Solve the differential equation $y' = \frac{x + y}{x}$.

$y = x \ln|x| + Cx$

The correct answer is Option (1) → $y = x \ln|x| + Cx$ ##

Rewrite the differential equation:

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

Assume that $y = vx$, this implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$

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

$ x \frac{dv}{dx} = 1 $

$ dv = \frac{dx}{x} $

$ \int dv = \int \frac{dx}{x} $

$ v = \log x + c $

$ \frac{y}{x} = \log x + c $

$ y = x \log x + cx $

The general solution is $y = x \log x + cx$.