The one-dimension Schrodinger's equation(spatial component of the S.E.) is

$$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ E-U\left ( x \right ) \right ]\psi=0$$

where E is the sum of the kinetic and potential energy of the system

If the potential energy U(x)=0,
the equation reduces to 

$$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ E^{\*} \right ]\psi=0
$$

where 

$$E^{\*}=\frac{1}{2}mv^{2}$$

But in the case where the potential energy of the system is non-zero

$$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ E-U\_{0}\left ( x\right ) \right ]\psi=0
$$

What can I say about E,specifically? 

Is $$E=\frac{1}{2}mv^{2}+U\_{k}\left ( x \right )$$ such that $$U\_{k}=U\_{0}$$ **necessarily**?

Asked because if $$U\_{k}=U\_{0}$$
then 
$$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ \left ( \frac{mv^{2}}{2}+U\_{k} \right )-U\_{0}\left ( x \right ) \right ]\psi=0$$
reduces to
$$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ \left ( \frac{mv^{2}}{2} \right ) \right ]\psi=0$$

which says that potential energies can never exists in the spatial component of the wave function

Clarification would help

Edit: $$\psi=\psi\left ( x\right )$$ to be clear