= Solution
Apply the <Euler-Lagrange equation> to the integrand of $F-\lambda G$, namely $py'^2-qy^2-\lambda ry^2$. With fixed zero endpoint values, it gives $(py')'+qy+\lambda ry=0$. Multiplying by $y$ and using <integration by parts> gives $F=\lambda G$, so each stationary <Rayleigh quotient> $F/G$ equals its eigenvalue.
In the half-line problem, $p=x$, $q=-x$, $r=1$, so the quotient is $R[y]=\int_0^\infty x(y'^2+y^2)dx/\int_0^\infty y^2dx$. For $y=e^{-\beta x}$, $\beta>0$, its numerator is $(\beta^2+1)/(4\beta^2)$ and denominator $1/(2\beta)$. Hence $R=(\beta+\beta^{-1})/2$, minimized at $\beta=1$:
$$
\boxed{R_{\rm exponential}=1.}
$$
For the compactly supported linear trial, $A>0$, the numerator is $A^2/2+A^4/12$ and denominator $A^3/3$. Thus $R=3/(2A)+A/4$, minimized at $A=\sqrt6$, giving
$$
\boxed{R_{\rm linear}=\frac{\sqrt6}{2}\simeq1.225.}
$$
The derivative jump at the cutoff is allowed in the quadratic-form trial space of the <Rayleigh-Ritz method>: the trial is continuous and has finite form energy.
The <exponential ground state of a linearly weighted Sturm-Liouville problem> proves which estimate is best. Vanishing endpoint terms give
$$
\int_0^\infty x(y'^2+y^2)dx=\int_0^\infty y^2dx+\int_0^\infty x(y'+y)^2dx\geq\int_0^\infty y^2dx.
$$
Equality occurs for $y=Ce^{-x}$, which also satisfies the differential equation with $\lambda=1$. Thus \b[the exponential trial gives the exact smallest eigenvalue], whereas the linear trial gives a larger upper bound. Finiteness at the singular endpoint supplies the regular branch with $xyy'\to0$; all the displayed trials have the required vanishing boundary terms.
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