Let y=y(x) be the solution of the differential equation:...

The given differential equation is a linear differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$, where

$P(x) = \dfrac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3+2)(2+e^{-2x})}$ and $Q(x) = 2 + e^{-2x}$.

We can rewrite $P(x)$ as:

$P(x) = \dfrac{3x^2(2+e^{-2x}) + 2e^{-2x}(x^3+2)}{(x^3+2)(2+e^{-2x})} = \dfrac{3x^2}{x^3+2} + \dfrac{2e^{-2x}}{2+e^{-2x}}$

The integrating factor (IF) is given by:

$IF = e^{\int P(x) dx} = e^{\int \left( \dfrac{3x^2}{x^3+2} + \dfrac{2e^{-2x}}{2+e^{-2x}} \right) dx}$

$IF = e^{\ln(x^3+2) - \ln(2+e^{-2x})} = \dfrac{x^3+2}{2+e^{-2x}}$

The general solution of the differential equation is:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substituting the values, we get:

$y \left( \dfrac{x^3+2}{2+e^{-2x}} \right) = \int (2+e^{-2x}) \left( \dfrac{x^3+2}{2+e^{-2x}} \right) dx + C$

$y \left( \dfrac{x^3+2}{2+e^{-2x}} \right) = \int (x^3+2) dx + C$

$y \left( \dfrac{x^3+2}{2+e^{-2x}} \right) = \dfrac{x^4}{4} + 2x + C$

Given $y(0) = \dfrac{3}{2}$, substituting $x = 0$:

$\dfrac{3}{2} \left( \dfrac{0+2}{2+e^0} \right) = 0 + 0 + C$

$\dfrac{3}{2} \left( \dfrac{2}{3} \right) = C \Rightarrow C = 1$

Thus, the particular solution is:

$y \left( \dfrac{x^3+2}{2+e^{-2x}} \right) = \dfrac{x^4}{4} + 2x + 1$

To find $y(1)$, substitute $x = 1$:

$y(1) \left( \dfrac{1+2}{2+e^{-2}} \right) = \dfrac{1}{4} + 2 + 1$

$y(1) \left( \dfrac{3}{2+e^{-2}} \right) = \dfrac{13}{4}$

$y(1) = \dfrac{13}{12}(2+e^{-2})$

Comparing this with $y(1) = \alpha(2+e^{-2})$, we get:

$\alpha = \dfrac{13}{12}$