I have the Schrödinger equation:
$$\dfrac{-\hbar^2}{2m} \nabla^2 \Psi + V \Psi = i \hbar \dfrac{\partial{\Psi}}{\partial{t}},$$
where $m$ is the particle's mass, $V$ is the potential energy operator, and $(-\hbar^2/2m) \nabla^2$ is the kinetic energy operator ($p^2/2m$).
The state function can be expressed as the product of space-dependent and time-dependent factors, $\Psi = \psi(r) \psi(t)$. If we substitute these into the above equation and divide by $\psi w$, we obtain a function on the left that depends on $r$ and a function on the right that only depends on $t$:
$$\dfrac{-\hbar^2}{2m} \dfrac{\nabla^2 \psi}{\psi} + V = \dfrac{i \hbar}{w} \dfrac{\partial{w}}{\partial{t}}.$$
Therefore, to be valid for all $r$ and $t$, each side must equal a constant, $E$:
$$\dfrac{-\hbar^2}{2m} \dfrac{\nabla^2 \psi}{\psi} + V = \dfrac{i \hbar}{w} \dfrac{\partial{w}}{\partial{t}} = E.$$
From this, we immediately have
$$w(t) = Ce^{-i(E/\hbar)t},$$
from which we can identify that $E = \hbar \omega$, where $\omega$ is the radian frequency of oscillation.
It's been a while since I've done PDEs, so I'm not sure how we "immediately have" $w(t) = Ce^{-i(E/\hbar)t}$. The form of $\dfrac{-\hbar^2}{2m} \dfrac{\nabla^2 \psi}{\psi} + V = \dfrac{i \hbar}{w} \dfrac{\partial{w}}{\partial{t}} = E$ reminds me of a separation of variables situation, but I'm not entirely sure.
Furthermore, I'm not sure how we identify that $E = \hbar w$ from $w(t) = Ce^{-i(E/\hbar)t}$.
I would greatly appreciate it if people would please take the time to explain this.
Note that we have performed a separation of variables and have two separate differential equations $$\frac{i \hbar}{w} \frac{\partial w}{\partial t} = E \\ \frac{-\hbar^2}{2m \psi} \nabla^2 \psi + V = E$$ The first equation is just a first-order ODE of the form $w’(t) = A w$, whose solution is a scalar multiple of $e^{At}$. Here $A = E/(i \hbar) = -i E / \hbar$, which is why the solution to the differential equation in time is as you have written. Note that the second equation above is referred to as the time-independent Schrodinger equation.
Intuitively, one can note that the solution to the differential equation in time is periodic with some frequency $\omega = E/ \hbar$. So the wave equation of the particle has frequency $\omega$, and rewriting the previous equation gives the relation $E = \hbar \omega$.