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

We begin with the given differential equation

$$\frac{dy}{dx}+\frac{y}{2}\sec x=\frac{\tan x}{2y},\qquad 0\le x<\frac{\pi}{2},\qquad y(0)=1.$$

To clear the denominators we multiply every term by $$2y$$, obtaining

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

Here we recognize that the derivative of $$y^{2}$$ is $$2y\dfrac{dy}{dx}$$, so we set

$$z=y^{2}\;\Longrightarrow\;\frac{dz}{dx}=2y\frac{dy}{dx}.$$

Substituting $$z$$ into the transformed equation gives

$$\frac{dz}{dx}+z\sec x=\tan x.$$ This is now a first-order linear differential equation in $$z$$ of the standard form

$$\frac{dz}{dx}+P(x)\,z=Q(x),\qquad P(x)=\sec x,\;Q(x)=\tan x.$$

The integrating factor is found from the formula $$\text{I.F.}=e^{\int P(x)\,dx}$$. Since

$$\int\sec x\,dx=\ln\!\bigl|\sec x+\tan x\bigr|,$$

we have

$$\text{I.F.}=e^{\ln|\sec x+\tan x|}=|\sec x+\tan x|.$$

Because $$0\le x<\dfrac{\pi}{2}$$, both $$\sec x$$ and $$\tan x$$ are positive, so $$|\sec x+\tan x|=\sec x+\tan x.$$

Multiplying the whole differential equation by this integrating factor yields

$$(\sec x+\tan x)\frac{dz}{dx}+z\sec x(\sec x+\tan x)=\tan x(\sec x+\tan x).$$

By construction, the left-hand side is the derivative of the product $$z(\sec x+\tan x)$$, that is,

$$\frac{d}{dx}\Bigl[z(\sec x+\tan x)\Bigr]=\tan x(\sec x+\tan x).$$

Integrating both sides with respect to $$x$$, we obtain

$$z(\sec x+\tan x)=\int \tan x(\sec x+\tan x)\,dx+C.$$

We now evaluate the integral on the right. We split it into two simpler integrals:

$$\int\tan x\sec x\,dx+\int\tan^{2}x\,dx.$$

Using the standard antiderivatives $$\int\tan x\sec x\,dx=\sec x$$ and $$\int\tan^{2}x\,dx=\int(\sec^{2}x-1)\,dx=\tan x-x$$, we find

$$\int \tan x(\sec x+\tan x)\,dx=\sec x+\tan x-x.$$

Therefore

$$z(\sec x+\tan x)=\sec x+\tan x-x+C.$$

To determine the constant $$C$$ we use the initial condition $$y(0)=1\;\Longrightarrow\;z(0)=y^{2}(0)=1.$$ At $$x=0$$ we have $$\sec 0=1$$ and $$\tan 0=0$$, so

$$1\cdot(\,1+0\,)=1+0-0+C\;\Longrightarrow\;1=1+C\;\Longrightarrow\;C=0.$$

Thus the relation simplifies to

$$z(\sec x+\tan x)=\sec x+\tan x-x.$$

Dividing by $$\sec x+\tan x$$ (which is positive in the given interval) gives

$$z=1-\frac{x}{\sec x+\tan x}.$$

Recall that $$z=y^{2}$$, so

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

This expression automatically satisfies the initial condition and is valid for $$0\le x<\dfrac{\pi}{2}$$.

Hence, the correct answer is Option D.