Explanation

(a)
Given that, $\frac{dy}{dx} = \frac{2x - y}{x + 2y}$ ….(i)
Let $y = vx\ \Rightarrow \ \frac{dy}{dx} = v + x\frac{dv}{dx}$
$\therefore v + x\frac{dv}{dx} = \frac{2 - v}{1 + 2v}$
$\Rightarrow \ x\frac{dv}{dx} = \frac{2 - v - v(1 + 2v)}{1 + 2v}$
$\Rightarrow \ \int_{}^{}{\frac{1 + 2v}{2(1 - v - v^{2})}dv = \int_{}^{}{\frac{1}{x}dx}}$
$\Rightarrow \log{k - \frac{1}{2}\log{\left( 1 - v - v^{2} \right) = \log x}}$
$\Rightarrow 2\log{k - \log{\left( 1 - v - v^{2} \right) = 2\log x}}$
$\Rightarrow \log{c = \log{\lbrack x^{2}\left( 1 - v - v^{2} \right)\rbrack}}$
$\Rightarrow c = x^{2}\left( \ 1 - \frac{y}{x} - \frac{y^{2}}{x^{2}}\ \right)$
$\Rightarrow x^{2} - xy - y^{2} = c$