What is the general solution of the differential equation e^y' = x?

The correct answer is Option (2) → $y = x \log x - x + c$ ##

The given differential equation is $e^{y'} = x$.

Taking log both sides we get

$\frac{dy}{dx} \log e = \log x$

$\frac{dy}{dx} = \log x \quad [∵\log e = 1]$

$dy = \log x \, dx$

$\int dy = \int \log x \, dx$

$y = x \log x - x + c$