= Solution
The leading <regular perturbation> equation has $c=0$. Its homogeneous solutions are $1$ and $(x+2)e^{-x}$, so
$$
y_0(x)=A+B(x+2)e^{-x}.
$$
The boundary data must also be expanded: $2e^\epsilon=2+2\epsilon+\cdots$, while the right endpoint value is independent of $\epsilon$. Thus $y_0(0)=2$ and $y_0(1)=3/e$, giving
$$
A+2B=2,\qquad A+\frac3eB=\frac3e.
$$
Since $2-3/e\ne0$, the unique values are $A=0$, $B=1$. Therefore
$$
\boxed{y_0(x)=(x+2)e^{-x}.}
$$
Its <derivatives> are $y_0'=-(x+1)e^{-x}$ and $y_0''=xe^{-x}$, so the leading differential equation and both leading endpoint values can be checked directly.
\Image[/past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2017-logarithmic-modes-and-perturbation.png]
{title=Logarithmic eigenmodes and the leading regular perturbation solution}
{description=The left panel shows three eigenfunctions after division by x and the change of variable t = log x. The right panel compares the derived leading term with numerical boundary value solutions at two small parameter values; these are original calculations.}
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