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

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

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

The given differential equation is

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

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

The integrating factor is $e^{\int \sec^2 x dx} = e^{\tan x}$

The solution is

$y e^{\tan x} = \int e^{\tan x} \tan x \sec^2 x \, dx + c$

Assume that $\tan x = u$ this implies $\sec^2 x \, dx = du$.

$y e^{\tan x} = \int u e^u du + c$

$= u e^u - e^u + c$

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