Find the general solution of the differential equation: $\cos ^{2} x \frac{d y}{d x}+y=\tan x$ where $0 \leq x < \frac{\pi}{2}$.

(N/A) We have the linear differential equation: $\cos ^2 x \frac{d y}{d x}+y=\tan x$.
Dividing throughout by $\cos ^2 x$,we get: $\frac{d y}{d x} + (\sec ^2 x) y = \sec ^2 x \tan x$.
This is a linear differential equation of the form $\frac{d y}{d x} + Py = Q$,where $P = \sec ^2 x$ and $Q = \sec ^2 x \tan x$.
The integrating factor $(I.F.)$ is given by: $I.F. = e^{\int P dx} = e^{\int \sec ^2 x dx} = e^{\tan x}$.
The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + C$.
Substituting the values: $y \cdot e^{\tan x} = \int (\sec ^2 x \tan x) e^{\tan x} dx + C$.
Let $t = \tan x$,then $dt = \sec ^2 x dx$. The integral becomes: $y \cdot e^{\tan x} = \int t e^t dt + C$.
Using integration by parts $\int u dv = uv - \int v du$ with $u = t$ and $dv = e^t dt$:
$y \cdot e^{\tan x} = t e^t - \int e^t dt + C = t e^t - e^t + C$.
Substituting $t = \tan x$ back: $y \cdot e^{\tan x} = \tan x e^{\tan x} - e^{\tan x} + C$.
Dividing by $e^{\tan x}$,we get the general solution: $y = \tan x - 1 + C e^{-\tan x}$.