Explanation

(a)
Given, $\frac{dy}{dx} = \frac{y(x + y)}{x(x - y)}$
Put $y = vx$
$\Rightarrow$ $\frac{dy}{dx} = v + x\frac{dv}{dx}$
$\therefore$ $v + x\ \frac{dv}{dx} = \frac{vx(x + vx)}{x(x - vx)}$
$\ \ \ \ \ \ \ \ \ \ x\ \frac{dv}{dx} = \frac{2v^{2}}{1 - v}$
$\Rightarrow$ $\frac{1}{2}\left\lbrack \frac{1}{v^{2}} - \frac{1}{v} \right\rbrack dv = \frac{dx}{x}$
$\Rightarrow$ $\frac{1}{2}\left\lbrack - \frac{1}{v} - \log v \right\rbrack = \log x + c_{1}$
$\Rightarrow$ $\frac{x}{v} + \log\left( \frac{y}{x} \right) + 2\log x = - 2c$
$\Rightarrow$ $\frac{x}{y} + \log{(xy)} = c$ [let $c = - 2c_{1}$]