If there are two similarly charged particles $Q$ and $q$ of masses $M$ and $m$, a Coulomb force is applied on each one (action-reaction). Suppose that $Q$ is stationary and we are shooting $q$ with an initial velocity $\vec{v_0}$ from $\infty$ into the field.
I have, as well, figured out that by rearranging Coulombs Law and the Fundamental Law of Mechanics we end up with the following differential equation:
$$x''(t) [x(t)]^2 = \frac{kQq}{m}$$
because:
coulombs law states that
$$F = \frac{kQq}{[x(t)]^2}$$
and 2nd Newton's Law:
$$F = ma = m x''(t)$$
By rearranging we get:
$$m x''(t) = \frac{kQq}{[x(t)]^2} \Leftrightarrow x''(t) [x(t)]^2 = \frac{kQq}{m}$$
We also know that its velocity is zero when $x = \frac{2kQq}{m{u_0}^2}$ (from Work-Energy Thm)
Is the general solution of this differential equation (2nd order non-linear ODE) the solution of the displacement over time function? If yes, what is it!?
Best Answer
This equation is said to "reduce to quadratures": you can essentially solve it exactly, in the sense that you get your solution as a well-defined integral. This integral is perfectly fine as a function, and it can be used if you so wish to calculate the solution numerically. Unfortunately, while you can do this integral exactly, the final step requires an inversion step which cannot be done using elementary functions.
To do this, one uses the Work-Energy Theorem, which is actually much more useful than you've realized so far. To apply it, simply multiply both sides of the equation by $x'(t)$ and integrate: $$ \begin{align} mx''(t)&=\frac{kQq}{(x(t))^2} \\\Rightarrow mx''(t)x'(t)&=\frac{kQq}{(x(t))^2}x'(t) \\\Rightarrow m\int x''(t)x'(t)dt&=\int\frac{kQq}{(x(t))^2}x'(t)dt \\\Rightarrow \frac12m(x'(t))^2&=\int\frac{kQq}{x^2}dx=\frac{kQq}{x(t)}+E, \end{align} $$ so that therefore $$ x'(t)=\sqrt{\frac{2}{m}}\sqrt{\frac{kQq}{x(t)}+E} $$ and by rearranging and integrating $$ \int \frac{x'(t)dt}{\sqrt{\frac{kQq}{x(t)}+E}}=\int\sqrt{\frac{2}{m}}dt, $$ and finally $$ t-t_0=\sqrt{\frac{m}{2}}\int_{x(t_0)}^{x(t)} \frac{dx}{\sqrt{\frac{kQq}{x}+E}}. $$
This integral can indeed be done exactly. Mathematica tells me that it equals $$ t-t_0=\sqrt{\frac{m}{2}}\left[\sqrt{\frac{kQq}{x}+E}\frac{x}{E}-\frac{kQq}{2E^{3/2}}\ln\left(kQq+2Ex+2x\sqrt{E}\sqrt{\frac{kQq}{x}+E}\right)\right]_{x(t_0)}^{x(t)}, $$ although depending on the sign of $E$ you may be able to achieve simpler expressions in terms of inverse hyperbolic functions. (The case where $E=0$ is also simple to handle.) The problem after that, though, is that this function is essentially impossible to invert analytically. Your solution is "the inverse of the function above", which definitely exists and is continuous and differentiable, but it does not have an analytical expression in terms of elementary functions.