= Solution
Introduce a <Lagrange multiplier> for the normalization and vary $J[y]=\int_0^1[p(y')^2+qy^2-\lambda wy^2]dx$. With variation $\eta$ satisfying $\eta(1)=0$, <integration by parts> gives
$$
\delta J=2[py'\eta]_0^1-2\int_0^1[(py')'-qy+\lambda wy]\eta\,dx.
$$
The upper endpoint term vanishes, and the lower term vanishes because $p(0)=0$ with the stipulated regular differentiable behavior. More generally the natural endpoint condition is $py'\to0$ there. The <fundamental lemma of the calculus of variations> gives $(py')'-qy+\lambda wy=0$, the stated <Sturm-Liouville problem>.
The <Rayleigh-Ritz variational principle> estimates the lowest eigenvalue by minimizing $R[y]=\int[p(y')^2+qy^2]dx/\int wy^2dx$ over an admissible finite-dimensional trial family. Restricting the family can only raise the minimum.
For the specific radial problem, $p=w=x$ and $q=0$. The trial $y=A(1-x^2)$ has $\int_0^1xy^2dx=A^2/6$. Normalization gives $\boxed{A=\sqrt6}$, with the negative sign an equivalent choice. The numerator is $\int_0^1x(2Ax)^2dx=A^2$, so
$$
\boxed{\lambda_1\leq R[y]=6.}
$$
This is the quadratic case of the <power-law variational bound for a circular membrane>. The exact first eigenvalue $5.7815$ is smaller because the polynomial trial is not the exact <Bessel function of the first kind> eigenfunction. Expanding a trial in the exact <orthonormal eigenbasis> expresses its <Rayleigh quotient> as a weighted average of eigenvalues, so no admissible trial can lie below the lowest one.
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