= Solution
Move the differential terms to the left in the sign convention of part (a). Then $\widetilde a=1+x$, $\widetilde c=0$, and $\widetilde b=(3+2x)-\widetilde a'=2(1+x)$. The <integrating factor> is therefore $w=e^{2x}$, up to a positive constant. The <Sturm-Liouville problem> is
$$
\boxed{-\frac d{dx}\left[(1+x)e^{2x}y'\right]=\lambda e^{2x}y,\qquad y(0)=0,\quad y'(1)=0.}
$$
The trial $y=xe^{-x}$ is admissible because $y(0)=0$ and $y'=e^{-x}(1-x)$ vanishes at one. Its <Rayleigh quotient> has numerator
$$
F[y]=\int_0^1(1+x)(1-x)^2dx=\int_0^1(1-x-x^2+x^3)dx=\frac5{12}
$$
and denominator
$$
G[y]=\int_0^1x^2dx=\frac13.
$$
Consequently the <Rayleigh-Ritz variational principle> gives
$$
\boxed{\lambda_{\min}\leq\frac54.}
$$
The numerator is positive for every nonzero admissible function, so the lowest <eigenvalue> is positive. The trial is not an exact <eigenfunction>: substituting it would require $x^2+2x-1=\lambda x$ for every $x$. Thus the upper inequality is actually strict. This example underlies the <exponential-polynomial trial space for a mixed Sturm-Liouville problem>.
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