 Solution

The leading regular perturbation equation has $c=0$. Its homogeneous solutions are $1$ and $(x+2)e^{-x}$, soThe boundary data must also be expanded: $2e^{ϵ}=2+2ϵ+\cdots$, while the right endpoint value is independent of $ϵ$. Thus $y_0(0)=2$ and $y_0(1)=3/e$, givingSince $2-3/e\ne 0$, the unique values are $A=0$, $B=1$. ThereforeIts derivatives are $y_0^{\prime }=-(x+1)e^{-x}$ and $y_0^{\prime \prime }=x{e}^{-x}$, so the leading differential equation and both leading endpoint values can be checked directly.