Find the general solution of the differential equation $x \frac{dy}{dx} = y(\log y - \log x + 1).$
Answer & explanation
Correct answer: option 2
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
$\int \frac{dv}{(v \log v)} = \int \frac{dx}{x}$
$\Rightarrow \log(\log v) = \log x + \log C$
$\Rightarrow \log(\log v) = \log Cx$
$\Rightarrow \log \left( \frac{y}{x} \right) = Cx$