Explanation

(a)
Given, $\frac{dy}{dx} = \frac{x - y}{x + y}$
This is a homogeneous equation
Put $y = vx\ \Rightarrow \ \frac{dy}{dx} = v + x\frac{dv}{dx}$
Given equation becomes
$v + x\frac{dv}{dx} = \frac{x - vx}{x + vx}$
$\Rightarrow x\frac{dv}{dx} = \frac{1 - v}{1 + v} - v$
$\Rightarrow \ \frac{1 + v}{2 - (1 + v)^{2}}dv = \frac{dx}{x}$
On integrating both sides, we get
$\int_{}^{}{\frac{1 + v}{2 - (1 + v)^{2}}dv = \int_{}^{}\frac{dx}{x}}$
Put $(1 + v)^{2} = t\ \Rightarrow 2(1 + v)dv = dt$
$\Rightarrow \frac{1}{2}\int_{}^{}{\frac{dt}{2 - t} = \int_{}^{}\frac{dx}{x}}$
$\Rightarrow \ - \frac{1}{2}\log{(2 - t) = \log{x + \log c}}$
$\Rightarrow - \frac{1}{2}\log{\left\lbrack 2 - (1 + v)^{2} \right\rbrack = \log{xc}}$
$\Rightarrow \ - \frac{1}{2}\log{\left\lbrack - v^{2} - 2v + 1 \right\rbrack = \log{xc}}$
$\Rightarrow \log{\frac{1}{\sqrt{1 - 2v - v^{2}}} = \log{xc}}$
$\Rightarrow x^{2}c^{2}\left( 1 - 2v - v^{2} \right) = 1$
$\Rightarrow x^{2}c^{2}\left( \ 1 - \frac{2y}{x} - \frac{y^{2}}{x^{2}}\ \right) = 1$ $\left( \because v = \frac{y}{x}\ \right)$
$\Rightarrow \ \frac{x^{2}c^{2}(x^{2} - 2yx - y^{2})}{x^{2}} = 1$
$\Rightarrow y^{2} + 2xy - x^{2} = c$