Explanation

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