Derivation by Schrodinger (post-hoc rationalization)

Consider old quantum theory, and let's focus on the case of a free particle. Probably, the reasoning of Erwin was as follows:

1. It seems that particles have an associated wave, whose parameters are given by the Einstein ($E=\hslash \omega$) and De Broglie ($p=h/\lambda$) relations, according to the experiments. Let's call this wave $\mathrm{\Psi }$. The oscillation/vibration takes place in a kind of internal space.

2. If the "internal space" were $\mathbb{R}$, then the associated wave might be $\mathrm{\Psi }(x,t)=A\sin (\frac{E}{\hslash }t)\sin (\frac{p}{\hslash }x)$. What PDE is satisfied by this function? A key observation was that $$-\frac{\hbar^2}{2m} \nabla^2 (\Psi)=\frac{p^2}{2m}\Psi$$

3. It would be great to find another differential operator, let's call it $\mathcal{E}$, such that $\mathcal{E}(\mathrm{\Psi })=E\mathrm{\Psi }$, since in that case we would have the PDE $$\mathcal E(\Psi)=-\frac{\hbar^2}{2m} \nabla^2 (\Psi),$$due to the classical fact that for the free particle case we have $E=\frac{p^2}{2m}$.

4. Unfortunately, when trying to extract the energy from $\mathrm{\Psi }$ we are led to the expression $$\frac{\hbar \partial}{\partial t} \Psi=E A\cos(\frac{E}{\hbar}t)\sin(\frac{p}{\hbar}x).$$ But we can apply ${\partial }_t$ twice, so $$-\frac{\hbar^2 \partial^2}{\partial t^2} \Psi=E^2 \Psi,$$and then $$\left(\frac{\hbar^2 \partial^2}{\partial t^2} +E^2 \right)\Psi=0$$

5. But, if we admit complex numbers, we can consider the decomposition $$\frac{\hbar^2 \partial^2}{\partial t^2} +E^2 =\left(\frac{\hbar \partial}{\partial t} -iE\right)\left(\frac{\hbar \partial}{\partial t} +iE\right).$$

6. Since $(\frac{\hslash \partial }{\partial t}+iE)(\frac{\hslash \partial }{\partial t}-iE)\mathrm{\Psi }=0$, we can consider another kind of waves, vibrating in a complex internal space, and defined by $$\tilde \Psi:=\left(\frac{\hbar \partial}{\partial t} -iE\right) \Psi=\cdots=\tilde A e^{-i\frac{E}{\hbar}t} \sin(\frac{p}{\hbar}x).$$ This expression is still a wave, and moreover, it satisfies: $$ \left(\frac{\hbar \partial}{\partial t} +iE\right) \tilde \Psi=0,$$ i.e., $$i\hbar \frac{\partial \tilde \Psi}{\partial t}=E\tilde \Psi.$$

7. So this is it, by renaming $\tilde{\mathrm{\Psi }}$ as $\mathrm{\Psi }$ we have: $$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \Psi.$$
   In conclusion, particles of energy $E$ and momentum $p$ have associated a wave of the form

and satisfying the PDE above, which was called Schrodinger equation.

Schrödinger was heavily inspired by an analogy with optics. He knew that classical mechanics (like $E=K+V$) is just an approximation for particles, in the exact same way that geometrical optics (treating light as straight rays) is just an approximation for wave optics (treating light as actual waves). He thought, "If Newtonian mechanics is just the 'ray' approximation of a deeper reality, what is the underlying wave theory?"