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

The given differential equation is $\frac{dy}{dx} + y\tan x = \sec x$

This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x) \cdot y = Q(x)$ where $P(x) = \tan x$ and $Q(x) = \sec x$.

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The integrating factor is $IF = e^{\int P(x)dx} = e^{\int \tan x dx}$

$\int \tan x \, dx = \int \frac{\sin x}{\cos x} dx$

Let $u = \cos x$, then $du = -\sin x \, dx$

$= -\int \frac{1}{u} du$

$= -\ln|u|$

$= -\ln|\cos x|$

$= \ln|\sec x|$

Therefore $IF = e^{\ln|\sec x|} = \sec x$

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Multiplying the entire equation by $\sec x$:

$\sec x \cdot \frac{dy}{dx} + y \sec x \tan x = \sec^2 x$

The left side is the derivative of $(y \sec x)$ by the product rule.

$\frac{d}{dx}(y\sec x) = \sec^2 x$

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Integrating both sides:

$\int \frac{d}{dx}(y\sec x) \, dx = \int \sec^2 x \, dx$

$y\sec x = \tan x + c$

$y\sec x - \tan x = c$

where $c$ is an arbitrary constant.