= Solution
\b[Variational formulation.] Take real $p,q,w$ with $p,w>0$ in the interval, and impose homogeneous <Dirichlet boundary conditions>, or other suitable conditions that cancel the boundary terms. The <functional>
$$
J[y]=\int_a^b\left[p(x)(y')^2+q(x)y^2-\lambda w(x)y^2\right]dx
$$
has <Euler-Lagrange equation>
$$
\frac d{dx}(2py')-2(q-\lambda w)y=0,
$$
which is equivalent to $-(py')'+qy-\lambda wy=0$. The same construction is constrained stationarity of the energy $\int[p(y')^2+qy^2]$ subject to $\int wy^2=1$, with $\lambda$ a <Lagrange multiplier>. For fixed zero endpoint values, the variation vanishes at both ends; homogeneous zero-flux conditions also cancel the boundary term $[2py'\delta y]_a^b$.
Multiplying the differential equation by $y$ and using <integration by parts> gives the <Rayleigh quotient>
$$
\boxed{R[y]=\frac{\displaystyle\int_a^b[p(y')^2+qy^2]dx}{\displaystyle\int_a^bwy^2dx}.}
$$
For a nonzero admissible trial <function>, $R[y]\geq\lambda_0$ by the <Rayleigh-Ritz variational principle>. The <Rayleigh-Ritz method> chooses trial functions satisfying the endpoint conditions, often with adjustable coefficients, and minimizes this quotient over that trial family. Expanding $y=\sum c_ny_n$ in a normalized <eigenbasis> shows why the lower bound holds: $R[y]=\sum\lambda_n c_n^2/\sum c_n^2$, a weighted average of <eigenvalues> no smaller than the least one.
\b[Radial quadratic trial.] The radial problem becomes
$$
-(r^2\psi')'=\lambda r^2\psi,\qquad\psi(1)=0,\qquad\psi'(0)=0,
$$
where the printed endpoint notation $y$ refers to $\psi$. Thus $p=w=r^2$ and $q=0$. At the origin the vanishing $r^2$ and regularity of $\psi$ make the boundary term vanish; the weight must be retained in both numerator and denominator.
For a quadratic $\psi=A+Br+Cr^2$, the derivative condition sets $B=0$ and the outer condition gives $C=-A$. Overall amplitude cancels in the <Rayleigh quotient>, so take $\widetilde\psi=1-r^2$. Then
$$
\int_0^1r^2(\widetilde\psi')^2dr=4\int_0^1r^4dr=\frac45,
$$
while
$$
\int_0^1r^2\widetilde\psi^2dr=\int_0^1(r^2-2r^4+r^6)dr=\frac13-\frac25+\frac17=\frac8{105}.
$$
Consequently
$$
\boxed{\widetilde\psi(r)=1-r^2,\qquad\widetilde\lambda_0=\frac{4/5}{8/105}=\frac{21}2=10.5.}
$$
\b[Why the estimate is higher.] For a regular radial <function> put $u=r\psi$. Then $u(0)=u(1)=0$, and direct differentiation transforms the equation to $u''+\lambda u=0$. Its first mode is $u=\sin(\pi r)$, corresponding to $\psi=\sin(\pi r)/r$ with its continuous value at the origin. Thus $\lambda_0=\pi^2$. The quadratic trial is not proportional to that <Spherical Bessel function>, so it has nonzero components in higher modes. Its <Rayleigh quotient> is therefore strictly larger than $\pi^2$, in agreement with $21/2>\pi^2$.
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