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