I know that the general one-dimensional Schroedinger equation is given by: $$-\frac{\hbar^2}{2m} \frac{\partial^2\Psi(x,t)}{\partial x^2} + U(x)\Psi(x,t) = i\hbar \frac{\partial \Psi(x,t)}{\partial t} $$
The source I am using mentions: "If the potential energy function is nonzero, these sinusoidal waves do not satisfy the Schroedinger equation". Why? I thought the reason why the general equation included $U(x)$ was exactly to address this problem. Then it continues by: "However, we can still write the wave function for state of definite energy $E$ in the following form" $$\Psi(x,t) = \psi(x)e^{iEt/\hbar} $$ I understand that this comes from $$ \Psi(x,t) = Ae^{i(kx - wt)} $$ But why a state of definite energy must have potential energy non-zero?