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

$\ln x + C$

The correct answer is Option (1) → $\ln x + C$ ##

We can write

The linear differential equation as

$\frac{dy}{dx} - \frac{y}{x} = 1$   $[\text{because } e^{\ln x} = x]$

The integrating factor (I.F.) as:

$\text{I.F.} = e^{-\int \frac{1}{x} dx}$

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

The solution as:

$y \times \text{I.F.} = \int Q \times \text{I.F.} \, dx + C$

$y \times \frac{1}{x} = \int 1 \times \frac{1}{x} \, dx + C$

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