Question

Sol. [Key Concept: The integrating factor collapses the LHS to $\frac{d}{dx}\lbrack y \cdot \frac{x^{3} + 2}{2 + e^{- 2x}}\rbrack$; integrate and apply $y(0) = \frac{3}{2}$.]

I.F. $= e^{\int\frac{6x^{2} + e^{- 2x}\left( 3x^{2} + 2x^{3} + 4 \right)}{\left( x^{3} + 2 \right)\left( 2 + e^{- 2x} \right)}dx}$

$$= e^{\int\frac{6x^{2} + e^{- 2x}\left( 3x^{2} - 2x^{3} - 4 \right) + \left( 4x^{3} + 8 \right)e^{- 2x}}{\left( x^{3} + 2 \right)\left( 2 + e^{- 2x} \right)}dx} $$Let $\left( x^{3} + 2 \right)\left( 2 + e^{- 2x} \right) = t$

$$\left( 6x^{2} + e^{- 2x}\left( 3x^{2} - 2x^{3} - 4 \right) \right)dx = dt $$I.F. $= e^{ln\left( x^{3} + 2 \right)\left( 2 + e^{- 2x} \right) - \int\frac{4e^{- 2x}dx}{2 + e^{- 2x}}}$

$${= e^{\mathcal{l}n\left| \left( x^{3} + 2 \right)\left( 2 + e^{- 2x} \right) \right| - 2\mathcal{l}n\left| 2 + e^{- 2x} \right|} }{= e^{ln\left| \frac{x^{3} + 2}{2 + e^{- 2x}} \right|} = \frac{x^{3} + 2}{2 + e^{- 2x}} }{\Rightarrow y \cdot \left( \frac{x^{3} + 2}{2 + e^{- 2x}} \right) = \int\left( \frac{x^{3} + 2}{2 + e^{- 2x}} \right)\left( e^{- 2x} + 2 \right)dx + C }$$$y \cdot \left( \frac{x^{3} + 2}{2 + e^{- 2x}} \right) = \frac{x^{4}}{4} + 2x + C$ at $y(0) = \frac{3}{2} \Rightarrow C = 1$

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

at $x = 1$

$${y \cdot \left( \frac{3}{2 + e^{- 2}} \right) = \frac{1}{4} + 2 + 1 }{\Rightarrow y = \frac{13}{12}\left( e^{- 2} + 2 \right) }$$So $\alpha = \frac{13}{12}$