In quantum mechanics, a particle is described by its wavefunction, $y(x)$, which is related to the probability of finding the particle at position $x$ (roughly speaking). This wavefunction satisfies the time-independent Schrödinger equation, $$-\frac{\hbar^2}{2m} y''(x) + U(x) y(x) = E y(x)$$ where $\hbar$, $m$ and $E$ are all positive constants (referring respectively to Planck's constant, the mass of the particle, and the energy of the particle), and $U(x)$ is the potential energy function.
Put the time-independent equation into the form of
$$y'' + P(x) y' + Q(x) y = 0$$
I know I'm supposed to show what I've tried, but I really don't know how or where to start.
Checking after the two hints in comment, one should get
$$y''(x)-\frac{2m}{\hbar}(U(x)-E)y(x)=0$$