= Solution
Multiplication by an <integrating factor> gives
$$
\boxed{p(x)=\exp\!\left(\int^xa(s)\,ds\right),\quad r=pc,\quad q=-pb,}
$$
since $p'=ap$ and hence $(pw')'=p(w''+aw')$.
For the singular-endpoint equation, take $p=x^{-1/2}$, $r=x^{1/2}/2$, $q=0$. The <indicial equation> is $\alpha(2\alpha-3)=0$. The constant branch violates the zero endpoint value unless its coefficient vanishes; the nontrivial admissible branch therefore has $\boxed{\alpha=3/2}$. More precisely, an <eigenfunction> has $w=Cx^{3/2}+O(x^{9/2})$ and $w'=(3C/2)x^{1/2}+O(x^{7/2})$ near zero.
Integrate the <self-adjoint> equation after multiplying by $w$, initially on $[a,1]$ with $a>0$:
$$
[wx^{-1/2}w']_a^1-\int_a^1x^{-1/2}(w')^2\,dx
+\frac\lambda2\int_a^1x^{1/2}w^2\,dx=0.
$$
At zero the boundary expression is $O(x^{3/2})$, the first <integrand> is $O(x^{1/2})$, and the second is $O(x^{7/2})$. Both <integrals> converge and the <boundary term> vanishes. Division by the positive <weighted norm> gives $\boxed{\lambda=I(w)}$.
The change of coordinate $z=x^{3/2}$, $w(x)=W(z)$ gives exactly
$$
W''+\frac{2\lambda}{9}W=0,\qquad W(0)=W(1)=0.
$$
Thus $\lambda_n=9n^2\pi^2/2$ and $w_n=\sin(n\pi x^{3/2})$. To prove a minimum over trial functions, rather than just identify the least <eigenvalue>, transform the quotient itself:
$$
I(w)=\frac92\frac{\int_0^1(W')^2\,dz}{\int_0^1W^2\,dz}\geq\boxed{\frac{9\pi^2}2}.
$$
The Dirichlet <Poincaré inequality> gives the last step, with equality precisely for a nonzero multiple of $W=\sin\pi z$. The trial class is the nonzero finite-energy class with zero <Sobolev traces>; endpoint values alone without an integrability requirement do not define a finite quotient. This is the <Power-coordinate reduction of a singular Sturm-Liouville quotient>.
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