= Solution
Use the normalization from part (ii), and write
$$
N_k=\int_1^e r\widehat y_k^2\,dx=\frac12,\qquad
I_k=\int_1^e r\widehat y_k e^x\,dx
=\int_1^e\frac{e^x}{x^2}\sin(k\pi\log x)\,dx.
$$
The <Sturm-Liouville eigenfunction expansion> coefficient is $c_k=N_k^{-1}\int_1^e ry\widehat y_k\,dx$. Although $\widehat y_k$ has zero endpoint values, $y$ generally does not. Retain the endpoint term in the <Lagrange identity for a Sturm-Liouville operator>:
$$
\begin{aligned}
\int_1^e\bigl(\widehat y_k\widehat Ly-y\widehat L\widehat y_k\bigr)\,dx
&=\bigl[x^{-1}(\widehat y_k y'-\widehat y_k' y)\bigr]_1^e\\
&=\alpha\widehat y_k'(1)=\alpha k\pi.
\end{aligned}
$$
The positive sign comes from subtracting the contribution at $x=1$. The differential equation also gives $\widehat Ly=re^x-(\beta-1)ry$. Substitution into the preceding identity yields
$$
I_k-(\beta-1+\widehat\lambda_k)c_kN_k=\alpha k\pi.
$$
Hence, away from <resonance in a differential equation>,
$$
\boxed{c_k=\frac{2(I_k-\alpha k\pi)}{\beta-1-k^2\pi^2}
=\frac{2\left[\alpha k\pi-\int_1^e e^x x^{-2}\sin(k\pi\log x)\,dx\right]}{1+k^2\pi^2-\beta}.}
$$
This is the effect of <boundary data in a Sturm-Liouville eigenfunction expansion>: omitting the endpoint contribution would discard the dependence on $\alpha$.
If $\beta=1+n^2\pi^2$ for a positive integer $n$, the <Fredholm solvability condition> is
$$
\boxed{I_n=\alpha n\pi.}
$$
When it fails, there is \b[no solution]. When it holds, $c_n$ is arbitrary and the other coefficients retain the displayed formula, giving \b[infinitely many solutions] differing by a multiple of $\widehat y_n$. Sufficiency follows by subtracting the smooth lift $h(x)=\alpha x(1-\log x)$, which has the required endpoint values and satisfies $Lh=0$, and applying the <Fredholm alternative> to the resulting homogeneous <Dirichlet boundary conditions>.
Finally, for $\alpha\ne0$, the displayed <Sturm-Liouville eigenfunction expansion> represents the solution on $1<x<e$ and in the <weighted inner product> sense. Every term is zero at $x=1$, so that series cannot give the nonzero prescribed value by substitution at the endpoint. The <boundary trace> must be imposed separately. Equivalently, expand $y-h$ with homogeneous <Dirichlet boundary conditions> and add $h$ back afterwards.
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