 Zero-eigenvalue solutions of the Laguerre differential equation

At eigenvalue zero the Laguerre differential operator on polynomials gives $x{y}^{\prime \prime }+(1-x)y^{\prime }=0$, equivalently $(x{e}^{-x}{y}^{\prime }{)}^{\prime }=0$. Thus $y^{\prime }=D{e}^x/x$ and the displayed solutions follow from the exponential integral. For any $x_0>0$, all solutions with $y(x_0)=1$ can be written $y(x)=1+D{\int }_{x_0}^x{e}^{t}dt/t$. The choices $D=0$ and $D=1$ are linearly independent despite sharing that one prescribed value.

The second branch grows as $e^x/x$ at infinity and has a logarithmic singularity at zero. It is consequently excluded from the regular weighted square-integrable eigenproblem with weight $e^{-x}$. Only the constant zero mode remains in that self-adjoint realization. A normalization at one interior point alone is not enough to select a unique second-order solution.