In general we have
Operator (Wavefunction) = Observable*Wavefunction
So the Schroedinger Equation is just a special case of the Eigenvalue Problem?
How are operators derived?
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No, it isn't.
The Schrodinger equation reads \begin{align} i\hbar\frac{d}{dt}\vert \psi\rangle &= \hat{H}\vert\psi\rangle \end{align} where the symbols have their usual meaning. On the other hand, an eigenvalue equation for an operator $\hat{A}$ reads \begin{align} \hat{A}\vert \psi\rangle = a\vert \psi\rangle \end{align} where $\vert \psi\rangle$ would be the eigenvector of the said operator with eigenvalue $a$.
As you can see the two forms of equations are different, $\hat{H}$ cannot be considered as an eigenvalue of any operator that is defined on the Hilbert space of states because it is not a complex number. Also, it is much less clear what would be the definition of $i\hbar\frac{d}{dt}$ as an operator on the Hilbert space. If you are insistent on treating it as an operator on the Hilbert space, its matrix elements in some basis would read $\langle i \vert \hat{H} \vert j\rangle=H_{ij}$ -- in other words, you would be forced to define it as the Hamiltonian itself. However, then the Schrodinger equation is still not an eigenvalue equation but just a tautological statement. See, Why can't $ i\hbar\frac{\partial}{\partial t}$ be considered the Hamiltonian operator?.
However, what is often referred to as the "time-independent Schrodinger equation" is indeed an eigenvalue equation. Namely, it is just the eigenvalue equation for the Hamiltonian of the problem.
\begin{align} \hat{H}\vert\psi\rangle = E\vert\psi\rangle \end{align}
However, it is not the same as the Schrodinger equation. The eigenvalue equation for the Hamiltonian would exist and can be solved regardless of the Schrodinger equation. It is the Schrodinger equation that posits that the Hamiltonian is the generator of translations in time and makes the eigenvalue equation of the Hamiltonian relevant to the question of time-evolution of states.