Explanation

(b)
Given, $\frac{dy}{dx} = \frac{x^{2} + y^{2}}{2xy}$
Put $y = vx$
$\Rightarrow$ $\frac{dy}{dx} = v + x\ \frac{dv}{dx}$
$\therefore$ $v + x\ \frac{dv}{dx} = \frac{x^{2} + v^{2}x^{2}}{2xvx}$
$\Rightarrow$ $x\frac{dv}{dx} = \frac{1 - v^{2}}{2v}$
$\Rightarrow$ $\frac{2v}{1 - v^{2}}dv = \frac{dx}{x}$
$\Rightarrow$ $- \log{(1 - v^{2})} = \log x + \log c$
$\Rightarrow$ $\log\left( 1 - v^{2} \right)^{- 1} = \log{xc}$
$\Rightarrow$ $\left( \frac{x^{2} - y^{2}}{x^{2}} \right)^{- 1} = xc$
$\Rightarrow$ $\frac{x^{2}}{x^{2} - y^{2}} = xc$
$\Rightarrow$ $x = c(x^{2} - y^{2})$