$\cfrac{{dy}}{{dx}} = y\tan x;y = 1$ when $x = 0$

.: We have, $\cfrac{{dy}}{{dx}} = y\tan x \Rightarrow \cfrac{{dy}}{y} = \tan xdx$ …(1)
Integrating (1) both sides,

we get
$\int {\cfrac{{dy}}{y}} = \int {\tan } xdx \Rightarrow \log y = \log |\sec x| + C$

When $x = 0,y = 1 \Rightarrow log1 = \log (\sec 0) + C \Rightarrow 0 = \log 1 + C$

$\Rightarrow C = 0\,\,\therefore \,\,\log y = \log |\sec x|$

Hence, $y = \sec x$

which is required particular solution.