Find the general solution of the differential equation $x \frac{dy}{dx} = y(\log y - \log x + 1).$

$\log(\frac{y}{x}) = Cx$

The correct answer is Option (2) → $\log(\frac{y}{x}) = Cx$ ##

Given differential equation is

$x \frac{dy}{dx} = y(\log y - \log x + 1)$

$\Rightarrow \frac{dy}{dx} = \frac{y}{x} \left( \log \frac{y}{x} + 1 \right)$

Put $y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$

$\Rightarrow v + x \frac{dv}{dx} = v(\log v + 1)$

$\Rightarrow \frac{dv}{v \log v} = \frac{dx}{x}$

On integrating both sides, we get

$\int \frac{dv}{(v \log v)} = \int \frac{dx}{x}$

$\Rightarrow \log(\log v) = \log x + \log C$

$\Rightarrow \log(\log v) = \log Cx$

$\Rightarrow \log \left( \frac{y}{x} \right) = Cx$