Let $$y = yx$$ be the solution of the differential equation, $$\frac{dy}{dx} + y\tan x = 2x + x^2\tan x$$, $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, such that $$y(0) = 1$$. Then

We begin with the given linear differential equation

$$\frac{dy}{dx}+y\tan x=2x+x^{2}\tan x,\qquad x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

The standard form is $$\frac{dy}{dx}+P(x)\,y=Q(x),$$ where here we have $$P(x)=\tan x \quad\text{and}\quad Q(x)=2x+x^{2}\tan x.$$

For such an equation the integrating factor is defined by

$$\text{I.F.}=e^{\displaystyle\int P(x)\,dx}.$$

Evaluating the integral, we note that $$\int\tan x\,dx=-\ln|\cos x|,$$ so

$$\text{I.F.}=e^{-\ln|\cos x|}=\frac{1}{\cos x}=\sec x.$$

Multiplying every term of the differential equation by this integrating factor gives

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

Because $$\frac{d}{dx}(\sec x)=\sec x\tan x,$$ the left-hand side is recognized as the derivative of the product $$y\sec x$$:

$$\frac{d}{dx}\bigl(y\sec x\bigr)=2x\sec x+x^{2}\sec x\tan x.$$

We now integrate both sides with respect to $$x$$:

$$y\sec x=\int\bigl(2x\sec x+x^{2}\sec x\tan x\bigr)\,dx+C,$$

where $$C$$ is the constant of integration. To evaluate the integral we observe that

$$\frac{d}{dx}\bigl(x^{2}\sec x\bigr)=2x\sec x+x^{2}\sec x\tan x,$$

which is exactly the integrand. Therefore

$$\int\bigl(2x\sec x+x^{2}\sec x\tan x\bigr)\,dx=x^{2}\sec x.$$

Substituting back, we obtain

$$y\sec x=x^{2}\sec x+C.$$

Multiplying by $$\cos x$$ gives the explicit solution

$$y=x^{2}+C\cos x.$$

To determine $$C$$ we use the initial condition $$y(0)=1$$. Since $$\cos 0=1,$$ we have

$$1=0^{2}+C\cdot1\;\Longrightarrow\;C=1.$$

Hence the required function is

$$y(x)=x^{2}+\cos x.$$

We now differentiate to find $$y'(x)$$:

$$y'(x)=\frac{d}{dx}\bigl(x^{2}\bigr)+\frac{d}{dx}\bigl(\cos x\bigr)=2x-\sin x.$$

First, evaluate at $$x=\frac{\pi}{4}:$$

$$y'\!\left(\frac{\pi}{4}\right)=2\left(\frac{\pi}{4}\right)-\sin\!\left(\frac{\pi}{4}\right)=\frac{\pi}{2}-\frac{1}{\sqrt{2}}=\frac{\pi}{2}-\frac{\sqrt{2}}{2}.$$

Next, evaluate at $$x=-\frac{\pi}{4}:$$

$$\sin\!\left(-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}},$$

so

$$y'\!\left(-\frac{\pi}{4}\right)=2\!\left(-\frac{\pi}{4}\right)-\!\left(-\frac{1}{\sqrt{2}}\right)=-\frac{\pi}{2}+\frac{1}{\sqrt{2}}=-\frac{\pi}{2}+\frac{\sqrt{2}}{2}.$$

We now form the required difference:

$$\begin{aligned} y'\!\left(\frac{\pi}{4}\right)-y'\!\left(-\frac{\pi}{4}\right) &=\left(\frac{\pi}{2}-\frac{\sqrt{2}}{2}\right)-\left(-\frac{\pi}{2}+\frac{\sqrt{2}}{2}\right)\\[4pt] &=\frac{\pi}{2}-\frac{\sqrt{2}}{2}+\frac{\pi}{2}-\frac{\sqrt{2}}{2}\\[4pt] &=\pi-\sqrt{2}. \end{aligned}$$

Thus

$$y'\!\left(\frac{\pi}{4}\right)-y'\!\left(-\frac{\pi}{4}\right)=\pi-\sqrt{2}.$$

None of the other stated combinations match this value, so the option that asserts this exact difference is the true one.

Hence, the correct answer is Option A.