Solve the differential equation: $(\cos^2 x) \frac{dy}{dx} + y = \tan x; \left( 0 \leq x < \frac{\pi}{2} \right)$
Answer & explanation
Correct answer: option 3
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
$y = \tan x - 1 + Ce^{-\tan x}$