By substituting $y = vx,$ the solution of the differential equation $\frac{dy}{dx} =\frac{x^2+y^2}{xy}, $is

$\frac{y^2}{2x^2}= log x + C$

The correct answer is option (2) : $\frac{y^2}{2x^2}= log x + C$

Substituting $y = vx $ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ in the given differential equation, we get

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

$⇒x\frac{dv}{dx} =\frac{1}{v}$

$⇒vdv=\frac{1}{x} dv$

On integrating, we get

$\frac{v^2}{2} = log \, x +C$

$⇒\frac{y^2}{2x^2}= log \, x + C$