{dx} = \frac{1 + v}{1 - v}$
$\Rightarrow$ $x\ \frac{dv}{dx} = \frac{1 + v^{2}}{1 - v}$
$\Rightarrow$ $\frac{1}{x}\ dx = \left( \frac{1}{1 + v^{2}} - \frac{v}{1 + v^{2}} \right)dv$
$\Rightarrow$ $\log_{e}x = \tan^{- 1}v - \frac{1}{2}\log_{e}{(1 + v^{2})} - \log_{e}c$ [integrating]
$\Rightarrow$ $\log_{e}x{= tan}^{- 1}\left( \frac{y}{x} \right) - \frac{1}{2}\log_{e}{\left\lbrack 1 + \left( \frac{y}{x} \right)^{2} \right\rbrack - \log_{e}c}$
$\Rightarrow$ $c\left( x^{2} + y^{2} \right)^{\frac{1}{2}} = e^{\tan^{- 1}\left( \frac{y}{x} \right)}$