The general solution of the differential equation $\frac{dy}{dx}+y\tan x=\sec x$

$y\sec x- \tan x = c$, where c is an arbitary constant

The correct answer is Option (1) → $y\sec x- \tan x = c$, where c is an arbitary constant

Given: $\displaystyle \frac{dy}{dx} + y \tan x = \sec x$

It is linear: $\displaystyle \frac{dy}{dx} + P(x)y = Q(x)$

$P(x) = \tan x$, $Q(x) = \sec x$

I.F. $= e^{\int \tan x\,dx} = e^{-\ln|\cos x|} = \frac{1}{\cos x} = \sec x$

Using the formula: $y \cdot \text{I.F.} = \int Q \cdot \text{I.F.} \, dx + C$

$\Rightarrow y \sec x = \int \sec x \cdot \sec x\, dx + C = \int \sec^2 x\, dx + C$

$\Rightarrow y \sec x = \tan x + C$