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If we have a Sturm-Liouville differential equation of the form $$ \frac{d}{dx}[p(x)\frac{dy}{dx}]+q(x)y=-\lambda w(x)y $$ and define the linear operator $L$ as $$L(u) = \frac{d}{dx}[p(x)\frac{du}{dx}]+q(x)u $$ then we get the equation $L(y)=-\lambda w(x)y$ which defines what is called the eigenvalue problem of the Sturm-Liouville differential equation.
My question: why is it called that way despite the fact that there is still a function $w(x)$ in the equation? I thought an eigenvalue problem would have the form $L(y)=-\lambda y$.
What's happening here?

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The operator can be written as $$ Ly = \frac{1}{w}\left[-\frac{d}{dx}\left(p\frac{dy}{dx}\right)+qy\right]. $$ This operator is defined on weighted $L^2$ space $L^2_w[a,b]$. With the correct endpoint conditions, $L$ becomes selfadjoint, and the eigenvalue problem is $Ly=\lambda y$. Absording the negative sign into the second derivative term is standard because that has the best chance of making $L$ a positive operator. For example, in the simplest case of $Ly=-y''$ on $[a,b]$ with endpoint conditions at $x=a,b$, $$ \langle Ly,y\rangle = \int_a^b (-y'')y(x)dx = -y'y|_a^b + \int_a^b(y')^2dx = \int_a^b (y')^2dx \ge 0. $$

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The function $w(x)$ is called weight function. You can find why they have used this weight function in the eigenvalue problem of the Sturm-Liouville differential equation from the following references:

$1.$ https://arxiv.org/ftp/arxiv/papers/0906/0906.3209.pdf

$2.$ "Differential Equations with Applications and Historical Notes" by George F. Simmons

I think this references will help you.

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