For example like in there Non homogeneous Sturm-Liouville problem and solution: Is the solution given in terms of $\lambda$, not particular eigenvalues?
$$\displaystyle\int_0^1r(x)\phi_n(x)\phi_m(x)dx=\delta_{mn}$$
Where $r(x)$ is weight function which is placed in S.L. as following:
$$\dfrac{d}{dx}(p(x)y'(x))+(q(x)+\lambda \underbrace{r(x)})y(x)=0$$
and where $\phi_{n,m}(x)$ are eigenfunctions of corresponding eigenvalues of homogeneous S.L. problem.
Best Answer
It comes from the Lagrange identity. If you have solutions $y_1$, $y_2$, with eigenvalues $\lambda_1$, $\lambda_2$, respectively, then $$ ((py_1')'+(q+\lambda_1r)y_1)y_2 = 0 \\ y_1((py_2')'+(q+\lambda_2r)y_2)=0 $$
So, if you subtract the second from the first, you get
$$ (py_1')'y_2-y_1(py_2')'=(\lambda_2-\lambda_1)ry_1y_2 \\ \frac{d}{dx}(py_1'y_2-y_1py_2')=(\lambda_2-\lambda_1)ry_1y_2 $$ The last equation is the Lagrange identity. Then, integrating over $[0,1]$ gives $$ (\lambda_2-\lambda_1)\int_{0}^{1}y_1y_2rdx = p(y_2y_1'-y_2'y_1)|_{0}^{1}. $$ So, if you have endpoint conditions of the following form for both $y_1,y_2$, then the evaluation terms vanish:
$$ A y_j(a)+B y_j'(a) = 0,\;\; A^2+B^2 \ne 0\\ C y_j(b)+Dy_j'(b) = 0,\;\; C^2+D^2 \ne 0. $$
And that gives orthogonality of $y_1,y_2$ with respect to the weight function $r$ provided that $\lambda_1\ne\lambda_2$. This is the history of how "orthogonality" with respect to the weight was formulated. This discovery eventually led to the general notion of "inner product" about 70 years later.