The differential equation $\frac{dy}{dx} +\frac{y^2}{x^2} =\frac{y}{x}$ has the solution

$x=y (log x + C)$

The correct answer is option (1) : $x=y (log x + C)$

We have,

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

Putting $y = vx $ and $\frac{dy}{dx} = v + x\frac{dv}{dx}, $ we get

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

$⇒x\frac{dv}{dx} = -v^2$

$⇒-\frac{1}{v^2}dv=\frac{1}{x}dx$

On integrating, we get

$\frac{1}{v}= log x + C $

$⇒x = y (log x + C)$