Solve the differential equation: $(\cos^2 x) \frac{dy}{dx} + y = \tan x; \left( 0 \leq x < \frac{\pi}{2} \right)$

$\tan x - 1 + C e^{-\tan x}$

The correct answer is Option (3) → $\tan x - 1 + C e^{-\tan x}$

Given:

$(\cos^2 x) \frac{dy}{dx} + y = \tan x$

Convert to linear form

Divide by $\cos^2 x$:

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

Integrating Factor

$e^{\int \sec^2 x dx} = e^{\tan x}$

Multiply throughout

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

Integrate

Let $t = \tan x$, so $dt = \sec^2 x dx$

$\int t e^t dt = e^t(t - 1)$

$y e^{\tan x} = e^{\tan x}(\tan x - 1) + C$

Final solution

$y = \tan x - 1 + Ce^{-\tan x}$