${\cos ^2}x\cfrac{{dy}}{{dx}} + y = \tan x\left( {0 \le x < \cfrac{\pi }{2}} \right)$

: The given equation is ${\cos ^2}x\cfrac{{dy}}{{dx}} + y = \tan x$

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

which is a linear equation of the type $\cfrac{{dy}}{{dx}} + Py = Q$

Where, $P = {\sec ^2}x$ and $Q = \tan x{\sec ^2}x$

$\therefore$ ${\rm{I}}.{\rm{F}}. = {e^{\int {{{\sec }^2}} xdx}} = {e^{\tan x}}$

$\therefore$
The solution is $y \cdot ({\rm{I}}.{\rm{F}}.) = \int {Q \cdot ({\rm{I}}.{\rm{F}}.)dx} + C$

$\Rightarrow y \cdot {e^{\tan x}} = \int {\tan } x{\sec ^2}x{e^{\tan x}}dx + C = I + C$

(2)
Now, $I = \int {\tan } x{\sec ^2}x{e^{\tan x}}dx$

Put $\tan x = t \Rightarrow {\sec ^2}xdx = dt$
$\therefore$ $I = \int {t \cdot {e^t}dt} = t \cdot {e^t} - \int {(1)} {e^t}dt$

(Integrating by parts)
$= t{e^t} - {e^t} = {e^t}(t - 1) = {e^{\tan x}}(\tan x - 1)$

From (2), $y \cdot {e^{\tan x}} = {e^{\tan x}}(\tan x - 1) + C$

$\Rightarrow y = (\tan x - 1) + C{e^{ - \tan x}}$, which is the required solution.