Find the general solution of the given differential equation. $\frac{dy}{dx} = ye^{\ln \frac{1}{x}} + 1$

$x\ln x + Cx$

The correct answer is Option (3) →$x\ln x + Cx$

$\frac{dy}{dx} = y e^{\ln\frac{1}{x}} + 1$

$e^{\ln\frac{1}{x}} = \frac{1}{x}$

$\frac{dy}{dx} = \frac{y}{x} + 1$

$\frac{dy}{dx} - \frac{y}{x} = 1$

$\text{IF} = e^{\int -\frac{1}{x}dx} = e^{-\ln x} = \frac{1}{x}$

$\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{1}{x}$

$\frac{y}{x} = \ln x + C$

$y = x(\ln x + C)$

The general solution is $y = x(\ln x + C)$.