"Where did we get that [Schrödinger's equation] from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger." Richard Feynman in The Feynman Lectures on Physics. Wrong! as the following shows.

The differential wave equation must satisfy

• The de Broglie-Einstein "postulates": l = h/p and n = E/h.
• It must be consistent with the equation: E = p²/2m + V, for total energy E of particle with mass M and kinetic energy p²/2m and potential energy V.
• It must be linear in the wave function, Y (x,t).

Then, we gather

1. From the de Broglie postulate, p = h/l , so that p²/2m = h²/(2ml ²).
2. From the Einstein postulate, E = hn.
3. Substituting these two previous results in E = p²/2m + V, we have: h²/(2ml²) + V(x,t) = h n.
4. Or, by k = h/l and w = 2pn, the previous result becomes (hk)²/2m + V(x,t) = (h-bar)w, where h-bar = h/2p.
5. To obtain the equation in (4) as a solution, we need to form a differential equation involving a wave function Y(x,t), but also the potential function, so the two different functands require a partial differential equation (pde).
6. The pde has as solution an equation involving a second degree term, k², and a linear term, w , so the pde must be second order and first order in the wave function for respective representation of them:

   	aD2x2Y(x,t) + V(x,t)Y(x,t) = bDtY(x,t)

   with a, b to be determined.
7. As a wave function test form, we consider:

   	Y(x,t) = cos(kx - wt) + gsin(kx - wt)

8. We differentiate this test form twice with repect to x and once with respect to t:

   	DxY(x,t) = -ksin(kx - wt) + kwcos(kx - wt);
   
   	D2xY(x,t) = -k2cos(kx - wt) - k2sin(kx - wt);
   
   	DtY(x,t) = wsin(kx - wt) - wgcos(kx - wt)

9. If we put the equations of (7) and (8) in the form of (6), we have:

   -ak2cos(kx - wt) - agk2sin(kx - wt) + V0cos(kx - wt) + V0gsin(kx - wt) = bwcos(kx - wt) - bwgsin(kx - wt)

   or

   [-ak2 + V0 + bwg]cos(kx - wt) + [-a k2g + V0g + bw]sin(kx - wt) = 0

10. For the last result to hold for all independent values of x,t , the coefficients of the cosine and sine terms must be zero, yielding two equations:

    -ak2 + V0 = -bwg; -ak2 + V0 = -bw/g.

11. Given the last two equations and equation (4) and three unknowns, we have a determinate problem. Subtract the second equation OF (9) from the first one, to find:

    0 = -bgw - bw/g Þ g = -1/g  Þ g2 = -1 Þ g = ±-1 = ±i

12. Substituting this in the first equation of (7), we have

    -ak2V0 = ±ibw

which can be compared with the equation of (4), (hk)²/2m + V₀ = hw, and this yields

a = -h2/2m, b = ih

since the sign doesn't matter.
13. Then we have our final equation:

    -h2/2m D2x2Y(x,t) + V0Y(x,t) = ih DtY(x,t)

    This pde satisfying our four conditions is Schrödinger's equation for a free particle with V(x,t) = V₀, a constant potential energy. Summarizing: We use two algebraic (hyperbolic) antitones to condition the derivation of the two differential terms of a pde for a (hyperbolic) wave equation.