Well, the Schrödinger equation indeed reads
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}).$$
The equation can be separated in the form
$$\psi(t,\vec{x})=T(t) \Psi(\vec{x}),$$
if the Hamiltonian is not time dependent, because then
$$\mathrm{i} \hbar \partial_t \psi = \mathrm{i} \hbar \dot{T}(t) \Psi(\vec{x})= T(t) \hat{H} \Psi(\vec{x}).$$
Now divide this equation by ##T(t) \Psi(\vec{x})##. Then you get
$$\frac{1}{T}(t) \mathrm{i} \hbar \dot{T}(t) =\frac{1}{\Psi(\vec{x})} \hat{H} \Psi(\vec{x}).$$
Now, since the left-hand side is independent of ##\vec{x}## and the right-hand-side is independent of ##t##, it must be a constant, which we call ##E##. This implies the equation for ##T##
$$\mathrm{i} \hbar \dot{T}=E T \; \Rightarrow \; T(t)=T_0 \exp(-\mathrm{i} \omega t), \quad \omega=\frac{E}{\hbar},$$
and for ##\Psi##:
$$\hat{H} \Psi(\vec{x})=E \Psi(\vec{x}),$$
i.e., ##\Psi## must be an eigenvector of ##\hat{H}## which of course is NOT ##\mathrm{i} \hbar \partial_t## but some differential operator in ##\vec{x}## like the standard case for the motion of a particle in some external potential,
$$\hat{H}=-\frac{\hbar^2}{2m} \Delta + V(\vec{x}).$$
So the general solution of the time-dependent Schrödinger equation can be written in terms of the energy-eigensolutions (i.e., the solutions of the time-independent Schrödinger equation)
$$\psi(t,\vec{x})=\sum_{E} a_E \exp(-\mathrm{i} \omega t) \Psi_E(\vec{x}), \quad \hat{H} \Psi_E(\vec{x})=E \Psi_E(\vec{x}).$$
The sum goes over all energy eigenvalues. If there are continuous energy eigenvalues ("scattering solutions"), there's also an integral over the corresponding values. Of course, there are cases, where the Hamiltonian has only a continuous spectrum, as in the case of the free particle.