Explanation

(b)
Given, $\frac{dy}{dx} = \frac{y^{2}}{xy - x^{2}}$
Put $y = vx$
$\Rightarrow \$ $\frac{dy}{dx} = v + x\ \frac{dv}{dx}$
$\therefore$ $v + x\ \frac{dv}{dx} = \frac{v^{2}x^{2}}{vx^{2} - x^{2}}$
$\Rightarrow$ $x\ \frac{dv}{dx} = \frac{v}{v - 1}$
$\Rightarrow$ $\left( 1 - \frac{1}{v} \right)dv = \frac{dx}{x}$
$\Rightarrow$ $v - \log v = \log x + \log k$ [integrating]
$\Rightarrow$ $\frac{y}{x} = \log x\ k\ \frac{y}{x}$
$\Rightarrow$ $\frac{y}{x} = \log{ky}$
$\Rightarrow$ $ky = e^{\frac{v}{x}}$