Explanation

(a)
Given, $\frac{dy}{dx} = \frac{x^{2} + y^{2}}{2xy}$
Put $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$
$\therefore v + x\frac{dv}{dx} = \frac{x^{2} + v^{2}x^{2}}{2vx^{2}}$
$\Rightarrow x\frac{dv}{dx} = \left( \frac{1 + v^{2}}{2v} - v \right)$
$\Rightarrow \ \frac{2v}{1 - v^{2}}dv = \frac{dx}{x}$
On integrating both sides, we get
$\int_{}^{}{\frac{2v}{1 - v^{2}}dv =}\int_{}^{}{\frac{1}{x}dx}$
$\Rightarrow \ - \log{\left( 1 - v^{2} \right) = \log{x + \log c}}$
$\Rightarrow \ - \log{\left( \ 1 - \frac{y^{2}}{x^{2}}\ \right) = \log{x + \log c}}$ ….(i)
This curve passes through (2, 1).
$\therefore\ - \log{\left( \ 1 - \frac{1}{4}\ \right) = \log{2 + \log c}}$
$\Rightarrow \ - \log{\left( \frac{3}{4}\ \right) = \log{2c}}$
$\Rightarrow \log{\left( \frac{4}{3}\ \right) = \log{2c}}$
$\Rightarrow c = \frac{2}{3}$
On putting $c = \frac{2}{3}$ in Eq. (i), we get
$\log{\left( \ \frac{x^{2}}{x^{2} - y^{2}}\ \right) = \log{\frac{2}{3}x}}$
$\Rightarrow 2\left( x^{2} - y^{2} \right) = 3x$