Explanation

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