McLaren Rulez said:
So is there a proof for this result? That the general solution of any linear DE is a superposition of the separable solutions. Thank you!
You're postulating too strong of a result.
To start, we must have results on existence and uniqueness of solutions to the initial or boundary value problem for a given differential equation. If the coefficients of the PDE are analytic functions (admit a locally-convergent power series expansion), then a major result is the
Cauchy–Kowalevski theorem. This result establishes the existence and uniqueness of an analytic solution to the initial value problem.
I'm no expert on the subject, but the online notes at
http://www.math.ucdavis.edu/~hunter/pdes/pdes.html seem to be a nice treatment. The heat and Schrodinger equation (at least in one space dimension) are discussed in Ch. 5.
Now, certain PDEs admit separable solutions. If we can establish a separable solution to an initial value problem that satisfies the restrictions of the Cauchy–Kowalevski theorem, then we can conclude that it is the unique solution.
Next, we can ask what PDEs will admit separable solutions. The time dependent Schrodinger equation
$$ i \partial_t \Psi(\mathbf{x},t)= \hat{H} \Psi(\mathbf{x},t)$$
$$ \hat{H} = - \Delta + V(\mathbf{x},t)$$
separates in space and time when the potential is independent of time, ##V(\mathbf{x},t)= V(\mathbf{x})##. Similarly, the time-independent Schrodinger equation
$$ \hat{H} \psi(\mathbf{x})= E \psi(\mathbf{x})$$
will admit separable solutions whenever there is some symmetry of the potential that would cause
$$\frac{d V(\mathbf{x})}{du} =0,~~~ u = f(\mathbf{x}).$$
This is a complicated way of saying that there is a change of variables such that the potential function is independent of at least some of the variables. In principle, there are even time-dependent potentials that are separable after a change of variables that mix space and time coordinates.