Let $$y = y(x)$$ be the solution of kthe differential equation :$$\frac{dy}{dx} + \left(\frac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3 + 2)(2 + e^{-2x})}\ \right),y = 2 + e^{-2x}$$, $$x \in (-1, 2)$$, $$y(0) = \frac{3}{2}$$. If $$y(1) = \alpha(2 + e^{-2})$$, then $$\alpha$$ is equal to :

The differential equation is linear in $$y$$:
$$\frac{dy}{dx}+P(x)\,y = Q(x),$$
where

$$P(x)=\frac{6x^{2}+\left(3x^{2}+2x^{3}+4\right)e^{-2x}}{(x^{3}+2)\,(2+e^{-2x})}, \qquad Q(x)=2+e^{-2x}.$$

1. Finding the integrating factor (IF)
We look for a function $$\mu(x)$$ such that $$\mu'=\mu P(x)$$. Observe that

$$\frac{d}{dx}\left[\ln\!\Bigl(\tfrac{x^{3}+2}{\,2+e^{-2x}\,}\Bigr)\right] =\frac{3x^{2}}{x^{3}+2}+\frac{2e^{-2x}}{2+e^{-2x}} =\frac{6x^{2}+\bigl(3x^{2}+2x^{3}+4\bigr)e^{-2x}} {(x^{3}+2)\,(2+e^{-2x})}=P(x).$$

Hence $$P(x)=\dfrac{d}{dx}\Bigl[\ln\!\bigl(\tfrac{x^{3}+2}{2+e^{-2x}}\bigr)\Bigr],$$ so

$$\text{IF}=e^{\int P(x)\,dx}=\frac{x^{3}+2}{\,2+e^{-2x}}.$$

2. Converting to exact differential
Multiplying the given equation by the integrating factor:

$$\frac{x^{3}+2}{\,2+e^{-2x}}\frac{dy}{dx} +\Bigl(\frac{x^{3}+2}{\,2+e^{-2x}}\Bigr)P(x)\,y =(x^{3}+2).$$

Because the IF was chosen correctly, the left-hand side is $$\frac{d}{dx}\bigl[\text{IF}\cdot y\bigr].$$ Therefore

$$\frac{d}{dx}\Bigl[\frac{x^{3}+2}{\,2+e^{-2x}}\,y\Bigr]=x^{3}+2.$$

3. Integrating
Integrate both sides with respect to $$x$$:

$$\frac{x^{3}+2}{\,2+e^{-2x}}\,y=\int (x^{3}+2)\,dx=C+\frac{x^{4}}{4}+2x.$$

Thus

$$y(x)=\Bigl(\frac{2+e^{-2x}}{x^{3}+2}\Bigr) \Bigl(\frac{x^{4}}{4}+2x+C\Bigr).$$

4. Using the initial condition
Given $$y(0)=\frac32$$.

At $$x=0$$: $$e^{0}=1,\;x^{3}+2=2,\;x^{4}/4+2x=0,$$ hence

$$\frac32 =\Bigl(\frac{2+1}{2}\Bigr)C =\frac{3}{2}C \;\Longrightarrow\; C=1.$$

5. Evaluating $$y(1)$$

At $$x=1$$:

$$x^{4}/4+2x+C=\frac14+2+1=\frac{13}{4}, \qquad x^{3}+2=3,\qquad e^{-2(1)}=e^{-2}.$$

Therefore

$$y(1)=\Bigl(\frac{2+e^{-2}}{3}\Bigr)\Bigl(\frac{13}{4}\Bigr) =\frac{13}{12}\,(2+e^{-2}).$$

The question writes $$y(1)=\alpha\,(2+e^{-2})$$, so

$$\alpha=\frac{13}{12}.$$

Option D which is: $$\dfrac{13}{12}$$