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For $\ell >0$, the logarithmic change of variable in a Sturm-Liouville problem sends this equation to $-Y^{\prime \prime }-\alpha Y=1$ on $(0,\ell )$. The normalized eigenfunctions are $\sqrt{2/\ell }\sin (n\pi t/\ell )$, with eigenvalues ${\lambda }_n=(n\pi /\ell {)}^2-\alpha$. If every ${\lambda }_n$ is nonzero, the Sturm-Liouville eigenfunction expansion givesIndeed the constant function has sine coefficients $2[1-(-1{)}^n]/(n\pi )$; division by each eigenvalue solves the equation in the operator domain. The series is uniformly convergent, with coefficients $O(n^{-3})$, while applying the differential operator is justified in the weighted square-integrable sense. At a zero eigenvalue the solvability condition at a Sturm-Liouville eigenvalue must instead be checked: an odd resonant index makes the constant forcing incompatible, whereas an even resonant index permits solutions with an arbitrary multiple of that zero mode.

The useful coordinate for this scale-dependent differential operator is $t=\ln x$. With $Y(t)=y(e^t)$, the chain rule givesThe interval becomes $0\le t\le 1$, and the endpoint values become $Y(0)=Y(1)=0$. This logarithmic change of variable in a Sturm-Liouville problem reduces the calculation to a constant-coefficient Dirichlet problem. Also $dx/x=dt$, so the transformation preserves the appropriate weighted inner product.

Use a logarithmic change of variable in a Sturm-Liouville problem: set $t=\log x$ and $Y(t)=y(e^t)$. The chain rule givesThe endpoint conditions become $Y(0)=Y(1)=0$. Multiplying $-Y^{\prime \prime }=(\lambda +1)Y$ by $Y^{\ast }$ and integrating shows that $\lambda +1>0$ for a nonzero solution. Put $\mu =\sqrt{\lambda +1}$. The first endpoint selects $Y=A\sin (\mu t)$ and the second requires $\mu =n\pi$, $n\ge 1$. HenceThese eigenfunctions are orthonormal in the specified weighted inner product, because $w(x)dx=dx/x=dt$ and