Where did $t-t_0 = \pm \int_{x_0}^x \frac{dx'}{\sqrt{\frac{2}{m}(E-U(x'))}}$? come from?

This came from the expression of conservation of energy: $E= \frac{1}{2} m v^2 + U(x)$, assuming $E$ is constant.

• First, solve for $v=\frac{dx}{dt} = \pm \sqrt{\frac{2}{m} (E-U(x))}$.
• Next, rearrange to write $dt = \pm \frac{dx}{ \sqrt{\frac{2}{m} (E-U(x))}}$.
• Next, integrate both sides: LHS from $t_0$ to $t$, and RHS from $x_0$ to $x$.

This gives the expression $t-t_0 = \pm \int_{x_0}^x \frac{dx'}{\sqrt{\frac{2}{m}(E-U(x'))}}$. You can plug in some given $U(x)$ and do the integral (analytically if you can, or numerically). You then get a function of $x$ on the RHS, which you can solve for in terms of $t$-- that's then your answer, $x(t)$.