= Solution
Multiply the original equation by $p(x)=\exp(\int^xu(s)ds)>0$. Since $p'=up$, it becomes $(py')'+qy+\lambda ry=0$ with $q=pv$, $r=pw>0$. This is <Sturm-Liouville form>. For two real <eigenfunctions>, multiply the equation for $y_n$ by $y_m$, subtract the equation for $y_m$ multiplied by $y_n$, and integrate. The <Lagrange identity for a Sturm-Liouville operator> gives
$$
(\lambda_m-\lambda_n)\int_a^br\,y_my_n\,dx=[p(y_my_n'-y_ny_m')]_a^b.
$$
Each separated homogeneous boundary condition makes the corresponding determinant zero, provided its pair of coefficients is not both zero. For distinct <eigenvalues>, this proves the <orthogonality of Sturm-Liouville eigenfunctions>.
For the constant drift, the <drift-removing substitution for a Dirichlet Sturm-Liouville problem> is $y=e^{-\alpha x}v$. It reduces the equation to $v''+\lambda v=0$ with $v(0)=v(\pi)=0$. Nonzero solutions occur exactly at
$$
\boxed{\lambda_n=n^2,\qquad y_n=e^{-\alpha x}\sin nx,\quad n=1,2,\ldots.}
$$
The zero and negative values give only the zero solution under both endpoint conditions. The self-adjoint expression is $(e^{2\alpha x}y')'+\alpha^2e^{2\alpha x}y+\lambda e^{2\alpha x}y=0$. Hence the weight is $r=e^{2\alpha x}$, and for the displayed normalization,
$$
\boxed{\int_0^\pi e^{2\alpha x}y_my_n\,dx=\frac\pi2\delta_{mn}.}
$$
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