Sturm-Liouville differential equation eigenvalue problem

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?

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 $$x=a,b$$, $$\langle Ly,y\rangle = \int_a^b (-y'')y(x)dx = -y'y|_a^b + \int_a^b(y')^2dx = \int_a^b (y')^2dx \ge 0.$$
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