# Isn't $E-U = K$ in Schrodinger's Equation?

I'm studying quantum mechanics in its most basic level (I don't even know if Physicists call this already quantum mechanics) and I have one doubt in Schrodinger's equation. The book presents the equation for the special case where the solution is of the form $\Psi(x,t)=\psi(x)e^{-i\omega t}$ and says that Schrodinger's equation (in one dimension) is

$$\dfrac{d^2\psi}{dx^2}+\dfrac{8\pi^2 m}{h^2}(E-U(x))\psi(x)=0$$

Where $m$ is the mass of the particle, $E$ it's total energy and $U$ it's potential energy function.

The first doubt that arises is the following: the book says that $E$ is the "constant total energy" of the particle. But wait, since $E = K + U$ and $U$ varies, clearly $E$ should vary. How can $E$ be constant if $U$ is not?

Also, when we write $E-U(x)$ isn't this simply $K$, the kinectic energy? Why do we bother then writing explicitly $E-U$?

I feel that the potential $U(x)$ on the equation and the one that is part of $E$ are different, but I'm not understanding how.