= Solution
For the regular Sturm-Liouville setting assume $p$ is positive and continuously differentiable, $q,w$ are <continuous>, and $w>0$. Apply a multiplier to $J[y]-\lambda(\int_a^bwy^2-1)$. Its <first variation>, divided by two, is
$$
[p y'\eta]_a^b-\int_a^b[(py')'+qy+\lambda wy]\eta\,dx.
$$
Endpoint-vanishing variations and the allowed fundamental lemma give
$$
\boxed{-(py')'-qy=\lambda wy}.
$$
If an endpoint value is free, its <variation> is arbitrary, giving the natural condition $py'=0$ at that endpoint. At a prescribed endpoint value the <variation> vanishes there, so no additional natural condition is imposed.
For eigenfunctions $u,v$ with different eigenvalues, multiply their equations by $v,u$ and subtract after integration. The resulting boundary expression $[p(uv'-u'v)]_a^b$ vanishes under common homogeneous Dirichlet or common natural boundary conditions. Hence $(\lambda_u-\lambda_v)\int_a^bwuv=0$, proving weighted <orthogonality>.
For the final Dirichlet problem the normalized stationary modes are
$$
\boxed{y_k(x)=\pm\sqrt2\sin(k\pi x),\qquad\lambda_k=k^2\pi^2-1,\quad k\geq1}.
$$
Solving $y''+(\lambda+1)y=0$ with both endpoint values zero gives these and excludes nonpositive $\lambda+1$. Distinct sine modes have zero integral product on $[0,1]$. Their Dirichlet values also make the same boundary expression vanish, which explains <orthogonality> even though their endpoint derivatives do not obey the natural conditions.
For global minimality, the Fourier sine expansion of an endpoint-vanishing <function> and Parseval's identities give $\int y'^2=\sum k^2\pi^2a_k^2\geq\pi^2\sum a_k^2=\pi^2\int y^2$. Equality holds exactly in the first sine eigenspace. Therefore
$$
\boxed{\min J=\pi^2-1=\lambda_1,\qquad y=\pm\sqrt2\sin(\pi x)}.
$$
Distinct modes are <orthogonal>; the two sign choices of the same minimum are not <orthogonal> to each other. This distinguishes the family of stationary modes from the two representatives of the minimizing eigenspace in the question's wording.
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