The general solution of the differential equation $x\, dy - y \, dx = \sqrt{x^2+y^2}dx $ is

$y+\sqrt{x^2+y^2}=cx^2$

The correct answer is option (1) : $y+\sqrt{x^2+y^2}=cx^2$

We have,

$x\, dy - y \, dx = \sqrt{x^2+y^2}dx $

$⇒\frac{dy}{dx} = \frac{y+\sqrt{x^2+y^2}}{x}$

This is a homogeneous differential equation.

Substituting $y = vx $ and $\frac{dy}{dx} = v + x\frac{dv}{dx}, $ it reduces to

$x\frac{dv}{dx} = \sqrt{v^2+1}$

$⇒\frac{1}{\sqrt{v^2+1}}dv=\frac{dx}{x}$

On integrating, we obtain

$log_e(v+\sqrt{v^2+1})= log_ex+logc$

$⇒v+\sqrt{v^2+1}=cx$

$⇒y + \sqrt{x^2+y^2}= cx^2 $, which is the required general solution.