The solution of the differential equation $\frac{d y}{d x}+\frac{1}{2} y \sec x=\frac{\tan x}{2 y}$, where $0 \leq x<\frac{\pi}{2}$ and $y(0)=1$, is given by

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

We have,

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

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

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

This is a linear differential equation with

Integrating factor = $e^{\int \sec x d x}=e^{\log (\sec x+\tan x)}=\sec x+\tan x$

Multiplying both sides by integrating factor $=\sec x+\tan x$ and integrating, we get

$u(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+C$

or, $y^2(\sec x+\tan x)=\sec x+\tan x-x+C$               .........(i)

It is given $y=1$ when $x=0$.

Putting $x=0, y=1$ in (i), we get $C=0$.

Putting $C=0$ in (i), we obtain

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

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