The solution of the differential equation 

$\frac{dy}{dx}+\frac{1}{2}y\sec x=\frac{\tan x}{2y},$where $0\le x<\frac{\pi }{2}$and $y\left(0\right)=1\text{, }$is given by 

We have,  

$\frac{dy}{dx}+\frac{1}{2}y\sec x=\frac{\tan x}{2y}$

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

$\Rightarrow \frac{du}{dx}+(\sec x)u=\tan x,$where $u=y^2$

This is a linear differential equation with

I.F.$=e^{\int \sec xdx}=e^{\log (\sec x+\tan x)}=\sec x+\tan x\text{. }$

Multiplying both sides by I.F .$= sec x + \tan x$ and integrating

or, $u(\sec x+\tan x)=\int \tan x(\sec x+\tan x)dx+C$

$y^2(\sec x+\tan x)=\sec x+\tan x-x+C$

It is given $y=1$ When $X = 0$ . Putting $x =0$,$y=1/h$ , we get Putting $C =0$ in (i) we obtain 

$y^2(\sec x+\tan x)=\sec x+\tan x-x$

$\Rightarrow y^2=1-\frac{x}{\sec x+\tan x}$