The solution of the differential equation $\dfrac{{dy}}{{dx}} + \dfrac{y}{2}\sec x = \dfrac{{\tan x}}{{2y}}$, where $0 \leqslant x \leqslant \dfrac{\pi }{2}$ and $y\left( 0 \right) = 1$, is given by:
A. ${y^2} = 1 + \dfrac{x}{{\sec x + \tan x}}$
B. $y = 1 + \dfrac{x}{{\sec x + \tan x}}$
C. $y = 1 - \dfrac{x}{{\sec x + \tan x}}$
D. ${y^2} = 1 - \dfrac{x}{{\sec x + \tan x}}$

Hint:
The given equation is a nonlinear differential equation. We will convert it into linear differential equation by dividing the equation by $y$. Then, make necessary substitution and compare with the standard linear differential equation. Find the integrating factor and apply initial value to get the value of $c$.

Complete step by step solution:
Given differential equation is a nonlinear equation.
$\dfrac{{dy}}{{dx}} + \dfrac{y}{2}\sec x = \dfrac{{\tan x}}{{2y}}$
Divide the equation by $y$ to convert it into a differential equation.
$\dfrac{{ydy}}{{dx}} + \dfrac{{{y^2}}}{2}\sec x = \dfrac{{\tan x}}{2}$
Let $v = {y^2}$, then
$
   \Rightarrow \dfrac{{dv}}{{dx}} = 2y\dfrac{{dy}}{{dx}} \\
   \Rightarrow y\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\dfrac{{dv}}{{dx}} \\
$
On substituting the values, we will get,
$
  \dfrac{1}{2}\dfrac{{dv}}{{dx}} + \dfrac{v}{2}\sec x = \dfrac{{\tan x}}{2} \\
   \Rightarrow \dfrac{{dv}}{{dx}} + v\sec x = \tan x \\
$
Here, we can observe that the given differential equation is of the type linear in $y$
That is the equation $\dfrac{{dv}}{{dx}} + v\sec x = \tan x$ is of the form $\dfrac{{dy}}{{dx}} + Py = Q$, where $P = \sec x$ and $Q = \tan x$
Then, integrating factor is given by \[{e^{\int {P\left( x \right)dx} }}\] which is \[{\text{IF}} = {e^{\int {\sec xdx} }}\]
We know that $\int {\sec xdx = \ln \left| {\tan x + \sec x} \right|} $
This implies, \[{e^{\ln \left| {\tan x + \sec x} \right|}} = \tan x + \sec x\]
We can solve the value of $v$ using the formula,
\[v = \dfrac{1}{{{\text{IF}}}}\int {{\text{IF}} \times Q\left( x \right) + c} \]
On substituting the values, we will get,
\[v = \dfrac{1}{{\left( {\tan x + \sec x} \right)}}\int {\tan x\left( {\tan x + \sec x} \right) + c} \]
Now, we will solve the integration part and then substitute its value.
\[\int {\tan x\left( {\tan x + \sec x} \right) = \int {\left( {{{\tan }^2}x + \tan x\sec x} \right)dx} } \]
Also, \[{\tan ^2}x = {\sec ^2}x - 1\]
Then, the value of integral is
$
   \Rightarrow \int {\left( {{{\sec }^2}x - 1 + \tan x\sec x} \right)dx} \\
   \Rightarrow \int {{{\sec }^2}x - \int {1dx + \int {\tan x\sec xdx} } } \\
   \Rightarrow \tan x - x + \sec x \\
$
Hence, the value of $v$ becomes,
\[v = \dfrac{{\tan x + \sec x - x}}{{\left( {\tan x + \sec x} \right)}} + c\]


Put $v - {y^2}$
$
  {y^2} = \dfrac{{\tan x + \sec x - x}}{{\left( {\tan x + \sec x} \right)}} + c \\
   \Rightarrow {y^2} = 1 - \dfrac{x}{{\tan x + \sec x}} + c \\
$
Put the initial condition $y\left( 0 \right) = 1$
$
  1 = 1 - 0 + c \\
   \Rightarrow c = 0 \\
$
Therefore,
\[{y^2} = 1 - \dfrac{x}{{\tan x + \sec x}}\]

Hence, option D is correct.

Note:
The differential equation of the form \[\dfrac{{dy}}{{dx}} + yP\left( x \right) = {y^n}Q\left( x \right)\] can be converted into linear form by dividing the equation by ${y^n}$ and then substituting ${y^{1 - n}}$ and then dividing by $1 - n$. Also, students must do the integration correctly.